Mechanical BE (Advanced Solid Mechanics) by IMTS INSTITUTE

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ADVANCED SOLID MECHANICS

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ADVANCED SOLID MECHANICS

CONTENTS: UNIT – I

01-103

ELEMENTARY PRINCIPLES – D’Alembert’s principle - Lagrange’s equation – Hamilton’s equation – Lagrangian and Hamiltonian TWO BODY CENTRAL FORCE PROBLEM Equations of motion and first integrals – Kepler’s laws – Scattering by a central potential – transformation from center of mass to laboratory frame. SPECIAL RELATIVITY IN CLASSICAL MECHANICS Relativistic LAgrangian and Hamiltonian for a particle – space, time and energy – momentum – four vectors – center of mass system for relativistic particles – invariance of Maxwell’s equations.

UNIT II

104-157

KINEMATICS OF ROTATION Orthogonal transformations – Euler poles – Rotating frames of reference and coriolis force

MECHANICS OF RIGID BODIES Angular momentum and kinetic energy – moment of inertia tensor – Euler’s equations of motion – Torque free motion – Motion of a symmetric top under gravity.

UNIT III

158-232

CANONICAL TRANSFORMATIONS Canonical transformations and their generators – simple examples – poisson brackets HAMILTON JACOBI THEORY

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Hamilton – Jacobi equations – Action angle variables – Application to kepler problem SMALL OSCILLATIONS Formulation of the problem – Transformation to normal coordinate – Linear triatomic molecule

UNIT IV

233-281

CLASSICAL STATISTICAL MECHANICS: Postulates – Liouville’s theorem – Micro canonical, canonical and grand canonical – examples – Partition function and entropy of ideal gas – Gibb’s paradox. QUANTUM STATISTICAL MECHANICS Liouville’s equation – Postulates of quantum statistical mechanics – BoseEinstein, Fermi-Dirac distributions

UNIT V

282-229

IDEAL BOSE GAS Equation of state – Bose-Einstein condensation – Landau’s theory of liquid Helium II – Black body radiation – Phonons IDEAL FERMI GAS Equation of state – free electron gas in metals – heat capacity – Pauli’s Para magnetism – Thermionic emission.

UNIT QUESTIONS

330-335

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UNIT I 1. Mechanics of a system of Particles (a) Conservation theorem for linear momentum: The net linear momentum of a system of n-particles is n

i 1

P   pi   mi vi ext

From Newton's second law, F



dP dt

i.e., the rate of change of linear momentum of a system of particles is equal to the net external force acting on the system. If

dP  0 . Integrating, P = constant. dt

Fext  0,

This gives the theorem for conservation of linear momentum of the system according to which "If the sum of external forces acting on the system of particles is zero, the total linear momentum of the system is constant or conserved." (b) Conservation theorem for angular momentum. The angular momentum of ith particle of the system about any point O, from definition is given by

J i  ri  pi ,

…(1)

where ri is the radius vector of ith particle from the point O and p i , its linear momentum. For a system of n particles, we have

J   J i   ri  pi i

dpi

dJ  dt Here,

Fi 

…(2)

 r  dt i



r F i

  

dri   pi  v i  pi  0  dt 

d pi = net force acting on ith particle. dt

Internal forces occur in equal and opposite pairs. Hence the net internal force acting on the system of particles is zero. Thus,

dJ  dt Here,

 ext

r F i

 ext

r F i

ext i

is the torque arising due to external forces only.

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 ext  0,

dJ 0 dt

J = constant.

Thus, if external torque acting on a system of particles is zero, the angular momentum of the system remains constant. This is the conservation theorem for angular momentum of a system of particles(c) Conservation of Energy: If the work done by a force is independent of path, the force is said to be conservative. If the forces acting on the system of particles are conservative, the total energy of the system of particles which is the sum of the total kinetic energy and the total potential energy of the system is conserved. This is the energy conservation theorem. On the other hand if the forces are non-conservative, the total energy of universe (mechanical energy + chemical energy + sound energy + light energy + heat energy etc.) remains constant.

Basic Concepts Degrees of Freedom. The number of mutually independent variables required to define the state or position of a system is the number of degrees of freedom possessed by it. For example, the position of a simple ideal mass-point can be defined completely by the three cartesian coordinates x, y, z. So it has three degrees of freedom. Extending this idea, for a system of N particles moving independently of each other, the number of degrees of freedom is 3 N. Constraints. Constraints are restrictions imposed on the position or motion of a system, because of geometrical conditions. Examples. (1) The beads of an abacus are constrained to one-dimensional motion by the supporting wires. (2) Gas molecules within a container are constrained by the walls of the vessel to move only inside the container. (3) The motion of rigid bodies is always such that the distance between any two particles remains unchanged. (4) A particle placed on the surface of a solid sphere is restricted by the constraint so that it can only move on the surface or in the region exterior to the sphere. For a particle constrained to move on a plane, only two variables x, y or r, 0 are sufficient to describe its motion and the particle is said to have two degrees of freedom. Thus, the constraint on the motion of tag. particle in a plane reduces the number of degrees of freedom by one.

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Very often, we can express constraints in terms of certain equations. For example, the equation of constraint in the case of a particle moving on or outside the surface of a sphere of 2

radius a is x + y + z > a if the origin of the coordinate system coincides with the centre of the sphere. (i) Holonomic and non-holonomic constraints. If the constraints can be expressed as equations connecting the co-ordinates of the particles (and possibly time) in the form f (r1, r2, r3, ..., rn, t) = 0

...(1)

then the constraints are said to be holonomic. Examples. (1) The constraints involved in the motion of rigid bodies in which the distance between any two particular points is always fixed, are holonomic since the conditions of constrains are expressed as

(ri  rj )2  cij2  0 (2) The constraints involved when a particle is restricted to move along a curve or surface are holonomic. Here the equation defining the curve or surface is the equation of constraint. If the constraints cannot be expressed in the form of Eq. (1), they are called nonholonomic constraints. Examples . (1) The constraints involved in the motion of the particle placed on the surface of a solid sphere are nonholonomic. The conditions of constraints in this case are expressed as 2

r - a > 0, where a is the radius of sphere. This is an inequality and hence not in the form of Eq. (1). (2) The walls of the gas vessel constitute a nonholonomic constraint. (3) An object rolling on a rough surface without slipping is also an example of nonholonomic constraint. (ii) Scleronomic and Rheonomic Constraints. If the constraints are independent of time, they are called scleronomic. If the constraints are explicitly dependent on time, they are called rheonomic. The constraint in the case of rigid body motion is scleronomous. A bead sliding on a moving wire is an example of rheonomic constraint. In the solution of mechanical problems, the constraints introduce two types of difficulties : (1) The co-ordinates ri are connected by the equations of constraints. Therefore, they are not independent. (2) The forces of constraint are not a priori known. In fact, they cannot be estimated till a complete solution of the problem is obtained.

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The first problem can be solved by introducing generalized coordinates, whereas the second is practically an insurmountable problem. We therefore reformulate the problem such that the forces of constraint disappear.

Generalised co-ordinates A system consisting of N particles, free from constraints, has 3N independent coordinates or degrees of freedom. If the sum of the degrees of freedom of all the particles is k, then the system may be regarded as a collection of free particles subjected to (3N—k) independent constraints. So only k coordinates are needed to describe the motion of the system. These new coordinates q1, q2, q2 ... qk are called the Generalised Coordinates of Lagrange. Generalised coordinates may be lengths or angles or any other set of independent quantities which define the position of the system. Definition. The generalised coordinates of a material system are the independent parameters q1, q2, q2 ... qk which completely specify the configuration of the system, i.e., the position of all its particles with respect to the frame of reference. Example. Consider the simple pendulum of mass m1 with fixed length r1. The single coordinate

i

will determine uniquely the position of ml since

the simple pendulum is a system of one degree of freedom. Since the only variable involved is q =

i .

 i , it can be chosen as the generalised coordinate. Thus

The two coordinates x1 and y1 could also be used to locate m1 but

would require the inclusion of the equation of the constraint

x12  y12  r12 .

Since x1 and y1 are not independent, they are not generalised coordinates. Generalised velocities: The generalised velocities of a system are the total time derivatives of the generalised coordinates of the system. Thus

qi 

dqi (i 1, 2,3,..., k ) . dt

Transformation equations The rectangular cartesian co-ordinates can be expressed as the functions of generalised co-ordinates. Let xi, yi and zi be the cartesian coordinates of ith particle of the system. Let t denote the time. Then, these cartesian co-ordinates can be expressed as functions of generalised co-ordinates q1, q2, q2 ... qk i.e.,

xi  xi (q1 , q2 ,..., qk , t )   yi  yi (q1 , q2 ,..., qk , t )  zi  zi (q1 , q2 ,..., qk , t ) 

…(1)

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Let ri be the position vector of the ith particle, i.e., ri = ixi + jyi + kzi, Then

ri  ri (q1 , q2 ,..., qk , t )

…(2)

Eq. (2) is the vector form of Eq. (1). The equations like (1) and (2) are called transformation equations. The functions and their derivatives in the above two equations are supposed to be continuous. The equations also contain the constraints explicitly. Configuration space In the case of motion of a single particle we can represent its trajectory in the three dimensional space by specifying its variables. For a system of N particles described by 3N space coordinates with (3N-k) equations of constraint in the real space, it is difficult to visualise the motion of the entire system. It is convenient to describe the motion of a system having k coordinates in a hypothetical k dimensional space. The instantaneous configuration of the system is described by the values of the k generalised coordinates q1, q2, q2 ... qk and corresponds to a particular point in a cartesian hyperspace, where the q's form the k coordinate axes. The point is called the system point and the k dimensional space is known as the Configuration space. At some later instant, the state of the system changes and it will be represented by some other point in the configuration space. Thus, the system point moves in the configuration space tracing out a curve. This curve represents “the path of motion of the system”. “The motion of the system”, as used above, then refers to the motion of the system point along this path in configuration space. Time can be considered formally as a parameter of the curve since each point in the configuration space has one or more values of time associated with it. Configuration space has nothing in common with the three-dimensional space which we can visualise physically. It is a purely geometric structure by means of which the laws of the variation of the state of a system can be formulated in geometrical language.

Principle of virtual work Consider a system described by n generalized coordinates qj (j=1, 2, 3, ..., n). Suppose the system undergoes a certain displacement in the configuration space in such a way that it does not take any time and that it is consistent with the constraints on the system. Such displacements are called virtual because they do not represent actual displacements of the system. Since there is no actual motion of the system, the work done by the forces of constraint in such a virtual displacement is zero. Let the virtual displacement of the ith particle of the given system be

 ri .

If the given

system is in equilibrium, the resultant force acting on the ith particle of the system must be zero, i.e., Fi = 0. It is, then, obvious that virtual work Fi . ri  0 for the ith particle and hence it is also zero for all the particles of the system.

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dW   Fi . ri  0

Thus

…(1)

The resultant force Fi acting on the ith particle is

Fi  Fia  fi

…(2)

Here, Fi is the applied force and f, is the force of constraint. Eq. (1) then becomes

 F . r   f . r  0 a i

…(3)

We now consider systems for which the virtual work done by the forces of constraints is zero, i.e.,

 f . r  0 i

…(4)

Then Eq. (3) becomes

 F . r  0 a i

…(5)

This equation is termed as principle of virtual work. 1.1 D’ Alembert's Principle Most of the systems we come across in mechanics are not in static equilibrium. Hence the principle must be modified to include dynamic systems as well. According to Newton's second law of motion,

Fi  pi

Fi  pi  0

…(6)

According to the above equation, a moving system of particles can be considered to be in equilibrium under the force ( Fi  pi ), i.e., the actual applied force Fi plus an additional force

 pi , which is known as reversed effective force on ith particle. Let us again assume that the forces of constraint do no work. Then, we can generalize Eq. (5) by the use of Eq. (6) to the form

 (F

 pi ). ri  0

…(7)

Eq. (7) is the mathematical statement of D’Alembert's principle. It is to be noted here that we have restricted ourselves to the systems where the virtual work done by the forces of constraints disappears. With this in mind we can drop the superscript a in equation (2) i.e., D’Alembert's principle may be written as

 (F

 pi ). ri  0 .

…(8)

1.2 Derivation of Lagrange's Equation of Motion

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Lagrange's Equations from D’Alembert's Principle Consider a system of particles whose position vectors are expressed as functions of generalized coordinates q1 , q2 , q3 ,..., qk ,...qf time t. Consider any particle of the system (ith particle) of mass mi and acted upon by an external force Fi. According to D’ Alembert's principle,

 (F

 pi ). ri  0

…(1)

Here p i , is the inertial force for ith particle and

 ri ,

is the virtual displacement of ith

particle due to action of force Fi .

ri  ri (q1 , q2 , q3 ,..., qk ,...qf , t ) .

In general,

ri 

…(2)

 ri ri r r r r  q1  i q2  ...  i qk  ...  i q f  i  t q1 q2 qk q f t

 k

ri r qk  i qk t

…(3)

 ri

The vertical displacement

in terms of generalized coordinates is given by

 ri   k

ri  qk qk

…(4)

pi  mi ri

Now,

Therefore Eq (1) becomes

 (F

 mi ri ). ri  0

…(5)

 m r . r   F . r

i i

Putting the value of

 ri

ri

i i

 qk   Fi .

Now,

d  ri  ri . dt  qk



ri .

ri  qk qk

…(7)

 ri d  r   ri .  i    ri . qk dt  qk  

ri d  ri   ri . qk dt  qk

Putting the value of ri .

…(6)

from Eq (4) in (6), we get

 m r . q i

 d  ri    ri .   dt  qk  

…(8)

ri from above equation in (7), we get qk

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d 

ri  d  ri   ri .  dt  qk k 

 m  dt  r . q i





    qk  

ri

 F . q i

 qk

…(9)

Differentiating Eq (3) partially with respect to qk ,

ri r  i qk qk

…(10)

Differentiating Eq (3) partially with respect to qk ,

ri  2ri  2ri  2ri  2ri  q1  q2  ...  qk  ...  qk qk q1 qk q2 qk qk qk t

…(11)

Also we have

d  ri  dt  qk

   ri    q1  qk

   ri  q1   q2  qk 

   q2  ...  qk 

 ri   qk

  qk 

 q f

 ri   q f

   r   q f   i  t  qk  

 ... 



 2ri  2ri  2ri  2ri q1  q2  ...  qk  ...  q1 qk q2 qk qk qk t qk

…(12)

Comparing Eqs. (11) and (12),

d  ri  ri   dt  qk  qk From Eq. (10),

…(13)

d  ri  d  ri  d   1 2   ri .    ri .   ri  …(14) dt  qk  dt  qk  dt qk  2 

Substituting (13) and (14) in Eq. (9),

d  1 2 ri  ri  qk   qk   Fi .  ri   ri . qk  qk  i k k 2 

 m  dt q i

d  1 ri  1 2 2   qk  mi ri    mi ri    qk   Fi . qk  qk  2  i k k 2 

  dt q i

d  

  dt q   2 m r k



2 i i

   1 2     mi ri    qk  qk  2  

ri

 F . q i

 qk

…(15)

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 2 mr

2 i i

 T  total kinetic energy of the system of particles ri

 F . q

and

…(16)

 Qk ,

…(17)

Here Q k ’s are components of generalized force. Eq. (15) becomes,

 d T

  dt q k





T    qk   Qk  qk qk 

…(18)

d T T   Qk . dt qk qk



…(19)

This is the general form of Lagrange’s equation. There are

such equations

corresponding to f generalized co-ordinates. When the system is wholly conservative,

Fi  Vi   Qk   Fi . i

Here

Vi ri

ri V r V   i . i   i  qk ri qk qk i qk

…(20)

 V    qV i

V  Vi  total potential energy of the system. i

Putting this value of Q k in Eq. (19), we get

d  T  T V    dt  qk  qk qk or

d  T  T V  0   dt  qk  qk qk

d  T   (T  V )  0   dt  qk  qk

…(21)

The potential energy V is the function of position co-ordinates qk and not of the generalized velocities qk . Therefore, Eq. (21) may be written as

d   (T  V )  (T  V )  0 dt qk qk But

…(22)

L  T  V , where L is known as Lagrangian function.

Eq. (22) becomes,

d  L  L 0 .   dt  qk  qk

…(23)

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This is Lagrange’s equation for a conservative system. Lagrange’s equation for systems containing dissipative forces. Consider a system of particles containing dissipative or frictional forces. The frictional force is proportional to the velocity of the particle, i.e.,

Fi( d )   iri , (d )

Here Fi

…(1) th

denotes the dissipative force, ri is the velocity of i particle and

i

is the

corresponding constant of proportionality. Forces of this type are derivable from Rayleigh’s dissipation function R defined by

R

1 i ri2  2 i

…(2)

i 1, 2,...n covers all the particles of the system.

Here

R  i ri   Fi( d ) ri Fi( d )  

R ri

from Eq.(1)

…(3)

The Lagrange’s equation in terms of r is given by

d  L  L (d )     Fi dt  r  r Here term

…(4)

R represents the dissipative force. ri

Lagrange’s equation in generalized coordinates qk is

d  L  L  Q(kd )   dt  qk  qk d

where Q k is component of generalized force. In order to find the Lagrange’s equation in generalized co-ordinates, we have to find the components of generalized force resulting from the dissipative force. (d )

If Q k

is the component of generalized force along qk , then

Q(kd )   Fi( d ) i

ri qk

   i ri

ri qk

[from Eq. (1)]

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   i ri Q(kd )    i





 qk

Q(kd )  

 ri ri      qk qk 

ri qk

 1 2 ( ri ) qk 2 1

 ( 2  r

2 i i

)

R qk

…(5)

Therefore the Lagrange’s equation for a system containing dissipative force is given by

d  L  L R  0   dt  qk  qk qk Hence the term

…(6)

R takes into account the dissipative forces. qk

Thus, if dissipative forces are acting on the system, we must specify two scalar functions – the Lagrangian L and Rayleigh’s dissipation function R- to derive the equations of motion. Application of Lagrange’s Equations In order to use Lagrange’s equations for the solution of a physical problem, one must use the following steps: (i)

Choose an appropriate coordinate system.

(ii)

Write down the expressions for potential and kinetic energies.

(iii)

Write down the equations of constraint, if any.

(iv)

Choose the generalized coordinates.

(v)

Set up the Lagrangian.

(vi)

Solve Lagrange’s equations for each generalized coordinate using, if necessary,

L T V .

the equations of constraint. (a) The Atwood’s machine: Let two small heavy particles of masses M1 and M2 be connected by a light inextensible rope of length l passing over a frictionless light pulley. It is found that the heavier particle descends while the lighter ascends, the system moving with a constant acceleration

The Atwood's machine is a conservative system with a holonomic constraint. There is only one independent coordinate x. The position of the other particle is determined by the constraint that the length of the rope between them is l. The P.E. of the system = V = – M1gx – M2g (l-x)

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The K.E. of the system = T =

1 ( M1  M 2 ) x 2 2

Hence, the Lagrangian function is given by L=T–V=

1 ( M1  M 2 ) x 2  M1 gx  M 2 g (l  x) 2

The Lagrange's equation for a conservative system is

d  L  L 0   dt  qi  qi Since the system has only one degree of freedom, there is only one equation of motion, involving the derivatives.



The equation of motion of the system is given by

d  L  L   0. dt  x  x From the expression for L we get,

L  ( M1  M 2 ) g ; x

L d  L  d  ( M1  M 2 ) x ;    [( M1  M 2 ) x]  ( M1  M 2 ) x . x dt  x  dt



We have, ( M1  M 2 ) x ( M1  M 2 ) g =0

x

M1  M 2 g. M1  M 2

(b) A bead sliding on a Uniformly Rotating wire in a Force Free Space Let OB be a straight frictionless wire fixed at a point O. Suppose the wire rotates about a perpendicular axis through O with constant angular velocity

 . Let r be the

distance of the bead from point O of the wire at time t. In this example, constraint is time dependent given by the relation





where



is the

angular velocity of rotation. Then the rectangular coordinates of the bead are given by x = r cos y = r sin



= r sin



= r cos

t

 t.

Here it is assumed that at t = 0 the wire is along the X- axis. Now,

x  r cos  r sin  y  r sin   r cos

Kinetic energy T is given by

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1 1 1 T  m( x 2  y 2 )  m[r 2  (r )2 ]  m(r 2  r 2 2 ) 2 2 2 The wire is rotating uniformly in a force-free space. Therefore the Lagrange's equation becomes

d  T  dt  r

 T 0   r

From the expression for T we get,

T T  mr and  mr 2 . r r

 Equation of motion is mr  mr 2  0 r  r 2 .

This shows that the bead moves outward because of centripetal acceleration.

Generalised momentum The generalised momentum conjugate to the generalized coordinate qk is defined as the quantity

L . It is represented by pk. qk

pk 

i.e.,

L qk

Cyclic or ignorable co-ordinates The Lagrangian L is written as function of qi and qi . If the Lagrangian of a system does not contain a given coordinate, say qk , then the coordinate is said to be cyclic. Since the coordinate qk is absent in the expression for the Lagrangian L, the partial derivative of L with respect to qk will vanish, i.e.,

L 0 qk

1.2.1 Lagrangian formulation of conservation theorems. (a) Conservation theorem for generalised momentum. According to Lagrange's equation,

d  L  L 0   dt  qk  qk If the generalized co-ordinate qk is cyclic, then

L 0 . qk

Thus, the Lagrange equation of motion reduces to

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d  L  dt  qk

 0 

L = constant qk

i.e.,

pk = constant

Thus, whenever a coordinate qk

does not appear explicitly in the Lagrangian function

pk is a constant of the motion. Hence we can state as a

L, corresponding linear momentum

general conservation theorem that the generalized momentum conjugate to a cyclic coordinate is conserved. (b) Conservation theorem for energy Consider a conservative system. The Lagrangian

does not depend upon time

explicitly. Then

L  L(q1 , q2 ,... qk ,... q1 , q2 ,... qk ,...).

Hence, the total time derivative of L  L (qk , qk ) is given by

dL L L  qk   qk dt k qk k qk Lagrange’s Eq. is

d  L  L 0   dt  qk  qk

…(1)

d  L  L   dt  qk  qk

We can rewrite Eq. (1) as

dL d  L   dt k dt  qk

 L d  L  qk    qk  qk    qk  k qk k dt  

d L   L   qk 0 . dt  qk 



L   qk

Now

L L T   1 2  (T  V )     mk qk  qk qk qk qk  2 

Eq. (2) becomes L 

…(2)

1  2  T   mk qk  2  

L  mk qk qk



L  constant. qk

m q k

L  2T  constant

2 k

 constant or

T  V  2T  constant

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15 (T  V )  constant.

 T  V  E  total energy = constant. Thus the energy conservation theorem states that, if the Lagrangian function does not contain the time explicitly, the total energy of the conservative system is conserved. (c) Conservation Theorem for Linear Momentum Consider a conservative system so that the potential energy V is dependent on position only and the kinetic energy is independent of position, i.e.,

V T  0 and 0 . qk qk

…(1)

Then, Lagrange equation of motion for such a coordinate is,

d  L  L 0.   dt  qk  qk d  L  V 0   dt  qk  qk

[using Eq.(1)]

d  L  V  Qk    dt  qk  qk

…(2)

i.e.,



pk 

Now, if we show that Q k represents the component of total force along the direction of translation of the system and

pk is the component of total linear momentum along the same

direction, then Eq.(2) will represent the equation of motion for total linear momentum. Generalized force is given by

Qk   Fi . i

ri qk

If n is the unit vector along the direction of translation, then

 ri  n  qk or

ri n qk

Thus

Qk   Fi . n  n . F i

which represents the component of total force along the direction of n. The kinetic energy,

T

. 1 m r  i i2 . 2 i

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16 .

T

 ri pk  .  .   mi r i qk  qk  qk .

 ri   mi vi qk   mi vi .n  n. mi vi i

which shows that pk represents the component of total linear momentum along the direction of translation. Now we can say that the equation

pk = Qk

is the equation of motion for total linear momentum of the system. If Qk = 0, pk = 0 i.e., pk = constant. This gives the conservation theorem for linear momentum.

It states that if a given

component of the total applied for linear momentum. It states that if a given component of the total applied force vanishes, the corresponding component of the linear momentum is conserved. (d) Conservation Theorem for Angular momentum: Consider a conservation system so that the potential energy V is dependent on position only and the kinetic energy is independent of position. i.e.,

V .

 0 and

 qk

T  0. qk

…(1)

Then Lagrange equation of motion for such a coordinate is

 d  L dt   q.  k

   L  0  qk 

 d  L dt   q.  k

   L   V  qk qk 

p k  Qk . Now if we show that with qk , a co-ordinate, the generalized force Qk.

is the component of total applied torque about the axis of rotation and the

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generalised momentum pk is the component of total angular momentum along the same axis, then Eq. (2) will represent the equation of motion for the total angular momentum. The generalised force Qk is given by The magnitude of change in position co-ordinate r; (qk) due to a change in rotation co-ordinate qk is given by Let n be the unit vector along the axis of rotation. Then Eq. (3) may be written as where x = 1| xt is the total torque. Eq. (5) shows that Q^ is the component of the total torque about the rotation axis. Further, Here L = ^ L, is the total angular momentum. Eq. (6) shows that

pk ist the component of total angular momentum along the axis of

rotation. Hence Eq. (2) Represents the equation of motion for the total angular momentum of the system. Thus the conservation theorem for angular momentum states that if the rotation coordinate qk is cyclic i.e., if the component of applied torque along the axis of rotation vanishes, then the component of total angular momentum along the axis of rotation is conserved Solved Examples Ex: 1(a) Set up the Lagrangian function for a simple pendulum and hence obtain the equation describing gits motion. Let the mass of the pendulum bob (idealized as a particle) be m and at nay instant t the string OB of the pendulum make an angle θ with its position of equilibrium OA. If l is the length of the string, then kinetic energy of the bob is given by 2

T = ½ mv



. . 1 1 m(l  )2  ml 2  2 2 2

(sin ce v = l  ).

Considering the horizontal plane passing through A as the reference level, the potential energy of the bob is given by

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18 Fig.

V  mg (CA)  mg (OA  OC )

 mg (l  l cos  )  mgl (1  cos  ). The Lagrangian function L is given by

L  T V 

1 2 .2 ml   mgl (1  cos  ). 2

The Lagrangian equation in the generalized co-ordinate



dt  L  L  0.   dt   .   

…(1)

Equation (1) gives

L .

 With

these

 ml 2  and

substitutions

L 

 mgl sin  ,

equation

(2)

equation

motion

written

d 2 .  ml    mgl sin   0 dt   ..

ml 2   mgl sin   0

  sin   0.

g l

…(3)

Ex: 2(a) Write the Lagrangian function and the equations of motion for a three-dimensional isotropic harmonic oscillator (a) in Cartesian co-ordinates, and (b) in polar coordinates. An isotropic harmonic oscillator may be supposed to be an oscillating particle acted upon by a force which is directed towards the fixed point (position of equilibrium) and whose magnitude varies linearly with the distance from the position of equilibrium. The force F may be represented as

F  kr ,

…(1)

where k is called force constant: But F  or

dV , where V is the potential energy dr

dV  F dr

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 kr dr

from equation (1).

The potential energy V is given by

V=  kr dr+c, where c is constant of integration



kr 2  c. 2

Taking the horizontal plane passing through the position of equilibrium as the reference level i.e., V = 0 at r =0; then from above equation we have c =0. Therefore the potential function or potential energy is given by

V

1 2 kr . 2

…(2)

(c) In Cartesian co-ordinates. The kinetic energy in the three dimensional isotropic harmonic oscillator in Cartesian co-ordinates is

T

. . . 1 m( x 2  y 2  z 2 ) 2

…(3)

The potential energy in Cartesian co-ordinates is

V

1 2 1 kr  k ( x 2  y 2  z 2 ). 2 2

…(4) from equation (2)

The Lagrangian function L is given by

L  T V 

. . . 1 1 m( x 2  y 2  z 2 )  k ( x 2  y 2  z 2 ). 2 2

…(5)

which gives

 L    kx , .  x  x  . L L   m y,  ky,  . y  y  . L L  m z,  kz.  .   z z  L

 m x,

…(6)

The Lagrange’s equations in Cartesian co-ordinates are

d  L  L  0,   dt   x.  x

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d  L  L   0, dt   y.  y   and

d  L  L  0.   dt   z.  z

.. d  .  m x  kx  m x  kx  0    dt    ..  m y  ky  0   .. m z  kz  0   

(b) In polar co-ordinates. Let instant. If

from eq. (6)

(r , ,  ) be the polar co-ordinates of the particle at any

( x, y, z ) the are Cartesian co-ordinates of the particle at the instant; then according to

transformation equations.

x  r sin  cos  ,   y  r sin  sin  , 

…(7)

z  r cos  , The kinetic energy of the particle is

T

. . . 1 m( x 2  y 2  z 2 ) 2

. . 1 .   m r 2  r 2  2  r 2 sin 2   2  , 2  

…(8)

The potential energy of the particle is

V

1 2 kr . 2

The Lagrangian in polar co-ordinates is written as

L  T V . . 1  .  1  m  r 2  r 2  2  r 2 sin 2   2   kr 2 , 2   2

…(9)

which gives

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 . . L 2 2 2  = mr   mr sin    kr , .  r  r  . . L L  2 2 2  mr  , = mr sin  cos    .     . L 2 2 2 L   mr sin   ,  0. .      L

and

 mr ,

…(10)

The Lagrange’s equations in polar coordinates are

 d  L  L  0,   .     dt   r  r     d L L   0,   .  dt           d  L  L   0.  dt   .     

…(11)

Making substitution from equations (10) in (11), we have . . d  . 2 2 2  m r   mr   mr sin    kr  0 dt  

m r  mr  2  mr sin 2    kr  0

…(12)

. . . d 2 2 2 mr    mr  sin  cos   2  0  dt  

…(13)

d 2 2 .  mr sin     0 dt  

and

…(14)

Equations (12), (13) and (14) are required Lagrange’s equation in polar co-ordinates.

1.3 The Hamiltonian Formulation Phase Space. In Lagrangian formulation there are 3n equations of motion of the 2

order for a system

containing n particles in the absence of holonomic constraints. For general consideration it is st

more convenient to write 6n partial differential equation of the 1 order in place of 3n equations of the 2

order. For the purpose there must be 6n degrees of freedom of 6n dimensional space

known as phase space-sometimes also called the  - space. In phase space momenta are also regarded as independent variables like space coordinates. A single particle in phase space is

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specified by six coordinates, 3 position co-ordinates, 3 position co-ordinates and 3 momentum coordinates. This six dimensional space is sometimes called the -space. The  - space is a superposition of -space. In brief phase space is a 6n dimensional space formed of co-ordinates q1, q2, …, qk, …, q3n; p1, . p2, …, pk, …, p3n. 1.3.1 The Hamiltonian Function H. The Lagrangian, in general is the function of .

q1 ,q 2 ,...,q k ,...,q1 ,...q 2 ,..., qk , and t, .

i.e.,

L = L(q1 ,q 2 ,...,q k ,...,q1 ,...q 2 ,..., q k ,..., t) L ( q k , q k , t) dL L . L L  q k   . qk  . dt t k qk k q

…(1)

The Lagrangian equation is

 d  L dt   q.  k

L

 qk



   L  0    qk

 d  L dt   q.  k

 .  

…(2)

With this substitution equation (1) becomes

 dL d  L  . dt k dt   q  k =

. ..  q k   L q  L .  t k q k 

  d  L .  L q k dt  . k   t    qk  

  d  L .  L L  q  k . k   t . dt   qk   

i.e.

But

L .

 qk

…(3)

= pk generalized momentum.

The equation (3) may be written as . d   L L  p q   k k dt  k  t

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. d  L   pk q k  L    . dt  k t 

…(4)

Now let us introduce a new function H known as Hamiltonian function defined as .

H   pk q k  L(q, q, t ).

…(5)

It will be seen later on, that the function H is equal to the total energy in most of the physical problems. From equation (5) it is seen that Hamiltonian H is the function of p, and t, i.e.,

…(6)

H = H (p, q, t).

1.3.2 Hamilton’s Equation As seen above the Hamiltonian function H is function of p, q and t, i.e.,

H = H (p, q, t). = H(p1, . p2,…, pk, …, q1, q2, …, qk,…t) dH =

H

 p k

dpk  k

H H dqk  dt. qk t

…(1)

Also from equation (5) of § 1.3.1, .

dH   q k dpk  pk d q k  dL. k

Since 

…(2)

L=L(q1 , q 2 , ..., qk , ..., q1 , q2 , ..., qk , ...,t), dL   k

 k

L  qk

L .

 qk

dqk   k

L .

 qk

dqk  pk d q k 

d qk 

L dt t

…(3)

   Since pk  L  .   q k   Substituting the value of dL from eqn. (3) in (2), we have .

dH   q k d p k  pk d q k   k

L .

 qk

  qk d pk   pk d qk  k

d q k   pk d q k  k

L dt t

L dt t …(4)

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24   .  Since p k  L form eqn. (2) of § 1.3.1 .    q k  

Comparing coefficients of dpk and dpk and dt in equations (1) and (4), we get

H

 ,  pk   . H   pk    qk 

…(5)

L H  . t t

…(6)

qk 



Equations (5) are referred as Hamilton’s equations or Hamilton’s canonical equations of motion. In the case of systems of n particles there are 6n partial differential equations of the first order in terms of rectangular coordinates in place of 3n Lagrangian equations of the 2

order.

Physical Significance of the Hamilton’s Function Since Hamiltonian function H is function of p, q and t, i.e.,

H=H(p1 , p2 , ..., pk , ..., q1 , q2 , ..., qk , ...,t),

the total time derivative of H is

dH H . H . H  qk   qk  dt t k q p k

  pk q k   q k q k  k

H t

[from equations (5) of § 1.3.2] =

H t

…(1)

But according to equation (6) of § 1.3.2,

Therefore

H L  . t t

…(2)

dH L  dt t

…(3)

If L is not an explicit function of time,

L  0 , thereby giving t

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H  0, dt i.e.,

…(4)

H = constant. Thus we may state that if the Lagrangian L is not an explicit function of time, the

Hamiltonian H is constant of motion. For conservative systems, the potential energy V does not depend upon generalized velocity, i.e.,

V .

 0.

…(5)

 qk Also we know that .

H   pk q k  L  k

L .

 qk

qk  L

 .       q k  . (T  V )   L k     qk 

(Since L=T-V)   V   T   qk  .  .   L k    qk  qk   .

T

  qk k .

  qk k

L

[from equation (5)]

 qk  1 .  m qk2   L .   q  2 k

1 .   Since T  m qk2   2 

  m qk2  L k

= 2T – L = 2T – (T-V) = 2T – T + V =T+V = K.E. + P.E. = E = Total energy of the system. Thus for conservative system, where the co-ordinate transformation is independent of time, the Hamiltonian function H represents the total energy of the system.

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Cyclic Co-ordinates and Routh’s Procedure. It is to be noted that the Hamiltonian equations are specially used to solve the problems involving cyclic co-ordinates. A co-ordinate qk is cyclic if does not appear in the Lagrangian. From Lagrange’s equation, we have

 d  L dt   q.  k

   L  0    qk

 d  L dt   q.  k

   L    qk

d L  pk   dt  qk    Since pk  L  generalised momentum corresponding to generalised co-ordinate q k  .    q k   .

pk 

or .

L qk

qk  

Also .

pk 

H qk

[from eqn. (5) of § 1.3.2]

L H  . qk qk

…(1)

Therefore if qk is cyclic, i.e. if qk is absent from the Lagrangian L, it will also be absent from the Hamiltonian H. Therefore if any co-ordinate qk is cyclic,

H =0. qk

pk = 0

[from eqn. (1)]

pk = constant =  (say).

…(2)

Then the Hamiltonian may be expressed as

H=H(q1 , q2 , ..., q k-1 , q +1 , ...q n , p1, p2 , ..., pk-1,  , p+1...pn ,t). .

Also

qk 

H pk

[from eqn. (5) of § 1.3.2]

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H 



[from eqn. (2)]

…(3)

To solve the problem involving cyclic and non-cyclic co-ordinates Routh devised a method by combining the Lagrangian and Hamiltonian procedures, being known as Routh’s procedure after his name. If q1 , q 2 , ...,qs are cyclic co-ordinates, then a new function as Routhian is defined by the relation .

R(q1 , q 2 , ...q k ,...,q n , p1, p 2 , ..., ps , qs+1 ..., q k ,... q n ,t) s

  pk q k  L.

…(4)

k 1

Here we have considered the problem involving n co-ordinates. The differential of R is k

k 1

L

dR   q k dpk  pk d q k  k 1

k 1

 qk

L L dqk  dt t k 1 qk n

d q k 

 s  n L . L .   dR   q k dpk  . d qk   . d q k   d qk  . k 1 k 1  q k 1  q k  s 1  p   k k k   s

L

 L L L dpk  dt  Since pk  . t  k 1 qk  qk  n



 dR   q k d pk   k 1

which yields

R  qk

k  s 1

 qk ,

R  qk

L .

 qk

d qk   k 1

L .

 qk

d qk 

L dt , t

 .  

…(5)

  q k for k = 1 to s

…(6)

 .   Since L  pk  .    q k   and

R .



 qk

L

R

 qk  qk



L

for k = s+1 to n.

…(7)

 qk

Equations (6) represent Hamilton’s equations with R as Hamiltonian while equations (7) point out that the co-ordinates form k = s+1 to n form Lagrangian equations with R as the Lagrangian function, i.e..,

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   R  0 for k = s+1 to n. .   q k 

…(8)

A co-ordinate which does not appear in the Lagrangian, will be absent from the Routhian R. Here d1, d2, ….ds are cyclic, therefore they will not be involved in the Routhian R and the momenta p1, p2, …ps conjugate to cyclic co-ordinates q1, q2, …, qs, respectively are constants hence may be replaced by a set of constants 1, 2, …s to be determined by boundary conditions. Keeping in mind this fact the Routhian R may be stated to involve (n-s) non-cyclic coordinates as variables and the generalized velocities corresponding to these (n-s) co-ordinates. Thus R may be represented as

R=R(qs+1 ..., qn , ps+1,... pn , 1 ,  2 , ...s ,t), thereby indicating that the Lagrange’s equations for non-cyclic co-ordinates can be solved as if the cyclic co-ordinates are absent and for cyclic co-ordinates we have Hamiltonian equations as usual. Thus according to Routhian procedure a problem involving cyclic and non-cyclic coordinates can be solved by solving Lagrangian equations for non-cyclic co-ordinates with R as the Lagrangian function and solving Hamiltonian equations for given cyclic co-ordinates with R as Hamiltonian function.

Applications of Hamiltonian equations. We have seen that dynamical problems can be solved easily using generalized co-ordinates and Lagrange’s equations. In fact there is hardly a mechanical problem where generalized coordinates are not applicable. For the solution of problems Hamiltonian equations are generally not so convenient as the Lagrangian equations and for their relation to more advanced mechanics, particularly to three fields; celestial mechanics, statical mechanics and quantum mechanics. Here we shall consider some examples concerned with the dynamics of the particles to illustrate the applications of Hamiltonian equations. Example 1. Motion of particle in central force field or planetary motion. or Write down the Hamiltonian and Hamilton’s equation for a particle in central force field in space. A particle moves in the XY-plane under the influence of a central force depending only on its distance from the origin. (a) Set up the Hamiltonian for the system.

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(b) Write Hamiltonian equations of motion. If (r, ) are the polar co-ordinates of the particle of mass m Lagrangian L is, L = T – V(r)

V(t) being potential energy due to central force.

. . 1 = m(r 2  r 2  2 )  V (r ). 2

…(1)

Then we have

L

  r   2 . . L p  .  m r      pr 

 mr

…(2)

H   pk q k  L .

 pr r  p   L .

=(mr) r  (mr 2  )  L . . . . 1  m r 2  mr 2  2  m(r 2  r 2  2 )  V (r ) 2



from (3) from (1)

. . 1 m(r 2  r 2  2 )  V (r )  T  V (r ) 2

… (3)

 total energy. .

r

From (2) we have

p pr . ,   2 m mr

With these substitutions the Hamiltonian takes the form [see equation (3)] 2 2 1   p    p   H  m  r   r 2   2    V (r ) 2   mr    m  

p 2  1  p  2    V (r ). 2m  r r2 

… (4)

This is the total energy expressed in terms of co-ordinates and momentum. The .

qk , are q k 

Hamiltonian

H .

equations,

corresponding

generalized

co-ordinates

 pk

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 pk 

H

 qk

In this case we have two co-ordinates r and ; therefore there will be four Hamilton’s equations. The two equations for qk are:

H pr   , pr m   . po  H   po mr 2  .

r

... (5) from (4)

The two equations for pk are

H p 2 V (r )  pr     , r mr 3 r   . H  po    0.  

… (6)

The last equation shows that the time rate of change of angular momentum is zero, i.e., angular momentum is conserved in planetary motion. Example 2: Write the Hamiltonian and equation of motion of a simple pendulum. 2 2

Referring to Ex. 1(a) in topic 1.2.1, the kinetic energy of the bob is T = ½ ml θ The potential energy is V = mgl (1-cos). The Lagrangian function L is given by L=T–V



1 2 . ml   mgl (1  cos  ). 2

Then we have

p 

L .



… (1)

 ml 2  .

… (2)

The Hamiltonian function H is .

H   pk q k  L .

 p   L . . 1   ml 2  2   ml 2  2  mgl (1  cos  )  2 

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

1 2 .2 ml   mgl (1  cos  ) 2

… (3)

= T+V = total energy. .

From equation (2),



p . ml 2

With this substitution equation (3) becomes 2

1  p  H  ml 2  2   mgl (1  cos  ), 2  ml 

Which yields

p 2   mgl (1  cos  ), 2ml 2

… (4)

p H H  2 ,  mgl sin  , p ml 

… (5)

The Hamilton’s equation for





p are

and

H , p

p  

H 

p   2 ml  p  mgl sin  . .



… (6)

These equations represent Hamilton’s equations for a simple pendulum. From first equation of (6), we have .

p  ml 2  , .

y  ml 2  .

With this substitution second equation of (6) becomes .

g l

  sin   0.

This represents the equation of motion of the simple pendulum and is exactly the same as obtained in Ex. 1(a) in topic 1.2.1 using Lagrange’s equations. Example 3. Write the Hamiltonian function and equation of motion for a compound pendulum. The Lagrangian L for compound pendulum is

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L = ½ I  2  mga cos  . Then, we have

… (1)

p 

L .



 I.

… (2)

The Hamiltonian function H is given by .

H   pk q k  L .

 p   L .

 I  2  ½ I 2  mgh cos 

[from (1) and (2)]

½ I  2  mgh cos  . .

From equation (2),



… (3)

p . I

With this substitution (3) becomes 2

p  H  ½ I     mgh cos  ,  I  

Which gives

p 2  mgh cos  , 2I

… (4)

H p H  ,  mgh sin  , p I  .

The Hamilton’s equations for .

 .



… (5)

and

p are

H , p

p  

H 

.  y    I  p  mgh sin  . .

… (6)

from eqn. (5)] From first equation of (6), we have .

p  I 

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p  I  .

With this substitution second equation of (6) becomes ..

I   mgh sin  ..



mgh sin   0. I

… (7)

But I = mk , k being the radius of gyration. With this substitution equation (7) or equation of motion is .



gh sin   0. k2

TWO BODY CENTRAL FORCE PROBLEM Introduction The development of any branch of physics is based on certain laws which, in turn, include certain concepts of fundamental nature. The development of Classical Mechanics is based on well known Newton’s three laws (i) the law of inertia or Galilio law (ii) the law of force (iii) the law of action and reaction. These laws include the concepts of absolute mass, absolute space and absolute time. The classical mechanics explains correctly the motion of celestial bodies like planets, stars and macroscopic and as well as microscopic terrestrial bodies moving with non-relativistic velocities (i.e. v < < c; c being the speed of light). The other class of mechanical phenomenon such as the non-relativistic motion of atoms, molecules, electrons, protons etc., is explained more correctly by Quantum Mechanics and the relativistic motion by Relativistic Quantum Mechanics. In this chapter we shall discuss the development of Classical Mechanics.

The Two-Body Central Force Problem (a) Reduction of Two-Body Problem to One-Body Problem A two-body system can be effectively reduced to one-body system by introducing the concept of reduced mass. Suppose a system is composed of two masses m1 and m2, then for an inertial observer the relative motion of these masses may be expressed by a fictitious particle of reduced mass . Let the instantaneous position of masses m1 and m2 be represented by r1 and r2 relative to an arbitrary origin 0. Let us assume that no external force acts on the system. Let F12 be the force experience by m1 due to m2 and F21 be the force on m2 due to m1 due to mutual interaction. Then equations of motion of particles are

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d 2 r1 F12  m1 2 dt i.e.

and

d 2 r1 F12  dt 2 m1

d 2r2 F21  m2 2 dt and

d 2 r2 F21  dt 2 m2

From these equations; we get

d 2r2 d 2r1 F21 F12  2   dt 2 dt m2 m1 But from fig. r2  r1  r

Fig.

d 2r F21 F12   dt 2 m2 m1 But according to Newton’s III law,

F21   F12  F

(say)

 1 d 2r 1  F  F   2 dt m m   2 1

 where

1 1   m1 m2 

Thus



F

m1m2 m1  m2

d 2r dt 2

is called the reduced mass of two particles.

…(1)

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This equation represents a one-body problem because it is similar to equation of motion of a single particle of mass



at a distance r from mass m2, considered as fixed origin of

inertial frame. Thus the two-body problem is reduced to one-body problem and the relative motion is represented by the motion of a fictitious particle of mass



placed (at m2), acted on by the

same force F.

1.4 (b) The equation of motion and first integrals Let us consider only the central force, where the potential V is the function of r only, so that the force is always directed along r. Let a single particle move about a fixed centre of force which we assume to be the origin of co-ordinate system. Using polar co-ordinates (r,

 ), the

kinetic energy of particle is given by

T  12  (r 2  r 2 2 ),  The potential energy

being reduced mass.

V = V(r).

The Lagrangian of the system is given by L=T–V

 12  (r 2  r 2 2 ) V (r ) As



…(1)

is cyclic co-ordinate, so that its conjugate angular momentum

p ,

which is

conserved, is given by

p  .

p 

so that

L   r 2 , 





…(2)

constant = J (say),

…(3)

d  r 2  0 . dt

Integrating, we get

 r 2 

where J is first integral and represents the magnitude of angular momentum. As



is constant, equation (2) gives

 

d 2 r  0 dt d dt

so that

1 2

r 2



1 2



r 2  0

= constant

…(4)

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The term

1 2

r 2

represents the areal velocity, i.e., the area swept out by the radius

vector per unit time. This can be verified from fig.

Fig. If vector r rotates by an angle

d

in time dt, the area swept out by r in time

dt  12 r   r d   dA (say), so that

dA 1 2 d 1 2  r  r dt 2 dt 2 From equation (4),

dA 1 2  r dt 2 Equation (1) gives

…(5)

constant

L  r r L V   r 2  r r

…(6)

The Lagrangian equation in terms of r is given by

d  L  L    0 dt  r  r

d V   r    r 2   0 dt r

…(7)

from (6) If we represent the force along r by F(r), then we have

F (r )  

V , r

so that equation (7) can be written as

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 r   r 2  F (r ) .

…(8)

This is the general equation of motion. But from equation (3),



J . r 2

2 

J2 ,  2r 4

So that equation (8) gives

J2  r  3  F (r ) r

…(9)

This is second order differential equation in r only Equation (9) gives

J2  r  3  F (r ) , r J2 V r  3  r r   J 2  V    r   r 2  r 1 2

r  

   1 J2  2 2 V  r   r 

…(10)

Multiplying both sides of this equation by r , we get

   1 J2  rr    2 2 V  r r   r 



r 2   

 d  1 J2  2 2 V  dt   r 

d dt

 d  1 2 1 J2  2  r  2 2 V   0 dt  r 

1 2

J2 r   V  cons tan t r 2 2

1 2

…(11)

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.  .  K .E.  T  12   r 2  r 2  2   

But

 .2 J 2   r  2 2   r   1 2

 12  r 2  12

J2 r 2

and potential energy = V and total energy

E  T V J2  r  V. r 2 . 2

1 2

…(12)

1 2

From (11) and (12), we have .

 12  r 2  12

J2  V  E  cons tan t , r 2

…(13)

i.e., the total energy of the system is constant, i.e., the total energy E, is constant of motion. This is another first integral of motion. From equation (13), we have .

 r2  E 

1 2

J2 V 2 r 2

. 2   J2 r   E   V  2 2 r   

2   dr J2   E   V  2 dt 2 r   

2  J 2  E  V     2  r 2   

 dt.

…(14)

Let the initial value of r be r0 ; then integrating equation (14), we get



dr 2  J 2  E  V     2  r 2   

  dt 

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t

…(15)

2  J 2    E  V  2 r 2   

This equation gives t as a function of r. However, from this equation we can find r as function of t and the constants. From equation (3), we have

J dt. r 2

d  If initially

  0 ,

then integration of above equation yields 

J dt 0 r 2 t

 d   0

J dt 0 r 2

  0  

 

This equation gives



J dt  0 r 2

…(16)

as function of t.

Equations (15) and (16) are the only integrations to be solved. Therefore, the problem have been reduced to quadratures with four constants of integration E, J , r0 ,0

(c) The differential equation for the orbit To obtain the equation of the orbit, we simply eliminate the parameter t, i.e., an equation, representing the dependence of r upon

 , is found.

From equation 3(b) of , we have .

J  r 2  d dt

J  r 2

Jdt   r 2 d The corresponding relation between the derivative relative to t and

d J d  2 dt  r d

…(1)



can be written as …(2)

The relations (1) and (2) are used to find the differential equation for the orbit from equation (9) of 3(b). From equations (2) the second derivative w.r. to t is given by

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d2 J d  J d   2 . 2 dt  r d   r 2 d  The equation (9) may be written as

J d  J dr  J 2  2  F (r )    r d   r 2 d   r 3 J d  J dr  J 2  F (r ).   r 2 d   r 2 d   r 3

…(3)

To simplify above equation we must remember that

d 1/ r  1 dr  2 r d d

…(4)

Using (4), equation (3) gives

J 2 d  d 1/ r   J 2  F (r ).    r 2 d  d   r 3

…(5)

1 u . r

Substituting Equation (5) gives



J 2u 2 d 2u J 2 3 1  u  F  . 2  d  u

 J 2u 2  d 2u 1  u   F  .  2   d u 

…(7)

This is the differential equation for the orbit if the force law F is known. Ex. 1

A particle describes a circular orbit under the influence of an attractive central force

directed towards a point on the circle. Show that the force varies as the inverse fifth power of the distance. Sol.

The equation of a circle of radius a passing through the origin, in polar co-

ordinates, is given by

r  2a cos . Putting

1 sec  r  , we have u  . u 2a

which gives,

du sec  tan   d 2a

and

d 2u sec  sec2   sec  tan  tan   d 2 2a

…(1)

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sec3   sec  tan 2   2a

…(2)

From equation (7) of § (c), we have

J 2u 2 1 F   u 

J 2u 2  sec3   sec  tan 2  sec    from (1) and (2)   2a 2a 



J 2u 2 sec3   sec  (tan 2   1)  2a 



J 2u 2 sec3   sec  sec2   2a 



J 2u 2 .2sec3  2a 



J 2u 2 .sec3  a



J 2u 2 .8a3u 3 a



8J 2a2



8J 2a2 1 . ,  r5



F(r) 

i.e.

 d 2u   d 2  u   

from (1)

.u 5

1 , r5

i.e., the force varies as the inverse fifth power of the distance.

1.5 (d) The Kepler problem: Inverse square law of force The inverse square law is most important of all the central force laws. It results in the deduction of Kepler’s laws of planetary motion. The Kepler’s laws of planetary motion are: (i)

All planets move in elliptical orbits having the sun as one focus.

(ii)

The areas swept out by the radius vector of planet relative to the sun in equal times are equal.

(iii)

The square of the period of revolution of any planet about the sun is proportional to the cube of the semi-major axis.

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Deduction of Kepler’s Laws Kepler’s first law. Dividing equation (14) by equation (3) of § 1.4 (b), we get

 J 2  2   E  V    2  2 r      

dr  r 2  d J

Jdr

d 

…(1)

2  J 2  E  V     2 r 2    

r 2

Under inverse square law of force, we have

F (r )  

k , r2

But

F (r )  

V r





where k is a constant.

V k  2 r r

or the potential energy V  

k . r

…(2)

Substituting this value of V is equation (1), we get

d 

2  k J 2  E      r 2 r 2    

r 2

Integrating, we get

 

where



Jdr 2  k J   E   r 2 r 2    

r 2

 ,

is constant of integration.

Substituting r=1/u, we get

  

Jdu 2  J 2u 2      E  ku   2    

   



du  2 E 2 ku 2  2  2 u  J  J 

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   

2   2 E  2 k 2   k    2  4    u  2   J   J     J

u

    cos 1

k J2

 2 E  2 k 2   2  4  J   J

J 2 1 k

    cos 1

 2 EJ 2   1  2  k 

J 2 1 k

 2 EJ 2   1  2  k 

  cos     

 2 EJ 2  J 2 1    1  cos      2 k  k 

u

k 

  2 EJ 2    1   1 cos         J2  k 2    

substituting and

…(3)

k

  J   2 EJ 2     1   ,  k 2    c

equation (3) gives

u  c[1   cos(   

1  c[1   cos(    r

which is the equation of the conic with



…(4)

…(5)

as eccentricity and one focus as the origin. Thus the

equation of the path of the two-body problem of reduced mass



is always a conic section,

which is the generalization of Kepler’s first law. The nature of the conic depends on the value of eccentricity given by eqn. (4). If

  

i.e., if

 2 EJ 2   1  1 or E  0, the conic is hyperbola.  2  k 

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  1,

i.e., if

 2 EJ 2   1  1 or E  0, the conic is a parabola.  2  k 

  1,

i.e., if

 2 EJ 2   1  1 or E  0, the conic is an ellipse.  2  k 

  0,

i.e., if

 2 EJ 2  k 2 or  1  0 E   , the conic is a circle.   2 2J 2  k 

In the case of elliptical orbits, when

      r=r2 =perihelion, then from eqn. (5), we have

1 . c 1   

r1  when

       r=r2 =aperihelion,

then eqn. (5) gives

1 , c 1   

r2 

The semi-major axis, which is one-half the sum of perihelion r1 and aphelion r2 is given by

a



r1  r2 2

1 1 1     2  c 1    c 1    

Substituting values of c and

a





k 

 2 EJ 2   1   1   J2    k 2  

k 2E

E

from eqn. (4), we get

k . 2a

…(6)

This shows that in the case of elliptical orbits the total energy depends solely on the major axis. Deduction of Kepler’s II law From eqn. of (b), we have

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dA 1 2 .  r   cons tan t , dt 2

…(7)

which represents the areas velocity, i.e., the area swept out by the radius vector per unit time is constant. This means that the areas swept out by the radius vector in equal times are equal times are equal which is Kepler’s II law. Deduction of Kepler’s III law. If T is the periodic time of describing the complete orbit, then the area of the orbit is given by

A



. T 1 dA dt   r 2  dt 0 2 dt

J dt 2

JT . 2

But area of the ellipse

(Since J  ur 2 ) …(8)

A   ab,

…(a)

where a and b are the semi-major and semi-minor axes of the ellipse respectively.

b  a 1  2

Also

 2 EJ 2   a 1  1   k 2    2 EJ 2   a  . 2   k  E

But

Therefore

k from equation (6). 2a

 kJ 2  ba  2   a k 

 J2   a1/ 2  .  k  Substituting value of b in equation (9) we get

 J2  A   a 3/ 2  .  k 

…(10)

Comparing (8) and (10), we get

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2 JT 3/ 2  J  a  . 2  k  2 J 2T 2 2 3 J   a . 4 2 k

Squaring,

T 2  4 2 a3

so that

T 2  a3 ,

 k

…(11)

i.e., the square of the period of revolution of the planet around the sun is proportional to the cube of semi-major axis, which is Kepler’s III law. Ex. 2. A charged particle is moving under the influence of point nucleus. Find the orbit of the particle and the periodic time in the case of an elliptical orbit. Sol. Let z be the atomic number of the nucleus and e the charge on the electron; then the central force between the nucleus and any electron in orbit is given by

F (r )  

k  ze2

so that and

ze2 ; r2

eccentricity

But



  1 

E



 2 EJ 3  2 EJ 2   1  .   2 4  k 2   z e 

k ze2  ; therefore 2a 2a

 2( ze2 / 2a) J 2    1    z 2e4    J2   1  , 2  a  ze   i.e.,

  1. Therefore equation (5) represents an elliptical orbit, i.e., the path of an electron moving

under the influence of point nucleus is an ellipse. The period T, from equation (11), is given by

T 2  4 2 a3 Here,

 k

k  ze2 .

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

T

 ze2

2   a3   . e  z 

Ex. 3. Use Hamiltonian equations to find the differential equation for planetary motion and prove that areal velocity is constant. Sol. The Lagrangian L is given by

L  T V .



1  .2 2 .2  k  r r   . 2   r L

…(1)

  r  .  L  p   r 2  . .  

But

and

 pr   r

…(2)

The Hamiltonian H is given by .

H  pr r  p   L . . . 1  .  k  pr r  p      r 2  r 2  2    .  r 2  . . . 1 . 1 k  pr r  p    r 2   2  2  2 2 r



pr 2



H



which yields

and



p 2 pr 2 p 2 k     r 2 2 2 r 2 r

p2 k pr 2   2 , 2 2 r r

p  H pr H  ,  2 , pr  p  r   p 2 H k   2  3 r 2 r r   H  0,   

from equation (2)

…(3)

…(4)

According to Hamiltonian’s equations

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H . r pr

   . H    pr  r 

and

…(5)

From (4) and (5), we get

H pr .  r pr 

   2 .  p H k    3  2   pr  r 2 r r 

and .

pr   r , so that

…(6)

pr   r,

From equations (6) and (7), we have ..

 r  

p 2 k  r3 r 2 2

 2 .  r   ..   k  r    3 r r2

 r   r  2 

i.e.,

r  r 2 

Since

p0   r 2 

k , r2

k . r2

…(8)

But

r 2   h(say ) r2

i.e.,

Let

d  h. dt

u

1 r

…(10)

Then

d h  2  hu 2 dt r

and

dr dr d d 1 2  .    .hu dt d dt d  u 



…(11)

1 du 2 hu u 2 d

 h

du d

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and

d 2r d 2u d  h 2 . dt 2 d dt

r  h 2 u 2

d 2u d 2

…(12)

Substituting values from (9), (10), (11) and (12) in (8), we get

 h 2u 2

d 2u 1 2 4 k 2  hu  u d 2 u 

d 2u k  u  2 2 d h



d 2u k u  2 d  h2

…(13)

This is the differential equation for planetary motion with u 

From equation (4).

1 . r

H  0. 

…(14)

From Hamilton’s equation, . H   p . 

…(15)

Comparing (14) and (15),

p  0

d  p   0 dt

i.e.,

p  co ns tan t

 r 2   cons tan t

1 2 . r   cons tan t 2

Since

  cons tan t

which represents that the areal velocity is constant.

1.6 Scattering in a central force field; Rutherford scattering. In order to discuss the scattering of -vely charged particle from the heavy nucleus, we assume; (i)

The heavy nucleus and the positively charged particle may be supposed to be point nucleus so that their dimensions are not taken into account.

(ii)

The nucleus of the atom is so heavy that it is at rest during collision.

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The mass of the +vely charged particle may be taken as constant because the velocity of +vely charged particle is very small as compared with the velocity of light.

Fig.1.6 Let a positively charged particle of charge

ze approach a heavy nucleus N of charge ze.

As both the charges are positive, there will be a force of positive, there will be a force of repulsion between them given by Coulomb’s inverse square law. This force of repulsion increases rapidly as the particle gets closer to the nucleus. The positively charged particle of initial velocity

0

repelled by the heavy positive nucleus and changes from a straight line to a hyperbola PAQ having one focus at N as shown in fig.. The asymptotes PO and OQ give the initial and final directions of the particle. As the initial and final directions of motion of the particle are not the same, the particle is said to be scattered. The perpendicular distance of PO from N=MN=p. This is the shortest distance from the nucleus to the initial direction of motion of the particle.

This distance is called the impact

parameter. Orbit of the particle. The equation of the orbit is given by

1  c[1   cos(   )] r

where

k

  J   2 EJ 2     1   k 2    c

…(10)

from equation (4) of § (d)

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Here

F

zze2 k  2 2 4 0 r r

In this case 

k 

zze2  zze 2 , so that c  , -ve sign indicating that the force is repulsive. 4 0 4 0 J 2

equation (1) becomes

1  zze2  [1   cos(   )] r 4 0 J 2 If the initial line set such that

…(11)

    , then equation (11) gives

1  zze2  [1   cos  ]. r 4 0 J 2

…(12)

Substituting value of k in equation (10), we have

 2 EJ 2 4 2   0    1  2 2     zz  e     or

  4 0 2 2 EJ 2    1     z2 z 2e4  

If the initial velocity of the particle is

E or

…(13)

0 , then we have

1 0 2 2

0 

 2 E  .

…(14)

According to the principle of conservation of angular momentum, we have .

0 p   r 2   J or

0 

J . p

…(15)

Comparing (14) and (15), we have

J  p or

Jp

 2 E 

 2 E  .

…(16)

Substituting value of J from (16) in (13), we get

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

2    4 0  .2 E p  2 E    1   z 2 z 2 e 4  

   8 Ep 2 0  1    zz e2  



    

   

…(17)

From this equation it is clear that

  1. Therefore equation (12) represents a hyperbola. Angle of Scattering. The angle between initial and final directions of the positively charged particle is called angle of scattering, i.e., the angle between the asymptotes is called the angle of



scattering. In fig. 1.6,

is the angle of scattering

  2  

From fig.1.6







 . 2 2

…(18)

The asymptotic directions are those for which r is infinite and then

 ,

so that

equation (12) gives

1   cos   0 or

cos   

…(19)



Using (18), equation (19) gives

1    cos        2 2





sin

cos ec

 2



  .

Squaring, we get cos ec

 2

2

  8 0 Ep 2  1  cot  1     zze2  2  

cot

 2





8 0 Ep zze2

…(20)

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  2cot 1

8 0 Ep zze2

…(21)

Above equation gives angle of scattering in terms of impact parameter p, energy E, the charge on the nucleus ze and the charge on the particle

ze .

Rutherford-scattering cross-section. The scattering cross section in any given direction is defined as

incident no. of particles scattered int o solid angle d per unit time   ( )d incident int ensity where

d is the element of solid angle in the direction of  and  ( ) is often known as

differential scattering cross-section. The incident intensity is defined as the number of particles crossing unit area normal to the incident beam in unit time. But the differential of solid angle

  d

d in the plane whose azimuth lies between  and

d  sin  d d .

where scattering is being considered through angle The

scattering

cross-section

through



and

  d.

angle

2

 ( )d    ( )sin  d d  2   sin  d . 00



any

plane

given

…(23)

If I is incident intensity, the number of particles scattered into solid angle

d per unit

time

 2 I   sin  d

…(24) from (22) and (23),

The cross-section of the particles having a collision parameter between p and P+dp is the area of the ring of radius p and width dp, so that

  p  dp  2 pdp Therefore the number of particles with impact parameter lying between p and p+dp =

I .2 pdp. As the number of particles scattered into solid angle

d lying between  and   d is

equal to the number of particles with impact parameter lying between p and p+dp, i.e., from (24) and (25), we have

2 Ip dp  2 I   sin  d.

…(26)

-ve sign is introduced because an increase dp in the impact parameter means less force is exerted on the particle; consequently, scattering angle decreases by an amount

d .

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Eq. (26) gives

    

Eq. (20) gives

p dp  

54 pdp sin  d

…(27)

zze2  cot . 8 0 E 2

zze2  1 cos ec 2 . d 8 0 E 2 2

zze2   cos ec 2 d. 16 0 E 2 Substituting values of p and dp in (27) we get

 zze2    zze2  2    cot    cos ec d 8 0 E  2  16 0 E  2        2sin cos d 2 2 2



 zze2  4     16   cos ec . 2  4 0 E 

…(24)

This is well known Rutherford scattering cross-section. It shows that the scattering crosssection and hence the number of particles scattered must be proportional to (i) cos ec square of the nuclear charge (ze), (iii) the square of the charge on the particle inversely proportional to the square of the initial kinetic energy, i.e.,

 2

, (ii) the

 ze  , and (iv)

1 E2

1.7 Laboratory and Centre of Mass Systems In a collision or scattering process at least two particles are involved. To describe such problem there are coordinate systems. 1. Laboratory coordinate system (or L-system): A frame attached with the laboratory is called laboratory coordinate system.

It is the usual coordinate system because

observations are taken in laboratory coordinate system. This system is abbreviated as Lsystem and in the Rutherford Scattering process the target is initially at rest. 2. Centre of Mass Coordinate System (C-system): A frame attached with the centre of mass of an isolated system of particles is called the centre of mass coordinate system or C-system.

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Let us consider an incident particle of mass m1 having an initial velocity u1 and the other particle of mass m2 at rest at O in L-system. Let r1 and r2 be the position vectors of m1 and m2 relative to a coordinate system fixed in laboratory. Let v1 and v2 be velocities of

m1 and m2 after collision in L-system. As there are no external forces, the linear momentum of the system is conserved. .

Then

Initially

m1 r1  m2 r 2  P(Cons tan t )

r1  u1 and r 2  0 Hence P  m1u1.

Thus

m1v1  m2v2  P  m1u1

…(1)

Let position vector of centre of mass be R. Then

RC 

m1r1  m2 r2 m1  m2

…(2)

Differentiating velocity of centre of mass is given by .

m r1  m2 r 2 m1v1  m2v2 VC  RC  1  m1  m2 m1  m2 .

Using (1), we get

VC 

m1u1 m1  m2

…(3)

Obviously the velocity VC of the centre of mass is along the direction of initial velocity u1 of incident particle. The centre of mass always lies on the line joining m1 and m2 to an observer fixed at centre of mass, the particle m1 would appear to be on the line making an angle

C

with the

direction of VC and relative to centre of mass the position vectors of two masses are (Fig. a)

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R1  r1  RC and R2  r2  RC

…(4)

These equations give .

R1  r1  RC .

R 2  r 2  RC

   or V2  V2  VC 

or V1  VC

…(5)

where V1 and V2 are velocities of two particles after collision in C-system From these equations, we get

mV 1 1  m2V2  m1v1  m2 v2   m1  m2 VC Using (3), we get

mV 1 1  m2V2  m1u1   m1  m2 

m1u1 0 m1  m2

…(6)

This result indicates that the two particles as seen from centre of mass appear to move is opposite directions in such a way that the net momentum in this system is zero. In other words, the velocity vectors of two particles lie along the line joining the particles (Fig. (b)).

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Relations between angles of scattering in Centre of Mass and Laboratory Frames Let

L

be the angle of scattering in L-system and

C

the angle of scattering in C-system

(Fig. (c)).

From Fig. tan

V1 sin C V1 cos C  VC

L 

…(7)

In this formula, we have to eliminate V1 and VC . For this we have

R1  r1  RC  r1  

m1r1  m2 r2 m1  m2

m2  r1  r2  m1  m2

As reduced mass

m1m2 and r1  r2  r , the relative position vector of two m1  m2



particles.

R1  .

R1 



r i.e., V1 

 m1

where V is the relative velocity of two particles. As collision is elastic, the relative velocity of two particles after scattering must be of same magnitude as initial relative velocity, i.e.,

V1 Also from (3),

VC  VC 

 m1

…(8)

m1 u1 m1  m2

 m2

Hence from (7), we get

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   u1  sin C  m1  tan  L        u1  cos C   u1   m1   m2  tan  L 

i.e.,

sin C cos C 

…(9)

m1 m2

Case (i). When m2  m1 ,

m1  1  tan  L  tan C i.e., in this case the scattering m2

angles are same in both the systems. Case (ii). When m1  m2 (as in the case of neutron-proton scattering)

tan  L  giving

L 

sin C   tan C cos C  1 2

C 2

Case (iii).

When m1  m2 or

m1  1, the denominator can never be zero, i.e., m2

tan  L   or  L   / 2. This implies that if a heavy particle is incident on a light particle initially at rest, the heavy particle will not bounce backward as a result of collision. Relations between scattering cross-sections. As

L

and

C

are not equal, the scattering

cross-sections will be different when expressed in two different systems. The relation between scattering cross-sections may be found by the fact that the number of particles within a given solid angle must be same in two systems. If

 C C 

and

 L  L 

are scattering cross-section

in C and L-systems, we have

2 I C C  sin C dC  2 I L  L  sin  L d L where I is intensity of incident particles. This gives

 L  L    C C 

sin C dC . sin  L d L

This is the required relation. The value of

d C may be evaluated using equation (9). d L

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SPECIAL RELATIVITY IN CLASSICAL MECHANICS Introduction The truth of the consequences of Newtonian mechanics is based on the approximation that the velocities of the particles of the system are small compared with velocity of light. When the velocities involved approach the velocity of light of the true description consistent with the experimental facts is provided by special theory of relativity. Our object is to show how the Lagrangian and Hamiltonian description can be altered to conform to the special theory of relativity. Basic Postulates of Special Theory of Relativity: A space system obeying Newton’s law of motion F=ma is called an inertial system. It appears that a system moving uniformly with respect to a ‘space system’ should itself be an inertial system. To show it, consider two co-ordinate systems-one inertial system S and the other S  moving uniformly with respect to S with velocity v (fig. 1.). A given point P is located by the radius vector r and

r with respect to the two systems which are related as r  r  vt ..

giving

…(1)

r   r,

i.e., the acceleration is same in both the systems proving that S  should also be inertial system. Such a transformation represented by eq. (1) is called Galilean transformation.

Fig. 1. Galilean Transformation is inadequate: The important point to note in such a transformation is that this shows the dependence of speed of light on direction. Thus, if we suppose that a light source placed at O, the origin of unprimed system, is emitting spherical waves traveling with the speed c, the velocity of a point on

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60 .

a given wave surface in unprimed system will be

r  cn, where n is the unit vector along r, the

position vector of the point under consideration; but according to eq. (1), in primed system, the .

corresponding wave velocity will be

r   cn  v which clearly shows that the magnitude of wave

velocity changes in the system moving with respect to the source of light and hence for such a system the waves will no longer be spherical. New transformation needed: Such a dependence of speed of light on direction, as mentioned above, was found to be objectionable since no experiment could detect it but, instead, experimental investigations-notably due to Michelson and Morley – indicated that the speed of light is the same in all directions and is independent of the relative-uniform motions of the observer, the transmitting medium and the light source. This led to the conception that Galilean transformation is not correct and should also be replaced by speed of light in all systems. Einstein predicted that the basic requirement for the design of such a transformation is the revision of the concept of time and simultaneity. Further, from the fact that the speed of light should have the same numerical value for all observers so long as they are all resident in inertial frames, he proposed, as a fundamental postulate of the relativity theory, that all phenomena of physics appear the same in all uniformly moving systems. The principle, on the other hand, can be illustrated further by starting that simply by a physical measurement, performed within a given co-ordinate system, one cannot infer that the coordinate system is intrinsically ‘stationary’ or ‘uniformly moving’ respect to other systems. However, a physical measurement in coordinate system can furnish the information that there exists a relative motion between different systems. Thus it all leads to a conclusion that ‘physical law must have the same meaning in all inertial frames,’ i.e., all physical laws must be phrased in an identical manner for all uniformly moving systems. This postulate is called the postulate of covariance of physical law.

According to the postulate, Newton’s equations of motion should also be

generalized to preserve their form in the new transformation – the so called Lorentz transformation. Postulates of Special Relativity: From above discussion, we arrive at the following two postulates of special relativity: (a) The Principle of Equivalence of Physical Laws: According to this principle, all phenomena of physics appear the same in all uniformly moving systems. This implies that the laws of physics are independent of the motion of the particular space-time coordinates to which they are referred; one can only detect the relative motion of one material body with respect to the other and not the absolute motion of bodies through space. (b) Principle of constancy of velocity of light: It states that the speed of light is same in all directions and is independent of the relative-uniform motion of the observer, source of light and the medium.

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Lorentz transformation: *(i) It must be linear, (ii) It should preserve the speed of light, and (iii) It should approach Galilean transformation in the limit of low speeds compared to speed of light. We consider two uniformly moving systems whose origins coincide at time t = 0. Further at t = 0, let a source of light, fixed at the origin of unprimed system, emit a pulse of light. Just as the disturbance in a pound resulting from dropping a pebble into the water is a system of circular ripples, so a pulse of light spreads out as a growing sphere. Thus an observer fixed in the unprimed system will see a spreading spherical wavefront propagating with the speed c, the equation of which can be written as

x 2 +y2 +z 2 =c2 t 2 .

…(1)

But since speed of light is invariant, observer in the primed system moving with respect to the source will also see the pulse of light propagating as the spherical wavefront from his own origin-equation being

x2 +y2 +z2 =c2 t2 ,

…(2)

which shows that time is not more scalar invariant but is a function of a particular coordinate system in uniform motion relative to other system or rather time interval between two events depends on the reference system of the observer. The need of transforming time scale in going from one system to another is also obvious from the fact that no rearrangement of space coordinates along can give wave pulses that are simultaneously concentric spheres in both the systems. Inspection eqs. (1) and (2) reveals that the desired transformation must be such that

x 2 +y2 +z 2 -c2 t 2  x2 +y2 +z2 =c2 t2

…(3)

Writing x1 , x2 and x3 for x, y, and z, the equation becomes 3

x i 1

2 1

 c2 t 2   x12  c2 t2. i 1

If we put x4  ict , a fourth imaginary coordinate, then 4

 1

 x 2   x 2 .

…(4)

This means that the square of the radius vector is invariant under a transformation of coordinates in the four dimensional x1 , x2 , x3 and x4 space. We know (cf. chapter 5) that in three dimensional case, if we rotate coordinate axes, then the components of a vector in new system are given by

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x    a  v xv , v 1

and the length of the vector remains unchanged, i.e., 3

x   x .    2

1

Thus basing on the analogy of above equation which represents spatial rotation of coordinate axes in three dimensional orthogonal space, we can say that eq. (4) represents orthogonal transformation of a vector in four dimensional space or in other words Lorentz transformations are merely the orthogonal transformation of four dimensional space-the latter also being recognized as world space or Minkowski’s space.

Lorentz transformation can,

therefore, be represented as 

x     a  v , xv , v 

where a  v is the linear transformation matrix. We are here concerned only with a transformation that involves uniformly moving systems whose axes are parallel and is called pure Lorentz transformation. Such a transformation does not involve any spatial rotation. Further general Lorentz transformation is a product of a space rotation and a pure Lorentz transformation. Let us consider that primed system is moving with velocity v along x3 axis. We shall not obtain, for pure Lorentz transformation, the matrix elements a  v of the transformation between x and

x ,



x     a  v xv

…(5)

v 

x1  a11 x1  a12 x2  a13 x3  a14 x4 x2   a 21 x1  a 22 x2  a 23 x3  a 24 x4

…(6)

x3   a 31 x1  a 32 x2  a33 x3  a34 x4 x4   a 41 x1  a 42 x2  a 43 x3  a 44 x4 . Obviously, x1  x2  and x3  being space co-ordinates will be real for which a14, a24, a34 should be imaginary (because x4  ict is imaginary). Similarly, for x4  to be imaginary it is essential that

a 41 , a 42 ,a 43 elements be imaginary while a 44 real. As the primed system is taken to be moving along x3 axis, then x1 and x2 , being in a direction perpendicular to that of system motion, will remain unaffected by the transformation. i.e.,

x1  x1.

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x2   x2 . As the components x3 and x4 undergo transformation and hence x1 and x2 should not appear in the transformation matrix of x3  and x4  . Thus eqs. (6) become

x1  x1

x2   x2

…(7)

x3   a 33 x3  a 34 x4 x4   a 43 x3  a 44 x4 . Therefore the matrix of transformation is

1  0 0  0

a 33

a 34

a 43

a 44

     

…(8)

Further these matrix elements have to obey the same sort of orthogonality conditions as for spatial rotation in three dimensional space, i.e.,

 a a v

  

…(9)

 1 if     0 if   . Putting

    3 and first v = 3 and then v = 4, we get a 332  a 34 2  1.

Also, with

  4

…(10)

and v = 3 and 4,

a 432  a 44 2  1. Further a 33a 43  a 34a 44  0.

…(11) …(12)

Thus orthogonality condition furnishes the three eqs. (10), (11) and (12) connecting the four matrix elements. The four unknown elements can be determined uniquely only when fourth relation between them is provided. We know that the origin of primed system (x3    is moving uniformly along x3 -axis; thus after time t its x3 coordinate will be vt, i.e.,

x3  vt  v.

x4 v  i x4  i  x4 ic c

which, with matrix relation for x3  , gives

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  a 34  i a 33  x4  0

a 34  i a 33 ,

which when substituted in equation (10), gives

a 33 

1 (1   2 ) i

a 34 

so that

(1   2 )

On solving eqs. (11) and (12), it is quite easy to show that

a 44 

(1   2 ) i

a 43 

and

(1   2 )

We can now write the Lorentz transformation as

1  x1     x  0  2   x3   0     x4  0  1  x    y   0  or   z   0    ict  0 

1-  -i / 1- 

i /

1-  1-  2

0 0

1-  -i / 1-  1/

   x1    x2      x3   x    4 

i / 1/

1-  1-  2

  x   y     z   ict    

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giving

x  x y  y z - vt

z 

1- 2 

1/ and

ict  x  y  z.

-i

1- 

t 

ict

1-  2

t-(v/c 2 )z

1-  2

Eqs. (13) are known as Lorentz transformation equations. transformation to real coordinate system is possible when 

                  

…(13)

It is obvious that the Lorentz

 1, indicating that one cannot have

relative velocity greater than c, the speed of light. The relativistic law of addition of velocities: Let us consider three frames of reference S,

S and S ; S moving with velocity v1 relative to S and S with velocity v 2 relative to S .

We want to find out the relative velocity of containing

S with respect to S. The transformation equations

S with S, and S with respectively. x  x

  y  y  z - v1t  z   2 2  ...(14) and 1-v / c  1  t-v1/c 2 .z  t   1-v12 / c2  

x  x

  y  y  z - v 2 t  z   2 2  1-v / c  2   ...(15) t-v 2 /c 2 .z  t   1-v22 / c2  

Substituting eqs. (14) in eqs. (15), we get

x  x y  y

z - v1t

z 

1-v

2 1

/ c2 

  v2   

1-v

2 2

  2 2  1-v1 / c   t-v1/c2 .z

/ c2 

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    1  v1v2 / c2   t  v1 +v 2    z   1-v 2 / c 2   1-v 2 / c 2     1    1   1-v22 / c2  t  v1 +v 2    v1v2   z 1  2  (1  v1v2 / c 2 )   c   

1-v

2 1

/ c 2 1-v 2 2 / c 2 

   v1 +v 2    z  t  2   (1  v1v2 / c )     2 2       v1v2   v1 +v 2    1 1  2    1+ 2  c  c2  v1v 2 / c      



where v 

z  vt

1-v

/ c2 

v1 +v 2 (1  v1v2 / c 2 )

and similarly,

t 

t   v+c2  .z (1  v2 / c 2 )

From these equations for

z and t we infer that two Lorentz transformations carried out

in succession are equivalent to one Lorentz transformation. velocity v of

We also note that the relative

S with respect to S is not simply the sum of v1 and v2 but rather v

v1 +v 2 . (1  v1v2 / c 2 )

…(16)

When v1 / c and v 2 / c are small compared with unity, v is very nearly equal to  v1  v2  but as one of the two approaches the speed of light, relation becomes significant.

Further, the

expression for v can be put in the form

 (1  v1 / c)(1  v 2 / c)  v  c 1  , 1  v1v2 / c 2  

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which shows that v cannot become equal to or greater than c, the speed of light, as long as the factors v1 / c and v1 / c are small enough. We, therefore, conclude that it is impossible to combine several Lorentz transformations in one involving a relative velocity greater than c. From eq. (16) it can be seen that if v1  c and v2  c , then

v

cc c 1  c2 / c2

that is, when velocity of light is added to the velocity of light, we obtain velocity of light. Kinematic effects of Lorentz transformation: It is not within our purpose here t discuss in detail the effect of Lorentz equations of length and time measurements in different frames of reference: therefore we shall not enter into physical discussions of these apparent paradoxes. Our chief aim to discuss the kinematic effects of Lorentz transformation is to show that no special privilege be given to any of the uniformly moving coordinate systems. all are completely equivalent to each other. The two consequences of Lorentz transformation, the so called Lorentz-Fitzgerald contraction of length and dilation of time scales, are being briefed as follows: Lorentz-Fitzgerald contraction: Let us consider that a rigid rod at rest is lying along Z-axis in the unprimed system. Its length in its own system is

l  z2  z1 An observer connected with the primed system that is moving uniformly with respect to unprimed system measures the length of the rod by locating simultaneously the position of both end points z1  and z2  in his system at the same time t. Thus from Lorentz equations, we have

z1 

z2 

z1   vt

1-  2

z2   vt

1-  2

so that apparent length is

 z1  z2  or

1-    z   z  2

1-   l 2

Thus, to the moving observer, rod appears to be contracted. The readers should note that we have used here inverse Lorentz equations. It would not be convenient to use direct equations (13) since events (location of end points) that are

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simultaneous in primed system (ends are measured at the same time

t ) will not be simultaneous

in unprimed system as they are at different points z1 and z2 . Time dilation: Suppose in unprimed system, at point z, a clock is located. At time t 2 on this clock an observer fixed in the moving primed system notes a time

t1  

t1  vz/c2

1- 

and at time t 2 the notes

t2  

t2  vz/c2

1-  2

so that the apparent time interval is

t2   t1  

t2  t1

1-  2

since  t2   t1    t2  t1  , it appears to the moving observer that the stationary clock is moving at a slow rate. The effect is called ‘time dilation’. Significant point to note is that if we fix the clock in the moving (primed) system and observer is in unprimed system even then the conclusion is the same, i.e., observer reports that the clock is running slow as compared to his own. Similarly, if the state of affairs be reversed in the case of length measurement of a rod, observer would again report a contraction in length of the rod. Thus we infer that simply by recording the time on a clock fixed in a certain system by an observer fixed in another system, it is difficult to distinguish which one of the systems is stationary or in motion; only a relative motion can be guessed.

These observed facts amount to the

‘principle of equivalence’. Relativistic Generalisation of Newton’s Law’s: Newton’s laws retain their form under Galelian transformation which is found to be incorrect and hence Newton’s laws are inaccurate representation of experimental phenomena. We want to extend the 3-vector law.

Fi 

d  dxi  m  dt  dt 

…(1)

to 4-vector form because above 3-vector form does not satisfy the principle of equivalence, i.e., its form changes under Lorentz transformation. The basic representation for such an extension are that: 1. a fourth component of force and momentum must be introduced, and

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2. the time is no more a scalar invariant and changes under Lorentz transformation. Thus differentiation cannot be performed with respect to time. It is therefore, necessary that a new parameter be found which is a true 4-scalar with respect to which differentiation in four dimensional form may be performed.

Such a 4-scalar is termed as proper time



discussed below. Proper time: Let the position vector of a point in four dimensional space be x - called position four vector. This particle during its motion, describes a path in world space called world line. The differential motion along the world line will involve a change in the position four vector x represented by a four vector dx . The dot product of two four vectors would be a world – or 4 – scalar. Therefore, dot product of dx with itself will be a 4-scalar and hence will remain invariant under Lorentz transformation. We thus define proper time or Lorentz invariant as

1 c2

(d )2  

  dx  . 2

…(2)

To explain the significance of proper time, let us consider a system in which the particle is momentarily at rest. Since the particle is at rest, so any change in the space co-ordinates of the particle in this system is zero i.e.,

dx  dy  dz   but if it rests for, say, time dt then

change in the fourth component will be will be (0, 0, 0,

icdt' . Thus components of the transformed vector dx 

icdt' ) and hence invariant is given by (d )2  

1 c2

  dx    dt 2

d  dt ,

or i.e.,

d is the time interval between two events* (or two world points) of the particle as read on

the clock moving with the particle.

d is therefore, called as particle’s proper time interval or

world time. Further,

(d )2  

1  2 2 2 dx    dy    dz   c 2 dt 2  2   c

 1 d  dt 1  2  c

 dt

1-v

 dt

1- .

1/ 2

2 2 2   dx   dy   dz              dt   dt   dt    

/ c2  …(3)

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If we consider two world points (or two events) and dx be their difference vector defined as

dx  x1  x2  , the subscript 1 and 2 denoting the two points, the magnitude of dx will be

 dx  

 r1  r2  c 2 (t1  t2 ) 2

  dr   c 2 (dt )2 , 2

Therefore,

(d )2  dt 2 

1 2 dr  . 2  c

From eq. (4) we note that

 d 

…(4) can be positive, negative or zero. If

is imaginary and is said to be a space- like interval. Since

 d 

is positive,

is 4-scalar invariant, time

or space like character of the proper time interval between two events is Lorentz invariant. Events separated by time like proper time interval can be ‘connected’ by signals traveling at a speed less than c, the speed of light, while the events that are separated by a space like interval cannot be bridged by any such signal.

Suppose spatial axes are so oriented that spatial

difference dr between two points is along z-axis, it can be written as dz.

Under Lorentz

transformation, the fourth component of difference vector dx will transformation as

v cdt  dz c cdt   1- 2  If dx is space like then

cdt  dz, and it is, therefore, possible to adjust

v v such that c dt equals dz and hence icdt' assumes a c c

value zero, i.e., the time like part of dx is zero. Thus, if the two events are separated by a space like interval in a given reference frame, it is possible to find a Lorentz system moving with velocity v which is less than c, in which the two events occur at the same time (simultaneously) but at different points in space. We shall now proceed to express velocity, force, momentum etc. in four world. (A) World velocity or Four velocity: We know the square of magnitude of a four vector when greater than or equal to zero is space like, while when less than zero is time like: we shall now show that in the case of world velocity, it is time like.

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World velocity u  is defined as the rate of change of the position vector of a particle with respect to its proper time, i.e.,

u 

dx d

with space components

ui 

dxi dxi   d dt 1- 2

…(5)

1-  2

and time components

ui 

ic dxi  d

1- 

…(6)

The magnitude of world velocity is

 u u    

   2  1-    ic

1-  2

   2  1-   ic

1-  2

c 2 1-vi 2 / c 2  vi 2  c 2   1- 2  1- 2   c 2 . Since square of world velocity is less than zero, it is time like and has a constant magnitude. Thus u

is invariant under Lorentz transformation. We can significantly conceive

here the usefulness of the four dimensional language in relativity: that all equations expressed in four vector language will automatically retain their mathematical form under a Lorentz transformation and hence satisfy the postulate of special relativity. (B) Four Force and Four Momentum: Now we proceed to generalize the Newton’s equation of motion

Fi 

d  mvi  . dt

…(7)

which is not invariant under Lorentz transformation. Its relativistic generalization should be a four vector equation, the spatial part of which would reduce to eq. (7) in the limit

 o.

We should

bring the following changes: (i)

Since t is not Lorentz invariant, it should be replaced by proper time τ.

(ii)

m can be taken as an invariant property of the particle.

(iii)

In place of vi , world velocity u  should be substituted, the space part of which, i.e,

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1-  2

reduces to vi for small velocities. (iv)

The left hand side force Fi should be replaced by some four vector K  , may be called Minkowski force. We require that K  should reduce to Fi in the limit of small velocities (

 o ); but it does not mean that spatial components of K  are the components of Fi

[as can be seen from equation (11) that K i , the spatial part of K  , is

reduces Fi to

1- 

which

  o ].

Thus taking into account all these changes, the relativistic generalization of eq. (7) is

d  mu   K  . d

…(8)

The spatial part of above equation can be written as

  2  dt 1-   

 K i 2  1-   mvi

 d dt  

 K i 2  1-   mvi

1- . 2

…(9)

If we continue to use the classical definition of force and define force as being the time rate or change of momentum in all Lorentz systems, then classical force should be defined by

 d F dt  

  2  1-  

 d dt  

  2  1-  

Fi 

mvi

…(10)

Eq. (9), then yields the relation between the space components of the force K  and Cartesian components of classical force F, i.e.

Ki 

1- 

…(11)

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Eq. (10) also yields the relativistic definition of classical linear momentum of the particle, i.e.,

mvi

pi 

1- 

…(12)

which reduces to mvi , the classical linear momentum as

 o.

The time like part of the four vector K  can be obtained from the dot product with world velocity

K  u  u

d d mu    d d

m   u u  2  

d  mc 2     d  2   0, since bracketed term is a constant. Thus

K  u  K i

1-  2

 K4

Fi vi icK 4  0 1- 2  1- 2 

icK 4 F .v   0. 2 1-  1- 2 

1- 

0

This gives the fourth component of Minkowski force as

K4 

i F .v c 1- 2 

…(13)

(C) Relativistic kinetic energy: If, as in the classical case, we define that the time rate of increase of T, the kinetic energy, gives the rate at which the force does work on the particle then:

†F .v 

dT dt

…(14)

and hence eq. (13) becomes

K4 

i dT . c dt 1- 2 

…(15)

From eqs. (8), we can write the fourth component of Minkowski force as

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d  mu4  d



1- 2 

 d dt  

 . 2  1-   mic

…(16)

From eqs. (15) and (16), we find that

i c

1-  2

 dT d   dt dt  

dT  dt

1-  2

 d dt  

  2  1-   mic

 , 2  1-   mc 2

which has the solution

T

mc 2

1- 2 

 T0 ,

…(17)

where T0 is a constant of integration. We can now check whether eq. (14) defining kinetic energy is consistent with the classical result. For small velocities

1- 

2 1/ 2

can be taken as

 1 2 2 1  v / c  and hence  2 

 1 v2  T  mc 2 1   T0 2   2c  Thus T will be classical kinetic energy if T0 is chosen

mc 2 . Notationally we prefer the choice

T0 = 0. Writing again eq. (15),

K4 

d d

 iT   c

 , 

we find that the fourth component of four momentum is

i p4  T c whereas the spatial part of four momentum is given by eq. (12). To express kinetic energy in terms of four momentum, we write the Lorentz invariant of four momentum, i.e., its magnitude

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 p p    



 iT    c  

mvi

1-  2

mvi

1-  2

 iT   c 

m2v 2 T 2  1- 2  c2

 p2 

T2 . c2

Also p p  m u  m c . 2

2 2

Therefore,

m2 c 2  p 2 

T2 c2

T 2  p 2 c 2  m2 c 4 ,

…(18)

which is the relativistic analogue of the relation T  p / 2m in non-relativistic mechanics except 2

that T here includes the rest energy. (D) Relativistic mass: We can preserve the classical definition of linear momentum as the product of mass times the velocity of the particle by defining the mass mr of a 75article, observed to be moving with speed v, by

mr 

1- 

so that the four momentum is

p  mr v,

…(19)

where mr is called relativistic mass of the particle and varies with its speed. Mass m0 is called the rest mass of the particle and is scalar invariant, i.e., unaffected by Lorentz transformation. Using eq. (19), the force, according to Newton’s second law of motion, is

 dp d d F   mr v   dt dt dt    d  m0 dt  

 . 2 2  1-v / c   m0v

  2 2  1-v / c   v

If the velocity varies with time, then

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1/ 2 3/ 2  dv 2v  dv   1 F  m0  1-v2 / c 2   v  1-v 2 / c 2   2   c  dt   2  dt

 m0

  dv 1-v 2 / c 2  v 2 / c 2  dt  1-v 2 / c 2 3/ 2   

dv / dt

1-v

…(20)

/ c2 

3/ 2

Readers may note that it is wrong to write

F

d dv  mr v   mr  dt dt

1-v



dv dt

because here mr has been taken as constant. Instead, it should have been taken as

F



dm d dv  mr v   mr  v r dt dt dt

1-v

 m0

 dv d v dt dt  

/ c2 

  2 2  1-v / c   m0

3/ 2 dv 2v dv  1  m0v  1-v2 / c 2   2   dt c dt  2

/ c2 

m0v 2 / c 2 dv dv  dt 1-v2 / c 2 3/ 2 dt

/ c2 

dv / dt

1-v

/ c2 

3/ 2

which is same as eq. (20). The variation of mass with velocity becomes quite significant as v→c. For v = 0.9 c, m = 2 m0 . That is, increase in the mass is 100%. (E) Mass-Energy Relation: Now we shall write kinetic energy expression in terms of rest mass, m0 . If a particle moves from rest (initial K.E. is zero) and acquires a velocity, v, in time, t, then taking into account the variation of mass with velocity, we write r

t .

T   F .dr   p .

dr dt dt

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  p .v dt  v dp 0

  v d 0  

  2 2  1-v / c   m0v

Integrating by parts, we get

m0v 2



1-v

/ c2 

m0v 2



1-v

 m0 

v dv

1-v

/ c2 

 m0c 2  1-v2 / c 2     0 v

/ c2 

m0v 2  m0c 2 1-v2 / c 2 



1-v

m0c 2

1-v / c 2



 m0c 2

…(21)

 mr c 2  m0c 2   mr  m0  c 2  mc 2

…(22)

Thus the kinetic energy of the particle is equal to the product of the increase in mass, Δm and the square of the speed of light. Note that taking T0  m0c , we can write eq. (21) as 2

T

m0 c 2

1-  2

 T0

which we have already obtained in eq. (17).

1.8 Lagrangian Formulation of Relativistic Mechanics: Now we want to set up a Lagrangian that will furnish the correct relativistic equations of motion.

In chapter 2, the concept of Lagrangian was introduced in two ways; namely D’

Alembert’s principle and Hamilton’s principle.

Since the D’ Alembert’s principle involved the

momentum pi which is now different from non-relativistic momentum mvi , approach to this new Lagrangian formulation is not possible by this way. Hamilton’s principle is still helpful in finding a function L for which the Euler-Lagrange equations, obtained from this variational principle, agree with known relativistic equations of motion. For example, we take the case of a single particle

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acted on by conservative forces independent of velocity and suppose a suitable relativistic Lagrangian for such a case is found to be

1-   V ,

L  mc 2

…(1)

where V purely depends upon positions; then introduction of the Lagrangian in Euler Lagrange equations

d  L  T  0,   dt  vi  xi

…(2)

should yield a relativistic equation. This in fact is true since from eq. (1) we have

1  .2vi / c 2 L  mc 2  2 vi 1- 2 



mvi

1-  2

which when substituted in eq. (2) gives

 d dt  

   V 1- 2   xi mvi

= Fi

 d  d  

  2  1-   mvi

1-  2

 Ki and is the relativistic equation of motion (cf. art 7.3 equation 9). (a) is correct. We have thus shown that the choice of L represented by eq. (1) is correct. We note that L is not longer (T-V), but

L  vi

mvi

1-  2

 pi ,

i.e., partial derivative of L with velocity is still the momentum. Switching to any desired set of generalized coordinates q j , canonical momenta can still be defined as

pi 

L .

qj

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which shows that if a coordinate is cyclic, its conjugate momentum will be conserved. Obviously, this is the same as in non-relativistic theory. Since the definition pf canonical momentum and the general form of Lagrange’s equations are the same as in non-relativistic case, the function .

H  q j pj  L j

would still represent a constant of motion provided L does not contain time explicity – a condition .

also imposed in non-relativistic case. As L is no longer (T-V) nor is

q

p j equal to 2T, we are

to prove in another way that H is also the total energy. In the case of a single particle, H is given by

H  i

mvi 2

1- 

 mc 2

1-   V , 2

which, on soling, reduces to

H

mvi 2

1- 

 V  T  V  E.

For velocity dependent potentials: We consider Lagrangian for a single particle in an electro-magnetic field. The force acting on a particle with charge q is given by

1   F  q  E  v  B  , c   which can be put in the form (cf. Chapter 2)

d   Fi     dt  vi Putting

 pi    

 d dt  

   q    q  v. A   c   xi  

  for relativistic case, we write 2  1-   mvi

   d     1- 2    dt  vi mvi

   q    q  v. A   c   xi  

…(3)

which can be derived from Hamilton’s principle provided L is given by

L  mc 2

1-   q  qc v.A 2

…(4)

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The canonical momentum will have additional terms due to the dependence of potential on velocity along with mui , i.e.,

pi 

L   vi vi



mvi

1- 



 mui 

 2  mc

1-   q  qc v.A 2

q Ai , c

…(5)

q A.i , c

This additional term is also present in the non-relativistic case.

Hamiltonian Formulation of Relativistic Mechanics: The Hamiltonian is expressed as .

H   qi pi  L

…(1)

  pi vi  L i

  pi vi  mc 2 i

1-   q  qc v.A 2

q   pi vi   vi Ai  mc 2 i i c q      pi  Ai vi  mc 2 c  i 

1-   q 2

Putting vi from eq. (5) art. 7.4 as

1-   2

vi 

q   pi  Ai  c  

…(2)

we get 2

q   H    pi  Ai   c  i 

 mc 2

1-   q 2

1-   2



1- 

2  q   2 2   pi  Ai   m c   q c   i  

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1-   2



2  q  2 2  p  A   m c   q c   

1-   2



m2v 2 2 2  (1   2 )  m c   q  

1-   m v 2



2 2

 

mc 2

1- 2 

 m2c 2 (1   2 )    q (1   2 ) 

 q

…(3)

…(3A)

 T  q

…(4)

From eq. (17) of art. 7.3, we know that

T

mc 2

1- 2 

 T0 .

…(5)

with the condition that T is zero when

 i  0 , we get

T0  mc 2

  T0  mc 2  

so that

putting for

mc 2 /

1-  2

   1 . 

…(6)

1-  in eq. (3A), we get 2

H  T  mc2  q.

…(7)

A Covariant Lagrangian and Hamiltonian Formulation: Although in art. 7.4 and 7.5, we have described relativistic Lagrangian and Hamiltonian formulation, but those descriptions were in a ‘certain sense’ in which time retained its nonrelativistic character.

A covariant formulation, however, demands that space and time be

considered as entirely similar coordinates in world space. The place of invariant parameter in world space is taken by proper time function of x instead of



instead of t. Therefore a covariant Lagrangian must be a

.  dx  xi , u     instead of x i and  instead of t. Hamilton’s principle  d 

in four dimensional language thus becomes

 I    L  x , u , d  0,

…(1)

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where L' is covariant Lagrangian. Covariant Lagrangian Formulation: We shall find L' for a particle in two cases; first when particle is entirely free and second when particle is acted on by external electromagnetic forces. (a) First Case: Euler-Lagrange equations deduced from covariant Hamilton’s principle represented by eq. (1) are

d d

 L  L 0     u  x    

…(2)

which for a free particle must reduce to

d (mu )  0. d The desired result can be achieved if covariant Lagrangian be chosen as

L  so that

 mu u , 

L L  0,  mu  p x u

which when substituted in eq. (2) yields the desired result for a free particle. (b) Second Case: For the second case when particle is acted on by external electromagnetic forces, suitable invariant Lagrangian is

L 

 q mu u  u A  c

…(3)

with canonical momenta

p 

L q  mu  A , u c

…(4)

so that eq. (2) takes the form

d d or

q     mu  A   c  x 

d  mu    d x 

 x

q   u A   0 c 

q  q dA  u A   c  c d

 q  d   u A    c  d

    u

 q    u A    c  

The right hand side of above equation is four force or Minkowski force K  . Therefore equation becomes

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d  mu   K , d and can be recognized as generalized Newton’s equation of motion. Here the momentum canonical to the time co-ordinate will not be simply

iT as in force c

free case but

p4 

i iE T  q   c c

…(5)

where E is the total energy. This can be obtained as follows: From eq. (4),

p4  mu4 

q A4 c

mic

q  i c



1-  2

  i  mc 2   q   c  1- 2    Refer to eq. (4) Art. 7.5, where

H

mc 2

 q .

1-  2

Since H represents the total energy E and taking T0  0 , we write

T

mc 2

1- 

we find

E  T  q , so that

p4 

i iE T  q   c c

This shows that p4 , the momentum canonical to time co-ordinate, is proportional to the total energy. Covariant Hamiltonian formulation:

In relativistic notation the covariant Hamiltonian for a

single particle is defined as

H   p u  L

…(6)

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with the corresponding eight equations of motion:

dp dH  dx dH     . and, dp d dx d

…(7)

From eqs. (3) and (4), we see that

L 

 q mu u  A u  c

p  mu 

and

q A c

which with eq. (6) yield

H   mu u  

q  q A u  mu u  A u c  c

  mu u   mc 2 ,  

since square of world velocity is

c 2 .

If we express covariant Hamiltonian in terms of canonical momenta

H 

 q     p  A  m   c 

…(8)

   mu 2  

   mc 2 ,  which is not identical with the total energy, With covariant Hamiltonian represented by eq. (8), space part of eqs. (7) will definitely lead to spatial equations of motion. For time like part, one of the equation are

dH   p4  q / cA4    u4 dp4 m or

p4 

i iE T  q   , c c

showing a fact that p4 is proportional to the total energy and this has already been noted in eq. (5). Covariant formulation forms a rather elegant method for the determination of physically interesting relations by equations of motion and quantities, viz, energy and momentum. Since the notion of force, in such a transformation, has completely disappeared, this method in no way reflects in no way reflects the Newtonian scheme.

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1.9 Four Vectors Having introducing the idea of four dimensional space, it is possible to extend ordinary vector analysis to four dimensions to generally valid laws in the form of equations between four dimensional vectors. These four-dimensional vectors are called four vectors or world vectors. As the four-dimensional coordinates are orthogonal, we must have ^ ^

i . i  1, j . j  1, k . k  1, p . p  0 ^ ^

^ ^

i . j  0, i . k  0, i . p  0, j . p  0, k . p  0

and

and so on. The world vectors are denoted by bars beneath their symbols. If A and B are two world vectors, then ^

A  i Ax  J Ay  k A2  p Ap B  i Bx  J By  k B2  p Bp The transformations law in Minkowski four dimensional space for the set of four coordinates

 x1 , x2 , x3 , p 

 x1, x2 , x3 , p

in one frame of reference to the set of four co-ordinates

in another frame moving with velocity v relative to the first frame is given by

x    av xv

…(1)

with the constraint

x12  x22  x32  p 2  x12  x2 2  x3 2  p2 Any set of four component A which transform under Lorentz transformations like the four-components

 x1 , x2 , x3 , p 

is called a four vector. The components A of four vector A

transform under Lorentz transformation according to  with A0  iA4  3

A    av Av

…(2)

v 0

the quantities a v being given by eqn. (1).

A.B  Ax Bx  Ay By  Az Bz  Ap Bp which may also be written as

 A B  A B  A B 1 1

 A3 B3  A0 B0

 A1B1  A2 B2  A3 B3  A4 B4  A.B  A4 B4

…(3)

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where A   A1 , A2 , A3 , A0  or  A1 , A2 , A3 , iA4  It can easily be verified that the scalar product given by eqn. (3) is invariant quantity, i.e.,

A B    A B  cons tan t    Since

…(4)

a A a B  A B       v

      a av A Bv         v A Bv 

(as av a =  v ) 

  A B 

(Here



is Kronecker delta symbol)

i.e., the scalar product of two four vectors in invariant quantity. The definition vector product of two world vectors is somewhat difficult, because the combination forms an antisymmetric tensor in three dimensions is represented by a vector. In four dimensions an antisymmetric tensor has six components while a vector has only four. Ex. 1. Show that the four dimensional volume element

" dx dy dz dt" is invariant under Lorentz

transformations. Sol: According to length contraction, we have

dx  dx 1 

v2 c2

dy  dy

dz  dz According to time dilation,

dt 

dt 1



v2 c2

Four-dimensional volume element in system

S

 dx dy dz dt

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      v 2   dt    dx 1  2   .dy.dz   2  c     v    1  c 2       dx dy dz dt = four dimensional volume element in system S.

Covariant Four-Dimensional formulation of the laws of Mechanics If the form of a law is not changed by certain coordinate transformation, the law is said to be invariant or covariant. If any physical law may be expressed in a covariant four dimensional form, then the law will be invariant under Lorentz transformations. In four dimensions the position vector will be termed as position four vector. The position four vector in four dimensions will be represented as ^

r= x1 x1  x2 x2  x3 x3  x4 x4 ^

where

x1 , x2 , x3 and x4 are unit vectors along x1 , x2 , x3 and x4 respectively. To differentiate

position four vector with ordinary position vector we have put r in place of r. Taking dot product of r with itself, we shall get a world scalar and hence Lorentz invariant, i.e., ^ ^ ^ ^ ^ ^ ^ ^  r.r   x1 x1  x2 x2  x3 x3  x4 x4  .  x1 x1  x2 x2  x3 x3  x4 x4    

 x12  x22  x32  x42  Lorentz invariant If d r represents the change in position four vector, then we have ^

d r  x dx1  x 2 dx2  x3 dx3  x4 dx4 taking dot product of this with itself, we shall get a world scalar and hence Lorentz invariant, i.e.,

d r.dr  dx12  dx22  dx32  dx42 =Lorentz invariant. or

d r  dx12  dx22  dx32  dx42

But

x1  x, x2  y, x3  z and x4  ict

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88 …(1)

Let us consider a system in which a particle is momentarily at rest. As the object is at rest, its displacement vanishes, i.e., dx = 0, dy = 0, dz = 0 and let the time be denoted by



, then

we have

d r  c 2d 2 2

...(2)

As d r is an invariant, we must have

dr  dx2  dy 2  dz 2  c 2dt 2  c 2d 2 2

…(3)

From equation (2) it is clear that d  d r / ic is an invariant since d r and c are invariant. The time d  , as measured in the rest frame is called the proper time. From equation (3) we have

d 2  

dt 2  2 c

1 dx 2  dy 2  dz 2  c 2dt 2  2  c

 2   dx 2  dy 2  dz 2   c              dt   dt   dt   

 v2  d  dt 1  2   c 

d 

dt  v2  1  2   c 

which is an expression for time dilation. As one of the components of a four vector is imaginary, the square of four vector may be +ve and –ve in like ordinary vector. If the square of magnitude of a vector is greater that or equal to zero, the four-vector is said to be space like; but if the magnitude is negative the four vector is said to be time like. As the magnitudes of the four-vectors are world between two world points which unaffected by Lorentz transformations. Let the now consider a difference vector between two world points which may be either +ve or –ve. Let R be the difference vector defined as

R  r2  r1 , where r1 , and r2 are the position four-vectors of the world points, respectively. Then the magnitude R is given by

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R  r2  r1  r2  r1  c 2  t2  t1  . 2

where r1 and r2 are the ordinary position vectors. Thus R will be space like if

 r2  r1 

 c 2  t2  t1 

…(4)

 c 2  t2  t1 

…(5)

and R will be time like if

 r2  r1 

Let the spatial axes be so oriented that the difference vector  r2  r1  is along X-axis and therefore equal to x2  x1. Then the Lorentz transformation of fourth component of R may be written as

c  t2   t1  

c  t2  t1    v / c  x2  x1   v2  1  2   c 

vx c2 t   v2  1  2   c  t

If R is space like, then

c  t2  t1   x2  x1 and hence it is possible to find a velocity v<c such that ic  t2   t1   vanishes. The vanishing of

ict  t2   t1  i.e, fourth component of R , means that if the distance between two events is spacelike, it is possible to find a system in which the two events are simultaneous. Now in order to illustrate the process of restating a physical law in a covariant four dimensional formulation, let us take the scalar wave equation of the type

2 

1  2 0 c 2 t 2

…(6)

Like three dimensional gradient, we may consider a four dimensional differential operator in which space, denoted by symbol

 and known as D’ Alembertian with components

    , , , x1 x2 x3 x4

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i.e.,

 i

Taking scalar product

^  ^   ^  j k x x1 x2 x3 x4

…(7)

 with itself we get a scalar quantity and hence Lorentz invariant, i.e,

 2 2 2 2  .  2  2  2  2   x 1 x 2 x 3 x 4  But x1  x, x2  y, x3  z, x4  ict

 2 2 3 1  2  2   2  2  3  2 2  = Lorentz invariant.  x x x c t  Thus equation (6) is invariant under Lorentz transformation. Clearly, any physical law may be invariant under Lorentz transformation, if the law may be expressed as a covariant four dimensional form.

Lorentz transformation of space and time in four-vector form Consider two systems S and

S , the latter moving with velocity v relative to former along

positive direction of X-axis. In Minkowski space let the co-ordinates of an event be represented by the quantities

( x, y, z, ict ) or x (  1, 2,3, 4) where

x1  x  x2  y  x3  z   x4  ict 

…(1)

Lorentz transformations of space and time with

x 

  v/c

are written as

x  vt

1    2

y  y

z  z vx c2 t  1   2  t

and

or equivalently

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x1  vt

    x1  i  x4   1      x2   x2   with  = x3   x3   vx1   t  2   c  x4       x4  i  x1    1   2   x1  

…(2)

1    2

Above transformations can be expressed as

x1    x1  0.x2  0.x3  i  .x4   x2   0.x1  1.x2  0.x3  0.x4   x3   0.x1  0.x2  1.x3  0.x4  x4   i  x1  0.x2  0.x3   .x4 

…(3)

These results can be written compactly by a single equation

x    v xv . Where

…(5) Equation (4) represents the Lorentz transformations of space and time in four-vector form. Similarly Lorentz transformations of any four-vector A (   1,3, 4) may be expressed as

A   av Av

…(6)

Four-vectors. 2

2

2 2

We have already seen that the quantity s  x  y  z  c t  x  y  z  c t 2

2 2

is invariant under Lorentz transformations. This quantity is minus times the interval between origin of four dimensional world and the world point

(x,y,z,t) . Let us put x=x1 , y=x 2 ,z=x 3 and

ict  x 4 , then the quantity s 2 may be written as

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i 1

s 2   xi 2  s2   xi 2 ,

…(1)

where the transformation relations between xi  and xi in the special case, when

s 2 moves with

a uniform rectilinear velocity v in the direction of x-axis, may be written from equations (4), i.e.,

v x1  i x4 c x1   , x2   x2 , x3   x3 2 2 1  v / c ) 

...(2)

v x4  i x1 c x4   2 1  v / c2 ) 

and

In a more general case these transformations can be written as a linear relation 4

xi     ik xk ,

i  1, 2,3, 4

…(3)

k 1

or, in summation convention,

xi  ik xk .

Equation (1), then gives s  xi xi  si  xi xi  ikil xl xk . 2

ikil   k l .

Obviously,

…(4)

These conditions are called the orthogonality conditions satisfied by coefficients of transformation

i j

for

s 2 to be invariant.

A vector with components x1 , x2 , x3 , x4 is called the four-dimensional radius vector. Similarly, a set of four quantities A1 , A2 , A3 , A4 , which transformations of the four-dimensional coordinates system transform like the component xi [like equations (2) in the special case or like equation (3) in the general case], is called a four vector

 Ai  .

Thus under a special Lorentz

transformation (4),

v A1  i A4 c A1  , A2  A2 , A3  A3 2 1  v / c2 ) 

and

v A4  i A1 c A4  2 1  v / c2 ) 

…(5)

or under a general Lorentz transformation, Ai  ik Ak

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It is obvious that by virtue of the orthogonality condition (4) the quantity invariant i.e.,

Ai Ai  Ai Ai

Ai Ai  Ai 2 is

…(6)

which is equal to the square of the magnitude of four-vector concerned. It is called the norm of four-vector Ai . The four-vector ( Ai ) could be called space-like, time-like or null, according as its norm is positive, negative or zero, respectively. Let us now combine two four-vectors ( Ai ) and ( Bi ) , linearly to form a third four-vector

Ci    Ai   Bi 

Ci such that then

CiCi   2 ( Ai Ai )   2  Bi Bi   2  Ai Bi   Norm of four-vector Ci

which is also invariant. The invariance of Ai Ai and Bi Bi thus gives invariance of Ai Bi . This quantity is called the scalar product of given four-vectors. A given four-vector ( Ai ) , in any system of co-ordinates, maybe spit into a spatial part

A( A1  A2  A3 ) and a temporal part A4 :

 Ai   ( A, A4 ),

…(7)

In the case of a general Lorentz transformation given by equations (11) and (12), the spatial and temporal parts of the vector ( Ai ) will change according to the formulae

A   

v v v (  1)  i A4 2 v c

…(8)

v    A    A4  i c  

…(9)

which are similar to equations (11) and (12) with r and t replaced by A and A / ic. The four-vector character of xi and the invariance of

d defined by equation (25) would

give the components of a certain four-vector (U i ) such that

Ui 

dxi . d

…(10)

This vector is called the velocity four-vector or the four-velocity of the particle.

The

splitting of this vector into spatial and temporal parts can be done by using equation (25) and splitting the position four-vector  xi  into spatial and temporal parts r and x4 . Thus

    dx4 dr dr / dt ic   (U i )   , ,  dt (1  u 2 / c 2 ) dt (1  u 2 / c 2 )   (1  u 2 / c 2 ) (1  u 2 / c 2 )     

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  u ic .  ,  (1  u 2 / c 2 ) (1  u 2 / c 2 )   

…(11)

It is clear from equation (10) that the four-vector (U i ) lies along the tangent to the time track. The norm of this vector, as defined by equation (10) is given by

U iU i 

u2 c2   c 2 2 2 2 2 1 u / c 1 u / c

…(12)

and hence (U i ) is a time-like four-vector. Equation (10) gives on differentiation,

Ui where

dU i  0, d

…(13)

dU i   Ai  is four-vector known as acceleration four-vector. This four-vector is evidently, d

orthogonal to (U i ) . Thus it lies in the direction of the normal to the same track. It may be written as

dU i dU dt  Ai   i  2 d (1  u / c 2 )1/ 2 For a particle in instantaneously rest system,

…(14)

u  0, i.e., U i   (0,0,0ic), and hence 0

0  dU i      Ai   d 

will have vanishing temporal part according to equation (13). Thus

 Ai 

 (a,0), the norm of

which is positive. Therefore, the acceleration four-vector is a space-like vector. The momentum four-vector or the four-momentum of the particle is defined as

( pi )  m0 (Ui ) the norm of which is given by

pi pi  m0 m0UiUi  m02c 2

…(15) …(16)

which shows that this vector is time-like one. It lies in the direction of the tangent of time track at

xi i.e., parallel to (U i ) . Its spatial and temporal parts are

  m0u im0c   ,  (1  u 2 / c 2 ) (1  u 2 / c 2 )    i    iE    mu, mc 2    p,  , c    c 

...(17)

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where p is ordinary three-momentum and E is the total energy of the particle. From this equation, we have

pi pi  p 2 

E2 c2

…(18)

combining equations (16) and (18), we get

p2 

E2  m0 2c 2 2 c

E 2  p 2c2  m02c4 .

…(19)

The energy expressed in terms of the momentum is called the Hamiltonian fuction H. Thus, we have

H c

p

 m0 c

2 2

m

2 2

 p2  1  2 2  .  m0 c 

For velocity small compared to c, p  m0c and

p2 H  m0 c  , 2m0 2 2

which, except for the rest energy m0 c , is the familiar classical expression for the Hamiltonian. In the case of photons, m0  0 and

p

v  c, equation (19) gives

E , c

…(20)

and hence equation (17) reduces to

 pi   ( p, ip).

The norm of which is pi pi  p  p  0 i.e., 2

 pi  , in this case, is a null four-vector.

The transformation equations for the components of four-momentum

 pi 

can be written directly

according to equations (8) and (9) on replacing A by p and A4 by iE/c. Thus

p  p  and

( p.v).v E (  1)   v 2 . 2 v c

…(21)

E    E  ( p.v). In special Lorentz transformations , these equations reduce to

px  px  px (  1)   v      px  2 E  c  

vE . c2 …(22a)

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py  py . pz  pz , E    E  vpx  Using these transformation relations, it can be proved in the straight forward manner that the equation (19) is invariant under Lorentz transformations i.e.,

E2  p c 2  E 2  p c 2 .

…(22 b)

Comparing equation (22a) and (22b) we find a striking analogy.

The quantities

px , p y , pz and E / c 2 transform exactly as the space-time co-ordinates x, y, z and t of a particle transform. The deep seated connection between r and p on one hand and t and E on the other, 1

are shown rather conspicuously in the quantum mechanical formulation of particle dynamics. . It is clear from equation (21) that if energy and momentum are conserved in an interaction according to one inertial observer, then necessarily energy and momentum are conserved in this interaction according to any other inertial observer also.

Furthermore, if

momentum four-vector is conserved then both the ordinary momentum and energy are conserved, since the conservation of a vector implies the conservation of all its components individually. The transformation equation of mass follows directly from the energy transformation. For this, we consider the mass energy relation which gives

m

where

m0 (1  u 2 / c 2 )

and

m 

E  mc 2 and E  mc 2

m0 (1  u2 / c 2 )

Thus the energy transformation relation gives

m   m(1  ux v / c 2 ).  v.v  m   m 1  2  .  c 

or in general

…(23) …(24)

Force four-vector or the four-force ( Fi ) is derived by differentiation the four-momentum with respect to



, i.e.,

 dp ( Fi )   i  d

d   d

 d      m0U i      d

 E   dp i dE    p , i  ,   c    dt (1  u 2 / c 2 ) c dt (1  u 2 / c 2 )  

  F i ( F .u ) .  ,  dt (1  u 2 / c 2 ) c dt (1  u 2 / c 2 )    where F 

…(25)

dp dE is ordinary force in three-dimensional mechanics an consequently  F .u. dt dt

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This equation may also be written as

( F .u )   ( Fi )   FM , i M  , c   where

FM 

F (1  u 2 / c 2 )

…(26)

, is known as the Minkowski force,

The scalar product of the four-force ( Fi ) and four-velocity (U i ) can be written directly,

U i Fi  U i

dU dm d  m0Ui   m0Ui i  UiUi 0 d d d

 c 2 .

…(27)

dm0 d

…(28) [using equations (12) and (13)]

Thus the scalar product is a direct measure of the rate of change of the proper mass of the particle with the proper time. In case proper mass is constant in proper time, equation (28) reduces to

U i Fi  0

i.e., four-force and four-velocity are mutually orthogonal. Moreover equation (28b), in this case, reduces to

( Fi )  m0 ( Ai ),

Let us now consider a particle at rest instantaneously in S-frame where it is subjected to a force F. Then u = 0 and equation (25) reduces to

( Fi )0  ( F ,0). In S' – frame this particle has the velocity u' = -v, and correspondingly.

  F .v   F i ( Fi )   ,   (1  v 2 / c 2 ) c (1  v 2 / c 2 )  The transformation relation gives

 F   F+

 F .v  v (  1)

 F   F+

 F .v  v (  1)

and

( F u    F .v  .

and

i  F .u  c (1  v / c ) 2

  i

 F .v  c …(29)

In special Lorentz transformations equations give

Fx =Fx ,

 F y  Fy ,  F z  Fz .

…(30)

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F u   F.v.

and

Let us consider now a more general case where a particle of rest mass m0 moves with velocity u and u' in frames S and S' under the action of force F = F' respectively. Then

F

d d  m u  and F    m u . dt dt

Let us assume that frame S moves relative to S' with a rectilinear constant velocity v in the direction of x-axis. Then m and m' are related according to equation (23) and the relationship of t and t' given by equation. Thus

   d d dt d  u v  u  v Fx    m ux    m ux     m 1  x2   x uv dt  dt dt  dt   c  x 1 2  c  

d 1 1  m(ux  v) vu  vu dt 1  2x 1  2x c c

  1    1  vu x  c2 

dm   dux  m dt   u x  v  dt   

1/ 2  du dm d  u2   1 x  m  ux  vm0 1  2   dt dt  c   1  vux  dt  c2 3/ 2  du dm vm0  u 2  du  1 x  m  ux  2 1  2  u  dt c  c  dt  1  vux  dt  c2

1  vu 1  2x c 1  vu 1  2x c



1 vu 1  2x c

1  du du y du   dm vm  u 2   dux x  ux  2 1  2   u x  uy  uz z  m dt c  c   dt dt dt    dt 

2 2 2  du du y duz  v dm  u x  u y  u z   dm vm  dux x m   ux   uy  uz  ux  dt c 2  dt dt dt  c 2 dt c2  dt 

du y v  dux  vux  du v dm  vux  dm u y v dm uz v  . 2  muz z 2  u y . 2  uz m 1  2   u x 1  2   u y m  dt  c  dt c dt c dt c dt c 2   dt  c  m

dux uv d dm  v d  1  ux  u y 2 (mu y )  z2 (muz )  dt dt  c dt c dt  1  vux c2

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d d mu y  mu  u v uz v dt  z  d y dt   mux   2  2 vu y dt c c 1  vu z 1 2 c2 c

 Fx 

uyv c  vux 2

Fy 

uz v Fz . c  vuz

…(31)

Similarly, we may have

Fy  

Fy Fz and Fz   . u v   ux v  x   1   1    c  c   

…(32)

These transformation equations can also be written as

v (u.F ) 2 Fy Fz c Fx   , Fy   , Fz    u v  u v  u v  1  x2   1  x2   1  x2  c  c  c     Fx 

…(33)

and the reverse transformation may directly be written by interchanging unprimed components and replacing v by –v*. §1.10 The Invariance of Maxwell’s Equations with Lorentz Transformations. Here we have to prove that Maxwell’s electromagnetic field equations are unchanged with Lorentz transformation. We consider two frames of reference S and and

S where S is at rest

S is in relative translatory uniform motion, moving with velocity v along ġ-axis. Any point in

S is expressed by co-ordinates

(x,y,z,t) and in S by (x,y,z,t).

In frame S Maxwell’s field equations in vacuum are written as div E = 0

(for a region in which no charge is present),

div H = 0, curl E=curl H 

 H c t 1 E 4 k  j c t c

1 E (for a region in which no current is present).  k c t In terms of components of electric fields E and H in directions of x, y and z-axes, we may write these equations as

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100

H x H y H z    0, x y z EZ E y  H x   , y z C t

Ex Ez  H y   , z x C t E y x



…(1)

Ex  H z  , y C t

H Z H y k Ex   , y z C t

H x H z k E y   , z x C t H y x



H x k Ez  . y C t

Here, we are dealing with fields, i.e., with physical entities which are functions of space and time, the field components relative to S i.e., Ex (x,y,z,t) , Ey (x,y,z,t)...H Z also be expressed in terms of

(x,y,z,t), can

(x,y,z,t) or any other set of variables which depend linearly on

(x,y,z,t). By this the physical significance of Ex ...H Z will not change. Only the functional form of Ex , E y ...H Z in its dependence on the various space-time points in the new notation is changed. In the same way the field equation take a new form. If the x-component of electric field in terms of

x,y,z,t as observed by the observer S is Ex (x,y,z,t) , then the new functional

dependence of E x is obtained from the relation

Ex (x,y,z,t)=Ex (x,y,z,t) In the frame of reference

…(2)

S we assume that to observer S the components of

electromagnetic field appear as Ex , E y , Ez , H x , H  y , H Z . In the same way as the observer S, E x is naturally a function of

x,y,z,t , but can also be expressed as a function E x

(x,y,z,t). i.e.,

Ex (x,y,z,t)=Ex (x,y,z,t)

…(3)

Using equations (2), we may write

Ex Ex x E x y E x z E x t     x x x y x z x t x

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Ex Ex x E x y E x z E x t     y x y y y z y t y

Ex Ex x E x y E x z E x t     z x z y z z z t z

…(4)

Ex Ex x E x y E x z E x t     t x t y t z t t t But using Lorentz transformations, we have

x  x

1  v

/ c2 

 a( say ).

y z t av x y z t  0,  0,  2  0,  1,  0,  0, x x x c y y y y x y z t x y z  0,  0,  1,  0,  av,  0,  0, z z z z t t t and

…(5)

t   a. t The similar equation as (4) may be deduced for other components. And then using

equations (5) we may have the following equations:

Ex Ex Ex E x v Ex Ex  a a 2 ,  av a , x x c t t x t

E y x

a

E y x

a

E y E y v E y E y ,  av a , 2 c t t x t

Ez Ez E z E z v Ez E z a a 2 ,  av a , x x c t t x t

…(6)

Ex Ex E y E y  Ez Ez Ex Ex   ,  ,  ; y y y y y y z z

E y z



E y z

and

Ez E z  . z z

Similar relations hold for H x , H y , H z . From Lorentz transformations, we may prove the following operational relations’

  av    a  2  , x x c t y y and

   , z z

…(7)

    av a . t x t

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Substituting equations (6) in the third and the second equations of the set (1), we have the set of equations

H x   Ez  E y    H z      av a y z c x t 

…(8)

H x  av H x  H y  H z      0. x c 2 t y z

…(9)

and

Eliminating

H x  , from these equations, we have x

E y   c H z  Ez   v H y   H x   a a  a  y c y z c z c t

…(10)

   v v  H x       a  Ez   H y    a  Ey   H z     . …(11)   y   c c c t   z   

If the components of fields in frames

S , i.e., E y , Ez , H  y , H Z are related with

E y , Ez and H x , H Z by the relations

v v     Ez   a  Ez   H y   .E y   a  E y    , H x   H x  c c    

…(12)

the equation may be written as

Ez  E y   H x    . y z c t

…(13)

which is similar to the third equation of set (1). Similarly,

Ex  Ez   H y    . z x c t E y 

and

x



Ex   H z   . y c t

Combining all these equations, we get curl E  

 H  c t

…(14)

Hence Maxwell’s third equation is invariant for Lorentz transformations. The equations (12) and (14) give us

v v     Ez   a  Ez  H y  , Ey   a  Ey  Hz , Hx  Hx c c    

…(15)

The remaining components also may be found in the following form by the similar method:

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kv  kv    H z   a  H z  E y  , H y   a  H y  Ez  and Ex   Ex  …(16) c c     Now using equations (15) and (16) with the divergence equation for E (i.e., first number of set (1)), we have

 E  E y  Ez    v  H Z  H y   v Ex  a x     0. …(17) a  a 2 y z  c  y z  c t  x By the similar method as of (14), we have curl H   

k E c t

…(18)

H Z  H y  k Ex    y z c t

So equation (17), yields

Ex  E y  Ez     0. x y z

Similarly, we may prove that

(taking kμ =1) …(19)

H x  H y  H z     0. x y z

…(20)

Thus all Maxwell’s field equations are invariant under Lorentz transformations. Starting from Ex , E y ...H Z  we may find the values Ex , E y ...H Z in the reverse process discussed above,

Ex  E x  H x  H x 

v

kv Hz c c Ey  , Hy   k 2   k 2  a 1v  a 1v   c   c  Ey  

Hy 

…(21)

kv Hy c c Ez  , Hz   k 2   k 2  a 1v  a 1v   c   c  Ez  

and

v

Hz

Hy

Hz 

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UNIT II MECHANICS OF RIGID BODIES Introduction This chapter is a problem involving motion of a particular types of body – the rigid body. By a rigid body we mean a rigid assembly of particles, with fixed inter-particle distances. Thus we always neglect the deformations that occur in actual bodies. In order to discuss the mechanics of a rigid body, we must have means of describing its position and orientation. A rigid body with N particles can almost have 3N degrees of freedom, three for each constituent particle but because of the large number of constraints, the number of degrees of freedom or independent coordinates required to describe its motion will be much less then 3N. 2.1 Generalised co-ordinates for rigid body motion. Euler’s Angles: (i)

Consider three non-collinear points p,q and r (fig. 2.1) in a rigid body. Each particle has

three degrees of freedom and so total of nine degrees of freedom is required. But due to the presence of constraints, the interparticle distance is fixed. Consequently, there will be three equations of constraints between them, viz,

Fig. 2.1

 pq 2  ( x1  x2 )2  ( y1  y2 )2  ( z1  z2 )2  constant

 qr 2  ( x2  x3 )2  ( y2  y3 )2  ( z2  z3 )2 

constant

 rp 2  ( x3  x1 )2  ( y2  y1 )2  ( z3  z1 )2  constant.

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Hence the total number of degrees of freedom of this system of non-collinear particle is reduced to 9-3=6. The position of each further particle, say P, requires three co-ordinates but three will be three equations of constraints for particle (because distance of P from p,q and r is fixed). Thus three co-ordinates for p less three eqs. Of constraints for p gives zero degrees of freedom. That is, any other particle, apart from p, q and r, taken to specify the configuration of the rigid body, will not add any new degrees of freedom. Thus the configuration of the body would be completely specified by p, q and r particles i.e. by six degrees of freedom. (ii)

We can also give an alternative way of showing that six generalised co-ordinates (degrees of freedom) are required to fix the configuration of the body. For assigning generalized co-ordinates, the fixed point in the body, which registers its translation and may with advantage be taken coincident co-ordinates relative to the co-ordinate axes of the external space. The remaining three co-ordinates then specify the orientation of the body relative to the external frame of reference. Orientation of the body can be described elegantly by locating a cartesian set of co-ordinates fixed in the rigid body (hereafter denoted by primed axes as show in fig 2.1(a,b)) called body set of axes.

Fig.2.1 (a)&(b) Thus configuration of a rigid body with respect to some Cartesian co-ordinate system in space (space set axes x,y,z) arbitrarily chosen is determined completely if the position of the origin, and orientation of a cartesian co-ordinate system fixed on the body (body set of axes

x' , y ' , z ' ) is known. To fix origin of co-ordinates, one can easily assign three co-ordinates which also register translation of the body. For specifying orientation of the body set of axes, one may take help of the direction cosines of body set of axes (primed system) referred to space set of axes (unprimed system) referred to space set of axes (unprimed system). Let

 i , i ,  i

be the direction cosines and i,j,k and i , j , k be the unit vectors along the '

axes x y z fixed in space and x , y , z axes fixed in the body then

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i '  1i  1 j   1k   j '   2i   2 j   2 k   k '   3i  3 j   3k  Therefore we get

i ' .i '  1  12  12   12   j ' . j '  1   22  22   22   k ' .k '  1   32  32   32  and

i ' . j '  1 2  1 2   1 2    j ' .k '   2 3   2 3   2 3   k ' .i '   31  3 1   3 1   As amongst nine direction cosines there are six relations connecting them, only three out of them are left unconnected. But these three uncorrected direction cosines are not independent of each other, and therefore we can not use them as degrees of freedom or generalized coordinates and consequently, we can not set up a Lagrangian in order to describe the rotation of the system in terms of these direction cosines. In fact, when we generate the primed axes (body set of axes) from the unprimed axes (space set of axes) through three successive rotations, the so-called Euler angles, then each transformation (rotation) is an orthogonal transformation and final configuration of primed axes will be shown to result from a composite orthogonal transformation. The most useful set of generalized co-ordinates for a rigid body are Euler angles, which are the angles or rotation about specified axes, executed in a specific sequence. 2.2 Euler’s Angles: Let us proceed to find three independent parameters which would completely specify the orientation of rigid body-the so called Euler’s angles. Let x,y,z be the orthogonal space set of axes with I,j, and k as unit vectors along these '

axes. Also x , y , z be the orthogonal body set of axes with i , j and k unit vectors along these axes. As shown in fig. (2.2), x-y plane is different from x  y plane; and also y-z plane is different '

from y’-z’ plane. In order to account for the rotatory motion, we shall carry out the transformation from space set of axes to body set axes. The transformation is worked out through three successive rotations performed in a certain specific order. That is, we rotate space set of axes

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x,y,z so as to coincide with body set of axes x , y , z through three successive rotatory operations that are worked out one after the other in a specific sequence.

Fig 2.2. Note that plane x  y is different from plane x-y. Also plane y′-z′ '

is different from plane y-z.

Fig. 2.2(a) Note that new plane y1, z1 contains body z'-axis, First Rotation : First of all the space set of axes is rotated about the space z-axis so that the y-z plane takes new position y1-z1. This new plane contains the body z' – axis. The rotation angle is . The new axes are x1 and y1 with unit vectors i1 and j1 as shown in fig 2.2. (a) The transformation to this new set of axes x 1,y1,z1 from x, y, z axes can be represented by the equations:

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i1  i cos   j sin  j1  i sin   j cos  k1  k  i1   cos  sin     (or) j1   sin  cos     k  0 0  1 

0  i    0  j  1   k 

Or simply the matrix of transformation is

 cos  sin   D    sin  cos  0 0 

0  0 1 

Second Rotation: Second rotation is performed about new x 1-axis so that z2 – axis coincides with body z axis. This also brings the plane x2-y2 in the plane x  y . The axes x2 and y2 '

obtained after rotation about x1 through an angle  still do not coincide with x and y axis. The '

rotation angles is . The new axes y2 and z2 with unit vectors j2 and k2 are shown in fig. 2.2 (b).

Fig. 2.2(b) Note that the z coincides with z and plane '

x2, y2 coincides with plane x , y

The transformation to this new set of axes x 2, y2, z from x1,y1,z1 set of axes can be represented by the equations.

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i2  i1 j2  j1 cos   k1 sin  k2   j1 sin   k1 cos  , So that the matrix of transformation will be

1  C  0 0 

0   cos  sin    sin  cos   0

As told earlier this transformation brings x2y2 plane into the plant x , y of the body set of axes. '

Third Rotation: Third rotation is performed about z2 axis i.e., about z axis so that the new axis x3 coincides with body

x ' axis and the axis y3 coincides with y ' axis. This completes the

transformation from space set of axes to body set axes. The angle is . The new axes are y3 (=

y ' ) and x3 (= x ' ) as shown in fig 2.2. (c),

Fig. 2.2(c)Space set x y z coincides x , y , z . Therefore

i3  i ' , j3  j' and k3  k ' . The transformation to this new set of axes x3 y3 z3 which coincides '

with body set of

axes x , y , z can be represented by the equations:

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i '  i2 cos   j2 sin  j '  i2 sin   j2 cos  k '  k2 So that the transformation matrix is

 cos  sin  0    B    sin  cos  0  0 0 1   Therefore, we have arrived at the body set of axes after three successive and sequential rotation of space of axes. The complete matrix of transformation A will be

 i2   i '   i3         j '    j3   B  j2   k '  k  k     3  2  i1  i       BC  j1   BCD  j  k  k     1 Let

A=BCD

 x' x     So that y '  A y     z' z      Or the inverse transformation from body set of axes to space set of axes will be given by

x  x'    1   y   A  y ' z  z'     '

Thus the matrix of transformation A furnishing x , y , z axes from the space set of axes xyz directly is the product of matrices taken in the indicated order of rotations, viz

 , ,  or

ABC A=BCD

 cos  sin  0    A    sin  cos  0  0 0 1  

1  0 0 

0   cos  sin    sin  cos   0

 cos  sin     sin  cos  0 0 

0  0 1 

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 cos  sin  0       sin  cos  0  0 0 1  

111

 cos     cos  sin   sin  cos  

 cos  cos     cos  sin  sin  A    sin  cos     cos  sin  cos   sin  sin  

sin  cos  sin   sin  cos 

  sin   cos   0

cos  sin 

sin  sin     cos  cos  sin    sin  sin  cos  sin     cos  cos  cos     sin  cos  cos   -1

The inverse transformation matrix from body set of axes to space set of axes is given by A =A because A is an orthogonal matrix. Therefore

 cos  cos     cos cos  sin  1 A   cos sin     sin  cos  cos   sin  sin  

sin  cos 

sin  sin     cos cos  sin    sin  sin   sin  cos    cos cos  cos    cos sin  cos  

A for transpose matrix of A Exercise: Prove that the resultant transformation A=BCD is an orthogonal transformation. Physically we except this, for, from an orthogonal system of axes, xyz, we come again to an orthogonal

x ', y ' z ' axes. We simply have to prove that AT = A-1. But A= BCD and each one of B,

C, D represents an orthogonal transformation for the simple reason that the transformation of axes is effected by a simple rotation. Therefore B, C, D satisfy orthogonality condition: T

-1

B =B ,C =C ,D =D T

-1

…(1)

Now A = (BCD) = D C B , Since, when transpose is taken of a matrix product, they can be multiplied only in the reverse order. Inserting (1) into this equation. T

-1

A =D ,C

-1

B , = (BCD) = (A)

-1

Proving that A represents an orthogonal transformation. Here again we have used the reversal property of an inverse of product of two or more matrices.

2.3 Rotating frames of reference and coriolis force Rotating Coordinate systems: Let us consider a reference frame xyz fixed in space and another reference frame

x yz

fixed in the body. Let at time t = 0 their origins and base vectors coincide.

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112

x yz will rotate with respect to frame Oxyz. We shall

establish a relation between the acceleration of body of motion in the two coordinate systems.

Fig. (2.3) Primed systems or body set of axes

x yz is rotating. Unprimed system or

space set of axes Oxyz is stationary.

Fig. 2.3(a). Rotation of a vector

Any point P(x,y,z) in the unprimed system (also called space set of axes or stationary coordinate system) is described by the position vector r. Motion of P is therefore described by

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r =r(t) At time t = 0, we have

r=ix+jy+kz,

Space set of axes

=i x  j y  kz

body set of axes.

Since x yz system rotates, unit vectors

i j, and k will also change with time

whereas unit vectors, I, ñ and k will remain constant in direction. As the point P moves, the velocity of P is different in the two systems for reason of their relative motion. We know that when a vector changes, then as shown in fig. (2.3 a), we have

dR=R sin  d = d R



dR d  R dt dt

   R, So that we can write

di    i, dt where



is the angular velocity of rotation. Therefore, the rate of change of vector r, when

primed system is rotating with angular velocity



with respect to unprimed system, is given by

dr d   i x  j y  kz  dt dt .

 x i  y  j  z k  x

di dj dk   y  z dt dt dt

 x i  y  j  z k  x   i   y   j   z   k 

 x i  y  j  z k    r  dr   dr  r =      dt unprimed  dt  primed

or the operator equation can be written as

d d     r, dt dt where

d is interpreted to the mean the time derivative of r in the rotating coordinate system. dt

We now apply it to the vector velocity of the body in the reference frame Oxyz (space set), i.e., to

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dr d r     r, dt dt to obtain the relationship, between acceleration;

d  dr   d 2 r   d    d r   r      2    r  dt  dt   dt   dt   dt 







d 2 r  d  d r  d r   r     r    2 dt dt  dt  dt



d 2 r d r d   2     r  r 2 dt dt dt



a  a    where we have set



d r d   r   r. dt dt





…(2)

d 2 r  a the acceleration in the primed system, and difference between dt 2

d d  and neglected since they differ by     r , which is a null vector. Additional dt dt





terms in (2) appear because of relative rotation of the two coordinate systems. The third term



 r

 is the famous centripetal acceleration as we can verify geometrically with reference

to fig. (2.3(b)). It is directed toward the axis and its magnitude is





    r   2 r sin  

v2 . r sin 

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115 Fig. 2.3(b).

The second term is present only when the point r is moving in the primed system and is called Coriolis acceleration.

The last term is the acceleration produced by the angular

acceleration of the primed system with respect to the unprimed system when

 is not constant.

If we assume that the Newton’s law of motion holds in the unprimed system, then

ma  F

d 2 r d  d r   m    r  2m    r  F, m 2 dt dt  dt 





in the primed system. Newton’s law, in the primed system, then assumes the form

 d 2 r d   d r   F   m    r  2m   r m 2 dt dt  dt   

d 2 r  Feff dt 2





…(3)

Let us put

d r  V  dt

and

so that to an observer in the rotating frame it appears as if the particle is moving under the influence of an effective force Feff , given by



 



Feff  F  2m  V   m     r  m



d  r. dt



2m  V  is called Coriolis force and it is obviously perpendicular to the plane containing  and

V  . When the particle is fixed with respect to the rotating coordinate system, V  =0 and the

Coriolis force is zero. For a uniformly rotating system,



is constant, and hence last term

 d   r  is zero. Therefore, acceleration a referred to a uniformly rotating system is m  dt  a 



 



F      r  2  V  . M

When the force F is due to earth’s attraction only then we put





F      r  ge M and





a  ge  2  V  .

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116

v2 since the magnitude of     r   g , the direction of g e is almost that of g. r 2.4 Angular velocity of a rigid body: The distance between any two points of a rigid body remains fixed. This means that the position vector of any point P relative to origin O of the body set of axes is constant in magnitude and therefore, it changes in direction only when the body is in motion (rotation). The relative th

velocity of any point P at the position of i particle is thus expressible in terms of the angular velocity  as

vi   Xri Here, we recall our discussion on the rate of change of a vector and the rotating coordinate systems where it was explicitly shown that the rate of change of a constant vector was absolutely expressed in terms of the angular velocity vector and it was applicable to any general vector G. Here, the body set of axes (primed axes) undergo rotation with respect to space set of axes and any changes in the components of G with respect to body axes will differ from the corresponding change with respect to space set of axes only by the effects or rotation of the body axes. Expressed in symbols, we have

 dG   dG        X G.  dt  space  dt body

…(1)

We can find out the axis about which the body and therefore the body set of axes through origin rotate. The procedure is to calculate the components of angular velocity vector with respect to the either system of axes. Now,   and  are, respectively, the angular speeds about the space zaxis, line of nodes and body z-axis or z’- axis. We shall denote them and ,  and . Components of angular velocity vector along the body set of axes: In consequence of the vector property of vectorially to from

,

we can combine these components

 . We shall obtain the components of 

along the body set of axes. Now



is along the space z-axis, therefore, its components along the body axes are found by applying complete orthogonal transformation A=BCD, since three orthogon transformations D,C and B are required to come to body axes.

   x '       y '        z '   

0   A 0      

Giving,

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117



  x '   sin  sin   y '   sin  cos    z '   cos  



….(2)









lies along the line of nodes. We perform orthogonal transformation B to come to body axes

arter rotation  has been performed ; and so the components of



along

x ', y ', z ' -axes are

obtained upon applying the final transformation B:

   x '          y '   B  0      0     z '      giving, 

  x '   cos   y '  sin    z '   



…(3)







Since

 is already parallel to z`-axis, no transformation is necessary. Its only component is z`-

component equal to  .

   x '      0     y '    0        z '        giving,

  x '  0   y '  0   z '   

….(4)





From eqs. (2,3,4), we write the components of

  ,  along x`,y`,z`-axis as follows:

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118 

  x '   sin  sin

  x ' 





  y '  sin  cos  



  z '  cos

cos , 

  y '  

sin 

  z '  0,



  x '  0

…(5)

  y '  0

…(6)





  z ' 

…(7)



Adding the components of the three angular velocities along individual axes we have the components of

       w.r.t. the body axes: x '  ( ) x '  ( ) x '   x ' 

…(8)



  sin  sin   cos 

 y '  ( ) y '  ( ) y '   y ' 

…(9)



  sin  cos   sin 

z '  ( ) z '  ( ) z '   z ' 

…(10)



  cos   Components of Angular velocity along the space set of axes: In an analogous way, we can obtain the components of  along the space set of axes. In going from the body set of axes to space set of axes, all the three rotational operations are to be performed in reverse order i.e.  back,

 back and  back. Therefore, components of 

-1

which is

along the z axis (the body z-axis) will be obtained by multiplying it with A or A . Therefore

   x  0         y   A   0          z      giving, 

  x   sin  sin    y   sin  cos    z   cos 





..(11)







Next,



is along the line of nodes and a rotation  about

z ' axis in reverse order would bring

x ', y ', z ' set to xyz. Therefore, transformation matrix D-1 (=DT because D is an orthogonal matrix) is to be applied. That is

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     x   cos   sin  0           1    y   D  0    sin  cos  0   0         0 1   0  0   0    z      giving, 

  x   cos  

  y   sin    z  0.

…(12)

Finally  itself lies along space z-axis and on transformation is required. Thus

   x   0       y      0      z         giving,

  x  0   y  0   z   . 

…(13)







From eqs. (11), (12) and (13), we write

x  ( ) x  ( ) x    x 



 sin  sin    cos 

 y  ( ) y  ( ) y    y 



  sin  sin    cos 

z  ( ) z  ( ) z    z 



…(14)

 cos    . Henceforth, we shall omit the prime sign to denote the body set of axes, since the body set of axes will be used exclusively. However, this sign may indicate a different set of axes within the body itself.

2.4.1 Angular Momentum of Rigid body: Moments and products of inertia:

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We shall be particularly interested in the rotational motion of rigid body for the simple reason that here only the salient features of rigid body structure have a great deal of influence on its motion. A rigid body can possess simultaneously the translational and rotational motion, and the equations of motion, governing the two types of motion, may involve both the translational and rotational co-ordinates. But we know that the break-up of the problems into two parts can be effected in most physical problems. This is possible as the Lagrangian L=T-V divides into two parts, one involving only the co-ordinates of the centre of mass concerned with translational motion and the other only the angle co-ordinates which may be chosen as Euler’s angles As the reader might be familiar that in rotational motion torque and angular momentum will come into play and a relation between them will be sought. Before proceeding to set up the general equations of motion, it will be necessary to consider carefully the meaning of rotation and in particular, the definition of angular velocity. Angular velocity: Consider the rotation of rigid body about an axis OQ within the body and take, for example, a point P of the body at a perpendicular distance PN from the axis. Then P moves in a circle with centre at N and radius PN. If r is the position-vector of P relative to the origin O, (may be taken as centre of mass) the PN =r sin , where  is angle of r with axis. In a time dt, P would have moved a distance r sin  d, along an arc of angle d. Linear speed therefore is r sin 

d dt

and it must be the magnitude of velocity r:

Fig 2.4 

| r | r sin 

d dt

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 r sin   Since velocity is a vector quantity, the form of its magnitude suggests that if as magnitude of a vector



along the axis of rotation, then

Xr



is taken

is a vector with the magnitude

as desired and we have 

r  v   X r. If either the axis of rotation (direction of

…(1)

 ) or the magnitude

d changes with time,  dt

will be a function of time. We can like wise consider changes in components of space set or body set of axes. If either set of components are represented by

,

along the

x ,  y , z then

  i x  j y  kz Vector property of



suggests that rotation along a certain axis can be decomposed into

rotation about orthogonal axes and vice-versa. Angular Momentum: If the rigid body is taken as a rigid collection of particles, then angular momentum is given by

L   m (ri X vi ) i

  mi (ri X vi ) i

  mi (ri X ( X ri ) i

Using eq. (1) and the fact that



is same for all the particles, we expand the vector triple

product to give

L  mi [ (ri .ri )  (ri . ) ri ] i

  mi [ ri 2  (ri . ) ri ] i

…(2)

In terms of components of vectors involved in this equation, we write

L  i Lx  j Ly  k Lz

 mi [(ix  j  y  k z )ri 2  (ri . ) (i xi  j yi  k zi )] Since (ri . ) is scalar and will be a multiplying factor in every component. Collecting coefficients of i,j,k and equating each, we obtain

i Lx  j Ly  k Lz  mi (ix  jy  kz )ri 2  ( xix  yiy  ziz ) (ixi  jyi  kzi ) 

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 i x  mi (ri 2  xi 2 )   y  mi xi yi  z  mi xi zi  i i  i   j x  mi xi yi   y  mi (ri 2  yi 2 )  z  mi yi zi  i i i  

…(3)

k x  mi xi zi   y  mi yi zi  z  mi  ri 2  zi 2  i i i   Equating coefficients of i,j,k we obtain the components of angular momentum L x, Ly, Lz, in terms of angular velocity components and we see that each is a liner function of all the components of angular velocity, (i.e., where in power of constant

coefficients).

x ,  y , z raised to unity occur in all the terms besides

emphasize

this

linearity

introducing

nine

quantities

I xx , I xy , I xz , I yy , I yz and I zx , I zy , I zz and writing

Lx  x  mi (ri 2  xi 2 )   y  mi xi yi  z  mi xi zi  i i  i   I xxx  I xyy  I xzz

Ly  x  mi xi yi   y  mi (ri 2  yi 2 )  z  mi yi zi  i i i  

…(4)

 I yxx  I yyy  I xzz .

Lz  x  mi xi zi   y  mi yi zi  z  mi (ri 2  zi 2 )  i i i    I zxx  I zyy  I zzz . Moments and products of inertia: The nine coefficients as introduced above, can be written as a 3x3 arrangement of rows and columns as in a matrix notation. We further enhance the linear nature of transformation of



- components into angular momentum components by means of the nine quantities:

 Lx   I xx I xy I xz   x        Ly    Lyx I yy I yz    y    L  I I  z   zx zy I zz   z  (or)

L  I .

Equation (6) implies that when I operates on angular velocity vector

…(5)

…(6)

,

a physically

different vector, the angular momentum, L, results. I is therefore a different physical entity, termed as the moment of inertia tensor. The components of I have been arranged in matrix form in eq. (5) in a way to facilitate understanding of linear transformation : otherwise I is basically a tensor.

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Now we come to the physical meaning of the components of I. the diagonal elements of matrix form,

I xx   mi (ri 2  xi 2 )   mi ( yi 2  zi 2 ) i

I yy   mi (ri  yi )   mi ( zi 2  xi 2 ) 2

...(7)

I zz   mi (ri 2  zi 2 )   mi ( xi 2  yi 2 ) i

are called the moments of inertia coefficients ; as the reader might recall form elementary physics that these are the moments of inertia of the body the x,y,z-axis separately. The less familiar quantities.

I xx  I yx   mi xi yi i

I yz  I zy   mi yi zi i

I zx  I xz   mi zi xi i

are termed as the products of inertia associated with the corresponding co-ordinate planes. Obviously I is a symmetric tensor and is additive in nature; the moments of inertia of a body are the sums of those of its parts. A mathematical structure having nine components in three dimensions is termed a tensor of rank two. In this sense, I is a tensor of rank two and we shall write such an entity in type I to emphasize its tensorial character. We add, however, a word of caution : not every nine quantities can be grouped as a tensor; what is further required is that the quantities must be attached to a co-ordinate system and connected by means of linear transformation to the corresponding quantities in a different co-ordinate system, derived from the first. We shall not take up this point here. We note that there exists a mirror symmetry of elements across the principal diagonal of matrix form of I, i.e. Ixy=Iyz etc. on this account, I is a symmetric tensor. In equation (2), (3) and (4), the matrix elements appear in a form suitable for the body composed of discrete particles. For a continuous body, the summation is replaced by a volume integration, with the particle mass becoming a mass density  (r ) , and so

L    (r )[r 2   (r.)r ] d r. Examples: We now obtain the moments and products of inertia in some simple cases of rigid bodies. It generally turns out that the products of inertia all vanish when one of the axes of the body lies along the axis of symmetry. (a)

Moments of inertia of uniform hemisphere about its axis of symmetry and about an axis lying in the base perpendicular to the symmetry axis. In spherical co-ordinates, the calculation is simple:

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I zz    ( x 2  y 2 )dV   02 0 / 2  0R  r 4 sin 3  d d dr  .

 2 R3  R5 4 2 .  MR 2 where M      mass of the hemisphere 5 3 5  3 

Similarly, moment of inertia about the x-axis:

Fig. 2.4(i)

I xx    ( y 2  z 2 )dV   02 0 / 2  0R  r 4 (cos 2  sin   sin 3  sin 2  ) dr d d R5 5 R5  5 R5  5 R5  5 

 02 0 / 2  0R (cos 2  sin   sin 3  sin 2  ) d d  02 0 / 2 sin   (1  sin 2  ) sin 3   d d    02  cos   / 2 d   02 0 / 2 cos 2  sin 3  d d  0  

2 2   2  2  3  0 cos  d  R5  2 1   2  .  02  cos 2  1 d   5  3 2  2    2  3  R 5  4  2 2     MR , 5  3  5 

R5 5

On putting

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2 R3 M . 3 Also I yy  I xx from symmetry. (b)

Moments and products of inertia of a uniform rectangular parallelepiped with respect to its edges. Consider the rectangular parallelepiped with edges of length a, b, c. The moment of

inertia about z-axis is found to be

 ba 3 ab 3  1 2 2 I zz   c0  b0  0a  ( x 2  y 2 )dx dy dz   c     M (a  b ) 3 3 3   Where

M   (abc)

From symmetry considerations, we find

1 I xx  M (b2  c 2 ), 3

1 I yy  M (a 2  c 2 ). 3

The products of inertia are also not difficult to obtain form the definition:

a 2 b2 1 I xy  I yx       xy dy dz    c   Mab 2 2 4 c b a 0 0 0

1 1 I xz  I zx   Mac, I yz  I zy   Mac. 4 4 Principal axes transformation: The symmetric nature of I bears an important advantage. We can choose a certain system of body axes with respect to which the off-diagonal elements should disappear and only the diagonal elements remain in the expression for I. such axes are called the principal axes of body and the corresponding moments of inertia as the principal moments of inertia. If we denote this form of inertia tensor by I` and I1, I2, I3 stand for the principal values,

 I1 0 0    I '  0 I 2 0  0 0 I  3 

…(1)

And therefore eq. (6.3-5) becomes

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 Lx   I1 0 0    x         Ly    0 I 2 0    y   L   0 0 I    3  z   z   I1 x  0 0     0  I 2 y  0   0  0  I 3 z   (or)

Lx  I1 x Ly  I 2 y Lz  I 3 z That is, each of angular momentum component along a principal axis is a function of corresponding angular velocity component only related to it via principal moment of inertia about that direction. The question of locating axes within the body is not difficult, once we know that I is not symmetric. Such a procedure leads us into the realm of diagonalisation of matrix, a process which merely means that by a suitable transformation of axes we shall kick out the off-diagonal elements, leaving only the diagonal ones. The diagonal form of inertia tensor in which I 1, I2, I3 appear as elements is very straightforward to obtain. We solve the following determinatal equation which will be cubic in I and therefore will furnish three values for I, viz, I1, I2, I3, which are desired principal moments of inertia:

I xx  I

I xy

I yy  I I yz

I xz

I yz

I xz  0.

…(3)

I zz  I

This is called the secular equation of inertia tensor and its solutions, the secular values or eigenvalues. In most of the easy problems in rigid dynamics, the principal axes can be obtained by inspection. For example, if the symmetry axis of the body is taken as axis of rotation and the origin of body axes lies on this, then the principal axes are the symmetry axis and any two perpendicular axes in the plane normal to the symmetry axis. In such cases, we shall have two equal roots of eq. (3), which shall be a clue to this. In the case of a sphere, every axis through centre is symmetry axis and therefore any three orthogonal axes through the centre are principal axes.

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Rotational kinetic energy of a rigid body: th

Refer of fig. (2.4 a) wherein the i particle of a rigid body is shown with position vector r i, referred to a fixed point O of the body. The body is therefore capable of rotation about the axis in the direction of unit normal n.

Fig. 2.4 (a) The kinetic energy of motion of a system of particles is defined by

1 T   mi vi 2, 2i

…(1)

Where the summation extends to all the particles of the body and v i is the linear velocity of the i

particle relative to O. Using the fact that

vi   X ri In a fixed origin, eq. (1) assumes the form

1 T   mi vi .vi 2i 1   mi vi .( X ri ) 2i 1   mi  .(ri Xvi ) 2i Where the last step follows from the inter-changeability of dot and cross in a scalar triple product, keeping the order of involved vectors intact. Rearranging slightly, we find that

1 T   .[  mi (ri X vi ) 2 i 1   . L. 2

…(2)

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Since the quantity within the larger bracket is identified as the angular momentum vector L about O. In terms of moment of inertia tensor I and



, eqn. can be rewritten as

L   mi  ri 2  (ri . ) ri  i

   mi  ri 2  ri (ri .)    i 

…(3)

 I Wherein the operator I is equivalent to the expression within the curly braces. This is legitimate since L is a function of all the components of



and is related to it through I alone. Substituting

eq. (3) into (1), we have

1 T  . I .  2 Let n be a unit vector in the direction of

…(4)



so that

  n and

the kinetic energy expression

looks the same as we are familiar from elementary mechanics:

T

2 2

n.I .n. 

1 2 I , 2

…(5)

provided we are able to show that the moment of inertia I about the axis of rotation

I  n.I .n

(see fig.. 2.4(i))

…(6)

agrees with its usual definition. We now wind up this section after expressing T in terms of components of moments of inertia tensor and the principal moments of inertia. Take up eq. (2):

1 T  .L 2 and break up into components of



and L:

1 [ix  j y  k z ]. [iLx  jLy  kLz ] 2 1  [x Lx   y Ly  z Lz ] 2 1  [x ( I xx x  I xy y  I xzz ) 2   y ( I yx x  I yy y  I yz z )

T

 z ( I zx x  I zy y  I zz z )] 

1 [ I xxx 2  I yy y 2  I zzz 2  2I xyx y  2I yz yz  2I zxxz ], 2

…(7)

using explicitly the symmetry property of inertia tensor viz I xy  I yx etc.

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We have already seen in the coordinate system of principal axes of the rigid body that L has the simple form

L  iI1x  jI 2 y  kI3z

…(8)

and therefore kinetic energy expression also assumes that simple form:

1 T  .L 2 1  [ix  j y  kz ].[iI1x  jI 2 y  kI 3 z ] 2 1 1 1  I1x 2  I 2 y 2  I 3 z 2 , 2 2 2

…(9)

Which is an expression involving the principal moments of inertia.

2.5 Moment of inertia of a rigid body: We shall show that n.I.n is the moment of inertia of the rigid body about the axis of rotation,

n.i.n  n{ mi (ri 2  ri (r.)}n i

By putting up the operator I. Simplifying, we get

n.i.n  n { mi (ri 2 n  ri (ri . n)} i

  mi {(ri 2 (n.n)  (ri . n) (ri . n)} i

  mi {(ri .ri )  (ri . n) (ri . n)} i

Since n.n  1, ri .ri  ri . Taking out common ri outside, we obtain 2

r.h.s.   mi ri .{ri  n (ri . n)} i

  mi ri (n X (ri X n)) i

As n X (ri Xn)  (n.n)ri  (ri .n)n Now, by virtue of the property of scalar triple product,

r.h.s.   mi (ri X n).(ri X n) i

  mi | ri X n |2  I i

Referring to fig. (2.4(i) we note that (ri x n) is the perpendicular distance of the j particle from the axis of rotation. Therefore, it is indeed the usual expression for moment of inertia of the body about the indicated axis

2.6 Equations of motion of a rigid body: Euler’s Equations:

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The motion of a rigid body with one point fixed will take place under the action of a torque N in such a way that its total angular momentum varies at the rate equal to N:

dL  N. dt

…(1)

Here the time-derivative refers to the space-axes, for, the equation holds only in an inertial system. In a coordinate system rotating with body, we have the following relation between the two time derivatives.

(d / dt )space  (d / dt )body   X L ...cf eq. (6.2  1) Eq. (1), in terms of body-axes is, therefore,

 dL    X L  N.    dt body

…(2)

It is found most convenient to choose the principal axes for which

L  I1xi  I 2y j  I3z k Where

…(3)

x ,  y and  z are the components of the angular velocity vector along the principal axes. From eq. (3) and remembering that the principal moments of inertia and the body base

vectors, i,j,k are constant in time with respect to the body co-ordinate system, we find that the time derivative of L, that is dL/dt, in rotating system is     dL     I1  x i  i2  y j  I 3  z k  dt body

…(4)

The x-component of equation (2) is obtained as 

Nx  N .i  I1  x  ( X L) x 

 I1  x  ( y Lz  z Ly ) 

 I1  x  ( I 2  I3 ) zx

…(5)

From eq. (3) By cyclic permutation of x, y, z we obtain the remaining two equations, viz, 

N y  I 2  y  ( I 3  I1 )z x

…(6)



N z  I 3  z  ( I1  I 2 )x  y

…(7)

These three equations are the Euler’s equations for the motion of a rigid body with one point fixed. Largange’s Method: An alternative approach to these equations is the Lagrange’s method using the Euler’s angles as the generalized co-ordinates. When one point of the body is fixed, only the rotational

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motion is possible and the Euler’s angles completely describe the orientation of the rigid body. In such a situation, the Lagrangian L, which is (T-V), should be a function of    only. Thus T will only involve these co-ordinates and likewise. V, the potential energy also. For conservative forces, the lagrangian can be written as 



L  T ( , , ,  ,  , )  ( , , ) 1  ( I1 x2  I 2 y2  I 3z2 )  V ( , , ), 2

…(8)

Where we have chosen preferentially the principal axes as the body set of axes; I 1, I2, I3 are the principal moments of inertia for the fixed point and the kinetic energy depends on

 , ,

via the

angular velocity components along the principal axes, x,y,z (cf. equations (2.4-8,9, 10). It will be remembered that the generalized forces corresponding to these angular co-ordinates will be the components of the impressed torque along the axis of rotation (cf, the section on generalized forces). The angle  happens to be the angle of rotation about the principal z-axis, so that 

angular velocity

Nz  

  

and we have generalized force or the z-component of impressed torque

V , 

And the Lagrange equation for  -coordinate can be written as

d  T dt   

 T V      

…(9)

=Nz For convenience, we reproduce here the angular velocity components expressed in terms of Euler’s angles and kinetic energy expressed in terms of

x ,  y and z

as referred to

the principal axes : 



x   sin  sin   cos  y   sin  cos   sin 



z   cos   and

1 T  ( I1x 2  I 2 y 2  I 3z 2 ) 2

From these, we obtain

z 



 1,

x 



 0,

 y 



 0,

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 y x z  0,  y ,  x ,   

T 





T x T  y T z    x   y  z 

= I1x .0  I 2 y .0  I 3z .1  I 3z

T T x T  y T z     x   y  z 

(and)

 I1x y  I 2 y (x )  I 3z .0  ( I1  I 2 ) x y . Substituting the values of

T T and in eq. (9), we obtain  

d ( I3z )  x y ( I1  I 2 )  N z dt I3z  x y ( I1  I 2 )  N z

(or)

…(10)

Which is identical with the z-equation obtained earlier. The other two equations can now be obtained simply by permutation of indices in eq. (7); we are justified in doing so due to the fact that identification of one of the principal axes as the z-axis is entirely arbitrary. It should be emphasized, however, that these two equations shall not be the generalized equations for  and  coordinates. Equations of motion about fixed axis: The equations of motion in the case of a rigid body with a fixed axis simplify further as only one component of angular velocity appears, namely along the axis of rotation, say z-axis ; then

z  ;

y  x  0.

Consequently from eq. (2.4). viz.

L  iLx  jLy  kLz

 i ( I xxx  I xyy  I xzz )  j ( I yxx  I yyy  I yzz )  k  I z xx  I zyy  I zzz  We obtain, on setting

z   ,  x   y  0 ,

L   (iLxz  jI yz  kI zz ),

…(11)

and eq. (1) the torque equation, becomes,

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dL d d  (iI xz  jI yz  kI zz )   (iI xz  jI yz  kI zz ) dt dt dt

di dj dk     (iI xz  jI yz  kI zz )    I xz  I yz  I zz  dt dt dt   Since Ixx etc., are constants when the point of the body is fixed. Also, since

x   y  0

di   X i  (ix  j y  kz ) X i dt  jz  j

dj   X j  (ix  j y  kz ) X j dt  iz  i

dk   X k  (ix  j y  kz ) X k dt =0 In view of these equations:

N  iN x  jN y  kN z 



 i ( I xz   I yz 2 )  j ( I yz   I xz 2 )  k ( I zz  ).

…(12)

Equating x, y, z components, we get three equations: Viz. 

N x  I xz   I xz 2 , 

N y  I yz   I xz 2

… (13)



N z  I zz . Form eqs. (13), it appears that only the last of them is consistent with the physical motion, while the first two determine the constraining torques which are required to prevent the axis of rotation from changing its direction. If the axis of rotation z happens to lie along one of the principal axes of the rigid body, e.g. the axis of symmetry, then we obtain the equations. Nx = Ny =0 Since Ixz = Iyz = 0 and 

N z  I zz  

…(14)

 I ,

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I being the only principal moment of inertia, that is, the moment of inertia about the axis of rotation. Kinetic energy, T, then has the simplest form

1 1 T  . L  I zz 2 2 2 1  I 2 2

…(15)

2.7 Torque Free Motion of a Rigid Body: Poinsot Solutions: One of the problems to which Euler’s equations can be conveniently applied is that in which there are no net forces or torques on the body; for example, the rotation of the earth can be described in this way as a fairly good approximation. In this problem, the centre of mass is at rest of movers uniformly so that the angular momentum is only due to motion about the centre of mass. Then we can choose the centre of mass as a fixed point and the origin for the principal axes in the body. Thus when N=0, Euler’s eqs. (2.6-5, 6,7) become 

I1  x   y  z ( I 2  I 3 ) 

I 2  y   z  x ( I 3  I1 )

…(1)



I 3  z   x  y ( I1  I 2 ). It is possible to integrate these equations in terms of elliptic functions with initial conditions of kinetic energy and total angular momentum, which are the integral or constant of the motion; for, there are no net forces acting on the system. The system is conservative and Euler’s dynamical equations furnish following two constants of motion:

(i)

Conservation of Kinetic Energy:

Multiplying eq. (1) by 



x ,  y , x

respectively and adding gives



I1x  x  I 2 y  y  I3z  z  x  y z ( I 2  I3  I3  I1  I1  I 2 ) = 0. But

 d 1 ( I1x 2 )  I1x  x etc dt 2

and therefore the left hand side is

d 1 1 1 ( I1x 2  I 2 y 2  I 3z 2 )  0 dt 2 2 2 The gives the conservation of kinetic energy, i.e.

T

1 1 1 I1x 2  I 2 y 2  I 3z 2  constant. 2 2 2

….(2)

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135

Conservation of angular momentum:

Another constant of motion is the angular momentum L, since we have presumed N=0, where

N

dL  0 and therefore L is conserved: dt

L  iI1x  jI 2 y  kI3z 

…(3)

We shall describe a very interesting solution of the problem due to poinsot. This is a geometrical representation of motion of rigid body based on the integrals of motion, eqs. (2) and (3) developing into the motion of inertia-ellipsoid. How this can be achieved is shown below: Inertia Ellipsoid Let us juxtapose eqs. (2) and (3) use alternative expression of T in terms of L: L= constant 2T=  L= constant Consider an angular velocity space in which every point corresponds to a set of values of

x ,  y , z

also,

.L  I xxx 2  I yyy 2  I zzz 2  2I xyxy  2I yzyz  2I zxzx …(5)

(for derivation of this see eq. (2.4 -7). Combine eq. (5) with eq. (4) to obtain

I xxx 2  I yyy 2  I zzz 2  2I xyxy  2I yzyz  2I zxzx  2T  constant.…(6) This is the equation of a quadric surface in - space, known as inertia ellipsoid. By proper choice of the axes in the body, the products of inertia may be made to vanish (see principal axes transformation) and the quadric surface becomes

I1x 2  I 2 y 2  I3z 2  2T

…(7)

In eq. (6) the axes of inertia-ellipsoid are not parallel to co-ordinate axes (i.e.

x ,  y , z .

But now the axes of the inertia ellipsoid lie along the principal axes of the body which are fixed in the body. The magnitude of the vector from the origin of co-ordinates in the -space to any point on the ellipsoid represents a possible value of the angular speed consistent with the given kinetic energy and angular momentum (since from eq. (4), for a given T and L magnitude, i.e, components

x ,  y , z



must have a constant

will change subject to this condition). The situation here

must be contrasted with that existing in the corresponding case of linear motion of a particle, where the magnitude of the linear velocity is completely dictated by the kinetic energy but here the angular momentum is important. It is apparent that in the direction of the minor axis of the ellipsoid (the principal axis of maximum moment of inertia), the angular speed is least, since

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136

1 2 I 2

Let us consider how the rigid body behaves on a fixed plane, under no torque acting, consider a neighbouring point on ellipsoid, referred by vector

  d .L  cons tan t , L  cons tan t Differentiating, we have

d (.L)  d .L  .dL  d .L d .L  0

(or)

Now, d  , since it connects two neighbouring points on the surface, lies in the tangent plane at the point in question, therefore from (8), tangent plane is perpendicular to L. But L is a constant vector from eq. (4). Therefore is this case the plane remains fixed in position relative to fixed system of axes. However, we now show that the tangent plane remains at a fixed distance from the origin of set of axes, as the motion proceeds. This plane on this score is called the invariable plane. L in eq. (4) represents the product of the component of



in the direction of L by the constant

magnitude of L; this equation requires that

.L   cos  L  2T (or)

 cos  



2  cons tan t. L

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Fig. 2.7 Motion of the inertia ellipsoid relative invariable plane. Refer to Fig (2.7)

 cos 

represents the perpendicular distance of origin O from the

invariable plane. It follows that the tangent plane remains at a constant distance from the origin as the motion goes on. If the origin is fixed, the tangent plane remains fixed both in orientation and position. The force-free motion of the rigid body can be then pictured as the motion of the inertia ellipsoid which rolls, without slipping, on the invariable plane, with the centre of the ellipsoid always at a constant height above the plane. Since the points of the body, which lie along the instantaneous angular velocity vector or the axis of rotation, have zero velocity, it follows that the point of contact between in poinsot ellipsoid (inertia ellipsoid) and the invariable plane (such a point is denoted by the radius vector



which is along the instantaneous axis of

rotation) has zero velocity momentarily, with respect to the plane. This explains why rolling occurs without slipping. The curve traced out by the point of contact on the ellipsoid is known at the polhode, while the similar curve on the invariable plane is called the herpolhode. The polhode is always closed curve, since the inertia ellipsoid would move in order to preserve the height of its origin from the invariable plane. In the case of a symmetrical body for which I1=I2=I, the poinsot ellipsoid is an ellipsoid of revolution. The angular velocity vector will thus remain constant in magnitude and be observed to process about the invariable line OO’ perpendicular to the invariable plane. The herpolhode is a circle on the invariable plane, centered at O, when vector  traces out a cone while performing processional motion; while the tip of



describes the circular polhode on the ellipsoid about the

symmetry axis (fig (2.7 a)).

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Fig. (2.7). I1=I2 for a symmetrical body. Physically this means that the direction of

ď ˇ

precesses in time about the axis of symmetry of the

body. When the ellipsoid is in motion, the vector*

ď ˛ traces out a cone in space, referred to as the

space cone. With respect to the body coordinate system, the angular velocity vector is also constant in magnitude and it like wise traces out a cone with respect to the rigid body, which is referred to as the body cone. The two cones are tangent to each other along the instantaneous axis of rotation. The motion of the symmetric body will thus appear as the rolling of the body cone on the space cone [fig. (2.7 b)] and conversely.

2.8 Force-free motion of symmetrical rigid body: We may also confirm the processional motion predicted by the poinsot method by analytical treatment. If we choose the symmetry axis as the principal z-axis, so that I1=I2 and eqs. (1) simplify to

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139



I1  x  ( I1  I 3 ) z  y 

I1  y   ( I1  I 3 )  z  x

…(9)



I3  z  0 We see at once from last of the eqs. (9) that

 z = constant,

…(10)

That is, the component of the angular velocity along the symmetry axis of the rigid body is a constant. If we now let



I1  I 3 z = constant, I1

…(11)

If n is the unit vector in the direction of

,

i.e.

  n ,

then define

 as n / I .

then we

can write the first two of eqs. (9) simply as

 x    y  

.  y   x. 

…(12)



Differentiating the first of eqs. (12) and eliminating 

y

between them, we obtain



 x   y 

 2 x (putting for  x )

…(13)



 x   2x  0

(or)

Which is a simple harmonic motion with an angular frequency, ,

x  A sin  t ,

…(14)

Where A is some constant and we have chosen initial condition, find

y

x  0,

when t=0. We can

by combining eq. (14) with eq. (12) ; the result is 

y 

x 

A cos  t.

…(15)

The solutions (14) and (15) show that the vector

 p  x i   y j ,

|  p |  A

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Fig. 2.8 has a constant magnitude and rotates uniformly about the z-axis with constant angular frequency -  as illustrated in fig. (2.8 a). Therefore, the total angular velocity

  x i   y j , z k   p  z k is also constant in magnitude and precesses about the z-axis with same constant angular frequency. This precessional motion is illustrated in fig (2.8 b), where we see that



sweeps

out a cone about the z-axis with the same constant angular frequency -. This motion, of course, takes place with respect to the principal axes of the body, which are themselves rotating in space with the angular frequency



which is large in comparison to . We also see. From equation

(11), that the more nearly equal I1 and I3 are, the slower will be processional velocity  as compared to

z .

The results can be applied to rotating earth, for it may be considered to be torque-free body. The earth is nearly symmetric about the polar axis and slightly bulged at the equator so that

I 3  I1. It is found that

I 3  I1 1  I1 300 and

z 

2 day

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Hence we would expect to find the period of procession of the axis of rotation about the symmetry axis to be

T

I 2 2  . 1  300 days  z I 3  I1

An observer on the earth would find the axis of rotation of the earth tracing out a circle about the north pole every 300 days (10 months). This resembles roughly with what is actually observed (nearly 14 months). The Motion of a symmetric top under the action of gravity: The symmetric top that we discuss now is approximated to a variety or rigid bodies, like the child’s top and gyroscope. It pivots around a fixed point O on the symmetry axis and we take the distance of O from G, the centre of mass, to be l. Taking body z`-axis, the symmetry axis, as the principal axis, we put I1=I2. Only force acting on the top is the force of gravity mg acting at G downward. (Fig. 2.8.1). The Euler’s angles are the most convenient set of generalized coordinates to describe the motion in this case. We find the Lagrangian L for the top:

L  T V 

1 1 I1 ( 2 x'  2 y ' )  I 3  2 z '  mgl cos  , 2 2

Which, on substituting for angular velocities in terms of Euler’s angles, from eqs. (8), (9) and (10) of art. 6.2* becomes

L

 2 I   1 I1 ( 2   sin 2  )  3 (   cos  ) 2  mgl cos  . 2 3

…(1)

Our aim in taking up the present discussion is to deal with the interaction of various types of motion that the top executes. As we know that, apart from translational motion, the top has following three types of motion: Precession

Which is rotation about space z-axis. Such a rotation corresponds to

Which is rotation about intermediate x1 axis or line of nodes. Such a

angle  Nutation

rotation corresponds to angle . Spin

Which is rotation about z` axis. Such a rotation corresponds to angle .

The spin velocity about z -axis is



 z ' given by ( ( cos   ) .

Therefore, we shall take up different cases, e.g. to see how the precession is affected when the notation is absent or when the notation is present. Obviously, notation corresponds to motion of body z axis between two  angles, say 1 and 2 in z y plane Fig. [2.8.2 b]. As the '

precession about z-axis is present, motion of z` axis from 1 to 2 or vice versa will not be '

confined in a plane z y but will be delineated, i.e. z`-axis will trace out a curve as shown in fig (2.8.3) while moving from 1 to 2 or vice versa.

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First integrals or equations of motion: Since the Lagrangian does not contain the Euler’s angles  and  and time t, i.e. they are cyclic, the corresponding momenta p p and total energy E are constant in time. 



x '   sin  sin   cos  y '   sin  cos   sin 



z '   cos  

Fig. 2.8.1. Coordinates for the symmetric top. These three integrals of motion are expressed by

p 

L 





 I 3 (   cos  )  I 3z '  I1a.

…(2)

Where a is a constant and

p 

L 





 I1  sin 2   I 3 cos  (   cos  )  I1b,

…(3)

Where b is a constant, and

T V 

I I1  2  2 (   sin 2 )  3  2 z '  mgl cos  2 2

…(4)

= E, Equations (2,3) may be solved for  and , yielding

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 

and



143

b  a cos  sin 2

…(5)

I1a  b  a cos   I 3  sin 2 

  cos  

…(6)

Substituting (5) and (6) in the energy equation, we obtain

 I a E' E  1 2I

1 2 1 2 2  I1   I1 sin   mgl cos  2 2 2

1  2 1  b  a cos    I1   I1    mgl cos  2 2  sin  

...(7)

*We find the P is the component of angular momentum along the body z`-axis and p is the component of angular momentum along the space z-axis or vertical. In principle, the essential features of motion of the symmetric top can described with the of equations (2,3,4) without solving equations of motion. This equation, when solved for  as a function of t, involves the use of elliptic integrals which tend to introduce complexity. Therefore some other ways must be found out which describe the phenomenon qualitatively. One such way is that of effective potential in equation (7) which, like the potential V(), is a function of  alone:

1  b  a cos   V '  ( )  mgl cos   l1   2  sin  

…(8)

This is a one dimensional problem in  co-ordinate. In fig. (2.8.2 a) we plot the effective potential as a function of . other values of



v ' () assumes infinite values for =0 or  and finite values for all

and, it follows that for some values of

minimum value. In figure

0

at which

,v'

assumes a

  b  a cos  cos    I1 sin   2

I1 (b  a cos  ) (a  b cos  ) 0 sin 3 

For a particular value of energy between the two values

between O and

v ' is minimum must therefore be a solution of the equation:

 b  a cos  dV '  mgl sin   I1a  d  sin   mgl sin  



 I a E' E 1

1 and 2 of  ,

2I3

…(9)

,  motion is a bound motion, confined

which are the roots of the equation

E '  V '( )

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1  b  a cos   E '  mgl cos   I1   2  sin  

(or)

144 2

…(10)

The variation in the angle  is referred to as the nutation of the symmetry axis of the top and is an up and down motion of the symmetry axis. If, however, the minimum effective potential equals the energy E , the angle  keeps fixed at value 0 and the top processes with the constant '

angular velocity given by 



b  a cos 0 sin 20

…(11)

about the vertical axis. Precession without Nutation: We shall calculate angular velocity of precession and spin, when nutation is absent i.e., z’-axis remains fixed at 0, From eq. (9), it is obvious that, for a given angular frequencies of precession

0 ,

0 given by*

 2



 I  2[ I 32  4mgl ( I1  I 3 ) cos 0 ] 0  3 2( I1  I 3 ) cos 0 

there will be two

Fig. 2.8.2 a

…(12)

Fig. 2.8.2 b

Thus corresponding to positive sign of the radical, precession will be fast while that will negative sign is called slow precession. Since  must be real, the quantity under the radical sign must be positive definite., i.e.,  2

I3   4mgl ( I1  I 3 ) cos 0  0 2

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 

145

4mgl ( I1  I 3 ) cos 0 I 32

…(13)

*From eqs (9) and (5), we find that 

mgl sin 0   0

I1a  I1b cos 0 0 sin 0



(or)

    mgl sin 0  0 [ I 3 (   0 cos 0 )  I1  0 sin 2 0 cos 0 sin 0



 I3 cos 2 0 ( 0   0 cos 0 )]  0 

(or)

 2



mgl sin 2 0   0 [ I 3   I 3 cos 2 0 ]   0 [ I 3 cos 0  I1 sin 2 0 cos0 I3 cos 2 0 ]  0  2



(or)

mgl sin 2 0  I3  0 sin 2 0   o [ I3 sin 2 0 cos 0  I1 sin 2 0 cos 0 ]  0  2



(or)

 0 [( I3  I1 ) cos 0 ]   0  I3  mgl  0

(or)

 0 [( I1  I3 ) cos 0  I3  0  mgl  0

 2



Which is quadratic on

 0 and gives the roots shown in eq. (12).

Thus this equation limits the values of



. Further, we can also solve eq. (12) for

for which steady precession may occur at the angle

0 in

terms of

z ',

the angular velocity about

z ' -axis

called spin angular velocity and is given by* 

0  Putting for

I 3z ' 

( I 32z '2  4mgl I1 cos 0 2 I1 cos 0

0

…(13A)

0 from eq. (5)

1/ 2 1 I 3z ' sin 2 0   4mgl I1 cos 0   1  1   (b  a cos 0 )    2 I1 cos 0   I 32 z '2    1/ 2 1 I 3 sin 2 0  '  '2 4mgl I1 cos 0    z   z     ….(14) 2 2 I1 cos 0  I 3    

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0   / 2 ,

146

physical motion is possible and as stated previously, for uniform

precession, quantity under radical must be positive, i.e.,

I32z '2  4mgl I1 cos   0

 4mgl I1 cos     I 32  

(z ' )min 

(or)

This is the minimum spin angular velocity uniformly at the angle

 0 and

…(15)

z '

below which the top cannot process

is given by equation (15). When

z '  (0' )min

quantity under



radical will not be zero and hence two values of 

0 

 0 are given by

(b  c cos 0 )  I 3 z '  sin 2 0 I1 cos 0

From eq. (2) we find that 



I 0   I3z '   I3  0 cos 0 On putting in eq. (12), we get  '



0 

( I 3z '  I 3  z cos 0 ) 

 '

[( I 3z '  I 3  0 cos 0 )2  4mgl  I1  I 3  cos 0 ) 2( I1  I3 ) cos 0  '



 2

[2( I1  I3 ) cos 0  0  I3z '  I3  0 cos 0 )2  I32z '2  I32  0 cos2 0

(or)

 2

2 I32z '2  0 cos2 0  4mgl ( I1  I 3 ) cos 0  2

 2

 0 .4 ( I1  I3 )2 cos2 0  I32 .z '2  I32z '2  0 cos2 0

(or)



 2

4 0 ( I1  I3 ) cos 0 I3z  4( I1  I 3 ) cos 0 I 3 cos 0  0 '



2I32 z ' . 0 cos 0  I32 z '2  2 I32z '  0 cos 0

(or)

 2

 I3  0 cos 2 0  4mgl ( I1  I3 ) cos 0 2



 2

 0 [4 ( I1  I3 )2 cos2 0  4 I3 ( I1  I3 ) cos 2 0 ] 

  0 4( I1  I3 ) I3z ' cos 0  4mgl ( I1  I 3 ) cos 0  0



 2



 0 I1 cos 0  I3z '  0  mgl  0

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147



It is quadratic in

 0 and gives eq. (13A).

and

0 



(b  a cos 0 )  mgl  sin 20 I 3z '

…(16)

 z ' ).

It is the slow precession which is ordinarily observed with a rapidly spinning top (large Nutational Motion:

Let us now investigate the nutational motion that we have mentioned earlier. At the angle

 0 the

effective potential V` (  ) has a minimum value ; it corresponds to a point of stable

equilibrium,

1 and  2 being the turning angles corresponding to the effective energy E'.It follows,

therefore, that situations exists for which the nutational motion of the axes of the top can be described as a frequency of these oscillations and given by

 1  d 2V '( )   *0       0 2   I1  d

…(17)

We find that   2  2 d 2V '( ) 2 2 |  I [ a  4 a  cos   3  cos    0 0 0 ].  0 1 0 0 d 2

The angular frequency of oscillation about

0

obviously depends on spin velocity  z . through a '



and on

…(18)



 0 . This motion is delineated by the processional angular velocity  0 which is dictated by

spinning velocity

 z ' . It is customary to

depict the motion of the top by tracing the curve of the

intersection of the figure axis (symmetry axis) on a sphere of unit radius. This curve is then the locus of the figure axis. The polar co-ordinates of a point on the locus are identical with the Euler angles

, 

for the body system. The bounding angles

1 and 2 on the unit sphere satisfy the

equation

The potential energy, in the neighborhood of stable equilibrium can be expended in Taylor series expansion i.e.

V ( x)  V ( x0 ) 

dV 1 d 2V |x0 ( x  x0 )  | ( x  x0 ) 2  .... 2 x0 dx 2 dx

Where V ( x0 ) is the value of potential energy at equilibrium position and is thus constant. Derivatives of V are taken at equilibrium position. We known that potential is extremum at equilibrium, so that

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148

dV |x 0 dx 0

E '  V '( ), in which P , P and E are determined from the initial conditions. If we

(10), when

multiply this equation by sin

 , it becomes a cubic equation in cos  . Referring to fig (2.8.2 a),

there are two real roots cos 1 ,cos 2 between -1 and +1. The third root can, in fact, be shown to be greater than 1, on a consideration of eq. (10) and is not of physical interest. Initially, when the top is spinning fast, nutation motion is absent and cos 1  cos 2  cos 0 . If initially



  0 , the

initial value cos 1 of cos  satisfies equation (10). During nutation, the processional velocity varies in accordance with eq. (5). If

| b | | a | or | P |  | P |, We can define angle  3 as follows:

cos 3  In view of equation (5), if sign. Also

P P



b a

  3 , 

…(19)

has the same sign as

 z ' and for   3 , it has an opposite

dV ' |  0 is negative, i.e. the slope of the potential curve in fig. (2.8.2 a) is negative, or, d

the angle between tangent to the curve and

 axis is greater than

Therefore,

V ( x) 

1 d 2V | ( x  x0 ) 2 2 x0 2 dx

The further terms are taken zero because higher power of ( ( x  x0 ) will be small enough (the deviations from equilibrium position are being taken as small) Comparing above expression with

V ( x) 

1 k ( x  x0 ) 2 2

We arrive at

k

d 2V |x dx 2 0

If the problem refers to rotation, then x will be replaced by

 , so that

d 2V k  2 | 0 dx

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149

  0 refers to equilibrium position. Thus frequency of oscillation will be given by 1/ 2

2  k  1  d V   0       2    I   I  dx  0 

90 , consequently

3  0 . If 3  1



then

 has same sign as  z '

throughout the nutation, and 

the figure axis traces out a curve shown in fig (2.8.3 a) on the unit sphere. If

3  1 ,  changes

sign during nutation. The corresponding motion is shown in fig (2.8.3 b)

Fig. 2.8.3 Fast top: The fast top is a top spinning very rapidly about the figure axis with angular velocity and is initially at rest at an angle

z '

1 and then released. For this, we have the initial conditions:

(i)

Initial angle,

  1 ,

(ii)

Initially at rest

  0,   0,

(iii)

Spinning with



z '



…(20)



   0  z '

And therefore equations (5,6,7) become

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     I 3  0 cos 1  b  I1  E '  mgl cos 1     

I a 3 0 I1



In motion,

…(21)



 and 

begin to differ from their initial zero values so that first two terms of eq. (7).

Will now have some value. Consequently, to keep E constant, the term

mgl cos  should

decrease. In other words, energy can be conserved only by a decrease in potential energy that is, by an increase in

 . The initial  3 is, therefore, the same as 1 , the minimum value  can have

b    from eq.21,  cos 1  When released in this manner, the top always starts to fall, and a   continues to fall until the other bounding angle axis then to rise again to

 2 is reached, precessing mean while. The figure

1 , the complete motion being shown in fig (2.8.3 c)

1 2 I1a  mgl 2 i.e. the rotational kinetic energy about the symmetry axis is much larger than the maximum 

allowable change in potential energy, * then we find

 is small and to a frist approximation, we

can set

d 2V ( ) | 0  I1a 2 [cf .eq.(18)] 2 d

0  a

(or)



I  3 0 I1 Let



…(22)

be the amplitude of this sinusoidal (nutational) motion in the unit of

place about the mean value

  angle

and take

 0 , obviously,

0  1   And therefore, at any time, angle

 is a function of time, varying sinusoidally about the value  :

  0   cos t  1     cos t.

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So that* 



a . (   cos t ) sin 1

Obtained on using eq. (5) * 1 can assume a maximum value

 / 2 at

which the top will fall and a minimum value zero at

which the top will be vertical. So maximum allowable change in potential energy will be

 E '  mgl (cos 00  cos  / 2)  mgl 

* with the value of  ,   

 

b  a cos  , then yields sin 2 

b  a (cos (1     cos t )) sin 2 [1     cos t ]

b  a [cos 1 cos (   cos t )  sin 1 sin (   cos t )] [sin 1  cos (   cos t )  cos 1 sin (   cos t )]2

The amplitude of oscillation is small, and we can write 

 is given by

The average of



  

 02

a (   cos t ) d (t ) sin 1  02 d (t ) 2



 a  a  .2   sin t  sin 1  sin 1 0 2



I Putting a  3 0 , we get I1 

  

I 3 0 . . I1 sin 1

…(23)

cos (   cos t )  1,sin (   cos t )     cos t 

 

b  a [cos 1  sin 1. (1  cos wt ) [sin 1  cos 1.(   cos wt )]2

b  a [cos 1  sin 1.(   cos wt )] sin 2 1  2sin 1.cos 1 (   cos wt )

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152

Neglecting the term cos (   cos wt ) in the denominator since being the square of 2

(   cos wt ) , it is very small.

b  a  cos 1  sin 1.(   cos wt )  a    sin 1 [sin 1  2 cos 1.(   cos wt )] 

But from eq. (20),

b  cos 1 a a sin 1.(   cos wt ) (   cos wt ) X sin 1 [sin 1  2cos 1.(   cos wt )] (   cos wt )



 

a sin 1.(   cos wt ) (   cos wt ) sin 1 [sin 1 (   cos wt )  2cos 1.(   cos wt ) 2]

We can neglect the terms containing square of 

  But

(   cos wt ) , so that

a sin 1.(   cos wt ), (   cos wt ) sin 1 sin 1.(   cos wt )

a (   cos wt ) sin 1

* 

mgl sin 1 I1 a 2

…(24)

Therefore, eq (23), becomes 

I mgl sin 1    3 0 . I1 sin 1 I1a 2 



I I2  3 2 0 . mgl. 1 2  I1 I 32  0



mgl 

I3  0

(Putting for a)

…(25)

Thus we conclude about Fast Top that: (i) Since I1a

mgl

The maximum notation angle is

2

given by

2  1  2

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1 and  2

We have mentioned earlier that turning points

1

and

153 are the two roots of eqs. (10) and therefore at fixed

 2 (i.e  is zero), we have

1  b  a cos  2  E '  mgl cos 1  mgl cos  2  I1   2  sin  2 

1  b  a cos (1  2 )  mgl cos 1  mgl cos (1  2 )  I1   2  sin (1  2 ) 

(or)

 mgl [cos 1 co s 2  sin 1 sin 2 ]

1  b  a cos 1 cos 2  a sin 1 sin 2   I1   2  sin 1 cos 2  cos 1 sin 2 

 mgl cos 1  mgl sin 1.2

1  b  a cos 1  a sin 1 sin 2   I1   2  sin 1  2 cos 1  b  cos 1  sin 1 2 1 2 a mgl sin 1.2  I1a  2  sin 1  2 cos 1 

(or)

    

 cos 1  cos 1  sin 1 2  1  I1a 2 .4 2   2 2  2 sin 1  4 cos 1 

 2 sin 1  2  2 I1 a   neglecting 4 cos 1 being small.  2 sin 1  2

Therefore,



mgl sin 1 I1a 2

 , the amplitude of natuation, is small as inferred from eq. (24).   0     cos wt , nutation is sinusoidal (simple harmonic)

(ii)

since

(iii)

since I 3  0  I1a 



mgl a

   is small as inferred from eq. (25). Thus the precession is slow. (iv)

Since

  a

Frequency of nutation is large.

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154

Thus the first top released from rest, precesses slowly and nutates simple harmonically with a large frequency and small amplitude. Sleeping Top: As a final example, we consider a top which is initially spinning with its symmetry axis vertical. In this case 



p  p  I3 (   )  I 3z ' ;

 and 

For

V '( ) 

are indistinguishable initially and lie along vertical. The effective potential then is

 I32z '2  (1  cos  )2 2 I mgl   cos   where   22 '2  2 2 I1  sin  I 3 wz 

For a rapidly spinning top ( 

1 ), the spinning motion is stable; if disturbed, it will exhibit a 2

small nutation about the vertical axis. The minimum spin angular velocity below which the top cannot spin stably about vertical axis when



1 is given by 2

1/ 2

min

 4mgl I1    2  I3 

If initially

z  z i  min, a

top will spin with its axis vertical continuously (therefore it is called a

‘sleeping; top). But when friction reduces is

min it will begin to wobble.

Few worked exampled 1. If, A,B,C; F,G,H are the moments and products of inertia of a rigid body about three mutually perpendicular and concurrent axes, prove that the moment of inertia of the rigid body about an axis making angles

,  ,

with the original axes is given by

I  A cos2   B cos2   C cos2   2F cos  cos   2G cos  cos   2H cos  cos  Take the three axes as the x,y,z axes of a Cartesian system. Consider a particle P of mass m at point r relative to origin O and n a unit vector whose components are cos cos  ,cos  and cos  . Let PM be the perpendicular drawn from P to the given axis. If

 is the angle between r and n.

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155

Fig.

PM  r sin  | rXn | but

r  ix  iy  ky

and

n  cos  i  cos  j  cos  k

Therefore

rxn  ( y  z cos   z cos  )i  ( z cos   x cos  ) j  ( x cos   y cos  )k PM 2 | rxn |2  ( y cos   z cos  )2  ( z cos   x cos  )2  ( x cos   y cos  )2

and

Moment of inertia of the whole body is given by I  m PM

 m( y 2  z 2 ) cos2   m( z 2  x2 ) cos2  )  m( x2  y 2 ) cos2 

2myz cos  cos   2mzx cos  cos   2mxy cos  cos   A cos2   B cos2   C cos2   2F cos  cos   2G cos  cos   2H cos  cos  , By the definition of coefficients of moments of inertia and products of inertia. 2. Find the kinetic energy of rotation of a rigid body with respect to the principal axes in terms of Eulerian angles and interpret the result when I1=I2. The kinetic energy is given by

1 T  [ I1x 2  I 2 y 2   I 3z 2 ] 2 Where

x ,  y , z , are

the components of angular velocity vector w.r.t. to the body set of axes

and and I1 , I 2 , I 3 are the principal moments of inertia along the body axes taken as the principal axes of the body. Putting the value of

T

z

etc. in terms of Eulerian angles,

 , , we get

   1  1 1  I1[ cos   sin  sin ]2  I 2 ( sin   sin  cos )2  I 3 ( cos   )2 2 2 2

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When I1=I2 this becomes

T

  1 2 2 1  I1[ sin    )2  I 3 ( cos   ) 2 2 2

This case happens when the rigid body is symmetric about one of the principal axes, namely the body z-axis 3. If T be the kinetic energy, G be the external torque about the instantaneous axis of rotation and



the angular velocity, then prove that

dT  G. dt Let

x ,  y , z , be

the components of angular velocity vector



along the body set of

axes (also the principal axes) Then

   x i   y j  z k

Also G lies along the same direction as

:

G  Gxi  Gy j  Gz k Now in these axes,

T So that

1 1 1 I1x 2  I 2 y 2  I 3z 2 2 2 2

dT  I1xx  I 2 y y  I3zz dt  x [Gx  ( I 2  I3 )yz ]  y [Gy ( I3  I1 )zx ]  z [Gz ( I1  I 2 )x y ]

Form the Euler’s equation of motion. Hence

dT  Gxx  Gy y  Gzz dt  G.. 4. A rigid body, symmetrical about an axis, has one point fixed on this axis. Discuss the rotational motion of the body when there is no other force acting except the reaction force at a fixed point. Since the body is symmetric and we choose it to be the z-axis of the body system, we have

I1  I 2 And the Euler equations therefore reduce to

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

I1  x  ( I 2  I 3 ) y z  0 

I 2  y  ( I 3  I1 ) z x  0

….(1)



I3  z  0 Third of the eq. (1) represents that can be put in the form, let 

 z is constant in magnitude. Then first two of the equations (1)

z cons tan t  

in these equations:

 I 2  I3    y  0  I1 

y

It is simple harmonic oscillation with frequency (angular)

 I1  I 3    , about the axis k,  I1 

 

So that the simple harmonic oscillation can be written as

 y  A sin  t With this choice, we can write

x  A cos  t. Thus

xi  y j  A cos  t i  A sin  t j is

frequency

a vector rotating round the fixed vector k z with

 . The angular velocity of the body is

  x i   y j  z k Which itself processes about the body z-axis with the constant angular velocity

.

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158

UNIT III Canonical Transformations 3.1 Introduction There are a number of problems in mechanics for the solution of which, it is often desired to change one set of position and momentum co-ordinates into another set of position and momentum co-ordinates which may be rather suitable. For instance we assume that q i and pj are the old position and momentum co-ordinates and Qi and Pj are the new position and momentum co-ordinates related by the transformations Pj = Pj (p1, p2, ...., pn, q1, q2, ...., qn, t) Qj = Qj (p1, p2, ..., qn, q1, q2, ..., qn, t)

}

... (1)

or in brief, Pj = Pj (pj, qj, t), Qj = Qj (p,qj,t),

... (2)

then if there exists a Hamiltonian H in the new co-ordinates, such that .

Pj  -

. H H and Q j  Q j Pj

... (3)

The transformation for which these equations (3) are valid are known as canonical or contact transformation. Through in the Hamiltonian formulation, as we have already seen that the momenta are independent variables similar to generalized coordinates, but the canonical transformations include the simultaneous transformation of the independent position co-ordinates and momenta qi, pj to the new set Qj, Pj. Here Q1, Pj are referred as canonical co-ordinates. Let H be the Hamiltonian in the old co-ordinates and L, L the Lagrangian in the old and new set of co-ordinates respectively, then by the definition of Hamiltonian, we have n

H   p j q j  L and H   Pj Q j  L j 1

... (4)

j 0

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159

Legendre Transformations The Legendre Transformation is a mathematical procedure used to change the basis .

from the (qj, q j , qj, t) set tot he (qj, pj, t) set. Let there be a function of only two variables f(x,y), so that the differential of f may be expressed as

f f dx  dy x y  u dx  v dy.

df 

Where u 

f f and y  x y

Let us now change the basis from x, y to the independent variables u, y so that the differential quantities may be expressed in terms of du and dy. Let there by another function g of variables u and y, such that g = f – ux.

... (3)

Therefore the differential of g is given by dg = df – u dx – x du.

... (4)

Substituting value of df from (1) in (4), we get dg = u dx + v dy – u dx – x du or

dg = v dy – x du,

... (5)

which has the required form. As g = g (u,y), therefore

dg 

g x du  dy u y

... (6)

Comparing (5) and (6), we get

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g g and v  u y

160



... (7)

Then above equations are in effect the converse of

... (8)

Thus by the help of equation (3) we can transform the basis from x,y to the independent variables u, y provided x and u satisfy equation (2). Generating function and the canonical transformations. An interpretation of Hamilton’s principle leads that the path of a system in configuration space between two fixed points when the conservative mechanical system moves from time t 1 to time t2 makes the time integral of the Lagrangian stationary if t2

  L dt  0 and   L dt  0

... (1)

But from equations (4) of 10.1, we have n

L   p j q j  H and L   Pj Q j  H j 1

j 1

which when substituted in (l) give









   p j q j  H  dt  0 and    p j Q j  H  dt  0 t t t2

 j 1



 j 1



... (2)

The equations (2) are simultaneously valid, if the integrands of the two integrals differ by a total time derivative of an arbitrary function say G, i.e., if . . n  n  dG  p j q j  H    Pj Q j  H    j 1   j 1  dt

... (3)

Then integrals (2), when combined, become t2



 



    pi q j  H    Pi Q j  H  dt  0 t1

  j 1

  j 1



or, with the help of equation (3),

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161

dG dt  0, dt t1

 t2

dG dt   G(t2 )  G(t1 )  0. dt t1



... (4)

The function G is called the generating function of the transformation. Now, consider the relation (3), in which the first bracket is regarded as a function of q j, pj and t and the second one as a function of Qj, Pj and t, showing that G, in general is a function of (4n+1) variables q j, pj; Qj, Pj and t, where j varies from 1 to n. It is, however, possible with the help of transformations (1) of 10.1, that G may be reduced to become the function of (2n+1) independent variables, one of which is t and the other 2n are from qj, pj; Qj and Pj. Let us assume in particular that G is a function of (2n+1) variables qj, Qj and t, i.e., say G = G1 (q1, q2, . . ., q1, Q1, Q2, ..., Qj, t),

... (5)

n n G . G . G dG   1 q j  1 Q j  1 dt t j 1 q j j 1 Q j

so that

Equating the two values of

... (6)

dG from (3) and (6) we get dt

n .  n   n  n G1 . G1 . G p q  H  P Q  H  q  Qj 1  j j   j j    j t ... (7) j 1 Q j  j 1   j 1  j 1 q j

n   G1  . G1 p  q   j  j   p j   q j  Q j j 1  j 1 

n   G1  G1 p  dq     p j    j j   q j  Q j j 1  j 1 

 .  G1   Q j   H  H    0,  t     G1   dt  0  dQ j   H  H  t    . dq j   since q etc.  j  dt  

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Treating qj, Qj and t as independent variables and equating the coefficients of like terms on either side, we find

pj 

G q j

... (8)

pj 

G1 Q j

... (9)

H H 

and

G1 , t

... (10)

where F1 is the function of qj, Qj and t, and hence it is acceptable in principle that the equation (8) may be solved to give Qj = Qj (qj, pj, t). which when substituted in (9) will yield Pj = Pj (qj, pj, t). Let us now choose G = G2 (qj, Pj, t). Comparing G1 and G2 we see than G2 can be obtained from G1 merely replacing Gj by Pj. Then keeping in mind Legendre transformations and equation (9) we can obtain G 2 from G1 by the relation G2 (qj, Pj, t) = G1 (qj, Qj, t) +



PjQj



G1 (qj, Qj, t) = G2 (qj, Pj, t) –

PjQj,

... (11)

Substituting this value of G1 in equation (7), we get .    d  p q   j j  H     Pj Q j  H   G2  q j , Pj , t    Pj Q j  dt  j  j  j 

 G2

  q j



qj   j

 G2 G Pj  2  Pj Q j   Pj Q j . ` Pj t j j 

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p q

163

H H 

G2 . G G q j  2 Pj  2  Pj Q j q1 t j P1 j

In above equations qj’s and pj’s are independent variables, therefore equating the coefficients of qj, pj on either side, we get

pj 

G2 q j

... (12)

Qj 

G2 Pj

... (13)

H H

and

G2 t

... (14)

Equations (12), (13) and (14) represent canonical transformations of generating function G2. Let us now choose G = G3 (pj, Qj, t). Comparing G1 and G2, we se that G3 can be obtained from G1 merely by replacing q1 by pj. Keeping in mind Legendre transformation and equation (8), we can write the relation

G2  p j , Q j , t   G1  q j , Q j , t    p j q j j

G1  q j , Q j , t   G2  p j , Q j , t    p j q j

... (15)

Substituting this value of G2 in equation (7), we get

    d     p j q j  H     Pj Q j  H   G2  p j , Q j , t    p j q j  j  j   j  dt    i

. . G3 . G . G p j   3 Q j  3  qj p j  q j p j p j t i Q j i i

 H   p j Q j H   j

. G3 . G . G p j  3 Q j  3   q j p j p j t j Q j j

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In above equations Qj’s and pj’s are independent variables, therefore equating coefficients Qj and Pj on either side, we get

 Pj 

G3 , Q j

... (16)

q j 

G3 , p j

... (17)

H H

and

G2 t

... (18)

Equation (16), (17) and (18) represent canonical transformations for the generating function G3. Let us now choose G = G4 (pj, Pj, t) Comparing G3 and G4, we se that G4 can be obtained from G3 merely by replacing Qj by Pj. Keeping in mind Legendre transformations and equation (16) we can write the relation

G4  p j , Pj , t   G3 ( p j , Q j , t )   Pj Q j

... (19)

G4  p j , Pj , t   G1 (q j , Q j , t )   p j q j   Pj Q j j

[(from equation (15)]

 G1 (q j , Q j , t )  G4 (q j , Pj , t )   p j q j   Pj Q j j

... (20)

Keeping this value of G1 in equation (7), we get .     d   p q   j j  H     p j Q j  H   G4  p j , Pj , t    p j q j   Pj Q j  j j  j   j  dt   G4 . G4 . G  Pj  Pj 4 p j Pj t j j .

  p j q j   q j p j   Pj Q j   Qj P j . j

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165

. . G4 . G . G p j  4 P j  4   q j p j   Q j P j . p1 t j P1 j j

In above equations pj’s and Pj’s are independent variable, therefore, comparing the coefficients of pj and pj on either sides, we get

q j 

Qj 

G4 , p j

H H

and

G4 t

... (21)

... (22)

... (23)

Equations (21), (22) and (23) represents canonical transformations for the generating function G4. From (10), (14), (18) and (23), it is apparent that these transformations are identical, since

G1 G2 G3 G4    , t t t t and hence they are expected to refer to the same transformation. These transformations give the necessary and sufficients conditions that the Lagrangian L and L in the old and new set of coordinates describe the same system. Conclusively, canonical transformations give a powerful device to find the solution of a mechanical problem by means of transforming the old set of co-ordinates qj, pj to new set of coordinates Qj, Pj all being cyclic, such that Qj = Qj (qj, pj, t), Pj = Pj (q1, pj, t).

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166 .

Now set of all co-ordinates Qj being cyclic are constant, say Pj =

Pj  

H  0 yields that all momenta Pj Q j

 j.

Evidently, if H is conserved,

cannot occur explicitly in H. As such

H (qj, pj) =

H  0 , i.e., t

H is a constant of motion which means t

H will no contain t and Qj, so that

H (Pj) =

H (  j ).

Hamilton’s equations for Qj becomes

H H   j , Pj  j

Qj 

where

j

are functions of

 j ’s only and so  j

are constant in time.

Solution of the above equation is therefore Qj =

 j t +  j, being constants.

Illustrations of canonical transformation. (i) G =

q ,P j

This generating function gives the following relations by (12), (13), 14)

pj 

G2   q j q j

q P

 Pj

Qj 

G2   Pj Pj

q P

 qj

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167

G2 H t

 G 2 is not a function of t, which follows that G generates the identity transformation. Similarly,

G   q j Pj

gives Qj = -qj, Pj = - pj and

H H

which represent the space inversion.

(ii) G   q j Q j j

The generating function corresponds to G1 and hence (8), (9), (10) give

pj 

G1   q j q j

Pj  

q Q j

G1   Q j Q j

H H

 Qj

q Q j

 q j

G1 H t

There is simply an interchange of coordinates and momenta i.e., the new coordinates are old momenta and the new momenta are the old coordinates.

(iii) G   f j ,(q j , t ) Pj   f j (q1 , q2 ,..., qn , t ) Pj j

where fj is the arbitrary set of functions. This is the transformation generated by G2 and given by (13), so that

Qj  

G2  f j (q1 , q2 ,..., qn , t ) Pj

which follows the new coordinates depend only on the old co-ordinates and time and do not involve the old momenta. This is known as a point transformation.

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f j  Q j   a j k qk . k

The form of the generating function is then

G2   a j k qk P1 j k

and the new momenta are to be determined by (12) as

pk 

G2   a j k Pj . q j

These equations when multiplied by ajk and summed over k, can yield Pj, i.e.

a

Pk   aik a jk Pj    ji Pj

j ,k

Pj   aik Pk

i.e.

showing the momenta also transform orthogonally. Corresponding transformation equations are

pj 

G1  Qj . q j

pj  

(iv)

Take

G1  q j . Q j

Problem of Linear harmonic oscillator.

G1 

1 2 q cot Q . 2

So that (8) and (9) give

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168

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169

G1  q cot Q, q

p

... (a)

G1 1   q 2 cos ec 2Q. Q 2

From (b),

q  (2 p) sin Q.

and then from (a),

q  (2 p) cos Q.

... (b)

... (c)

Squaring and adding (c) and (d), we get 2

p + q = 2P.

In a more general case, take G1 

1 mq 2 cot Q ; m,  being constant. 2

Then (8) and (9) give

p

G1  m q cot Q. q

... (e)

G1 1 m q 2 p  Q 2 sin 2 Q  2P  q   sin Q  m 

(f) gives

and then (e) gives

p

 2m p  cos Q

... (f)

... (g)

... (h)

Since the generating function does not contain f explicitly, the value of H (Hamiltonian) is not affected by the transformation and so H is to be expressed in terms of new Q and P by means of (g) and (h). Taking k as force constant of the linear restoring force acting on a particle, the potential energy is given by

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170

1 2 mq 2 kq 2 p 2 kq 2 V  kq and H     2 2 2 2m 2



p 2 m 2 k   q 2 if  2  2m 2 m

with the help of (g) and (h), the value of H becomes,

H   P cos2 Q   P sin 2 Q   P .

... (j)

This H is cyclic in Q and conjugate momentum P is constant.

Also (j) gives

F







... (k)

when H = E = constant of energy. The equation of motion for Q then yields.

Q

H  p

which gives on integration

Q  t  



being a constant of integration to be determined by initial conditions. As such (g) and (h) reduce to

 2E  q  sin (t   ), 2   m 

p

 2mE  cos (t   ),

... (l)

... (m)

which gives customary solution for a harmonic oscillator. Integral Invariants of Poincare

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171

If we define a Phase space analogus to configuration space in Lagrangian formulation as a 2n-dimensional Cartesian space formed of co-ordinates q1, q2, ..., qn, p1, p2, ...pn then a complete dynamical specification of a mechanical system is given by a point in this space. Poincare pointed out the expressions which are in this space. Poincare pointed out the expressions which are invariant under Canonical transformation just as the form of Hamilton’s equations are preserved under Canonical transformation. According to Poincare the integral

J1    dq j dp j j

taken over an arbitary surface

S of 2n dimensional phase space (q 1, pj) is invariant under

Canonical transformation. We known that position of a point in two dimensional surface S may be completely specified by two parameters u and v such that qj = qj (u, v) and pj = pj (u, v) According to Jacobian (a special type of determinant discussed in advanced calculus) transform the area-element dqj, dpj into the area element du dv such that

dq j dp j 

i.e.,

 (q j , p j )  (u, v)

 (q j , p j ) (u, v)



du dv

... (3)

qi pi u u qi Pi v v

... (4)

Now J2 is invariant means

  dq s

dp j    dQ j dPj , s

... (5)

which with the help of (3) yields

  s

(q j , p j ) (u, v)

du dv   s

(Q j , Pj ) (u, v)

du dv

... (6)

S being arbitrary, the integrals are equal if the integrands are identical i.e. if

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

 (q j , p j )

 (u, v)

172

 j

 (Q j , Pj )  (u, v)

du dv

Thus J1 is invariant if the sum of the Jacobians is invariant. The transformation from q, p to Q, P can be affected by the generating function of the type G (q, p, t) so than

pj 



and

p j v



G2 G , Qj  2 , q j Pj

  G2  u  q j

  G   2 v v  q j

p j

  2 G2 Pk  2 G2 qk     k q P u j k  k q j Pk u

  2 G2 Pk  2 G2 qk     k q j qk u  k q j Pk v

…(7)

Making substitutions of (7) in L.H.S. of (6) we find with the help of (4).

 (qi , p j )



 (u, v)

 j

 j

q j  2 G2 Pk   u k q j Pk u

 2 G2 qk k q q u j k

q j  2 G2 Pk   v k q j Pk v

 2 G2 qk k q q v j k

 G2 j qk

 q i



qi p j u u qi Pj v v

q j qk u u q j qk v v

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 j

q j Pk u u q j Pk v v

 G2 j qk

 p k

173

The terms of the first series on right being antisymmetric under the interchange of indices, j, k, since then two columns are interchanged, vanish and hence we get

 j

 (qi , p j )  (u, v)

 j



 q k

G2 Pk

q j Pk u u q j Pk v v

Also we have

0  j



 P k

G2 Pk

Pj Pk u u Pj Pk v v

since the value of the summation is not affected by the interchange of dummy suffixes and so j=k gives two columns identical there by vanishing the determinant. Adding the last two equations, we get



 (q j , p j )  (u, v)

 j

 j



 P k

G2 Pk

Pj Pk u u Pj Pk v v



 2 G2 Pk i P u P  j k

 2 G2 qk Pj k P q u u j k

 2 G2 Pk k P P v  j k

 2 G2 qk Pj k P q v v j k



 q k

G2 Pk

q j Pk u u q j Pk v v

... (8)

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174

  2 G2 Pk  2 G2 qk  But      k Pj qk u   k Pj Pk u  Q j  2 G2 Pk  2 G2 qk    G2  and      v v  Pj  k Pj Pk v k Pj qk v    G2   u u  Pj

Q j

... (9)

Equation (8) with the help of (9) yields,

 k

 (q j , p j )  (u, v)

Qk u Qk v

 k

Pj u Pj

 k

 (Qk , Pk )  (u, v)

... (10)

v

which establishes the desired invariance. Similar invariance of integrals though rather complicated, but can be exhibited as

J2    

  dq

dp j dqk dpk ,

S being 4-dimensional surface of 2n-dimensional phase space. And in general

J 2    dq1 dq2  dqn , dp1 , dp2  dpn shows that the volume in phase space is invariant under canonical transformation. Conditions for transformation to be canonical To show that the transformation Pj = Pj (pj, qj, t), Qj = Qj (pj, qj, t) is canonical if the expression

 n    p j dq j   Pj dQ j  j 1  j 1  that if G = G (qj, Qj),

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175

G  q j   G  Pj   Q j  Pj 

then

... (1)

As G = G(qj, Qj), we have

dG  

G G dq j   dQ j q j n Q j

  Pj dq j   Pj dQ j , j

  Pj dq j   Pj dQ j  dG.

i.e.,

Since the right hand side is an exact differential, therefore in order that the given transformation be canonical, the left hand side of the above relation,

  Pj dq j   Pj dQ j . j

10.6.

Bilinear Invariant as condition for a transformation to be Canonical If a transformation from q, p to Q, P be canonical, then the bilinear from

 ( p j dq j   q j dp j )

…(1)

Remains invariant The condition (1) is equivalent to

 ( p j dq j   q j dp j )   ( p j dQ j   Q j dp j ), j

…(2)

i.e., if (2) is true then the transformation from old set of co-ordinate q,p to the new set of coordinates Q, P will be canonical and in that case Hamiltons equation’s can be expressed as

H H dt , dp j   dt , p j q j

…(3)

H H dt , dPj   dt , Pj Q j

…(4)

dq j  Qj  Thus for arbitrary

 q j , p j ,

we have

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 H    Pj  dq j  dt     q j   j j  P j  

176

 H  dp j  q j 

   0 

…(5)

 ( p j dq j   q j dp j )   Hdt  0

(or)

…(6)

With similar expressions for Q, P, i.e,

 ( Pj dQ j   Q j dPj )   H dt  0

…(7)

Subtracting (7) from (6) we find

 ( p j dq j   q j dp j )   (Pj dQ j   Q j dPj ), j

Which establishes the desired invariance

An important remark In case of canonical transformation

H ( p j q j  H ( p j Qj ) when integrated between the limits t0 and t1 gives,



{  p j q j  H ( p j p j )dt   {  p j Q j  H ( p j Q j )dt t0

 [G( p j q j )]t1  [G( p j q j )]t0 . Comparison of coefficients will give

H ( p j q j )  H ( p j Qj ) 3.2 Solved Examples Ex. 1. Show that the transformation

1 q P  ( p 2  q 2 ), Q  tan  1 2 p Is canonical.

In the existing case the transformation will be canonical if the expression

( p dq  P dQ) is an

exact differential. Here

p dq  PdQ

1 p dq  q dp  p dq  ( p 2  q 2 ). 2 p2  q2

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1  P dq  ( p dq  q dp ) 2 1  ( p dq  q dp) 2 1  d ( pq ) 2 = an exact differential Hence the given transformation is canonical. Ex. 2. Prove that the transformation

1  Q  log  sin p  , P  q cot p, is canonical q  The transformation will be canonical if the expression

p dq  P dQ Is an exact differential. Hence

  1 p dq  P dQ  p dq  q cot p  log sin p  q   p dq  q cot p

 sin p  1 d  1/ q sin p.  q 

cot p q cos p dp  sin p dq sin p. q2  p dq  cot p (q cot p dp  dq ) p dq  q 2

 q c ot 2 p dp  ( p  cot p )dq  q (1  cos ec 2 p) dp  ( p  cot p )dq  d {q ( p  c ot p)} Which is exact differential Hence the given transformation is canonical. Ex. 3. If the transformation equations between two sets of co-ordinates are

p  1(1  q1/ 2 cos p) q1/ 2 sin p, Q  log (1  q1/ 2 cos P) (i)

the transformation is canonical, and (ii) the generating function of this transformation

G3  (eQ  1)2 tan p (i)

Here

p dq  P dQ  p dq  2(1  q1/ 2 cos p)q1/ 2 sin p.

cos p dq  2q sin p dp 2q1/ 2 (1  q1/ 2 cos p)

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 p dq  sin p cos p dq  2q sin 2 p dp 1  ( p  sin 2 p) dq  q (1  cos 2 p) dq 2

1  p {q( p  sin 2 p)} 2

…(1)

Which is an exact differential Hence the given transformation is canonical (ii) We have Q  log(1  q

1/ 2

cos p)

eQ  1  q1/ 2 cos p

(or)

1/ 2

(or)

 eQ  1  eQ  1 giving p    , cos p  cos p 

So that p  2(1  q

1/ 2

cos p)q1/ 2 sin p becomes

 eQ  1  eQ  1 p  2 1  .cos p  sin p  cos p  cos p

…(2)

2eQ (eQ  1)tan p. Now, from relations (16) and (17) of

q



10.3, we have

G3 G and P  3 p Q

Which become with the help of (1) and (2), as 2

And

 eQ  1  G3 Q 2 2     (e  1) sec p p cos p  

…(3)

G3  2eQ (eQ  1) tan p Q

…(4)

Integration gives

G3   (eQ  1)2 sec2 p dp taking constant of integration to be zero in a particular case  (eQ  1)2 tan p And

G3  2  eQ (eQ  1) tan p dQ

…(5)

Constant of integration being zero in a particular case

 (eQ  1)2 tan p

…(6)

Equations (5) and (6) both give G3  (e  1) tan p Q

Ex. 4. find the values of

 and  so that the equations

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Q  q cos  p, P  q 2 sin  p Represent a canonical transformation. Also find the generating function G3 for the case. The give transformation will be canonical if the expression

p dp  P dQ is an exact differential

Or if

p dq  q sin  pd (q cos  p)

Or if

p dp  q sin  p{ q 1 cos  p dq   q sin  p dp} is an exact differential

Or if

( p   q2 1 sin  p cos  p)dq   q 2 sin 2  p dp.

is an exact differential

…(1)

Is an exact differential Now, we know that

Mdx  Ndy is an exact differential if

M N  y x Thus the expression in (1) will be exact differential if

  ( p   q 2 1 sin  p cos  p)  (  q 2 sin 2  p) p q (or)

1   p2 1  (cos2  p  sin 2  p)  2 p  p 2 1 sin 2  p

(or if)

1   q2 1  (1  2sin 2  p)   q 1 sin 2  p

(or if)

1   q2 1  2 q2 1 sin 2  p  2 q2 1 sin 2  p

This being an identity, equating the coefficients of like terms, We get 1  2 q

2 1

Which is satisfied for Hence

 0 thereby giving 2 q 2 1  1

  1and 2 1  0

1 2

  ,  2

Now to find G3, we have from (16) and (17) 

q

 G3 G3 and P  p Q

…(2) 1/ 2

 Q  Here Q  q cos  p gives q     cos 2 p  

And



p  q1/ 2 sin 2q 

1 as   ,   2 2

Q .sin 2 p  Q tan 2 p cos 2 p

Equation (2) become 1/ 2

 Q  G3    p  cos 2 p 

and

G3  Q tan 2 p Q

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1/ 2

 Q  And therefore G3       cos 2 p 

dp and alsoG3    Q tan 2 pdQ of which the second

integral gives the from of G3 as

1 G3   Q 2 tan 2 p 2 Ex.5. show that if there exists a generating function G, such that

G  L  L; L, L being the t

Lagrangian in old and new set of co-ordinates, then the transformation is canonical. According to Hamilton’s principle, the path of a conservative mechanical system between fixed points from time t1 to time t 2 makes the time integral of the Lagrangian stationary it

 tt L dt  0 and   tt L dt  0 2

Subtraction gives

  tt ( L L) dt  0

i.e.,

  tt

(or)

  tt

dG dt  0 dt dG dt   {G(t2 )  G(t1 )}  0 dt

Which by definition of generating function follows that the transformation is canonical Ex.6. Prove that the following relations for a canonical transformation:

q j Qk



Pk Qk Q p j P p j Qk ,  k ,  k ,  p j pk p j Qk q j Pk q j

Relations (16) and (17) of

qj  



q j Qk

Since



10.3 give

G3 G3 and pk   p j Qk



 2G3 P  2Gk and k   Qk p j p j p j Qk

 2G3  2G3  Qk p j p j Qk 

q j Qk



Pk . p j

Similarly, from relations (21) and (22) of



10.3, we can show

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q j Pk



181

Qk p j

From relations (8), (9), we can show

p j Qk

From relations (12), (13) , we can show



p j Pk

Pk q j



Qk q j

Ex. 7. If f be a function depending on position, momenta and time, and H be the Hamiltonian then prove that

 f H f H    q  p p j q j j j 

n df df   dt dt j 1

  

Let qj and pj be the position and momentum co-ordinates and t denote time; then

f  f (q j , p j , t ) n

So that

df   j 1

(or)

n n f f f dq j   dp j   dt q j t j 1 p j j 1

n f f   t t j 1

 f  f   qj pj    p j  q j 

…(1)

But, we have from Hamiltonian equations, 

qj 

H  H ,p p j q j

Substituting these values in (1), we get n df f   dt t j 1

 f H f H   q j p j p j q j

  

Ex. 8. Show that the generating function for the transformation.

p

1 q , p  PQ 2 isG  , Q Q

From relations (8) and (6) , we have

Here



p

G G ,P   q Q

p

1 q and P  2 Q Q

…(1)

Equations (1) become

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1 G q G  and 2   Q q Q Q Integration gives G  

dq dQ and alsoG    q 2 , taking constant of integration to be zero in Q Q

particular cases. Both of these integrals give

q

Ex. 9. Show that

G

q Q

Q j generates the exchange transformation in which position co-ordinates

and momenta are interchanged Here the generating function G is of the first type as

G  G1 (q j , Q j , t ) and then from results (8) and (9) we have

pj 

G1 G , Pj  1 q j Q j

…(1)

Now, when G1  q j , Q j , we get

G1 G  Q j and 1  q j q j Q j From eqns. (1) and (2), it is clear that

p j  Q j , and Pj  q j Which follows that the position co-ordinates and momenta are interchanged Ex. 10. show that

G3    Q j p j generates the identity transformation i

From relation (16), (17) and (18), we have

qj  

G3 G G and , Pj   3 and also H  H  2 p j Q j t

…(1)

And from the given generating function, we get



G3 G G  Q j and  3  p j and also 3  0 p j Q j t

…(2)

It is clear from equations (1) and (2) that

q j  Q j and , Pj  p j and also H  H These show that the new and old position co-ordinates and momenta are similar. Hence, are similar. Hence, the given function generates the identity transformation.

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Ex.11. Find the values of q, p ; the position and momentum co-ordinates, for a harmonic oscillator described by the Hamiltonian H 

1 2 ( p   2 q 2 ) and generated by the generating function 4

1 G1   q 2 cot 2 Q 2 1 G1   q 2 cot 2 Q 2

Given that

So that equations (8) and (9) give

p

G1   q cot 2 Q q

P

and

From (2), we have

…(1)

G1  q 2 cos ec 2 2 Q Q

…(2)

 P  q   sin 2 Q and then from (1), we get     P  p   cos 2 Q   

Now substituting the values of p and q in H  We find H 

1 2 (P   2q2 ) 2

1  P p 2  cos 2 2 Q  sin 2 Q   2   

 P  vP, 2 Where

v   / 2 , the frequency of the oscillation this shows that the Hamiltonian cyclic

in Q and the conjugate momentum P is therefore a constant. It is observed from (3) that P is actually a constant energy divided by v, say

P So that

E , where vP  H  E (total energy constant) v Q

H v P t

dQ   v dt

i.e.,

Q  v(t  t0 ) 

 (t  t0 ) 2

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184

  P  q  (t  t0 )  a sin  (t  t0 ),  sin 2 2   

thus

 2 E  (2 E )  P  a            

where

p  a cos  (t  t0 )

and

Ex. 12. Using the invariance of bilinear form show that the

1 , P  qp 2 p

Q Here

 P  P2  q  2qp  p

i.e.,

dP  P2 dq  2qp dp

and

Q  

1 p p2

i.e.,

dQ  

1 dp. P2

transformation is canonical

 1   1    PdQ   QdP  ( P 2  q  2qp  p)   2 dp    2  p  ( p 2 dq  2qp dp)  p  p    pdq   q dp 

2q 2q  p dp   p dp p p

  q dq   q dp Which shows that the transformation is canonical. Problems 1. (a) what is a canonical transformation? (b) Prove that the following transformations are canonical (i) when Q=p, p =-q, (ii) when Q=q an p, p = log sin p

2. Prove that transformation

 P  q  sin Q, p  (mP k ) cos Q is canonical and  k

that its generating functions is 3.

Show that

p

show

1 kq 2 cot Q . 2

p j generates an exchange of position co-ordinates and

momenta

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185

G2   q j p j generates the identity transformation. j

5. Show that

G2   p j Q j generates the inversion of phase –space i.e. both the

new

momenta and co-ordinates are the old momenta and co-ordinates with

signs reversed

6. Define canonical transformations and obtain the transformation equations

corresponding

to all possible generating functions. 7. Write note on canonical transformations and the relation between old and

new

Hamiltonians. 8. Show that the transformation

1  Q  log  sin p  , p  q cot p is canonical q  Find the generating function F(q, Q), 9. (a) What are canonical transformations? Discuss briefly why they are considered useful? (b) Show that the transformation

Q  log (1  q cos p)

P  2 q (1  q cos p)sin p is canonical. Find the generating function f (q, Q) 10. Test whether the following transformations are canonical (i)

Q  P, p  q

(ii)

1  P  q cot p, Q  log  sin p  q  q  2 P sin Q

(iii)

p  2 P cos Q Q  log (1  q cos p)

(iv)

P  2 q sin p (1  q cos p) 11. (a) Discuss the conditions for a transformation to be canonical. (b) Find the values of

 and  so that the equations

Q  q 2 cos ( ,P) p  q sin  p represents a canonical transformation what is the form of generating function F3 in this case. 12. What

are

functions.

canonical Obtain

transformations. an

expression

Using for

the

G1 

1 m q 2 cot q as 2

displacement

generating

linear

harmonic

oscillator

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13. What is a canonical transformation? Show that the problem of a linear

harmonic

oscillator can be solved using a canonical transformation. 14.

Derive the equation of canonical transformations of a system and discuss the significance

the

generating

function

the

transformation

solving

the

harmonic oscillator problem. 3.3 Poisson Brackets Introduction. It is observed that the form of the equations of motion for a conservative mechanical system remains unaltered when the set of position and momentum co-ordinates q, p are replaced by an other set of position and momentum co-ordinates Q,P related to the original set with the help of canonical transformation. Though this replacement changes the functional form of the Hamiltonian, but its value is unchanged, unless the canonical transformation is a department of time. It f be a dynamic variable defined by the function

f (q, p, t ) , then, we have to study how

this transforms under canonical transformation. In case the value of f does not change, we most have

f (q, p, t )  f {q(Q, P, t ), p{Q, P, t )}  f (Q, P, t ), say

…(1)

In view of the relation (1), the Hamiltonian does not act as dynamical variable in respect of time dependent transformations. Since it is rather difficult to conceive the effects of finite canonical transformations on dynamical variables, therefore we consider infinitesimal canonical transformation (abbreviated as ICT) in which the new set of variables Q, P differ from the old one, i.e., q,p by such quantity that their products are negligibly small. We can thus write

q j  Q j   q j , p j  Pj   p j

…(2)

Now, if G be the generating function such that

G  G(q j , Q j , t ) Then from equations (8) and (9) and (10) of

pj 

…(3)



10.3 we get

G G G , P and H  H  q j Q j t

…(4)

From equation (3), we get

dG 

G G G dp j  dQ  dt q j Q j t

Which with the help of eq. (4) becomes

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187

dG  p j dq j  Pj dQ j  ( H  H )dt

…(5)

or, making substitutions from (2), this takes the form

dG  (q, Q, t )  p j dq j  ( p j   p j )d  q j   q j   p j dp j  ( p j   p j ){dq j  d ( q j )}  {H (Q, P, t )  H (q, p, t )}dt

d{G(q, Q, t )}  {H (Q, P, t )  H (q, p, t )}dt

(or)

 p j dp j  p j dq j  p j d ( q j )   p j dq j Neglecting the product

 p j d ( q j )

d{G(q, Q, t )}   p j  q j   q j dp j   q j dp j  p j d ( q j )

(or)

{H (Q, P, t )  H (q, p, t )}dt   p j dq j   q j dp j  d ( p j  q j )  {H (Q, P, t )  H (q, p, t )}dt

 p j dq j  dq j dp j  {H (Q, P, t )  H (q, p, t )}dt

(or)

 d{G(q, Q, t )  p j  q j }

….(6)

Since on the left-hand side of (6), only the derivatives of q j , p j and t appear, therefore the right –hand side must also be a function of q,p,t and let it be

 (q, p, t ), where  is a

small parameter and the negative sign has been taken for better adjustment in the following discussion:

  p j dq j   q j dp j  {H (Q, P, t )  H (q j , p j , t )}dt

  (q, p, t )

…(7)

From equation (7), we find

 qj  

  ( q, p, t ) p j

 p j   and

  ( q, p, t ) q j

H (q, p, t )  H (Q, P, t )   The function

 (q, p, t ) as

…(8)

…(9)

  ( q, p, t ) t

…(10)

 (q, p, t ) introduced here is known as the generator of the ICT. If we take

a dynamical variable, then

 (q, p, t )

and

 (q, p, t ) generate

the same ICT,

provided Q, P and q, p are related by a finite canonical transformation. Now the function

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 (q, p, t ) being dependent of position co-ordinates, momentum and time, represents the state of the system at an instant t. If we now propose to find the change in functional form caused by an infinitesimal canonical transformation, then taking

 (q, p, t ) as

a dynamical variable and

 (q, p, t ) as

the

generator of the ICT we have from the equations (8) and (9) of the previous article,

 qj  

  ,  p j   p j q j

The transformation being infinitesimal, the change in the functions form may be defined as

 (q, p, t )   (q, p, t ) (q, p, t ) Now if we expand

 (q, p, t ) in

…(2)

a Taylor series; retaining linear terms and using (2), we

get

 (q, p, t ) 

 (q, p, t )  (q, p, t )  qj   pj. q j q j

…(3)

With the help of (1), the relation (3) becomes

       ( q, p, t )  (q, p, t )   (q, p, t )  (q, p, t )  …(4) p j q j p j  q j 

 (q, p, t )   

The bracketed quantity in (4), i.e., dynamical variables

 (q, p, t )

is generally denoted by

    is termed as passion – bracket of the two  q j p j q j p j

and

 (q, p, t ) under the infinitesimal canonical transformation. It

 ,  p,q        ,  p ,q q j p j q j p j

…(5)

In this notation (4) can be written as

    , 

…(6)

Properties of Poisson –brackets Proof I. The Poisson – bracket of two dynamical variables is-variant under infinitesimal canonical transformation Form relation (1) , if f is a dynamical variable and f its new value, then we have

f (Q j , Pj , t )  f (q j , p j , t ) So that

f f q j f q j   Q j q j Q j p j Q j

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189

Now if we assume that ICT is generated by

q j Q j Where

  jk  

 (q, p, t ) then from (2), (8) and (9), we have

p 2  2  and k   p j qk Q j q j pk

…(2)

 jk is the Kronecker delta, defined as unity if j=k and zero if j  k If the derivatives with respect to Pj are considered, we get similar expressions. Now (1), if

f is successively replaced by

 and 

and the terms of order

 2 are neglected, then we find

           Q j Pj Q j Pj q j p j q j p j Which justifies the omission of argument (6) and follows that the Poisson bracket of two dynamical variables say

 and 

is a dynamical variable whose value is independent of the

variables used to define the state of the conservative system. Proof II. The Poisson’s bracket does not obey the commutative law of algebra, i.e.

 ,     ,  We have from the definition of Poisson-bracket,

 ,  

     q j p j q j p j

           q j p j q j p j     ,   Prop

III.

The

Poisson

bracket

obeys

the

distributive

law

algebra,

i.e.

  ,     ,     ,   We have

  ,   

           q j p j q j p j       q p j j 

      p q j j 

      p q j j 

  

                q p q p   q p q p j j   j j j j  j j

  

  ,     ,   Prop. IV.

 ,     ,      , 

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 ,  

We have



190

         q j p j q j p j  q j

                p j   q j q j  p j  p j

                 X    q j p j q j p j   q j p j q j p j     ,      ,  

 ,  ,      ,   ,    ,   ,    0 which is known as Jacobi’s identity

Prop. V. We have

 ,   ,  



  ,      ,   q j p j q j p j



 q j

           .    q j p j q j p j  p j q j p j

          q j p j q j p j 

  2    2  2    2    2     2  q j p j q j q j p j q j p j q j p j q j  p j   q j

  2    2  2   2       2 p j q j q j p j 2   p j q j q j p j p j



 2    2     2    2        q j 2 p j p j p j 2 q j q j  q j 2 p j p j p j 2 p j q j 



 2 p j q j

       2     q j q j q j p j  p j q j

   x       q j p j q j p j 

Similarly,

  ,   , 



 2    2     2    2         q j 2 p j p j p j 2 q j  q j 2 q j p j p j 2 q j q j 



2  p j q j



 2    2     2     2         q j 2 p j p j p j 2 q j q j  q j 2 p j p j p j 2 q j q j 

       2     q j p j q j p j  p j q j

          q j p j p j p j 

And

 ,   ,  

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

 2 p j p j

191

      2       q j p j q j p j  p j p j

        q  p q j p j j j 

  

Adding all together, we find that the terms on right hand side cancel and we are left with

 ,  ,      ,   ,    ,   ,    0  Pj , pk   0, q j , qk   0

Prop IV

The definition of Poisson break gives

( , ) 



     q j p j q j p j

( p j , pk ) 

p j pk pk p j  0 q j p j q j p j

p j pk p   jk  1 when k  j and 0 when k  j , k  0, 0 q j q j q j

 q j , qk   0

And similarly, Prop. VII.

q j , pk    jk ,  jk being Kronecker delta ( , ) 

We have



     q j p j q j p j

(q j , pk )  

q j pk pk q j  q j p j q j p j

q j pk p  0, as k  0  p j q j q j

  jk sin ce

pk  1 when k  j p j  0 when j  k

Prop. VIII

   , q j   . and  , p j    p p j

We have



( , ) 

    .  q j p j q j p j

( , q j ) 

 q j q j   . q j p j q j p j

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

And similarly

192

q j q j  as 1 0 p j q j p j

( , p j )  

 q j

Canonical equations in terms of Poisson brackets. So far we have not varied the time. Now making a consideration of the variation of the time let us assume that the transition from the old set of coordinates (q j, pj, t) to the new set (Qj, Pj, t+dt) takes place in two steps. In the first step the position co-ordinates and momenta are changed to canonical co-ordinates in the second step the time is changed. Let

d ,(Q, P, t ) be the whole change in the value of  (Q, P, t ) when time is varied,

then we have

d (Q, P, t )   (Q, P, t ) 

 (Q, p, t ) dt t

…(1)

Now according to relation (4) of §11.2, if

  Q, P, t   ( , H )dt , 

since

H be the generator of ICT, then

 being a time parameter may be taken as dt.

(1) becomes d   , H    dt where 





If we consider a particular case in which

 is

 . t

…(2)

one of the co-ordinates say

  Q j then

  0 so that (2) gives dQ j dt





 Qj , H .

…(3)

Similarly if we take   Pj , then

dPj dt



 Pj , H



Combining the equations (3) and (4) with equations (3) of

….(4)



10.1, the canonical

equations in terms of Poisson brackets can be expressed as

Qj 

. H H  (Q j H ) p j     Pj , H  . . P Q j 

…(5)

The equations (5) are referred to as the equations of motion in Poisson bracket from. Note : Equations (3) and (4) of this section may easily be found by the following considerations: The rate of change of

 (Q, P, t ) with t is

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193

d  .  .  Qj  pj dt Q j Q j .   H  H  H  by canonical equation Q j  Q j Q j Pj Pj Pj .

Qj  

And

H Q j

  , H  by the definition of Poisson bracket.

   , H  .

Thus,

  Q j , Pj in succession, we find

If we put

Q j  [Q j , H ] and P j  [ Pj , H ]. Ex. 1. If

 ,

(a)

    ,    ,    ,  , t  t   t 

(b)

d d d  ,    ,    ,  . dt  dt   dt 

We have

be the Poisson bracket, then prove that

 ,  

    ,  q j p j q j p j



          ,     t t  q j q j q j p j 



 q j

                t  p j q j p j  t  q j

    p j

            p q j p j  t  j 

                           q j  t  p j q j p j  t    q j p j  t        ,     ,  t   t

            q j  t  p j 

 . 

similarly the other result can be proved.

Some applications of Poission brackets to Mechanics:

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194

Consider a function G  G(q j , p j , t ) depending on coordinates

momenta and time as the Hamiltonian, we have

[q j , H ] 

H .  qj p j

[ pj, H ]  

and

…(1)

H .  pj . q j

…(2)

The total time derivative of the function by

dG G H . G .   qj pj dt t q j p j



G  [G, H ]. t

…(3)

Since the function depending on dynamical variables which remain unchanged during the course of motion of the system, are said to be the integrals of motion, therefore the condition that the function G be an integral of motion is that

dG G  0 i.e.,  [G, H ]  0. dt dt

…(4)

In case this integral of motion G does not involve an explicit time dependence, (4) requeces to

[G, H ]  0 which means that the function G not involving explicit time dependence is an integral of motion of its Poisson bracket if the Hamiltonion H varnishes. Replacing G by q j and p j in succession in (3), we find .

q j  [q j , H ] and p j  [ p j , H ]

…(6)

thereby exhibiting the whole set of Hamilton’s canonical equation in terms of Poisson’s brackets. Again if we replace G by H in (3), we get

dH H   [H , H ] dt t i.e.,

dH H  . dt t

…(7)

This result has already been discussed in Chapter VII [2] Poisson’s theorem. The Poisson bracket of two integrals of motion is itself an integral of motion. According to Jacobi’s identity for three arbitrary functions v, u, w of q,p we have

[u,[v, w]  [v,[w, u]]  [w,[u, v]]  0.

…(8)

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195

Assuming u and v any two integrals of motion and putting w = H in (8), we get

[u,[v, H ]]  [v,[ H , u]  [ H ,[u, v]]  0. Treating u, as two constants of motion, we have

[v, H ]  0  [ H , u]. So that the first two terms in the L.H.S. of (2) vanish, and we are left with

[ H [u, v]]  0. which allows that the dynamical variable

[u, v]  cons tan t. Hence the Poisson bracket of two constants of motion is itself a constant of motion. This proves Poisson’s theorem. In case u and v depend on time explicitly, the equation (3) reduces to,

d  u, v  [u, v]  [u, v], H ] dt t  u   u    , v   u,   [u,[v, H ]  [v,[ H , u ]]  t   t  by using Jacobi identity for writing the last two terms.

 v   u   u,  [v, H ]  v,   v,[u, H ]  t   t  (by combining 2

and 3 terms)

 v   du   u,  [v, H ]  v,  [u, H ]  t   dt 

 v   du   u,   v,  by (3)  t   dt 

 0, since So that

dv du 0 . dt dt

[u, v]  cons tan t ,

Which also proves the Poisson’s theorem. [3] Quantum Poisson Brackets. Consider the Poisson bracket of the form [u1 , u2 , v1 , v2 ]. Expanding it is two different forms without disturbing the order of u1 relative to u2 , we have

[u1 , u2 , v1 , v2 ]  u1.[u2 .v1 , v2 ]  [u1 , v1 , v2 ]u2 by prop. IV of Poisson brackets.

 u1 u2 , v1  u2  v1 u2 , v2   u1 , v2  v2  v1 u1 , v2  u2

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196 …(10)

[u1 , u2 , v1 , v2 ]  [u1.u2 , v1 ]v2  v1[u1 , u2 , v2 ]

and



 u2 u2 , v1   u1 , v1  u2  v2  v1 u1 u2 , v2   u1 , v2  u2



…(11) Comparison of (10) and (11) yields

u1v1 u2 , v2   v1u1 u1 , v2   u1 , v1  u2v2  u1 , v1  v2u2 (u1v1  v1u1 ) u2 , v2   u1 , v1  u2v2  v2u2 

i.e.

…(12)

Theorem u1v1  v1u1 and u2v2  v2u2 in classical mechanics but in quantum mechanics, if the variables are not pure numbers but have a different significance, then these results are not true. The equation (12) still holds if u1 and v1 are quite independent of u2 and v2 , so that we have

u1v1  v1u1 u2v2  v2u2    (cons tan t ) u1 , v2  u2 , v2  Dirac’s theory gives



…(13)

ih , h being Plank’s constant and i  (1). So in quantum 2

Mechanics we get

uv  vu 

ih [u, v]. 2

…(14)

Tracing q j and p j as certain kind of operators in quantum mechanics, we get from (14)

  q j qk  qk q j  0  p j pk  pk p j  0   ih q j pk  pk q j   jk   2

…(15)

These equations constitute the fundamental quantum conditions and

 q j q j   0,

 p j p j   0, and  q j pk    jk are sometimes known as the quantum Poisson brackets. 3.4 Hamilton- Jacobi method: Canonical transformations provide a general procedure for easy solution of mechanical problems. These are two ways of effecting such transformations:

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197

One way to obtain the solution of a mechanical problem is to transform old set of co-ordinates into new set of co-ordinates that are all cyclic and consequently all momenta are constants. In this way the new equations of motion can be integrated to give a solution and is adopted when Hamiltonian H is conserved This way was only a sort of co-ordinate transformation and not a method in itself; that is, we switch H (q1, pi) to K(Qi, Pi) and then apply Hamilton’s equations of motion, which are then integrated to yield result.

(ii)

Another way to obtain the solution is to seek canonical transformation from coordinates and momenta (q,p) at time t to a new set of constant quantities which may be 2n initial values (q0, p0) at t=0. The transformation equations will be, therefore,

q  p (q0 , p0 , t ) p  p(q0 , p0 , t ) Since these equations of transformation give the value of q, p in terms of their initial values (q0,p0) at time t=0, they are then the desired solution of a mechanical problem. Thus greatest advantage so such a transformation is that we are doubly benefited i.e. in obtaining transformation equations we arrive at the solution as well. The way is due to Jacobi. Thus Jacobi’s way is a transformation as well as a method in itself. That is, in it we do not first transform and then apply Hamilton’s equations of motion but while performing transformation, we are arriving at the result as well. This procedure is more general because it can be applied, in principle, to the cases for which Hamiltonian involves the time. Hamilton-Jacobi Partial Differential Equation: Under the second kind transformation, as we know, the new set is of constant co-ordinates (initial values q0, p0). If we requite that the transformed Hamiltonian K is zero, then new equations of motion (involving co-ordinates of transformed as Pi, Qi) are

K   0 Pi   K Pi   0  Qi 

Qi 

…(1)

Which obviously ensures that the new co-ordinates Qi, Pi, are constant in time i.e. both the coordinates cyclic in the technique The new Hamiltonian K is related to old Hamiltonian H and to the generating function F by the equation

KH

F , Pi

Which will be zero only when

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198

F 0 t

…(2)

The generating function F is to be the function of one old co-ordinate and one new coordinate because it has only 2n independent variables and performs the transformation between old and new set. Let us take F to be a function of (q,p,t).* with such a generating function, we can write the equation of transformation as

pi 

F ( q , P, t ) qi

With this substitution Hamiltonian H (qi, pi, t) becomes

H (q1 , q2 , ...qn ; p1 , p2 ,... pn ; t ) 

 F  H  qi , t  and equation  q i  

F (q, P, t ) 0 t

Is then

 F F  F (q, P, t ) H  q1 , q2 , ...qn ; , ;t   0 qi qn  t 

…(3)

Eq. (3) is a partial differential equation in (n+1) variables (n for q’s and one for t and p is constant, not a variable) while situation has reduced the number of variables by n. Eq. (3) is called Hamiltonian’s principal function and is denoted by S. Thus eq. (3) is also termed as H-J equation for Hamiton’s principal function. Solution of Hamilton Jacobi Equation: Solution of Hamilton Jacobi equation is to obtain transformation equation in which co-ordinate of new set be constant quantities. Since these co-ordinates, being constant quantities, can be taken as initial values, the solution of Hamilton-Jacobi equation is the desired solution of the mechanical problem. We can express the transformation equation and hence the solution as:

q  q (q0 , p0 , t ) (or) Which

q  p ( i ,  i , t )

 i and  i are related to initial values of q and p at time t0. Now we will proceed to obtain solution of Hamilton Jacobi eq. (3). If we integrate eq. (3)

to obtain its solution S, the Hamilton’s principal function, then it will merely provide dependence on the old co-ordinates and time. It will not be apparent form the solution that now the new momenta Pi appear is S. Eq. (3) has the form of a first order partial differential equation in (n+1) variables. Consequently, a complete solution must involve (n+1) independent constants of integration

1 ,  2 ...,  n1. As eq. (3), Whose solution is S, is a partial differential equation, only derivatives of

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199

S with respect to qi and t will solution of eq. (3), where differential of

 is

 will be zero and that of S will remain) Out of

any additive constant (since the

(n+1) constants of integration one

should therefore be an additive constant to S but since this additive constant will have no effect in transformation, we can take S to be involving only n constants (1 ,  2 ,.... n ).* Hence a complete solution of eq. (3) can be written as

S  S (q1 , q2 ,....qn , 1 , 2 ,... n , t )

…(4)

Where none of n constants is solely additive, We can take these n constants as the new momenta pi, i.e.

1  Pi Therefore n transformation equations can be written as

p1 

S (q1 , 1 , t ) qi

Which are n equations and t=t0 give the n values of

…(5)

 i , in terms of the initial values

of q and p.

Thus constants of integration are evaluated in terms of the specific initial conditions of the problem i.e

i  i (qi , pi , t ) (or)

Pi  Pi (qi , pi , t )

…(6)

Which is one of the transformation equations. The second transformation equations which provide the new constant co-ordinates appear as

Qi  i 

S (qi ,  i , t ) Pi

…(7)

S (qi ,  i , t )   i Which are n equations and at t=t0 give the n values of

i

in terms of the known initial values of

qi . Eq. (7) can then be ‘turned inside out to furnish qi in terms of i and i and t: qi  qi (i , i , t )

…(8)

Which solves the problem by giving the co-ordinates as function of time and initial conditions (since

i and i are known at t=t0 in terms of the initial values of pi, and qi). Discussion about the Hamilton’s Principal Function which is the solution of H-J

equation : From the discussion of the solution of Hamilton Jacobi equation, we recognize S, the Hamilton’s principal function, as the generating function which gives rise to a canonical

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transformation involving constant momenta and constant co-ordinates. As we have explained in earlier steps of this methods, with the achievement of transformation equations, solution of mechanical problem is automatically obtained or in other words, while solving the HamiltonJacobi equation, we are, at the same time, obtaining a solution to the mechanical problem. Choice of quantities

i

 i as the new momenta is arbitrary to some extent; instead, we can choose n

which are independent functions of

 i   i (1 ,  2 ,... n )

 i the

constants of integration i.e.

this choice expresses Hamilton’s principal function as S (qi ,  i , t ).

To have more information about the significance of the function S, we consider its total time derivative, i.e

S  S (qi , i , t )

dS S  S  s  qi   i i  dt i qi t i

So that

…(9)

But from eq. (5) we have



S pi qi

And from eq (3),

H

S 0 t

(or)

S  H dt

Also

 i  0 since  i



is constant.

Putting these in eq. (9), we get  S   pi q i  H dt i L

So that

S   L dt  a

constant

…(10)

The expression differs from Hamilton’s principle in a constant showing that this time integral is of indefinite form. Thus the same integral when in definite form shapes the Hamilton’s principle which results in the solution of mechanical problems, while in indefinite form it shapes the Hamilton’s principal function that provides an alternative way of solving the problems. In actual calculations, result expressed by eq. (10) is of no practical use since the integration cannot be performed until qi and pi are known functions of time i.e. until the problem is solved. Hamilton –Jacobi Equation for Hamilton’s characteristic function:

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Conservation systems: For conservative systems in which H does not depend on time t explicity, we have H as constant,

 i say,

the total energy E of the system. Hamilton- Jacobi equation for Hamilton’s

principle function S (q,t) becomes

 S  S H  qi 0   qi  t

…(1)

Since the explicit dependence of S on t is involved in the last term, we can separate the variables. Let us assume the solution of the form

S (qi , i , t )  W (qi , i )  1t

…(1)

From which it follows that

S W S  and  1. qi qi t Putting these in eq. (1), we find that

 W  H  qi   1  qi  Which is independent of time. Out of

…(3)

 i integration

constants one constant

 i is

a constant of

motion being equal to H. It is perhaps in almost all the cases, the total energy, but not always Physical significance of W (Hamilton’s characteristic function): we write

W  W (qi , i ) So that

dW W  W d i  qi   i i dt qi  i dt  i

Since

W  qi qi

 i ' s are constant, Now W S   pi qi qi

So that  dW   pi qi i dt 

(or)

W   (  pi qi )dt i

= action A.

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Thus W is identified as action A. Function W is called Hamilton’s characteristic function. Type of canonical transformation generated by W: Let us first consider a canonical transformation in which the new momenta Pi are all constants of the motion

 i and

where

1 in

particular is the constant of motion H. That is, H

( (qi , pi )  1 . Taking generating function to be W(q, p) = F, equations of transformation with W(q,p) as the generating function will be

and

pi 

W ( q, P ) qi

…(4)

Qi 

W W (q, P)  Pi  i

…(5)

on using eqs. (15) and (16). Further it is taken that

H (qi , pi )  i Which on using eq. (4) gives partial differential equation of W of the form.

 W  H  qi   i  qi  Which is exactly of the same form as eq. (3) and therefore W is the generator of a canonical transformation considered, quite different from that due to S, the Hamilton’s principal function, is the condition H= 1 (a new momenta) which determines W. Now in such a transformation, the new Hamiltonian is

W t W  i  (qi ,  i )   i t

KH

Thus old and new Hamiltonian are equal. As Qi , new coordinates, do not occur in K, they can be recognized as all being cyclic. Thus W, the Hamilton’s characteristic function, generates a canonical transformations in which all new co-ordinates are cyclic. This canonical transformation then exactly resembles with that discussed. Using Hamilton’s eq. of motion, the new co-ordinates are given by 

Qi 

K ( 1 )  1if i  1  i



(or)

Q1  1

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and

Qi  t  1

and

Qi 

giving

Qi  i 



W i

203   W W  fromQ    i  Pi i  

K (1 )  0 if i  1  i

W when i  1  i

...(6)

Therefore, out of all new co-ordinates Qi, only Q1 emerges out as involving time and is thus not a constant of motion. It is not necessary always to take

 i and constants of integration

 i in W as the new constant momenta (see solved problems) Kepler’s Problem Solution by H- J Method the problem, in general, consists in finding the path of a charged particle under the action of a central force. To be more close to the Kepler’s problem, we choose the motion in an inverse square field (cf. chapter on motion under a central force). Let us suppose that a particle with charge e is moving about a stationary nucleus with positive change ze, where z is atomic number. Denoting the conjugate momenta along r and



directions respectively by

pr and p

Hamiltonian function becomes

H

1  2 p 2  e2 z  pr  2   2m  r  r

As the system is conservative, Hamiltonian will represent the total energy

…(1)

1 of

the system, i.e.

for such a system

H  1  E (say)

pr 2 

Thus

S 2me2 z  2mE  qi r

Since no explicit depence on t is involved in H, we can separate the variables. Further for such cases we shall proceed with Hamiliton’s characteristic function W (spatial part of Hamilton’s principal function S) taking as the generating function for the transformation. We write

pi 

Giving and

S W  qi qi

S W   r r   S W  p     

pr 2 

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204

Eq.(2) is, therefore,

1  W  2me2 z  W     2   2mE  r  r  r    2

Separating the variables W  Wr (r )  W ( ), we get

1  W  2me2 z  Wr    2 mE     2  r  r  r    2

…(3)

Since in a central force motion angular momentum is conserved

p  constant.

Therefore, we can take

p

dW   2  a constant d

W   2  a constant of integration

(or)

…(4)

This when substituted in eq. (3), gives

 Wr 2me2 z  2 2    2mE   2  r r r  

…(4a)

 2me2 z  2 2  Wr    2mE   2 dr  a constant of integration r r  

(or)

This gives the transformation function as 1/ 2

 2me2 z  2 2  W  W  Wr   2    2mE   2  r r  

dr  a constant of integration …(5)

Further, we know that

Q1  And

W  t  1. 1

W W  1. when i  1. Hence  2 . 1  2

Form eq. (5), on calculating

2 

W    1

2

we find

 2 dr r2

 2me2 z  2 2  2 mE   2   r r  

…(5a)

The integral is evaluated to give 1/ 2

1 me2 z  m2e4 z 2 2mE     2  r  22   24 2 

cos (   2` )

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Where

2'  2   / 2

(or)

1/ 2  1 me2 z   2 E 22  '  1  1  cos (    )    2  r  22   me4 z 2   

…(6)

Eq. (6) is of the form

1 1  1 cos (  2 ')  r l Which is the equation of a conic with eccentricity

l 

And If

…(7)

 where

 22 me2 z

 2 E 2   1  4 22  .  me z 

…(8)

E  0,1, path of the particle is an ellipse; E  0,1, path is parabola

E  0,1, path is a hyperbola In Kepler problem we are interested in elliptical path. From eq.(7), the relation between the major axis of the conic and the energy when it is ellipse can be expressed as

2a 

2l 1 2

[We know that when conic equation is

1 1  (1 cos  ) r p Semimajor axis is given by

a Hence

p for ellipse. 1 2

p  l , see Art.4.4]

Which on using relation (8) gives

ze2 E 2a Readers may compare the results [equations (7) and (9)] with those derived in the chapter on ‘Motion under a central force.

3.5 Action (Phase interals) and angle variables:

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We shall now extend Hamilton – Jacobi procedure so as to adapt it particularly to the solution of problems in periodic motions. In this technique we do not choose

 i as

the new

momenta but, instead, we use suitably defined constants J i which form a set on n independent functions of the

 i ' s and are called action variables

Suppose we have separated the variables in Hamilton-Jacobi partial differential equation, obtaining

pi 

Wi (qi , 1 ,  2 ,.... n ) qi

…(1)

Which gives

pi  pi (qi , 1 ,  2 ,..... n ) We shall define the phase integral or action variable conjugate to the coordinate qi by the integral

Ji 



…(2)

pi dqi

Where the integration is to be carried over a complete period of oscillation or rotation cycle of q i . In case qi is a cyclic coordinate, its conjugate momentum pi is constant and this integral is to taken from 0 to 2 giving

J i  2 pi

…(3)

For all cyclic co-ordinates If we compare eq. (2) with the action A defined as 

A   pi qi dt ,    pi dpi i

It becomes quite obvious why, J i is designated as action variable. Eq. (2) can also be written as

Ji 



Wi (qi, 1 ,... n ) dqi qi

…(a)

Since the definite integral evaluated here is not a function of qi i.e. qi simply appears here as a variable of integration; each action variable J i is function of

1.... n constant of integration. Thus

we have

J i  J i (1...,  n ) We can solve above equations for

…(4)

 's

getting

i  i ( J1..., J n ) By means of these relations we can express the Hamilton’s characteristic function W as

W  W (q1 ,....qn J1 ,....J n )

…(5)

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207 '

While Hamiltonian appears as a function of J i s only

H  1  H ( J1 ,....J n ) The generalized coordinates conjugate to J i are called angle variables expression eq. of transformation Qi 

i 

i and

are given by the

W Pi

W (qi ,  i ) J

…(6)

Since Ji has the dimensions of an angular momentum (equation 3), coordinate conjugate to it should be an angle and hence the name angle variable. Now the equations of motion for the angle variable are 

i 

W ( J1 ,.....J n )  vi ( J1....J n ) J

...(6a)

having a solution

i  vi  i

...(7)

where vi ' s are a set of constant functions of the action variables. Eq. (7) shows that

i ' s are

linear functions of time eqs. (6) and (7) can be combined and solved for q’s to give

qi  qi ( J1...J n , 1...n , t )

...(8)

the greatest advantage of action and angle variables lies in the fact that one can obtain the frequencies of periodic motion without finding a complete solution for the motion of the system as can be seen from the following: let us denote change in an angle variable

i with the completion of one cycle by qi one

of the co-ordinates, by i given as

i  Putting

i 

Wi dq j qi



...(9)

W in eq. (6), we get J i

i 



 2W dqi q j J j



 J j

W

 q

dqi 

 J j

 p dq i

...(10)

on using transformation equation,

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Thus

i 

208

 J i   ij J j

 1 wheni  j  0 wheni  j The above expression shows that only when qi goes through a complete period,

i

change by

unity. Then from equation (7),

i  vi t As we have just stated i  1 when qi goes through a complete period i.e. when t   i . where thus

 i is the period associated with qi 1  vi i

vi 

(or)

i

 i

...(11)

from above expression vi is to be identified as the frequency associated with the periodic motion of qi and hence we calculate the frequency of periodic motion without finding a complete soluction for the motion of the system. 4. Determation the frequency of the Kepler problem for which Hamiltonian is given by

1  2 P 2   H  Pr  2   2m  r  r Apply the method of action angle variables We have solved the Kepler problem by H.J. method. In the value of H, instead of  , we have taken

ze2 . therefore, on replacing ze2 by  we can use the relations, derived there, for the

solution of the problem.

2  0  

 2 dr  m  2 2  r 2  2mE  2  2  r r  

Further replacing E by

1

we get

2    

 2 dr  2m  2 2  r 2  2m1   2  r r  

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  2  

209

 2 dr  2m  2 2  r 2  2m1   2  r r  

The roots of the quadratic equation

2m  2 2 2m1   2 0 r r 2m1r 2  2m r   22  0

or are given by

r



 1  2 21 2 2      21 21  m 

 2  2      1  1  1 22   . 21  m    

…(1)



This gives the minimum and maximum values of r. For applying the angle variable method, we required one complete cycle of coordinate r which consists of the motions from rmin to rmax and back to rmin . The angle variable are represented by eq. (3a) i.e.,

Ji 

 p dq . i

For our present case, the angle variables are

and

J 

 p d

Jr 

 p dq . r

Angle variable J will be

J 

 p d  

W d 

   2 d (cf . Art.3.17  1) 2

   2 d  2 2 , 0

and J r is given by

Jr 

 p dq   r

W dr r

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  2

r max

r min

1 r

210

 2m  2 2   2   2m1  r r  



 2m r

 2m r   2 2  dr

2m1r 2  2m r   2 2  2 . r

 2m r

Putting, in the last integral, r 

dr   

 2m r   2 2 

 2 2 2 

dr r

 2m r 1

 2m r   2 2 

1 u

du u2

(4m1r  2m )dr

 2m r

(2m1r  2m )dr

 2

and

 2m r   2





2m

 2m   1

dr    m 2  21 2 2    r     21  4m12     2

1 du u2 2 2 2  1  2m1 2m 2  2  2  2  u  u u  

 2 

 2m1r 2  2mr   22   r min.

r max.

 2m         m 2  21 2 2    . log r   r        21  4m12  1    21   2

2 2 2 

r max.

    r min.

 2m1  2mu   22u 2 

…(2)

The first term of eq. (2) vanishes for both the limits by virtue of eq. (1) and the second is equal to

   2m        . log  r   21   1     

 / 21 [ 1 (1 21 22 / m 2 )1/ 2 ]

     m 2  21 2 2     r     2 2  4 m  1  1       / 2 [ 1(1 2  2 / m 2 )1/ 2 ] 1 1 2 2

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1/ 2 1/ 2  2m      21 2 2     21 2 2   .   log 1    l og  1    21  m 2  m 2    1    21  

   2  2 1/ 2  1 2   1   2m  2  m  2    1      log  2 1/ 2  1     21 2    2 1  m 2      1

 2m   2m      .log(1)     . i  1   1 



2 m

 2m1 

and third term of eq. (2) becomes

2 2 2 

 2m

 2 2 

 2mu   2 2u 2 

 2 2 

 2m1 2m   2 u  u2   2 2  2 

du 2  2m m2  2  m   1     u   22  2 4   2 2    

du  2m1 2 2  m2 2 m  u  2   2 2 2  

  m   u  2  2     2 2 sin 1  2 2 2   2m1 2  m     24    u 2 2  m  2 2 sin 1   2m1 2 2  m2 2

    

rmax .

        rmin .

rmax .

   r . min

   2 2  m r  2 2 sin 1  2 2 2   r 2m1 2  m  

rmax .

   rmin .

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Substituting the values of rmax and rmin . from equation (1), we get

 2 2 sin 1 (1)  sin 1 (1)   2 2 . Thus we find that

Jr  0  

Therefore

 2m1 

2 m

 2m1 

J  J r 

Now putting

2 m

1  E,

 2 2

 2 2 .

2 m

 2  m1 

the total energy, we find from above equation that

E  1 

2 2 m 2

 J  J r 

…(3)

If v and vr be the required frequencies then by eq. (6a) of art. 3.19, we have

 H   2 2 m 2 v    J    J  J 2  r   H   2 2 m 2 vr    J r r   J  J 2  r 

   

   

 2 m 2

 J  J r  4 m 2

 J  J r 

        

…(4)

Thus we find that vr , v i.e., the two frequencies are commensurable and hence the given system is degenerate one (if a system possesses two or more commensurable frequencies, it is termed as degenerate). The cause of this degeneracy here is the nature of central force (  / r form) due to vr , v assume the same value.

SMALL OSCILLATION There are mechanical problems which involve continuous systems, and which can be treated as limiting case of a system of mutually interacting particles. The motion of a given particle is influenced by the motion of al the other particles in such a system and the overall motion of the particles. Under restrictive condition of small amplitude of oscillation, the entire system develop superposition of distinct modes of motion, referred to as normal modes of motion. In each such mode, the system oscillates with its own characteristic frequency and assumes a configuration

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213

peculiar to the frequency which distinguishes it form other normal modes. Such a system of mutually interacting particles is called a coupled system. Stable and unstable Equilibrium: While dealing with the small oscillations executed by various systems, we shall frequently come across the condition of equilibrium. A brief account of the equilibrium state will be given here.

Point of stable equilibrium:

Point or unstable equilibrium:

dV/dx is positive at a neighbouring point A.

dV/dx is negative at a neighbouring point Fig. 3.5

In Fig. (3.5 a and b), potential energy as a function of one dimension, say x, for a particle executing bound and unbound motion, respectively, has been plotted. At x 0, the slope of potential energy curve,

dV , is zero. Consequently, force is zero there. dx

That is ,

F ( x)  

dV 0 dx

And a particle placed at such points with zero velocity will continue to remain at rest. When the force acting on a particle vanishes, the particle is said to be in equilibrium. Point x 0 in Fig. (3.5 a) is at the minimum of potential energy, and represents a state of equilibrium. Now, force also vanishes at points at points like x0 in fig (3.5 b), at which potential energy the total energy and is  1 2 maximum (because at these points, E  V  m x  0 ; that means force m x  0) . Then the 2

question arises : what is the distinction between the equilibrium points at which the potential energy has a minimum value and those at which it has a maximum value?

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214

Fig. 3.5.1 At the former point x0 in fig (3.5 a), at which V is minimum, if the particle is displaced then the force

F ( x)  

dV dx

Will tend to return it and it will oscillate about the equilibrium point, performing bounded motion. These point are called points of stable equilibrium. On the other hand, if the particle is displaced slightly from point of equilibrium x 0 in fig. (3.5 b), then it will be acted upon by a force

F ( x)  

(dV ) dV  dx dx

Which will tend to push the particle away from equilibrium point, when released. For this reason such points are called the points of unstable equilibrium.

Stable Equilibrium (i)

If a slight displacement of the system from its position of equilibrium results only in small bounded motion about the point of equilibrium, then the system is said to be in a stable equilibrium. The examples are a bar pendulum at rest, a suspension galvanometer at its zero position.

(ii)

Further, the state of stable equilibrium also corresponds to minimum of potential energy because

Unstable Equilibrium

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If a slight displacement of the system from its equilibrium results in unbounded motion, then the system is said to be in an unstable equilibrium. The example is a rod standing on its one end, or an standing on its one end.

(ii)

Such a state is not characterized by minimum of potential energy. Therefore, if the system disturbed form

Potential energy about a point of stable equilibrium: Suppose the particle is slightly displaced from the point of stable equilibrium executing small oscillations. Then the potential energy function can be expressed in the form of a Taylorseries expansion. That is,

1 (d 2V ) V ( x)  V ( x0 )  x0 ( x  x0 )2  ...., 2 2 dx Since

(dV ) x0  0 dx At a point x0 of stable equilibrium. Since zero level of potential energy is arbitrary, we can choose it at the point x0 making

(V ) x0  0 . Then, to a first approximation, we can write

V ( x) 

1 (d 2V ) x0 ( x  x0 )2 2 2 dx 

1 k ( x  x0 )2 2

Where

 d 2V  k   2  x0  dx  Bounded motion is possible only when the potential at equilibrium is minimum. Suppose the system is disturbed from equilibrium energy dE above the equilibrium energy. Since the potential energy is minimum at equilibrium, any deviation from the equilibrium position will result in an increase in potential energy and the kinetic energy would then decrease (as the energy is conserved). Due to the decrease in kinetic energy, the velocities would also decrease and approach zero in the course of time; the motion is thus bounded. Such a conclusion can be drawn form the shape of the curve of fig (3.5.1 a). equilibrium by an increase in energy above the equilibrium energy, then the potential energy would decrease. Consequently, kinetic energy and hence velocities would increase infinitely with time, indicating unbounded motion. The same conclusion can easily be drawn from the shape be of the curve of fig. (3.5.1 b).

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215

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216

3.6 Formulation of the problem: Lagrange’s equations of motion for small oscillations We are going to make a strong limiting assumption about the motion, namely, the displacements of the particles are restricted to small deviations from position of stable equilibrium. We shall then consider only conservative system in which the potential energy is a function of position only. Also, the constraints that depend on time will be excluded from our formulation. A system of particles is said to be in stable equilibrium if all the particles are and remain at rest. In the conservative force fixed, therefore, generalized forces acting on each particle must vanish:

 V Qj     q  j

  0

…(1)

This equation yields the values qoj of the generalized co-ordinates that the particles have in equilibrium configuration. If we assume the equilibrium to the stable, the potential energy must be a minimum when evaluated at these q0 j . The displacements of the generalized co-ordinates from their equilibrium value will be denoted by u j and so

q j  qoj  u j

…(2)

Since the qoj are fixed, we use the u j as generalized co-ordinates for the system. Since we shall be interested only in motions for which displacements are shall, every relevant dynamical quantity may be appropriately expanded in a Taylor series about the equilibrium, retaining only first order non-vanishing terms. The potential energy, when expended thus, becomes,

V (q1, q2, ........qn )  V (u1 , u2 ......, un )  V (q01 , q02 ,.....q0 n )  V  j  q  j

 1   2V u    i j , k  q q 2 0  j k

…(3)

  u j uk  ... 0

It contains the constant level of potential energy at equilibrium, which may arbitrarily be chosen as reference level with the potential energy zero there. The second sum term is linear in u j and vanishes because of stability requirement, eq. (1). We are therefore left with the quadratic term as the first approximation to V :

V

1   2V    u j uk 2 j , k  qi qk 0

…(4)

1   V jk u j uk 2 j,k The coefficients

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ADVANCED SOLID MECHANICS   2V V jk    q q  j k   2V   q q j  k

217   0   Vk j 0

are constants, characteristic of equilibrium configuration and are symmetric in the indices. They can be arranged in the form of a symmetric matrix V. To verify that type of potential in fact establishes are interaction among the particles, we calculate the generalized force on the j

particle

Qj 

V q j



V u j

…(5)

  V jk uk j,k

Indeed, the force tending to change any generalised co-ordinate depends on the displacement of all the others and we have the correct form of the potential energy given by (4) We recall now, from that kinetic energy expression in generalized co-ordinates is a homogeneous quadratic function of generalized velocities:

T 

  1  m jk q j q k 2 j ,k

  1  m jk u j u k 2 j ,k

…(6)

Where the coefficient m jk are given by

 r m jk  mkj   mi  i  q i  j

  ri       qk 

 m jk (q1 , q2 ...qn )

….(7)

And are functions of the q j . As before, we can expand these coefficients in a Taylor series about the equilibrium values of the q’s and obtain

 m jk  m jk (q1 , q2 ......qn )  m jk (q01 , q02 ,....q0 n )     ui  ... i  ql 0

…(8)

Since T is already quadratic in the small quantities uj, we can not use the higher order terms and therefore we choose the lowest order approximation to T by taking

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m jk (q1 ,......qn )  m jk (q01 , q02 ,....q0 n )  mkj

218

…(9)

Which are symmetric with respect to the indices J and k. We shall like to denote the constant values of m jk at equilibrium by T jk to conform to the nation used for potential energy coefficients

Vij Therefore

T

  1  T jk , u j u k 2 j,k

…(10)

Near the equilibrium configuration of the system, Lagrangian may, therefore, be written as

L

  1  (T jk , u j u k  V jk u j uk ) 2 j,k

…(11)

Lagrange’s equations of motion in our nation appear as

d  L dt   u  j

 L   0,  u j 

And with L given by equation (11), finally we get the equations of motion for the coupled system :

 (T jk u k  V jk uk )  0,{ j  1,...n}

…(12)

The coordinates uk may be used to define a column matrix u and in matrix notation equations (12) appear as

    T11 T12 ...T1n   u1   V11        T21 .... ... ...   u 2    V21  ... .... .... ...   ...   ...      Tn1 ... ... Tnn      Vn1  un 

V12 ...V1n  u1    .... ... ...  u2  0 .... .... ...  ...    ... ... Vnn   un 



(or)

T u +Vu = 0

…(13)

Either set of equations (12) or (13) represents the equations of motion near equilibrium. Each of net n equations will generally involve all the coordinates and it is this set which must be solved to find the motion of the system. This is somewhat tedious. A great simplification would result if we transform the set into another form of n equations, each of which involves only a single variable. This is basically the reason for the introduction of normal coordinates and normal frequencies. The most direct and natural way to attack the equations (12) is to try a solution

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219

uk  Ak eit

…(14)

 Cak  it

Where C is a scalar factor common to all co-ordinates. This is harmonic oscillator solution which an equation of type (13) or (12) must have ; amplitudes are represented by the factor Cak. Why we this form of amplitude will become obvious later. We taken real part of solution (14) to correspond to a physical soluction*.

*It is worthwhile to mention here both the real and imaginary parts are separately solution of the same differential equation. To clarify this, we break up uk as

uk  Rk iI k

…(A)

When we substitute the general (14) into the equations of motion (12), we obtain n equations for the unknown ak.

 (V jk   2 Tjk )ak  0

( j  1, 2,..., n)

…(15)

For an n-particle system, these are the equations written in full expansion below:

( j  1), (V11   2 T11 ) a1  (V12   2 T12 ) a2  ...  (V1n   2 T1n )an  0 ( j  2), (T21   2 T21 ) a1  (V22   2 T22 ) a2  ...  (V2 n   2 T2 n )an  0 ....

....

...

( j  n), (Vn1   2 Tn1 ) a1  (Vn 2   2 Tn 2 ) a2  ...  (Vnn   2 Tnn )an  0 These are n simultaneous linear homogeneous equations for the n unknowns a j and can be solved for (n-1) unknowns of these in terms of the remaining one, if and only if the determinant of the coefficients vanishes:

V11   2 T11 V12   2 T12

... V1n   2 T1n

V21   2 T21 V22   2 T22

... V2 n   2 T2 n

...

....

Vn1   2 Tn1

...

... Vnn   2 Tnn

0

The equations, in effect, is of nth degree for the frequency

1 ,.........n 2

…(16)



and will possess n roots

for which our solution, equation (14), represents a possible solution of the

equations of motion (12). These values of

 are the normal frequencies of the system mentioned

earlier. For each of these frequency values, equations (15) can be solved for (n-1) ak’s in terms of the remaining ak. Where Rk and Ik are real quantities. Also, let use suppose the u k satisfies a linear differential equation with real coefficients of the type 



Au k  B u k  Cuk  0

…(B)

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220

Differentiating (A) and substituting in (B), we have 



A R k  B R uk  CRuk  0 

and



A I k  B I k  CI k  0

On separating real and imaginary parts of the resulting equation and equating each to zero. From equation (A) we see then that, if uk is a complex solution of a linear differential equation with real coefficients, the real and imaginary parts are separately solution of the same differential equation. The standard convention is to select the real part as the solution. Eq. (13) may also be written as

(Va   2 Ta)  0

…(17)

Where a is now a column matrix of n components:

 a1  a  a   2 ....    an  3.6.1 Properties of T, V and

:

In the first place, these frequencies

l will turn out to be real positive quantities and on

this account, the motion given by eq. (3.6-14) will be completely oscillatory about the position of stable equilibrium. However, the characteristic that renders the frequencies as described above so useful is that the numbers ak corresponding to a given

 , which can be obtained by solving

the set of n homogeneous linear equations, represent the behaviour of the physical system at that particular frequency; therefore we say that the set ak is a normal mode of the frequency. Each frequency will posses its own mode of vibration that can exist independent of the other modes; the word ‘normal’ it used here in this sense. To distinguish between different modes and to identify a particular mode, we put a second label subscript to ak ; for example aki will denote the kth normal mode, corresponding to lth frequency We shall proceed to show that

l

l 2 are

all real. Let akl be the kth component of lth

eigenvector al . Then, one of the equations obtained from equation (15) for a frequency n

V k 1

akl  l 2  T jk akl

 l  T jk akl k

l is

…(1)

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221

l has been written for l 2 If the frequency is

m 2 ,then the complex conjugate of equation (2) can be written as n

V k 1

a jm *  m *  T jk a jm *.

…(2)

Multiply equation (1) be a jm * and sum over j. also multiply equation (2) akl and sum over k. Then subtract the resultant equation to get. n

k 1

  (V

akl a jm *  V jk a jm * akl )

 (1  m *)   Tjk akl a jm * k j

Since the left hand side of above equation is zero, we are left with n

(1  m *)   T jk akl a jm *  0

…(3)

k 1 j 1

Let us first consider the case l=m. The eq. (3) becomes n

(l  l *)   T jk akl a jl *  0

…(4)

k 1 j 1

We can easily show that

 T jk akl a jl *  0

k j

And is positive definite * and consequently, from eq. (4) we arrive at the result

1  l * Which implies that Now, since

…(5)

1 or l 2 are all real.

1 ’s are all real, all components akl ' s

will also be real

a jm *  a jm *To show it, let us put

akl  akl  i bkl Then



T jk akl a jl *   T jk (akl  i bkl ) (a ji  i b jl )

k, j

  (T jk akl a jl  T jk bkl b jl )  i  (T jk bkl a jl  T jk a kl b jl ) k, j

k, j

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The second term on right hand side is zero because T jk  Tkj i.e.T jk is symmetrical. The first term on right hand side is positive definite; for, if akl be replaced by qk and a ji by q j in

 T jk akl a jl Then it becomes

 T jk qk q j Which is the expression of kinetic energy and therefore positive definite. Further, from eq. (3), we have

(l 2  m 2 )  Tjk a jm akl  0

…(6)

j ,k

Now multiply (1) a ji and sum over j. We obtain

l 2 

 V jk a jl akl j ,k

…(7)

 T jk a jl akl j ,k

This equation says that

l 2 is a positive quantity **

Therefore, the eigen frequencies are all real and the motion for a given

l 2

will be

completely oscillatory about the position of stable equilibrium. Where the potential is not minimum, numerator in eq. (7) might assume negative value, and motion with imaginary frequency would give rise to an unbounded exponential rise of u j with time, in accordance with solution . The motions, then, obviously unstable which we do not desire in our discussion. If

l 2  m 2 ,

we obtain from eq. (6),

 T jk a jm akl  0, l  m

…(8)

j ,k

Which is called the orthogonality property. Further, the coefficients a jl are not completely determined form eq. (1), by virtue of the fact that this is a set of homogeneous linear equations. This uncertainty be removed be requiring that

 T jk a jl akl  1 j ,k

lm

…(9)

The two equations (8) and (9) can be combined into on by means of kronecker delta symbol

 T jk a jm akl   ml

 ml

…(10)

j ,k

Which are then termed as the orthonornality conditions

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223

  1  T jk u j u k the denominator is twice the value the kinetic energy would have if the 2 j ,k

velocities were the a jl and is therefore positive, and since by equation V 

1  V jk u j uk the 2 j ,k

numerator is twice the value the potential energy would have if the displacements were the a jl ; this energy is positive because we have assumed the equilibrium potential energy to be a minimum. Neither numerator nor denominator can be negative and the denominator cannot be zero, hence

l 2 is finite and positive.

We may rewrite eq. (9)

 T jk a jm akl  0

…(11)

j ,k

In matrix form as

aT Ta  1*

…(12)

If in eq. (1) we introduce the diagonal matrix

 characterized as

ml  l  ml We get n

 V jk akl   T jk akl ml

k 1

Which can be written in matrix form as

Va  Ta  Multiplying above equation by

aT , we get

aT Va  aT Ta On

aT Va  

…(13)

On using eq.(12)

3.7 Normal co-ordinates and normal frequencies of vibration We now look for a solution of equations of motion where only one single frequency (normal) is involved in the solution and in such a case, the co-ordinate appearing in it will be called the normal co-ordinate. Therefore, the generalized co-ordinates, each one of them executing oscillations of one single frequency, are called normal co-ordinates.

 a b   q1  T *(q1 , q2 )      A MA  b c  q2  (or)

aq12  2bq1 q2  AT MA

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224

 q1  A     a column matrix  q2 

Where

a b  M    matrix b c AT  (q1 , q2 )  transpose matrix of A The transformation of equations of motion into such co-ordinates (normal co-ordinates), denoted by

 and

therefore of Lagrangian, is effected by means of linear transformation. We

define the new set of co-ordinates

 k related

to the original co-ordinates u j by the defining

equations:

u j   a jk k

…(1)

If we express u j s as the elements of the single column matrix (u) and

 respectively then we can

write eq. (1) as

u  a

…(2)

In order to find out Lagrangian in new co-ordinates, we shall first express kinetic and potential energy in terms of

 . The potential energy is

1  V jk u j uk 2 j ,k 1   u j V jk uk 2 j ,k Which is quadratic in u and therefore can be expressed as V

1 V  uT Vu 2 Putting for u from eq. (2), we get

1 V  (a )T V a  2 1   T aT Va  2 1   T  2 On applying eq. (13) of art. 3.6.1. Further above equation is quadratic in

1  l l 2 2l 1   l 2 l 2 2l

 so that

V

…(3)

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225

The kinetic energy is given by   1   T jk u j u k 2 j k   1    u j T jk u k 2 j k 1 T   u Tu 2

T

1 T T    a T a 2 

1 T    2



1  2  l 2 l

(from eq. 2)

(using eq. 12 art 3.6.1)

…(4)

Therefore, the Lagrangian in new co-ordinate system will be

L

1 n  2 1 n 2 2 l l  l  2  2 l 1 l 1

L

  l ,

Giving 

 l



l 1

n L   l 2l l l 1

Which when substituted in

 d  L dt     l

   L  0  l 

Gives n



 ( l 1

 l 2l )  0

Therefore the equations of motion in new coordinates are 

 1  121  0 

 2  2 22  0 ... ... ... ... 

 l  l 2l  0

…(5)

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226

1 corresponds

only to

1 ,2 to 2 etc.

coordinate executes only one single frequency oscillation and therefore

thus each

1 ,2 , etc. are termed as

normal co-ordinates. The solution of equation 

 l  l 2l  0 Will be

l  A1 cos l t  Bl sin l t ,

if l 2  0

l  A1t  Bl

if l 2  0

l  Al e  t  Bl e  t , l

In this first case, since

if l 2  0

l 2 are

…(6)

real and positive, all co-ordinates always remain finites for any

time t and we then say that the equilibrium is stable for the case

l 2  0 . But for rest of the two

cases, we find that the co-ordinates become infinite as the time advances and consequently such solution refers to unstable equilibrium. Normal modes of vibration:

  2 v

Writing The solutions become

1  A1 cos 2 v1t  B1 sin 2 v1t 2  A2 cos 2 v2t  B2 sin 2 v2t ...

... ... ...

...

n  An cos 2 vnt  Bn sin 2 vnt

…(7)

Where A’s and B’s are 2n arbitrary constants to be determined from the initial conditions. Suppose we choose these constants such that all, except A 1 and B1 are zero. Then only co-ordinate

1

will very periodically with time, while all the other co-ordinates will remain zero for

all times. Such a situation corresponds to a normal mode of vibration, i.e. we say that the system is vibrating in a normal mode. Obviously, and of the co-ordinates can be given liberty to very with time by such a choice of co-ordinates and therefore, there will be n normal modes of vibration and n normal frequencies v1 , v2 ,.....vn corresponding to each normal co-ordinate

1 ,2 ,3 ,....n .

We may write equation (7) as

l  Al cos (l t   l ) Where

 l is the phase factor. Then from equation (1), the old co-ordinates are given by u j   a jk Ak cos (k t   k ) k

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3.8 Linear Triatomic Molecule: We shall consider a linear triatomic symmetric molecule of the type YX2 (e.g. CO2), shown in Fig.3.8. Y is the central atom and we shall neglect the interaction between this and the end atoms (X). We further assume that there exists an elastic bond between the central atom and the end atoms, of force constant k. let the mass of central atom be M. we denote the displacements of atoms form equilibrium configuration by the generalized co-ordinates q1, q2 and q3. then

Fig. 3.8 Kinetic energy

1 2  2 1  2 T  m(q1  q3 )  M q 2 2 2

 m 0 0   2T  (q1 , q 2 , q 3 )  0 M 0  0 0 m   

(or)



   q1     q2     q3   

giving

m 0 0    T  (Tij )   0 M 0  0 0 m   Potential energy:

V 

(or)

…(1)

1 1 k (q2  q1 )2  k (q3  q2 )2 2 2

1 k (q12  2q2 2  q32  2q1q2  2q2 q3 ) 2

 k  k 0   q1     2V  (q1 , q2 , q3 )  k 2k  k   q2  0  k k  q    3

Giving

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 k  k 0   V  (Vij )   k 2k  k  0  k k   

…(2)

We then write the secular equation

| V   2T | 0

(or)

k   2m

k

2k   m

k

0 k  0

…(3)

k  m 2

Which gives on development, gives

(k   2 m)[2k   2 M ) (k   2 m)  k 2 ]  k .k .(k   2 m)  0  0 (k   2 m)[2k   2 M ) (k   2 m)  2k 2 ]  0

 2 ( 2 m  k )[k ( M  2m)   2 m M ]  0 This cubic equation gives three values of

:

1  0 Normal frequencies :

2 

k m

…(4) 1/ 2

 k  2m   3   1   M  m 

The first case refers to translator motion (1  0) of the atoms and the rest two to the oscillatory motion. In order to calculate the corresponding normal co-ordinates

1 ,2 and 3 ,

we shall

proceed with

q j ,  a jk k k

(or)

 q1   a11 a12 a13  1        q2    a21 a22 a23  2  q  a a    3   31 32 a23  3 

…(5)

Therefore problem is to calculate components of eigenvectors a1, a2 and a3. Let us deal with the first case :

1  0

and calculation of the components a11, a21 and a31

of eigenvector a1 : For this purpose, we shall apply relation

(Vij   2 Tij ) a j  0, (i  1, 2, 3) j

From eq. (3) we write for a1 as

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 k  12 m  k 0   a11     2k  12 M  k   a21   0  k   k k  12 m   a31  0 Putting

12  0  k  k 0   a11      k 2k  k   a21   0 0  k k  a     31 

giving

ka11  ka21  0 ka11  2ka21  ka31  0  ka21  ka21  0 yielding

a11  a21  a31   (say)

(or)

…(6)

    a1         Which shows that displacements of all the atoms is in the same direction and are equal in

amount (see Fig. 3.8.1 a)

Fig. 3.8.1 Let us now deal second case;

2  (k / m) and calculate the components a12, a22 and

a32 of eigenvector a : For this purpose, we can write

 k  2 2 m  k 0   a12     2 2 k  2 M  k   a22   0  k   k k  2 2 m   a32  0 Which gives on putting

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2  (k / m) a22  0 and

a12  a32  

So that components of a2 are

    a2 :  0      

…(7)

Similarly, for third case :

k 

2m  

3   1   and components of eigenvector a3 are to M   m 

be calculated. Putting

32 

k  2m  1   m M 

We get

 2mk  M  k    0 

k k

M m

k

 0    a13    k   a23   0   a  2mk   33    M 

yielding

2mk a13  ka23  0 M kM ka23  a23  ka33  0 M 2m ka23  ka33  0 M 

From these equations if follows that

a13  a33   and

a23  

2m 2m a13    M M

So that the components of a3, will be

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     2m  a3 ;     M      

…(8)

Eqs. (6) and (8) define the required components of eigenvectors if

,  ,

are known for which

we apply the following orthogonality T

…(9)

a T a=I because

    m 0 0    2 m (aij )    0    and (Tij )   0 M 0   M  0 0 m          We write eq. (9) as

        m 0 0     0     0 M 0      0 0 m 2m        M         m   0     M     m 2m      M 

(or)





  1 0 0  2m   0     0 1 0  M      0 0 1   

m

 m  1 0 0     0  2 m    0 1 0   m  m   0 0 1

    2 (2m  M ) 0  0 1 0 0     2    0 1 0 0 2 m 0      2m    0 0 1  2  0 0  2 m 1     M    



1 (2m  M )1/ 2



1 (2m)1/ 2



…(10)

1 1/ 2

  2m    2m  1  M     

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Now using eqs. (6), (7), (8) and (10) with eq. (5), we can write for the normal co-ordinates and

1 ,2

3 associated with normal frequencies 1 , 2 , and 3 , respectively, that is:   1   (2m  M )    q1   1     q2    (2m  M ) q    3   1   (2m  M )  

1 (2m)

1 (2m)

  1     2m     2m  1  M        1  2m / M     2   2m        2 m  1  M    3       1    2m     2m 1  M       

Which completes the discussion It can be seen that : Is case II, with

2 , we find that

a22  0, a12  a32 Which indicates that central atom does not take part in motion and atoms oscillate with equal amplitude but opposite in phase, Fig. (3.8.1 b) We find that in case III,

a13  a33   a23  

2m  M

Which indicates that end atoms vibrate in phase with equal amplitude whereas the central atom vibrates in opposite phase with a different amplitude. Fig. (3.8.1 c) Since in case III, the molecule is asymmetrically stretched oscillating dipole moment will be associated with the motion and corresponding absorption band will appear in the infrared region.

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UNIT IV Classical Statistics Mechanics: Basic Concepts The experimental data of modern physics may be divided, roughly speaking, into two main classes. On one hand, we have experiments in microscopic (i.e., small scale) physics where dimensions are small, for example, the determination of properties of individual isolated small systems, such as nuclei, atoms or molecules. On the other hand, we have experiments in macroscopic (i.e. large scale) physics which determine properties of bulk matter, namely, macroscopic constants such as specific heat, dielectric constant etc. When one is dealing with macroscopic system, he is not concerned with the detailed behaviour of each of the individual particles. If the macroscopic parameters of an isolated system do not vary in time, then the system is in equilibrium.

A macroscopic system at equilibrium has certain properties; for

example, the energy, heat capacity, entropy, volume, pressure etc. interest in thermodynamics.

Which are of particular

Thermodynamics is concerned with the relationships between

certain macroscopic properties (the thermodynamical variables, or functions) of system in equilibrium. If the state of the substance is in the form V = V (T,P) and we know the entropy, S, at one pressure, we can calculate the entropy at other pressure and at the same temperature from the relation:

 S   V        P T  T  P In thermodynamical sense, the system is a collection of macroscopic bodies such as solids, liquids or gasses, among which thermal, physical or chemical changes can occur.

In a

‘mechanical’ macroscopic system, it means a collection of microscopic bodies (atoms, ions or electrons). Our intuition tells us that it should not be necessary to measure the macroscopic properties of a system but it should be possible to calculate them if the properties of the constituent molecules and laws of force between them are known. Statistical mechanics is a branch of science which establishes the relation between macroscopic behaviours (bulk properties) of the system in terms of its microscopic behaviour (individual properties). Statistical mechanics as its name implies is not concerned with the actual motions or interactions of the individual particles but explores the most probable behaviour of assembly of particles. If we attempt to determine the actual behaviour of a gas consisting of say 23

molecules whose position and velocities are known at any initial time, we have to solve 10

equations of motion. Moreover, it is not possible to have a complete knowledge regarding the

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positions and velocities of all the molecule. Under such circumstances, the method of statistical mechanics is used. It enables to predict the average properties of the system without a detailed knowledge about the initial condition of its component. Radioactive decay is a good example of the statistical nature of the phenomenon. In radioactive decay one cannot say which atom of the radioactive material will decay first and when. Applying the principle of statistical mechanics, it can be said that a certain average number of atoms will decay at any given instant of time. The methods of statistical mechanics are important in such cases. The study of statistical mechanics can be classified mainly in the following two categories: (1) Classical statistics or Maxwell-Boltzmann statistics. (2) Quantum statistics or Fermi-Dirac statistics. The classical statistics or Maxwell-Boltzmann statistics utilizes the classical results of Maxwellâ&#x20AC;&#x2122;s law of distribution of molecular velocities and Boltzmann theorem relating entropy and probability. The quantum statistics are developed by Bose, Einstein, Fermi and Dirac. One form is known as Bose-Einstein statistics, while the other is known as Fermi-Dirac statistics. The essential difference between these two types is that Bose-Einstein statistics is used in case of indistinguishable identical particles having zero or integral spin while Fermi-Dirac statistics is used in case of indistinguishable identical particles having spin half. Bose-Einstein statistics holds good for photons while Fermi-Dirac statistics holds for elementary material particles. The classical statistics is only a limiting case of quantum statistics.

Phase Space: In classical mechanics, the position of a point particle is described in terms of three Cartesian coordinate x, y, z and the particle is said to have three degrees of freedom. The state of the .

motion of particle is described in terms of velocity components the differentiation with respect to time). .

x, y, z (where the dot indicates

Sometime it is more convenient to use momentum

coordinates px , p y , pz ( m x, m y, m z where m is the mass of the particle) instead of velocity coordinates.

Thus the position of a single particle can be specified in terms of Cartesian

coordinates x, y, z and its corresponding momentum components px , p y , pz .

In order to

facilitate the application of the laws of mechanics, it is convenient to invent a scheme which can be used to describe the state (position and momentum) of the particle at a particular instant. For this purpose, we imagine a six dimensional space in which the six coordinates x, y, z and

px , p y , pz are marked along six mutually perpendicular axes in space. The combined position

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and momentum space is then called as phase space. A point in the phase space represents the position and momentum of the particle at some particular instant. Division of phase space into cells: The meaning of a point in phase space can be understood with the help of uncertainty principle. The phase space is divided into six dimensional cells whose sides are dx, dy, dz, dpx , dp y , dpz .

Such cells are called phase cells.

Further we

approach close to the limit of a point in phase space reducing the size of a cell. The volume of each cell is given by

d =dx dy dz dpx dpy dpz But according to uncertainty principle.

dxdpx  h,

dydpy  h,

Thus

d  h3

dzdpz  h

A point in the phase space is actually considered to be a cell whose minimum volume is 3

of the order of h . So the particle in the phase space cannot be considered exactly located at point x, y, z, px , p y , pz but can only be found somewhere within a phase cell centred at that point. The generalization of the above remark can be made for a complex system. For example the state of an assembly of N systems can be described in terms of 6N coordinates i.e., 3N position coordinates and 3N momentum coordinates. Now the state of the particle can be defined in 6N dimensional space (conceptual space) known as phase space or  space. Let the system has f degrees of freedom (number of independent coordinates needed for the description of the system). The instantaneous configuration of the system can be described by a set of generalized position coordinates q1 , q2 , q3 ...q f and corresponding momentum coordinates p1 , p2 , p3 ... p f i.e., by total of 2f parameters.

The state of the particle can be described by imagining a

conceptual space having two dimensions. This 2f dimensional coordinate momentum space so formed is called as phase space of  space. The set numbers (q1 , q2 , q3 ...q f ; p1 , p2 , p3 ... p f ) is regarded as a point in phase space of 2f dimensions. Thus the dynamical state of a system may be specified by locating a point in a 2n dimensional space (in general); the coordinates of the point being the values of the n coordinates and n momenta which specify the state. The space is referred to as phase space and the point is called a phase point. Ensembles: If we have a collection of particles, we shall refer to a single particle as a system and to the collection of particles as a whole as an assembly. The collection of a large number of assemblies is called as an ensemble. The numbers of the ensemble are identical in features (like

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volume, energy, total number of particles etc.) and are known as element. The element differ in their states i.e. they differ in coordinates and velocities. Thus an ensemble is defined as a collection of a very large number of assemblies which are essentially independent (i.e. in calculating the possible eigenstate of an assembly we do not have to worry about any interaction between the assembly of interest and the other assemblies) of one another but which have been made macroscopically as identical as possible. By being macroscopically identical, we mean that each assembly is characterized by the same values of set of macroscopic parameters which uniquely determine the equilibrium state of the assembly. In an ensemble, the system play the same role as the non-interacting molecules do in a gas. The macroscopic identity of the systems constituting an assemble can be achieved by choosing the same values of some set of macroscopic parameters. These parameters uniquely determine the equilibrium state of the system.

Accordingly three types of ensembles i.e.,

microcanonical, canonical and grand canonical are most widely used. Their descriptions are as follows:

4.1 Fundamental Postulates of Statistical Mechanics The application of statistical mechanics to gases, certain fundamental postulates are made. These are: 1. Any gas under consideration may be considered to be composed of large number of molecules which are constantly in motion and behave like very small elastic spheres. 2. All the cells in phase space are of equal size. 3. All accessible microstates corresponding to possible macrostates are equally probable. This is the most fundamental and hence important postulates of statistical mechanics. For example, if w consider as earlier, the distribution of 4 distinguishable particles a, b, c, d in two cells i and j, then Probability of microstate (a b c d, o) =

1 1  4 2 16

 1  1  1  3   2  2  16

Probability of microstate (a b c, d) = 

 1  1  1  2  2   2  2  16

Probability of microstate (ab, cd) = 

 1  1  1  3    2  2  16

Probability of microstate (a, bcd) = 

Thus, we can conclude that the probability of all accessible microstates of the system is equal. From this postulate it follows that the probability of occurrence of a given macrostate is proportional to the number of microstates, corresponding to the given macrostate. Thus the

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probability P(E) of system possessing energy E is proportional to the thermodynamic probability W(E) i.e., P(E)  W(E) P(E) = cW(E) or Where c is a constant of proportionality. 4. The equilibrium state of a gas corresponds to the macrostate of maximum probability. 5. The total number of molecules is constant. This is in accordance with the principle of conservation of matter. If n1 is the number of molecules in cell 1, n2 is cell 2, n3 in cell 3 etc. If n is the total number of molecules, then

n1  n2  n3  ......  ni  .......nk   ni  n  a constant

…(1)

6. The total energy of the system is constant. This is again in accordance with the conservation of energy of the system. If E 1 is the energy of each particle in cell 1, E2 the energy of each particle in cell 2 etc. and E the total energy, then

E1n1  E2 n2  E3n3  ...   Ei ni  E  a constant

…(2)

Uses of Ensemble: In practice, we consider the three main types of ensembles and each ensemble corresponds to a different experimental situation. In fact, the three ensembles, which we have introduced and discussed above are only examples of the infinite number of ensembles that can be considered. These are particularly useful for two main reasons: (a) Firstly, they correspond approximately to the types of thermodynamic measurements most frequently made in practices. (b) Secondly, in large number of systems, it is useful to find that the values of thermodynamic quantities that are not very sensitive to the method of measurement. For example, in the measurement of specific heat of a liquid of known mass and thus known particle number, at temperature far below the boiling point, it matters very little whether the liquid is isolated at constant temperature so that the number of particles is fixed as in a microcanonical ensemble or in equilibrium with its vapour so that the number particles can fluctuate as in grand-canonical ensemble.

Comparison of Ensembles Property

Microcanonical

Canonical

Grand-canonical

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Isolated system, that

Systems

is, energy and number

thermal contact with

exchange both energy

heat reservoir. Hence

and

system can exchange

particles.

particles

the

system are constant.

are

Systems

matter

can

i.e.

energy but no matter i.e. particles Probability

density

2. Probability

(E) = constant, in the

distribution

range E to E +dE = o,

(E) =Ae

E/kT

 ( E )  e( n  E ) / kT 

outside this range. None Fluctuations in energy 3. Fluctuations Z=Δ  4. Partition

Z =  e-  E d

function

Fluctuations in both energy and matter i.e., particles. 

Z   en / T Z n

n 0

4.2 Liouvilles Theorem: We have seen that the dynamical state of a system at some instant of time can be represented by a point in the phase space. This point will not be stationary but will move along a definite trajectory which is determined from the equation of motion .

qi 

H

and

 pi

pi  

H  qi

Where H=H(q1 ,..., qf ;p1 ,..., pf ) is the Hamiltonian of the system. As a result of this motion, the density  of system in phase space changes with time. We are interested in finding

  at a given point in phase space. To deduce use is made of t t

Liouville’s theorem. This theorem is primarily concerned with defining a fundamental property of the phase space-the space of position and momenta coordinates, in which the system, represented by a point, moves in time. The theorem consists of two parts: (1) The principle of conservation of density in phase space. This part states the conservation of density in phase space i.e. the rate of change of density of phase point in phase space is zero or

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d 0. dt (2) The principle of conservation of extension in phase space. This part gives the conservation of extension in phase space i.e.,

d ()  0 or the volume at d

the disposal of a particular number of phase points is conserved throughout the phase space. First Part: Consider any arbitrary hyper volume

d    q1  q2 ...... q f ,  p1  p2 ....... p f in the phase space located between q1 and q1   q1......q f

and q f   q f ;

p1 and

p1   p1 ,.... p f and p f   p f . The number of systems (phase points) in this volume element ( q1... q f ;  p1... p f ) change as the coordinates and momenta of the systems vary with time. Let  be the density of phase points. Now the number of phase points in this volume element at any instant t is given by

 N     q1... q f ,  p1... p f . The change in number in phase points in the volume element per unit time is given by . d d ( N )  ( )    dt dt

(Where  

d ) dt

   q1... q f ;  p1... p f

Fig. 4.1 In a time dt, the change in number of phase points within the volume of phase space is given by

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  dt  q1... q f ;  p1... p f This change is due to the number of systems entering and leaving this volume in time dt. Consider two faces of hyper volume normal to q-axis with coordinates q1 and

q1   q1 as

shown in fig. (4.1). In this figure only two axes are shown. .

q1 is the component of velocity of phase point at q1 ,...q f , p1 ,...pf , then the number of

phase points entering the first face in time dt will be .

 q1 dt  q2 ... q f  p1... p f .

…(1)

Again the phase points leaving the face (q1   q1 ) in time dt will be .     .  q1   q1  q1   q1  dt  q2 ... q f  p1... p f      q  q  1  1 

…(2)

This is due to the fact that density  changes with change in position and momentum coordinates and at opposite phase coordinate q1 changes to q1   q1 , therefore density  changes to

.            q1  . Similarly the velocity q1 changes to    q1   

.  .    q1    q1  . q1      q 1       Neglecting higher order terms, we have .  .    .  q   q    1  q   q  dt  q ... q ; p ... p 1 2 f 1 f 1  1  q1 q1      

…(3)

Subtracting equation (3) from equation (1), we have .  .    q  dt  q1... q f ; p1... p f    1  q1  q1 q1   

…(4)

Similarly for p1 coordinate, we have .  .    p 1   dt  q1... q f ; p1... p f    p1  p1 p1   

…(5)

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The net increase in time dt of number of systems in this volume of phase space is then obtained by summing the net number of system entering the volume through all the faces labeled by q1 ,...q f , and p1 ,...pf . Hence . . f   d    qi  pi ( N )      dt qi i 1   qi  

   . .     q1   p i      qi pi    

dt  q1... q f ; p1... p f Now

…(6)

d  ( N )  dt  q1... q f  p1... p f dt t

.   f       qi pi    .  .   Therefore dt  q1... q f  p1... p f       q1  pi  t qi   qi pi   i 1   qi    

dt  q1... q f  p1... p f . . f       pi  qi      t qi i 1   pi  

   . .     q1   p i      qi pi    

…(7)

The result, however, can be immediately simplified. Inaccordance with the equations of motion in canonical form, we have .

qi  .

H

and

 pi

pi  

H  qi

 qi  pi 2 H   , qi pi qi pi

Now

Since the order of differentiation is immaterial .  .  q  p i  i    q  p i 1 i  i f



   0.  

…(8)

Substituting equation (8) in equation (7), f   .  .       qi  pi      pi   t q , p i 1  qi

…(9)

This result is known as Liouville’s theorem. Equation (9) can be written as

 dqi  dpi     0    pi dt  t q , p i qi dt

…(10)

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and is identical with the equation of continuity in hydro-dynamics. If  is a function of q, p and t, and q, p are functions of t, the total differential coefficient of  with respect to t is given by

d    dq  dp   .  . dt t q dt p dt Generalising this for all fp’s and fq’s, we get

d    dqi  dpi   .  . dt t dt dt i qi i pi

…(11)

Comparing equations (10) and (11), we get

d  0. dt

…(A)

Inaccordance with Gibbs, this form of expression may be called the principle of the conservation of density in phase space. Therefore, the density of a group of points remains constant along their trajectories in the phase space. If at any time the phase points are distributed uniformly in phase space, they will for ever have uniform density.

Second Part: For this part, we have prove that

d    0. We know that dt

 N  , or

d d d  N        . dt dt dt

Since the number of phase points

N

in a given region of the phase space must remain fixed,

as the system can neither be created nor destroyed, we have



d d        0 dt dt

We have proved that



…(12)

d    0. dt …(13)

d  0. Hence it follows that dt

d    0. dt

…(14)

Since  ≠ 0, hence

d    0. dt

…(B)

Following Gibbs, this equation gives the principle of conservation of extension in phase space.

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Alternative proof of the second part: Let q1 ,...q f , and p1 ,...pf be the position and momenta coordinates which constitute the phase space. Let

 p1. p f  q1. q f  V1

is an element of volume. Let

 V2

is the element of

volume after an interval of time dt. The new coordinates, therefore, are

p f  p1    p  dt . p  dt   f  1  t   t   and

q f  q1    q  dt . q  dt    1  f t   t  

…(15)

because the rate of change of momenta and position coordinates are

pi

 t i

after a time dt the change will be

pi

and

qi

 t

and

qi

 t .dt and  t dt. i

Let the volume element be  2 , then .

2   ( p1  p1 dt )  ( p f  p f dt )  (q1  q1 dt )... (q f  q f dt ), Where

…(16)

. p1 . q  p1 and 1  q1 . t t

Equation (16) can be written as .    p  2   p1... p f  q1... q f 1  1 dt   p1   

.    p f 1  dt    p f  

. .     q  q f 1  1   dt ... 1  dt  .  q1   q f     

…(17)

Equation (17) can be simplified by considering only two terms i.e. . .     p  p 1 2    2   p1  p2  q1  q2 1  dt 1  dt   p1  p2     . .     q  q 1 1  dt 1  2 dt   q1  q2     . .    p  p 1 2    p1  p2  q1  q2 1  dt  dt   p1 p2   

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. .    q  q 1 1  dt  2 dt  .  q1 q2    2

[neglecting the terms involving (dt ) ] . . . .    p  p  q  q 1 2 1   p1  p2  q1  q2 1  dt  dt  dt  2 dt   p1 p2 q1 q2    . .  2   p  q i   p1  p2  q1  q2 1     i  i 1  pi qi  

  dt     

…(18)

Hence the general form of equation (17) will be . .  f   pi  qi    2   p1... p f  q1... q f 1     i 1  pi qi   . .  f   q  p i  1 1    i   i 1  qi pi  



 2  1 dt

  dt     

  dt  ;    

.  f  .  p  q i  1    i  i 1  pi qi  

  ,  

…(19)

dt can be taken out of the summation as it does not have the suffix I and is independent of summation. We know that .

qi 

. H H and p i   pi qi

 qi p 2 H  i  [Since the order of differentiation is immaterial] qi pi qi pi .  .  p  q i   i   p qi i 1  i f

   0.  

Substituting in equation (19) we have

 2  1 dt

 f   1  (0)   0.  i 1 

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 2  1 dt

245

represents the rate of change of volume of phase space, we have

d   0 dt or

 d  f  pi qi   0.   dt  i 1 

4.3 Microcanonical ensemble: The microcanonical ensemble is a collection of essentially independent assemblies having the same energy E, volume V, and number N of systems, all the systems are of the same type. The individual assemblies are separated by rigid, impermeable and well insulated walls (fig. 4.3(a)) such that the values of E, V and N are not affected by the presence of other systems. We can not actually specify the macroscopic energy of an assembly exactly.

Fig. 4.3(a)

Consider a closed system for which the total energy H (q p) = E remains constant, according to the equation E (q1 ,...q f , p1 ,... p f )  cons tan t.

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Fig. 4.3(b) The locus of all the phase points having equal energies in phase space is called an energy surface or ergodic surface.

We can now imagine a family of such energy surfaces

constructed in phase space and let us consider two neighbouring surfaces with energies E and E+E (Fig. 4.3(b)). Each surface divides the phase space in two parts, one of higher and the other of lower energy, hence they will never intersect each other. As they include some phase volume in between them, they will contain certain number of phase points. The number of phase points between them will be constant. A very useful ensemble can be obtained by taking the density as equal to zero for all values of the energy except in a selected narrow range of E and E+E. Using the terminology of Gibbs such an ensemble, specified by  = constant

(in the range E to E+E )

 = 0.

(Outside this range)

may be called a microcanonical ensemble. We observe the following properties: (i)

As  is a function of energy, this ensemble is in statistical equilibrium.

(ii)

The average properties predicted by such ensemble will not vary in time being in statistical equilibrium.

(iii)

As  is constant within the energy shell, the distribution of phase points is uniform (by Liouville’s theorem).

An ensemble of this kind can be regarded as obtained from an originally uniform ensemble by discarding all systems having phase points with positions that do not fall within the limits in the phase space that correspond the energy range E and E+E.

4.4 Canonical ensemble; The canonical ensemble is a collection of essentially independent assemblies having the same temperature T, volume V, and number of identical particles N.

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Fig. 4.4 To assure ourselves that all the assemblies have the same temperature, we could bring each in thermal contact with a large heat reservoir at temperature T or we could simply bring all of the assemblies in thermal contact with each other. Fig. 4.4. represents symbolically a canonical ensemble. The individual assemblies are separated by rigid, impermeable, but diathermic walls.

Since energy can be exchanges

between the assemblies, they will reach a common temperature. Thus in canonical ensemble, system can exchange energy but not particles.

4.5 Grand canonical ensemble: The microcanonical ensemble is a collection of independent assemblies having the same energy E, volume V and number N of system and the canonical ensemble is a collection of independent assemblies having the same temperature T, volume V and number of identical systems N. Thus in going from microcanonical ensemble to canonical ensemble, the condition of constant energy has been relaxed. This simplifies the calculations in thermodynamics where the exchange of energy takes place. Now the next logical step is to abandon the condition of total number of particles.

Actually in chemical process this number varies and in various

physical problems, e.g., radioactive decay, it is difficult to keep the number of particles constant. Such an ensemble in which exchange of energy as well as of particles takes place with the heat reservoir is known as grand canonical ensemble.

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Fig. 4.5 The grand canonical ensemble is a collection of essentially independent assemblies having the same temperature T, volume V and a chemical potential . We will discuss the chemical potential in a later article. In the grand canonical ensemble we then effectively have a collection of assemblies, each occupying a separate volume V, but which can exchange energy and molecules as well with each other. Fig. 4.5. represents a grand canonical ensemble. The individual assemblies systems are separated by rigid, permeable, diathermic walls. The grand canonical ensemble will thus correspond to the situation when we know both the average energy and the average number of particles in assembly, but are otherwise completely ignorant about the state of the system. Uses of the Ensemble: We have introduced the three main types of ensembles and each ensemble corresponds to a different experimental situation. If fact, the three ensembles which we have introduced are only examples of the infinite number of ensembles that can be considered. These three are particularly useful for two main reasons: Firstly, they correspond approximately to the types of thermodynamic measurements most frequently made in practices.

Secondly, in large

assemblies, it is useful to find that the values of thermodynamic quantities are not very sensitive to the method of measurement. For example, in the measurement of specific heat of a liquid of known mass and thus known particle number, at temperatures far below the boiling point, it matters very little whether the liquid is isolated at constant temperature so that the number of systems is fixed as in a macrocanonical ensemble or in equilibrium with its vapour so that the number of systems can fluctuate as in grand canonical ensemble. Example. Let us derive the relation

 C1  E   , using the idea of ensemble.  K 

  exp. 

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The probability that the phase points of the system S 1 lie in the element of volume dV1 is 1 (E1)dV1. Similarly the probability that the phase points of the system S n lie in the element of volume dVn is n (En) dVn. Thus the probability that the ensemble lies in the element of volume dV is

 (E)dV=1 ( E)dV1.2 ( E2 )dV2 ..n ( En )dVn Where dV  dV1dV2 dV3 ...dVn and

 (E)=1 ( E1 ) 2 ( E2 )...n ( En ).

…(1)

Equation (1) can be written as

log  ( E )   log n ( En )

…(2)



 1 1 1    1 2 dE1  dE2  ...  dE1  dE  ...    E1 E2 2 E2 2  1 E1

Since



depends only on E quantities like

…(3)

1  n for different values of n are equal  n En

i.e.

1 1 1 2 1   ...  cons tan t   ( say ). 1 E1 2 E2 K

…(4)

Integrating the equation, we get

 C1  E    K 

1  exp.  where C1

…(5)

and K are constants.

Using equation (1) we have

  C1  ...  Cn    E1  ...  En    K  

  exp.  or

 C1  E    K 

  exp. 

…(6)

Density of Distribution in Phase Space: The use of ensembles in statistical mechanics is guided by the following factors: (1) There is no need to maintain distinctions between individual system as we are only interested number of systems at any time which would be found in different states that correspond to different regions in the phase space. (2) The number of elements in an ensemble is so large that there is a continuous change in their number is passing from one region of phase to another.

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The condition of an ensemble at any time can be specified by the density  with the phase points are distributed over the phase space. It is called the density of distribution or probability density or distribution function. The density of distribution  if the phase points is a function of f position

coordinates

and

momentum

coordinates

i.e. q1 , q2 , q3 ...q

and

p1 , p2 ,... p f

corresponding to the 2f axes in phase space. The density of distribution is also a function of time because at any time, the density will change due to the motion of phase points in phase space. Hence

   (q1 , q2 , q3 ...q f ; p1 , p2 , p3... p f ; t )

…(1)

   (q, p, t )

…(2)

or simply as

The physical significance of the density of distribution is that it denotes the number of system N which are found at any given time in a given small volume of the phase space. The so called hyper-volume  of the phase points is

  q1 , q2 ,...q f ; p1 , p2 ,...p f .

…(3)

The number of systems N lying in the specified region can be obtained by multiplying the density of distribution and the hyper-volume in the phase space i.e.,

N   (q, p, t )q1 ,......q f .p1 ,.......p f , or simply,

…(4)

N  q1 ,......q f .p1 ,.......p f . f

In brief

N   qi .pi

…(5)

i 1

It can be seen that the density of distribution will be analogous to the probability that a number of systems or ensemble is found in a given state. By integrating over the whole of the phase space, we can write

N    dq1 ,......dq f .dp1 ,.......dp f .

…(6)

Equation (6) gives

 ( q, p, t ) N



 ( q, p, t )

 ...  (q, p, t )dq ,......dq 1

…(7) f

as the probability per unit extension in the phase space that the phase point for a system chosen at random from the ensemble would be found at time t to have the specified values of q’s and p’s. It is sometimes convenient to regard  as normalized to unity inaccordance with the equation

1   ...  (q, p, t )dq1 ,......dq f .

…(8)

The quantity  itself then gives directly the probability per unit volume of finding the phase point for system picked at random from the ensemble in different regions of the phase space.

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General Discussion of Mean Values (Ensemble Averages) The average value of variable u in an ensemble is defined by multiplying each possible value ur by the number nr of the system in the ensemble which exhibits this value, adding the resultant product for all possible values of the variable u and then dividing this sum by the total number of systems in the ensemble. Let us be the variable which can assume any of the M discrete values

u1 , u2 ,.....uM . with respective probabilities

(u1 ), (u2 ),.....(uM ). Then mean (or average) value of u is denoted by u and is defined by

u

 (u1 )u1   (u2 )u2  .....   (uM )uM  (u1 )   (u2 )  ..... (uM )

…(1)

or in shorter notation by M

u

  (u )u i

i 1 M

  (u ) i

i 1

More generally, if f(u) is any function of u, then the mean value f(u) is defined by M

  (u ) f  u 

f (u ) 

i 1

…(3)

  (u ) i 1

This expression can be simplified. Since (ui) is defined as a probability, the quantity M

 (u1 )   (u2 )  ...   (uM )    (ui )

…(4)

i 1

represents the probability that u assumes any one of its possible values and this must be unity, i.e. M

  (u )  1 i 1

…(5)

This is so called ‘normalization condition’ satisfied by every probability. Hence equation (3) becomes M

f (u )    (ui ) f  ui 

…(6)

i 1

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If an ensemble consists of n systems, the variable u can be expressed as a function of all position and momentum coordinates of the systems of the ensemble.

If the probability

distribution function is continuous in all position and momentum coordinates then equation (2) may be written as

u

 u(q, p) (q, p)d    ( q, p ) d 

d   dq1 ,......dqM , dp1 .......dpM .

Where

But according to normalization condition

 (q, p)d   1 u   u(q, p) (q, p)d .



This expression gives the ensemble average of the quantity u. If f(u) and g(u) are any two functions of u, then M

f (u )  g (u )    (ui )  f  ui   g  ui   i 1

i 1

   (ui ) f  ui    (ui ) g  ui  or

…(7)

f (u)  g (u)  f (u)  g (u) Furthermore, if c is any constant, then

cf (u)  cf (u). Suppose we wish to determine the average distance from the origin of a set of points 

distributed along a line. Let x(i) be the distance of the i points from the origin, x , the average distance from the origin, and N, the total number of points, then N

x

 x(i) i 1

…(9)

If the line is divided into cells; and the distribution is given in terms of the number of points in each cell, then

x

 N x(i) i

…(10)

N th

Where Ni is the number of points in the i cell, and x(i) the location of the i cell. If our distribution is given in the form of a continuous function n(x), we have

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x or

noting that













253

xn( x)dx

…(11)

n( x)dx  N , we have 

 x 

xn( x)dx

  

…(12)

n( x)dx

Macroscopic behaviour as an average over microscopic behaviour: If one observes a macroscopic body for a long time, then the thermodynamic properties (macroscopic) remains practically constant. However, since it never stays constant from the microscopic point of view, one can never say precisely in which microscopic state the system is found. One can only define the probability for the set of all possible microscopic states of the system. Consider a physical quantity x for the system under study, x is a dynamical quantity from microscopic point of view and is a function of microscopic states. The microscopic value of x is represented by x(q , p) = x(P) where P is the phase point.

The observed value xobs , in

macroscopic sense should be the average of the microscopic x, i.e.

xobs  x

…(1)

Suppose M is the set of all microscopic stats which can be realized by the system under a given macroscopic condition. The probability d that these microscopic states are realized is

 of the phase space is

defined that one of the microscopic states in the volume element realized

d 

  ( P)d , Where d   dq ,......dq .dp ,.......dp 1

…(2)



Where

 belongs to M and  is probability density or the distribution function. When the

integration extends over the whole phase space we have

 d   1.

…(3)

When the distribution functions are given, the average value (1) can be obtained by multiplying all the possible values of x(P) by their respective probabilities and integrating over all states i.e.,

x   x ( P )  ( P )d 

…(4)

Density of Phase Points in a Classical Ensemble:

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According to the principle of conservation of density in phase space

(d  / dt )  0, i.e.,

the rate of change of density of phase point in phase space is zero. This shows that density is constant in time. Therefore the density must be functions of constants of motion of the system. Thus

   ( E, N ),

…(1)

because for an ensemble, total energy E and number of particles N are constants of motion Let the ensemble consists of n systems S1 , S2 ,...Sn containing N1 , N 2 ,...N n particles respectively. The probability that the phase points for the ensemble lie in the element of volume

d is

given by

 ( E, N )d   1 ( E1, N1 )d 1, 2 ( E2 , N2 )d 2 ,...n ( En , Nn )d n

…(2)

d   d 1d 2 ...d n

Where

 ( E, N )  1 ( E1 , N1 ) 2 ( E2 , N2 )...n ( En , Nn )

and

Expression (2) is based on the fact that the probability that the phase point for the system S j will lie in the element of volume

d  j   j ( E j .N j ) d  j . The general solution of eq. (2) is of the form

log  ( E, N )  a  bE  cN

…(3)

Where a, b and c are some constants independent of E and N. Here we consider the following three cases: (i)

When E and N (both) are constants. Eq. (3) takes the form

log   cons tan t

i.e.,

  cons tan t

…(4)

As B is constant in this case, the phase points move in hyper-volume in such a way that the density of phase point in this volume remains constant.

This is the case of microcanonical

ensemble. (ii)

When N is constant. In this case, eq. (3) can be written as

log   bE  cons tan t

i.e.,

  e( C  E ) / C 1

…(5)

Where C1 and C2 are constants of ensemble. In eq. (5), E is chosen negative because  decreases as E increases. Here the constant C1 is equivalent to Gibb’s free energy while C2 is equivalent to kT where k is Boltzmann’s constant and T is absolute temperature. This is the case of canonical ensemble. (iii)

When E and N (both) are variable. In this case

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  e ( C C N  E ) / C 1

…(6)

C1 , C2 and C3 are constants. C1 is equivalent to (-pV), C2 is equivalent to kT and C3 is equivalent to chemical potential. This is case of grand canonical ensemble.

4.6 Partition Function and their Properties: Partition Function:

The calculation of partition function for the system at a specified

temperature T (i.e. when the system is in contact with a heat reservoir at this temperature – a case of canonical ensemble) and also for the system which is isolated and has fixed energy (because even for the such a system such a system the mean values of the macroscopic parameters of the system are related to its temperature T as if it were in thermal contact with a heat reservoir at this temperature-a canonical ensemble case dealt as an approximation to microcanonical ensemble)results in the evaluation of a host of other physical quantities, e.g., mean gas pressure, total mean energy of the gas, heat capacities of the gas. Therefore, central idea is to evaluate the partition function, which we have done in art. 3.1-2, eq. (4). When the assembly contains n distinguishable particles, then classical partition function is

Z

1 e E ( q , p ) / kT d  3n  h

The summation has been changed into integration as the assembly is having a continuous energy distribution. On the other hand, if the assembly consists of n indistinguishable particles, then n! permutations between the n particles of the assembly becomes meaningless because they will refer to the same state of assembly. Therefore classical partition function is further divided by n!. This partition function viz.

Z

1 e E ( q , p ) / kT d  3n  n !h

is called semi-classical partition function. In art. (5.11-2), we shall show that quantum partition function for canonical ensemble will come out to be

Z   e  Er , r

where r signifies the state of the system and



1 . kT

Effect of shifting the zero level of energy on the partition function, Mean Energy, Specific heat and Entropy:

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On Partition Function: changes by an amount

256

Suppose the standard state (or reference level of energy)

0 ,

the energy of each state r becomes Er  Er   0 .

The

partition function becomes

Z    e  ( Er  0 ) r

 e  0  e  Er r

 e 0 Z log Z   log Z   0 ,

which shows that the partition function is changed. (iv) On Entropy: To see the effect on entropy, we first arrive at the expression of S in terms of Z. We have shown that (also derived in chapter 2) Helmholtz energy F is related to Z by the expression

F  kT log Z

  log Z

(where   kT ).

Also entropy (statistical) is related to F by the expression

 F 

      V 

  log Z  . 

Now thermodynamical energy is related to



by the expression

S  k k

  log Z  

    k log Z   . (log Z )    

1 Z    k log Z   . Z   

 1 1 Z   k log Z   .  Z  2  1/    

 1 Z   K log Z   , Z   

1 sin ce   .



As we know that

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257

1 Z Z  

S  k (log Z   E ).

In the present case when the energy is shifted by

0,

we find that the new entropy is

S   k (log Z    E )  k (log Z   0   E   0 )

 k (log Z   E ), which means the entropy remains unchanged. Boltzmann’s Entropy Relation: In thermodynamical theory, the entropy is defined as the thermal property of the substance which remains constant during as adiabatic change. It is a function of the state. The entropy does not depend on a particular reversible process by which the substance is brought from one state to another. According to the principle of increase of entropy, whenever any physical or chemical process takes place, the entropy, whenever any physical or chemical process takes place, the entropy of the universe increases; while according to the principle of degradation of energy, whenever an irreversible process takes place, a certain quantity of energy of the universe is converted from a form in which it was available for work into a form in which it is unavailable for work. Thus the ‘available energy’ of the universe is tending towards zero, and after an infinite period of time, this will tend to ‘heat death’. Now the question is whether the heat death will be the ultimate fate of the universe. According to G.N. Lewis the increase of entropy corresponds merely to a loss of information with regard to a state of a system and is thus purely subjective concept. The second law of thermodynamics, in terms of entropy, states that for any thermodynamical process occurring in a closed isolated system, dS0, where the quality sign holds if the process is reversible and the inequality sign holds if the process is irreversible. In other words the entropy increases during spontaneous process taking place in a closed, isolated system until it reaches a maximum value at the final equilibrium state. This law also gives the conversion of heat energy into mechanical energy and vice versa. The heat energy is contained in the disorderly motion of the molecules of a gas and mechanical energy means an orderly motion of gas molecules. It is the law of nature that disorder is orderly state to transform to disorderly state but it is highly improbable that a highly disordered motion of molecules of a single body by itself be converted into an orderly motion. Thus the whole of the mechanical energy be converted into heat, but the whole of the heat cannot be converted into work.

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We have noted above that entropy is a measure of degree of randomness in the assembly.

The entropy of a physical assembly in a definite state depends solely on the

probability of that state. A relation between entropy which is a thermodynamical quantity and probability, a statistical quantity, was established by Boltzmann which is known as Boltzmann entropy relation. Let S be the entropy,  is the maximum probability of the physical system assembly in a definite state, then we can write …(1)

S = f()

Consider two systems A and B having entropy values S1 and S2.

The entropy of

combined system will be …(2)

S=S1+S2 [because entropy is additive]

and the probability of a composite system which consists of two entirely independent systems is equal to the product of the probabilities of these two systems i.e.,  = 1 2 [probability is multiplicative]

…(3)

where 1 and 2 are the probabilities of different systems i.e., the probabilities of the system A is 1 and that of B is 2. Again S1= f(1) and

…(4)

S2= f(2) Substituting equations (I), (3) and (4) in equation (2), we get

f (12 )  f (1 )  f (2 )

…(5)

This functional equation can now be used to determine the form of the function f in the following manner. Differentiating equation (5) with respect to

1 , keeping 2

2 f (12 )  f (1 ) where

constant, we get …(6)

f  stands for differentiation.

2 , keeping 1

Differentiating equation (5) with respect to

1 f (12 )  f (2 ).

constant, we have …(7)

Dividing equation (6) by equation (7)

2 f (1 )  1 f (2 )

f (1 )

2



f (2 )

1

In general

1 f (1 )  2 f (2 )  2 f  3   ...  k (say)

…(8)

where k is some constant. From equation (8), we have

f (1 ) 

1

; f (1 ) 

2

; f (3 ) 

3

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Integrating, we have

f (1 )  k log 1  C, f (2 )  k log 2  C, or in general

S  f ()  k log   C,

…(9)

where C is the integration constant and can be determined by initial conditions. In case of a

 1

perfect order there is only one manner in which the perfect order can be obtained i.e.,

and

S=0 and therefore C=0. Now equation (9) becomes

S  f ()  k log .

…(10)

Partition Function and its Correlation with Thermodynamic Quantities: (a) Partition Function: Let us consider an assembly of ideal gas molecules obeying classical statistics e.g. Maxwell th

Boltzmann distribution law. Following this distribution law, let nr molecules occupy r state with energy between Er and Er +d Er , and degeneracy, g r . Then

nr  gr e .e Er / kT

 gr Ae  E r

…(1)

Where

A  e and  

1 kT

So that total number of molecules in the assembly

N   nr r

 A gr e  E r or

N   g r e  E r  Z A r

…(2)

Where Z is called the Boltzmann partition function or simply the partition function. The term

g e 

 Er

represents the sum of all the g r e

 E r

terms for every energy state of the given molecule.

Consequently the quantity, Z, indicates that how the gas molecules of an assembly are distributed or partitioned among the various energy levels and is called a partition function. Taking the weight, g r , of an individual level as unity, we can write the equation (2) as

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Z   e  E r

…(3)

(b) Correlation with Thermodynamic Quantities. (i)

With Entropy, S: Entropy is related to the weight,

 , of most probable configuration by

S  k log max ,

…(4)

Where for a classical system having total number of molecules as

N (  nr ), r

We have

g r nr   N ! r nr !

log   log N !  (nr log g r  log nr !). r

Using Stirling approximation

log n!  n log n  n, We get

log   N log N  N   (nr log g r  nr log nr  nr ). r

According to Maxwell-Boltzmann distribution law, for a most probable configuration, we have

nr  gr e e  E r

 gr Ae  E r so that

log max  N log N  N  [nr log gr  nr log

 g Ae

 E r

n ] r

 N log N  N   nr log g r   nr log g r r

 nr log A+ nr  Er   nr . r

Taking

n

 N total number of particles, and

n E r

 E total energy of the gas

molecules of the assembly, we get

log max  N log N  N log A   E

 N log

N  E A

 N log Z   E

(Using eq.2)

…(5)

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Putting in eq. (5), we get that entropy is

S  k log max

 kN log Z  k  E  kN log Z  k .  kN log Z 

1 .E kT

E . T

…(6)

For an ideal gas

E

3 NkT 2

so that

S  kN log Z 

3 Nk . 2

which gives the entropy of the assembly of ideal gas molecules. (ii)

With Helmholtz Free Energy, F:

Helmholtz free energy F is given by

F  E  TS E   E  T  kN log Z   T 

(Using eq. 6)

 E  NkT log Z  E   NkT log Z . (iii)

…(7)

With total Energy, E:

Average energy of a system of N particles (assembly) is given by

nr Er E  r E  N  nr r

 g Ae  .E   g Ae   Er

 Er

(Using eq.1)

g E e 

 Er

 Since

(Using eq.2)

…(8)

Z   g r e  E r r

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  g r e E r / kT r

we have for isothermal isochoric transformation

Er  Z   E r / kT     gr  2  e  T kT  V r



1 kT 2



g r e E r / kT Er

 Z   E r / kT kT 2     g r Er e  T V r

 ZE or

kT 2  Z  E   Z  T V

NE 

(Using eq. 8)

NkT 2  Z    Z  T V

or total energy, E is given by

E  NE 

NkT 2  Z    Z  T V

    NkT 2  (log Z )   T V (iv)

With Enthalpy, H: Enthalpy is given by

H  E  pV  E  RT

(for ideal gas

PV  RT )

    NkT 2  (log Z )   RT  T V (v)

…(10)

With Gibb’s Potential, G: Gibb’s potential is given by

G  H  TS E      NkT 2  (log Z )   RT  T  kN log Z   T  T V  (Using eqs. 5 and 10)

    NkT 2  (log Z )   RT  NkT log Z  E  T V

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    NkT 2  (log Z )   RT  NkT log Z  T V     NkT 2  (log Z )   T V

(Using eq. 9)

 RT  NkT log Z (vi)

…(11)

With Pressure, P, of the Gas: Pressure is given by

 F  P     V T     NkT  (log Z )   V T

…(12)

Using eq. (7). (vii)

With Specific heat at Constant Volume, CV : It is given by

 E  CV     T V



 T

  2   NkT T (log Z )V 

   2  Nk  2T (log Z )  T 2 2 (log Z )  T  T V Example 1: Show that if the partition function is given by Z, then mean energy,

E

…(13)

E , is given by

1  (log Z ), Where   kT 

Refer to the case (iii) of relation of Z with total energy E. We have shown there in eq. (8) that

Z E   g r Er e  E r

…(1)

Further

Z   g r e  E r r

Z   g r Er e  E r  r

 Z E

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264

1 Z Z 



(Using eq. (1))

 (log Z ) 

Which is desired.

4.7 Gibbs Paradox: For equation (4a), we have already stated that it does not satisfy the additive property of the entropy and which can be made more obvious if we take two systems denoted by indices a and b at the same temperature T a = Tb = T. If the particles in the two systems are different, the entropy of the joint system ab will be (using equation 4a).

Equation (4a)

  4 m 3/ 2  E 3/ 2  V      3   n   3n    n log  .   2 h3     Sab = Sa + Sb

 na k[log Va 32 log T  C ] nb K[log Vb 32 log T  C ], Where C is a constant term including constant factors e.g., k, m of equation (4a). Now if the particles in the two systems are the same and if, for convenience, we take

Va  Vb  V ; na  nb  n, we have to consider 3n particles in a volume 2V. We get

Sab  2nk[log 2V 32 log T  C ]  2nk[log V 32 log T  C]  2nk log 2  Sa + Sb  2nk log 2, which shows that by joining two moles of two different gases by removing a partition between them, the entropy of the joint system increases by an additional factor 2nk log e 2 which can not be accounted. This is called Gibbs paradox. The explanation is quite simple.

After the removal of partition, a molecule of each

system, being indistinguishable, can be found anywhere within the volume 2V instead of V and hence external parameter becomes 2V and the possible states of energy are to be calculated with the total volume 2V instead of simply V. This explains the occurrence of the additional factor nk log 2. But if we take two systems as the same e.g., two molecules of the same gas, in that case the molecules will be indistinguishable (as in quantum mechanics). Indeed, if we treat the

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265

gas by quantum mechanics, the molecules are to be regarded as completely indistinguishable, i.e. in that case one can not 1 mole

1 mole

V Fig. 2.

Mixing of two gases. say or observe and label the individual particle. So applying this idea of indistinguishability to classical approximation, we can correct the equation (4a). We note that if there are n molecules in a system, then they can not be distinguished in n! ways and hence to arrive at the correct result, we must divide equation (4a) by n! which leads to equation (5) which embodies the additive property of the entropy and then

 4 m 3/ 2  V  E 3/ 2  5n Equation (5)  n log       2   3h   n  n   2 Sab =2nk[log 2V  log 2 32 log T  C]  Sa + Sb . Gibbs paradox is thus resolved by a concept of quantum mechanics. While dealing with grand canonical ensemble we shall observe that entropy, deduced by the application of this ensemble, bears the additive property.

Quantum Statistical Mechanics 4.8 Equation of Motion and Liouville’s Theorem The Poisson bracket a, b is defined by

 a b a b   .  q  p  p  q i i   i i

a, b    i

…(1)

The canonical equations of motion and total time derivative dD/dt can be written as

qi , H  

H .  qi , pi

 pi , H   

H .  pi , qi

…(2)

dD D D dqi D dpi    dt t i qi dt i pi dt



 D H D H  D     t pi qi  i  qi pi

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266

D  D, H  . t

…(3)

The last result is the equation of motion in terms of Poisson bracket. The Liouville theorem D/dt = 0, can now be expressed as



D  D, H  . t

…(4)

The implicit dependence of D(q,p,t) on time is thus given by the Poisson bracket. We shall deal

D =0. Therefore, the t

only with those ensembles for which D does not depend on time explicitly, stationary ensemble described by D(q,p) or (q,p) is the same for all times.

  0, t

  q, p  , H   0.

…(5)

4.9 Postulatory Foundations of Quantum Mechanics: Postulate I: In classical mechanics, the sate of a system is specified by its generalized coordinates q and generalized momenta p at a particular time t. In quantum mechanics, the complete knowledge of a dynamical state (values of p and q at a time t) cannot be determined. The state of the system is completely determined by the state function of wavefunction

 * denotes

The wavefunction, in general, is complex, finite single valued and continuous. the probability of finding the system in a volume dq at time t, where of

(q, t ).

 * is the complex conjugate

 . All the possible information about the system can be derived from this wavefunction. In

general the state of a system cannot be measured directly.

We measure certain physical

quantities such as energies, momenta etc which are called observables. Postulate II: Any observable or dynamical variable can be represented by an operator 

say



A ( p, q, t ). An operator A is a mathematical operation which may be applied to a function

f ( x) which changes the function to another function g ( x) . This can be represented as 

A f ( x)  g ( x).

…(1) 



Moreover, the operators must be so chose that for any pair of operators A and B

h           A B  B A   A, B  ,   2 i    

  

 



…(2)

 





where  A B  B A  is known as commutator of the two operators and A , B is an operator     associated with the classical poisson bracket

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 a b a b  [ a , b]     . pi qi  i  qi pi If for arbitary functions

1

2

and

…(3) 

and arbitary constants c1 and c2 , the operator A

satisfies the condition, 



A (c1 1  c2 2 )  c1 A 1  c2 A 2 ,

…(4)

the operator is known as linear operator. 

For any two well-behaved function  1 and  2 , an operator A satisfying the equation



    2 *  A 1 dq    A 2  * 1dq













…(5)



is said to be Hermitian. Let

n

be a well-behaved function of the state of the system and this operates on the



operator A such that it satisfies the equation 

A n  n n

n

then we say that the numbers

…(6) 

are eigen values of the operator A and the operand



the eigen-functions of A ; and the eigen-value belongs to each other. A set of the functions

n

n

and the eigen-functions

n

are 

of operator A

is said to be complete with respect to a class of

functions if any  belonging to the class can be expressed as a linear combination of  n .

   cn n

…(7)

where cn are constants. Postulate III: For an assembly of N identical system, with time t, and if a measurement of value a of

N   , all in the state  (q, t ) at the

a( p, q) is made at t on all these systems, then the average

a( p, q) is found from the following equation 

  *(q, t ) A( p, q) (q, t )dq a(t )    *(q, t ) (q, t )dq where

…(8)

 * stands for the complex conjugate of  . Postulate IV: The time variation of a quantum mechanical state is determined by the

equation

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268

h (q, t ) 2 t

…(9)

(q, t )   (q)e(2 i / h) Ent

…(10)



where H is the Hamiltonian operator. Equation (9) is called as Schroedinger time dependent equation. Some useful theorems based on these postulates are given below:

Theorem I: The eigen functions which, after being operated by the same Hermitian operator give two eigen-values, will be orthogonal functions. 

Let  1 and  2 be the two wavefunctions and A is the Hermitian operator, then 



A 1  a1 1 and A 2  a2 2 ,

which yield 

 2  A 1dq  a1  2  1dq,

…(11)



 ( 1  A 2 )* dq  a2 *  1 2 * dq

…(12)

Now

 (



 A 2 )* dq   1 ( A 2 )* dq 

  ( A 2 )* 1dq 

  ( 2 *) A 1dq,

…(13)

and therefore

(a1  a2 *)  2 * 1 dq  0.

…(14)

We also know that eigenvalues are always real, hence

(a1  a2 )  2 * 1 dq  0.

…(15)

As (a1  a2 )  0, hence



* 1 dq  0.

Hence the two wavefunctions are orthogonal. Theorem II:-

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1

and

2

269

are linearly independent eigen functions of an operator belonging to the

eigen value a (same for both), then a linear combination of these functions will also be eigan 

function of the operator A and it will have an eigen value a. In this case 



A 1  a 1 and A 2  a 2 .

The linear combination of these wave-functions can be written as

  c1 1  c2 2 , 

then



A  A  c1 1  c2 2   c1 A 1  c2 A 2  a  c1 1  c2 2  

Thus  is an eigen function of operator A which has a eigen value a. Theorem III:In the expansion

   cn n , the quantity cn * cn represents the probability that a n



measurement of R will yield the eigen value

n

corresponding to the eigenfunction  n .

We observe that

  * d    c * 1

* c2 * 2 * ... c1 1  c2 2  ... d

 c1 * c1  c2 * c2  ...   cn * cn   cn we can interpret

…(1)

cn ds the probability that the system described by  is in the nth state. Now 



 R    * R d 

   cn * n * R  cm m d n

   cn * n * mcm m d   cn n 2

…(2)

This represents the expectation value of the physical quantity R when the system is described by

. Let

 ( n)

denotes the probability that an individual measurement yields the value

n .

Now by definition, we have

 R    (n)n

…(3)

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Comparing eqs. (2) and (3), we get

 (n)  cn Thus

 ( n)

…(4) 

is the probability that a measurement R will yield the value

n .

Indistinguishably of particles and its consequences The consequences of indistinguishability of particles in quantum statistics may be explained in the following manner: Let there be two points of collection and only two cells to be occupied. In classical statistics, each particle has a recognizable individuality, hence the particles are distinguishable and each particle is as likely to be in one cell as in the other. This means that in classical statistics, each or both of the particles can occupy any one of the two cells. Thus, in this case, we have four possible arrangements, each of which is counted for assigning a probability to the distribution. This is as shown in Fig. (a)

Fig. (a,b,c) In Bose-Einstein statistics, the particles loose their individuality, hence they are indistinguishable. So we must concentrate our attention on the cell rather then the individual particles. Each cell may have number of particles, so that in this case, we have only three possible arrangements, as shown in Fig. (b) In Fermi-Dirac statistics, since the particles are indistinguishable, again we have to concentrate our attention on the cell rather then on individual particles. But in this case, the particles obey Pauli’s exclusion principle according to which it is not possible to have more than one representative point in any one cell. Obviously, each cell will contain at the most one particle. Therefore, in this case, only one arrangement is possible, as shown in Fig. (c) 4.10 Bose – Einstein Distribution Law

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Bose used Planck’s hypothesis, according to which radiation in a temperature encloser are composed of light quantas known as photons each of energy, E= hv. These photons in the encloser are indistinguishable. Basic postulates: In Bose-Einstein statistics, the conditions are: 1.

The particles of the system are identical and indistinguishable

Any number of particles can occupy a single cell in the phase space.

The size of the cell cannot be less than -34

10 4.

h3 , where h is a Planck’s constant (h=6.63 X

Joule. Sec)

The number of phase space cells is comparable with the number of particles, i.e. the occupation index f ( Ei ) is

ni 1. gi

Bose-Einstein statistics is applicable to particles with integral spin angular momentum in units of

h . All particles which obey B.E. statistics are known as Bosons. 2

Consider a system consisting of n independent and indistinguishable particles. The particles have definite energies (or momenta) and occupy definite positions. Therefore, they can be represented by phase points in the phase space. In order to determine the energy (or momenta) distribution of these particles in the most probable state known as equilibrium state, we divide the available volume in the phase space into large number (say k) of compartments (quantum groups or energy levels), each compartment representing a small interval of energy (or momentum) as shown in Fig. (a) Further we divide each compartment into elementary cells, each 3

of size h where h is Planck’s constants Fig. (b). We suppose that the size of the compartment is very large as compared to the size of the cell so that each compartment contains a very large number of elementary cells.

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Fig. (a)

Fig. (b)

Let there be n1 , n2 ....n1...nk particles having mean energy values E1 , E2 ,......Ei ,...Ek respectively in compartments numbered as 1,2… i… k containing

g1 , g2 ...gi ...gk cells

respectively in them. The total number of particles in the system is

n  n1  n2  ...  n1  ...  nk Now consider ith compartment. It has ni indistinguishable particles distributed among its g i cells. Suppose, that ni particles are arranged in a row and distributed among, g i quantum states with

( gi  1) partitions in between. The total number of possible arrangements or particles and partitions is equal to the total number of permutations of (ni  gi  1) objects in a row. Therefore, the total possible ways of arranging ni particles with g i -1 partitions

 (ni  gi  1)! As the particles are indistinguishable, the rearrangement of particles among themselves will not give rise to any new distribution. The number of such meaningless permutations is ni !.. Hence above result would be divided by ni !. Moreover, this also includes permutations of

( gi  1) partitions among themselves. These permutations also do not produce different states and hence are meaningless. The above result is further divided by ( ( gi  1)!. Hence, the actual number of ways in which ni particles are to be distributed in g i cells in the ith compartment is given by

Wi 

(ni  gi  1)! ni !( gi  1)!

…(1)

Similar expressions will be obtained for other compartments. Therefore, the total number of different arrangements for all the n particles of the system gives thermodynamic probability.

W( n1 ,n2 ,.....ni .....nk ) 

(ni  gi  1)! (n2  g2  1)! (n  gi  1)! (n  g k  1)! X X ... X i X ... k ni !( gi  1)!)! n2 !( g 2  1)!)! ni !( gi  1)!)! nk !( g k 1)!)! (ni  gi  1)! i 1 n !( g  1)!)! i i k



…(2)

Where  denotes multiplication of terms stated above for various values of I from i=1 ot k Most Probable Microstate The most probable microstate corresponds to the state of maximum thermodynamic probability. In equation (2), ni and gi both are very large numbers. Hence we may neglect 1 in the above expression.

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W( n1 ,n2 ,.....n1 ,....nk  

(ni  gi )! ni ! gi !

…(3)

Taking natural logarithm on both sides, we have k

In W  [ In (ni  gi )! In ni ! In gi !] i 1

As ni and g i are very large numbers, we van use Sterling approximation

In n! (n In n)  n Applying sterling approximation, we get k

In W  [ (ni  gi ) ln (ni  gi )  (ni  gi )  ni ln ni  ni  gi ln gi  gi ] i 1

 [ (ni  gi ) In (ni  gi )  ni In ni  gi In gi ]

…(4)

i 1

Here g i is not subject to variation whereas ni varies continuously To get the state of maximum thermodynamic probability, we differentiate equation (4) and equate it to zero; i.e.

 ( InW )  0 k   1 1    ni In  ni  gi    ni  gi   ni   ni In ni  ni  ni  ni  ni  gi  i 1  

 ( In W )

 k

   ni In  ni  gi    ni In ni i 1

( gi is a mere number  gi  0) k



 ( In W )    In

(or)

i 1



(or)

i 1



 ni  gi    n   k ni

 

 i 1

 ni   In   ni  0 n  g   i i  

  ni  In   ni  0   ni  gi  

…(5)

In addition, our system must satisfy two subsidiary or auxillary conditions: (i)



Conservation of total number of particles, i.e. n=a constant

n   ni = constant i

i.e.

 n   n i 0

…(6)

i 1

(ii) Conservation of total energy of the system i.e. E=a constant.

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274

E   ni Ei =constant



 E   Ei  n i  0

i.e.

…(7)

Now we shall apply the Landrangin method o undetermined multipliers. For the let us multiply eqn. (6) by

 and eqn. (7) by  , and add the resulting expression to eq. (5) so that we get k   ni      In       Ei   ni  0 i 1   ni  gi  

The Variations

…(8 )

 ni are independent of each other. Hence we get  ni In   ni  gi

      Ei  0 

(or)

 ni In   ni  gi

      Ei 

(or)

ni  e   Ei  e (   Ei ) ni  gi

(or)

ni  gi  e   Ei ni

(or)

1

(or)

ni 

gi  e   Ei  1 ni

   Ei

1

…(9)

This equation represents the most probable distribution of the particles among various energy levels for a system obeying Bose-Einstein statistics and is therefore, known as Bose-Einstein’s Distribution law for as assembly of bosons.

Substituting of

ni 



1 , the equation (9) becomes kT

gi g   Ei /ikT  Ei e .e  1 e .e 1 

…(10)

B.E. energy distribution function. The energy distribution function f ( Ei ) is the average number of particles per quantum state in the energy level Ei . It is given by

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275

ni gi

Substituting the value of ni / gi from equation (10), we get

f ( Ei ) 

1  Ei / kT

…(11)

1

e e

For continuous distribution of energy, the distribution function is written as

f (E) 

1  Ei / kT

e e

…(12)

1

B-E energy distribution law for continuous variation of energy. When the energy levels of the system are very closely packed, they form a quasicontinuous spectrum. In such a case, if g (E) dE is the number of energy states between the energy range E and (E+dE), then

gi  g ( E )dE If n(E) dE is the number of particles whose energy lies between E and (E+dE), then ni =n (E) dE. Substituting in eqn. (10), we get

n( E )dE 

g ( E ) dE e e 1  Ei / kT

…(13)

The quantity g(E) denotes the density of states. For the system consisting of free particles with no spin g(E)dE is given by

 2m  g ( E )dE  2 V  2  h 

3/ 2

…(14)

E 2 dE

Substituting this value in eqn. (11), we get

 2m  n( E )dE  2 V  2  h 

3/ 2

1 2

E dE e .e E / kT  1 

…(15)

This is Bose-Einstein energy distribution law for continuous distribution of energy among free particles with no spin. 4.11 Fermi – Dirac Distrubution Law Consider a system consisting of n independent and indistinguishable particles. The particles have definite energies (or momenta) and occupy definite positions. Therefore, they can be represented by phase points in the phase space. In order to determine the energy (or momenta) distribution of these particles in the most probable state known as equilibrium state; we divide the available volume in the phase space into large numbers (say k) of compartments (quantum groups or energy levels), each compartment representing a small interval of energy (or

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276 3

momentum). Further each compartment is divided into elementary cells; each of size h , where h is a Planck’s constant. Further, we suppose that the size of the compartment is very large as compared to the size of the cell so that each compartment contains a very large number of elementary cells. Let the compartments be marked 1,2,………i,……..k and their mean energy values be represented be E1, E2,………Ei…………..Ek containing g1,gy….gi…gk cells respectively in them. The total number of particles in the system is n=n1+n2+………+ni…………….+nk Our aim is to find the Fermi-Dirac distribution law i.e., the distribution of n i Fermions in gi cells of the ith compartment. For this, the basic postulates are: Basic postulates (i)

The particles of indistinguishable so that there is no distinction between the various ways in which ni particle are chosen

(ii)

The particles obey Pauli’s exclusion principle according to which there can be either no particle or only one particle in a given cell. Therefore, the number of cells must be much greater then the number of i.e., gi>>ni.

Out of ni number of particles in the ith compartment with mean values E i, the first particle can be placed in any one of the available gi states, i.e., this particle can be assigned to any of the gi sets of quantum numbers. Thus, the first particles can be distributed in g i different ways, in accordance with paili’s exclusion principle and the remaining (g i-ni) will remain vacant. The second particle can be arranged in (gi-1) different ways and the process continues. Thus, the total number of different ways of arranging ni particles among available gi states with energy level Ei is given by

 gi ( gi  1) ( gi  2) ( gi  3)..............[( gi  (ni  1)] 

gi !  gi  ni !

…(16)

Further, since the particles are indistinguishable , it would not be possible to detect any difference when ni particles are reshuffled into different ways occupied by them in the energy level Ei Thus, out of these ni permutations of the indistinguishable particles among themselves will be meaningless. Therefore, the total number of different and distinguishable ways is



gi ! ni ! gi  ni  !

…(17)

Therefore, the thermodynamic probability for the microstate (n1 , n2 ,.....ni ,...nk ) of the system is given by

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277

gi ! gk ! g1 ! g2 ! X X ..... X X .... X n1 ! g1  n1 ! n2 ! g 2  n2 ! ni ! gi  ni ! nk ! g k  nk ! gi ! n1 ! gi  ni !

 i 1

….(18)

Where  denotes multiplication of terms for various values of

i form1to k

Most Probable Microstate Taking natural logarithm on both sides of eqn. (18) k

In W   i 1

 In gi ! In ni ! In ( gi  ni )!

As ni and g i are very large numbers, we can use Sterling approximation In n!=(n In n)-n Applying Sterling’s approximation, we have k

 gi In gi  gi  ni In ni  ni  ( gi  ni ) In ( gi  ni )  ( gi  ni )

In W   i 1

 i 1

 gi In gi  ni In ni  ( gi  ni ) In ( gi  ni )

Here gi is not subject to variation whereas ni varies continuously. Differentiating both sides, we have k





 ( In W )    ni  ni  In ni  ni  ( gi  ni )  ni  In ( gi  ni )  ni  ni ( gi  ni ) i 1   k

 i 1

 In ( gi  ni )  In ni  ni

To get the state of maximum thermodynamic probability

 ( InW )  0 k

  In ( g



i 1

(or)

 i 1

 ni )  In ni  ni  0

 gi  ni   In   ni  0 ni  

(or)

 ni    In   ni  0 gi  ni  i 1 

(or)



i 1

 ni   In   ni  0  gi  ni 

…(19)

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In addition, our system must satisfy the two auxillary considtions: (i)

Conservation of total number of Particles, i.e., n=a constant

n   ni =constant



i 1

 E    ni  0

i.e.,

…(20)

i 1

(ii) Conservation of total energy of the System i.e. E=a constant

E   ni Ei constant



i 1

 E   Ei  ni  0

i.e.,

…(21)

i 1

We shall now apply Langrangian method of undetermined multipliers. For this, let us multiply eqn. (12.31) by

 and eqn.

(12.32) by

 and

add the resulting expression to eqn.

(12.30)

 i 1

  ni    In       Ei   ni  0   gi  ni  

As the variations

 ni are independent of each other, we get  ni In   gi  ni

      Ei  0 

(or)

ni  e (   Ei ) gi  ni

(or)

gi  ni  e   Ei ni

(or)

gi  1  e   Ei ni

(or)

gi  e   Ei  1 ni

(or)

ni 

   Ei

1

…(22)

This equation represents the most probable distribution of the particles among various energy levels for a system obeying Fermi-Dirac statistics and is therefore known as Fermi-Dirac distribution law, for an assembly of fermions,

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

The parameter

279

1 , where k is Boltzmann’s constant, has the same role as in case kT

of M-B distribution law. Substituting in equation (22), we have

ni 

   Ei / kT

1



 Ei / kT

e e

…(23)

1

Fermi-Dirac energy distribution function Since there can be a maximum of one particle quantum state, the function f ( Ei ) is the ratio of the number of quantum states of energy Ei occupied by electrons to the total number of quantum states available in the energy level Ei. Therefore, the value of f ( Ei ) for the Fermi-Dirac distribution at a particular energy Ei is the probability that under equilibrium a quantum state of that energy is occupied by a particle. Form equation (24) f ( Ei ) is given by

f ( Ei ) 

ni gi



1  Ei / kT

e e

…(25)

1

For continuous distribution of energy at energy E, the distribution function is written as

f ( Ei ) 

1  Ei / kT

e e

…(26)

1

F-D energy distribution law for continuous variation of energy If the energy levels are very close together, then the distribution of energy of the particles may be considered continuous. For this distribution the number of particles n(E)dE whose energies lie between E and E+dE, then ni  n( E )dE and given by

n( E)dE  f ( E) g ( E)dE Where

…(27)

g ( E )dE is the number of quantum state of energy between E and E+dE.

Substituting the value of

n( E )dE 

f ( E ) in eqn. (27), we have

g ( E )dE e .e Ei / kT  1

…(28)



For particles like electrons of spin angular momentum  orientations. For a system of such particles

1 h , there are two possible spin 2

g ( E )dE is given. Substituting, we get

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 2m  n( E )dE  2 X 2V  2  h 

3/ 2

280 1

E 2 dE   Ei / kT e .e 1

…(29)

This is the Fermi-Dirac energy distribution law for continuous variation of energy among free particles with spin

1 . 2

Comparison of the two Statistics Bose-Einstein

Fermi-Dirac

1. Particles are indistinguishable

1. Particles are indistinguishable.

2. Only quantum states are taken into

2. Only

consideration.

states

are

taken

into

consideration.

3. There is no restriction on the number of

3. There is restriction on the number of

particles in a given quantum state. 4. Phase space in known V = h

particles in a given quantum state. 4. Phase space is known V = h

5. Number of distinguishable ways are

5. Number of distinguishable ways are given

given by

W  i

(ni  gi  1)! ni !( gi  1)!

W  i

6. Maximum probability distribution

 7. At

quantum

high

distribution

6. Maximum probability distribution

1 e

(   Ei )



  1

temperatures,

Bose-Einstein

approaches

gi ! ni !( gi  ni )!

7. At

Maxwell-

  1

approaches

Fermi-Dirac Maxwell-

Boltzmann distribution. 8. Applicable to electrons and antisymmetric

particles known as bosons. 9. The energy at absolute zero is taken to

e

temperatures,

distribution

Boltzmann distribution. 8. Applicable to photons and symmetrical

high

1 (   Ei )

particles known as fermions. 9. Due to Pauli’s exclusion principle, all the

be zero.

electrons cannot occupy the lowest energy level. Even at absolute zero temperature, some of the electrons are distributed at higher energy levels. Hence, at absolute zero, the energy is not zero. Therefore, at the complete degenerate state, the energy is in dependent of temperature. 10.

ni 

e

i (   Ei )

 1

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10.

ni 

281

gi e(   Ei )  1

11. The two distributions can be represented by a single equation

gi  e(   Ei )    ni Here If

  1

  1

gi  1 or ni  gi i.e.,  can be neglected. In the case of diluted systems, the ni

two statistics lead practically to similar results. It means that at high temperatures, the two statistics give practically the same results.

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UNIT V Ideal Bose Einstein Gas Equation of State for ideal Quantum Gases The nonrelativistic energy of an elementary particle (no internal degrees of freedom) is just the kinetic of its translational motion,

  p 2 / 2m.

Under normal conditions (big box) the

translational levels are closely spaced and so the discrete spectrum can be treated approximately as continuous (quasiclassical). Then gives 3/ 2  2V  g ( )d   g s  3   2m   1/ 2 d  ,  h 

g s  2s  1,

where g s is the degeneracy associated with the spin s of the particle. The of quantum states of energy between



and

  d .

The

…(1)

g ( ) is the number

g ( ) is the density of states. In fact,

it is the density of quantum states because it refers to one-particle system and not to the states of N particle system. When multiplied by the distribution function

f ( , T ,  )  exp[(   ) / kT ], f ( , T ,  ) 

1 exp[(   ) / kT ]  1

f ( , T ,  ),

(Classical),

(Quantum),

it gives the density of occupied quantum states. The total number of particles in a system is then given by

N   dN   f ( , T ,  )g ( )d  . In general, if Rn is the value of the quantity R in the quantum state n, then we can write

R   f ( n , T ,  ) Rn n

  gi f ( i , T ,  ) Ri ,

(all

i

different),

R   f ( , T ,  ), R( )d .

(2)

Thus the sum over the quantum states can be transformed under appropriate conditions (closely spaced energy levels) to an integral by the substitution

 (...) or  g (...)   (...)g ( )d . i

( states )

( levles )

(3)

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Using upper sign for FD statistics and lower sign for BE statistics,

N   dN g s

 2 V  1/ 2 d 3/ 2 (2 m ) 0 exp[(   ) / kT ]  1, h3

E    dN g s

 g  kTg s

 2 V  3/ 2 d  3/ 2 (2 m ) 0 exp[(   ) / kT ]  1, h3

(4)

(5)

 2V 3/ 2 (2 m )  1/ 2 In 1  exp[(    ) / kT ]d . (6) 3  0 h

Partial integration of the grand potential gives  2 2 V  3/ 2 d  3/ 2 g   gs (2m)    PV , 0 exp[(   ) / kT ]  1 3 h3

(7)

Where we have used (4.66),  g   PV . From (5, 7), for any statistics

PV  Putting x 

 kT

2 E. 3

(8)

, write (7) as P  gs

 4 V x3/ 2 dx 3/ 2 5/ 2 (2 m ) ( kT ) 0 exp[( x   ) / kT ]  1. 3h3

(9)

The equation of state of the gas is determined by (9) in terms of the parameter μ. For e

 / kT

 1,





x3/ 2 dx e x   / kT  1 

1

  x3/ 2 e x   / kT  1 dx 0





x3/ 2e / kT  x 1 e / kT  x  dx

   e / kT   x3/ 2e x dx e  / kT  x3/ 2e2 x dx   0  0



 g   PV   g s

3 1/ 2  / kT e 4

 3/ 2V h

1  / kT   1 25/ 2 e  ,

(2m)3/ 2 (kT )5/ 2 e  / kT 1 25/ 2 e  / kT  .

(10)

(11)

The first term in (11) gives just the classical value for g s  1. The next term is the correction to it. We can write

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g  g

 gs

284

 3/ 2V 5/ 2

(2m)3/ 2 (kT )5/ 2 e2  / kT .

(12)

 1  h3 N  2 , 3/ 2  g s V (2 mkT ) 

2  / kT

(13)

we have, to the same approximation, for the equation of state

  Nh3 PV  NkT    5/ 2 3/ 2  g s 2 (2 mkT ) V 

(14)

Thus, the departure from the MB statistics is large when for a given density the temperature is lowered. This is a measure of the degeneracy. In the FD case, the pressure increases due to the correction term, implying the appearance of an effective repulsion between the particles. In the BE case, it leads to a decrease in pressure, implying an effective attraction between the particles.

5.1 Ideal Bose Einstein Gas In Bose-Einstein distribution, we consider a system of identical (indistinguishable), independent, noninteracting particles of integral spin (bosons) that have symmetrical wave function. Energy and Pressure of the Gas: Consider a perfect Bose-Einstein gas of n bosons. Let these particles be distributed among states such that there are n1, n2, n3…, no. of particles in quantum states whose approximate constant energies are ε1,ε2,…εi… respectively. For a perfect Bose-Einstein gas, consisting of material particles, the formula for the most probable distribution is

ni 

   i

1

…(1)

Where g i is a measure of degeneracy. Since the gas is ideal, the interaction between the particles is assumed to be negligible so that the energy may be regarded as entirely translation in character. The results thus obtained will be applicable to a monatomic gas. Equation (1) can also be written as

ni 

 i

1

Where, for convenience, we have put

…(2)

D  e .

We know particles in a box of ‘normal’ size, the translational levels are closely spaced so that we can integrate over phase space instead of summing over particle state. The number of particle states

g ( p)dp lying between momentum p and p  dp is determined from

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285

4 3 p 3 g ( p)  g s 3 , h /V 4 V 2 p dp, h3

g ( p)dp  g s

giving

…(3)

where g s ( 2s  1) is the spin degeneracy factor (arising due to the spin, s, of the particle). Equation (2) can now be written as

dn( p)   gs . Since

  d



g ( p)dp De  p

/ 2m

1

4 V p 2 dp 2 h3 De p / 2 m  1

…(4)

p2 we can write number of particles lying between energy range  and 2m

from equation (4), i.e.

4 V 2m (m / 2 )1/ 2 d  dn( )  g s 3 . h De / kT  1 Where



1 has been substituted. kT

dn( )  g s .

4 mV h3

 2m 

 1/ 2 d  De / kT  1

…(5)

Let us put



x

dx 

d , kT

So that we express

dn( )  g s .

4 mV h3

 gs

 2m 

x1/ 2 dx(kT )3/ 2 De x  1

1  2 mkT      h2 

3/ 2

.V .

x1/ 2 dx . De x  1

From thermodynamical properties of diatomic molecules, we note that translational partition function is

 2 mkT  Zt    2  h 

3/ 2

.V .

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It follows then

dn 

2 g s Zt x1/ 2 dx .  De x  1

On integration total number of particles is given by

n   dn 

2 g s Zt







x1/ 2 dx . De x  1

…(6)

and energy

E    dn, 

2 g s Zt



.k .T 



x3/ 2 dx . De x  1

…(7)

We shall now evaluate the integrals of equations (6) and (7).

For a Bose Einstein

system, α should be positive otherwise the value of

ni  gi ( Dei  1)1 , would become negative at sufficiently low energies which is physically unacceptable. When α is positive D>1. The integral of equation (6) is





1

x   e x  x1/ 2 dx 1/ 2 e  x 1   dx. De x  1 0 D  D 



  x1/ 2 0



 e x  e x e2 x  2  ...  dx. 1  D  D D 

1  1/ 2  x 1  1/ 2 2 x x e dx  x e dx  ... D 0 D 2 0



1 1  1  3/ 2  3/ 2  ...  . 2D  2 D 3 D 

…(8)

and similarly, integral of equation (7) is





x3 / 2 dx 1  3 / 2  x 1    x e dx  2  x3 / 2 e2 x dx  ... x De  1 D 0 D 0



3  1 1  1  5/ 2  5/ 2 2  ...  4 D  2 D 3 D 

…(9)

Using (8), we get from equation (6),

n

g s Zt D

1 1   1  3/ 2  3/ 2 2  ...   2 D 3 D 

…(10)

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287

and from equations (9) and (7),

E

3kTg s Zt 2D

1 1   1  5/ 2  5/ 2 2  ...   2 D 3 D 

…(11)

Substituting value of g s Z t from equation (10) in equation (11), we get

E

3nkT  1 1 1 1   1  5/ 2  5/ 2 2  ... 1  3/ 2  3/ 2 2  ...  2  2 D 3 D  2 D 3 D 



3nkT  1 1 1 1   1  5/ 2  5/ 2 2  ... 1  3/ 2  3/ 2 2  ...  2  2 D 3 D  2 D 3 D 



3nkT  1 1 1 1  1  5/ 2  5/ 2 2  ...  3/ 2  3/ 2 2  ...  2  2 D 3 D 2 D 3 D 



3nkT 2

1 1   1  25/ 2 D  35/ 2 D 2  ...

1

…(12)

If we take only first term in equation (10) then

D

g s Zt , n

which when substituted in equation (12) gives

 3 1  n E  nkT 1  5/ 2   2  g s Zt 2 

2   1  n  .   ...  5/ 2     3  g s Zt  

For one gram molecule of the gas, since nk = R, we write energy expression as 2   3 1  n  1  n  E  RT 1  5/ 2   5/ 2   ...     2  g s Zt  3  g s Zt   2  



3 RT. 2

…(13)

where η has been substituted for bracketed terms.

Gas Degeneracy: The pressure of the gas can be calculated from the relation

 E  P     V  S

…(14)

which means, in order to calculate pressure, we must first set up a relation in energy E and volume V. Let us consider a particle of mass m enclosed in a container of volume abc. Then wave function ψ of the particle must satisfy the Schroedinger equation

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8 2 m    2 ( E  V *)  0, h 2

Where

…(15)

V *( x, y, z ) represents the potential energy of the particle which for a single particle in

the box should be zero. Therefore



h2   2  2  2    8 2 m  x 2 y 2 z 2

   E . 

…(16)

To solve this equation, let us write  in terms of variable,

  X ( x) Y(y) Z(z),

…(17)

so that equation (16) assumes the form



h2  1  2 X 1  2Y 1  2 Z       E, 8 2 m  X x 2 Y y 2 Z z 2 

…(18)

which means

 1 2 X  Ex  2 2 8 m X x  h 2 1  2Y   2  Ey  2 8 m Y y  2 2  h 1 Z  2  E z  8 m Z z 2  

…(19)

We consider first of equations (19). i.e.



h2 1  2 X  Ex 8 2 m X x 2

the general solution of which can be written as

X ( x)  A sin( Bx  C),

…(20)

in which the constants A, B and C are to be determined by the application of boundary conditions. We know that the probability of finding the particle at any point

 ( x, y, z )

at the point. Therefore

( x, y, z ) is equal to

X ( x) . which is a function of x-coordinate only, would

determine the probability of the particle being found somewhere along the X-axis. Since the particle is within box, its probability

X ( x) , of being found at the walls ( x  0 and x=a) is

equal to zero.

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Fig. Thus

X(x)=0 at Applying

x=0 x=a



X ( x)  0 at x  2, to equation (20), we get

0  A sin C, giving

sin C  0 or C  0. Now applying

X ( x)  0 at x  a to the same equation we find

0  A sin( Ba  C ) giving Ba  C  rx , where rx is an integer. or

B

rx , since C is zero. a

Therefore, equation (20) finally assumes the form.

X ( x)  A sin

rx a

…(21a)

Similarly, for the rest two equations of relation (19). We can find

Y ( y )  A sin Z ( z )  A sin

ry b rz . c

…(21b)

…(21c)

Substituting (21a), (21b) and (21c) in equations of relation (19), we find

rx 2 2 h 2 Ex  2 2 , 8 a m Ey 

ry 2 2 h2 8 2b 2 m

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rz 2 2 h2 Ez  2 2 , 8 c m giving

E  Ex  Ey  Ez



2 h2 2  rx 2 ry rz 2    .  8 2 m  a 2 b 2 c 2 

If container is cubic,

a  b  c  1, so that 2 2 2 h2 2  rx  ry  rz  E 2   . 8 m  l2 

Further for cubic l, volume is,

V  l3 l 2  V 2/ 3 ,

so that and taking

rx 2  ry 2  rz 2  r 2 , we have

E

h 2 2 r 2. 8 2 mV 2 / 3

…(22)

Thus we have established a relation in energy and volume. Now from equation (14), we obtain the pressure of ideal Bose Einstein gas. Thus

 E  P     V 

  V

 h2 2 r 2   2 2/ 3   8 mV 

2 h 2 2 r 2  . 2 3 8 mV 5/ 3 

2E , 3V

which is in agreement with classical result. Using equation (13), pressure of an ideal Bose gas is

 2 3 1  n V . RT 1  5/ 2  3V 2  2  g s Zt

2   1  n     ...  5/ 2   3 g Z    s t

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291

RT . . V

…(23)

We find from equations (13) and (23) that there are deviations in energy and pressure form the values for ideal gas behaviour.

An additional factor η occurs which can be held

responsible for these deviations. The deviations from perfect gas behaviour exhibited by Bose Einstein gas is termed as ‘gas degeneracy.” The gas degeneracy is obviously a function of 1/D. Also,

1 D

n g s Zt

n  2 mkT    g sV  h 2 

3/ 2

…(24)

which shows that for particles of small mass at low temperatures and small volume or high pressure, the gas degeneracy will be more marked. The deviations due to gas degeneracy are rather small as compared to those due to van der Waal’s forces and that is why it is not possible to observe this effect under normal conditions. Degeneracy for Helium: Helium also obeys Bose-Einstein statistics. For helium molecular weight is twice that of hydrogen, it can exist in gaseous state at much lower temperatures )(the boiling point being 4.2 K at atmospheric pressure) and the molar volume is 345 c.c. We find

1  0.134, D Which is applicable as compared to the case of molecular hydrogen and that is why there is some possibility of observing gas degeneracy in the case of helium.

Bose Einstein Condensation: From relation (24),

1 n  D g sV

 2 mkT    2  h 

3/ 2

We note that at temperatures nearing zero, i.e.,

, T  0, 1/D will have a large value which means

the degeneracy will be more marked. It means the gas will deviate highly from its perfect gas behaviour. The reason is as follows: While arriving at the Bose-Einstein distribution, we have assumed that, because of the closeness of the energy levels, we could always replace the discrete distribution by a continuous distribution giving 

n   dn( ) 0

(see equation 6)

i.e. we have changed summation into integration. As long as the change in occupation number,

ni from state to state is small compared to the number of particles in the states, this is a valid FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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procedure. If, however, the temperature in an ideal Bose gas is lowered to zero, the particles will begin to crowd into a few levels and the above condition will be violated. This means that when we are working at low temperatures, we must be careful in replacing the summation into integration. Now from equation (5) the number of particles lying between energy range

  d i



and

is given by

dn( )  g s g ( ) 

Where

g ( )d  . De / kT  1

4 mV h3

We note that for ground state unity because there is one state at of states at

  0,

(2m )

   0  0, g ( )  0,

  0.

while actually it should have been

Therefore above distribution fails to give the number

while this state, called the ground state, is very important at low temperatures

because a large number of particles occupy it at such temperatures. We further note that if

  0,

the above distribution holds good as

g ( )  0. Therefore above distribution can still be

applied for all states except ground state which should be treated separately. For a single state, we write

ni  For ground state

   i

1

i  0  0 ni  n0 

and gi  1. Therefore number of particles in ground state

 1

Therefore, the total number of particle states we write

n  n0   dn  n0   g s  n0  from equation (6). Value of x is

2 g s Zt





Further

x1/ 2 dx , De x  1

…(25)

 / kT ,

 2 mkT  Zt    2  h 

and



4 mV (2m )d  . h3 De / kT  1

3/ 2

.V .

n  n0  g s Zt F3/ 2 ( ),

…(26)

Where

F3/ 2 ( ) 







x1/ 2 dx De x  1

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

 2D

293

2  1 1  1  3/ 2  3/ 2 2  ...   2 D 3 D 



…(27)

When

 0 F3/ 2 (0) 

So that



D  e  1,



1 1  1  3/ 2  3/ 2  ...  2  2 3 

=2.612 Let T  T0 When

 0

D  1 so that from equation (10) we get

1 1   n  g s ( Zt )T T0 1  3/ 2  3/ 2  ...  3  2 

 gs (Zt )T T0 .F3/ 2 (0). Therefore,

g s ( Zt )T T0 

n F3/ 2 (0)

…(28)

Putting the value of g s from equation (28) in equation (26), we find

n  n0  n.

F3/ 2 ( ) F3/ 2 (0)

 Zt T T

 2 mkT    .V .F3/ 2 ( ) h2    n0  n. 3/ 2  2 mkT0    .V .F3/ 2 (0) 2  h  3/ 2

T   n0  n    T0  or

3/ 2

F3/ 2 ( ) . F3/ 2 (0)

T  n  n  n0  n.    T0 

3/ 2

F3/ 2 ( ) . F3/ 2 (0)

…(29)

From equation (27) we find that with D>1, F3/ 2 ( ) will have less value than when

D  1, i.e.,

F3/ 2 ( )

  0.

 F3/ 2 (0).

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n given by equation (29) acquires its maximum value when   0 .

This means that Thus for

294

  0 , the maximum number of particles occupying states above ground state is given

T  n  n    T0 

3/ 2

, When T  T0

…(30)

and the rest of the particles, given by

n0  n  n  n     ,

…(31)

T  T0

When

must condense into ground state. Thus from equation (31) we note that, as the temperature is lowered, beginning at T  T0 the molecules fall rapidly into the ground state. There is a sort of condensation into this state. This phenomenon is known as Bose Einstein Condensation. The temperature T 0 at which the Bose Einstein condensation begins to occur depends upon the density of the gas. If we consider liquid helium to be a gas, we would obtain a value of about 3.14 K for T 0. It is found experimentally that liquid helium does undergo a rather unusual transition at 2.19K. Below this temperature, the liquid helium displays the properties of a superfluid. It is generally agreed that this transition in liquid helium is associated with a Bose-Einstein condensation. For temperature above T0, α must decrease in order to keep

n  (n0  n)  n.

But

if α becomes significantly different from zero, then n0 becomes very small. We therefore assume for temperatures above T0 that n0  0. Therefore for T  T0

n0  0, and

n  n  g s Zt F3/ 2   . From equation (29) when

n  n , we find that

T  n  n   T0  or

3/ 2

F3/ 2   F3/ 2  0 

T  F3/ 2    F3/ 2  0   0  T  T   2.612  0  T 

3/ 2

…(32)

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which when substituted in the expression for n gives the total number of particles states equal to

T  n  g s Zt 2.612  0  T 

3/ 2

…(33)

Thermal Properties of Bose Einstein Gas: From the discussion of degeneracy of Bose Einstein gas, we infer that every thermal property of the gas should be studied separately for temperatures greater than T0 and for the temperatures less than T0 . We, therefore, again calculate the energy of the Bose Einstein gas; calculation of specific heat and entropy will then follow: (i)

Energy: From equation (7),

E

2 g s Zt kT







x3/ 2 dx De x  1

2 g Z kT 3  4  s t . 4 3  





x3/ 2 dx De x  1

 3/ 2.k.T .g s .Zt F5/ 2   ,

…(34)

where

F5/ 2   



3 

 3  . 4 

 4  3 



x3/ 2 dx De x  1

1 1   1  5/ 2  5/ 2  ...  ,  2 D 3 D 

…(35)

from equation (9), Further from equation (6),

n  g s .Zt







x1/ 2 dx . De x  1

 g s .Zt F3/ 2   .

…(36)

From equations (34) and (36).

F   3 E  nkT 5/ 2 . 2 F3/ 2   Case I. T  T0 ,

 0

Putting the value of

and

D  1, we should replace n by n i.e,



F  0 3 nkT 5/ 2 . 2 F3/ 2  0 

…(37)

n from equation (30),

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T  3 E  nkT   2  T0  From equation (35), with

296 3/ 2

F5/ 2  0  . F3/ 2  0 

D  1, we have F5/ 2  0   1 

1 1  5/ 2  ... 5/ 2 2 3

 1.342 Therefore,

( E )T T0

T  3  nkT   2  T0 

3/ 2

T  3  nkT   2  T0  Case II. For T  T0

  0,

and

1.342 2.612

3/ 2

0.5134

D  1. From equation (13),

 3 1  n E  nkT 1  5/ 2  2  2  g s .Zt

n T   2.612  0  Substituting g s .Zt T  ( E )T T0

…(38)

2   1  n  ,   ...  5/ 2     3  g s .Zt 

3/ 2

from equation (33), we have

3/ 2 3   3 2  T0   T0  5/ 2 5/ 2  nkT 1  2  2.612    3   2.612     ... 2 T  T   

 3 T   nkT 1  0.4618  0  2 T  

3/ 2

3   T0   0.0226    ... T  

…(39)

Specific Heat – Energy expressions can be used to calculate the specific heat Cv by the

(ii)

relation

 E  Cv     T v Case I.

For T  T0 :

 E  Cv    .  T v Using equation (38),

Cv 

3 nk 5 3/ 2   T  0.5134 2 T03/ 2 2

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T  15   0.5134  nk   4  T0 

…(40)

Case II: For T  T0 , using equation (39), we get 3/ 2 3  3   T0   T0  Cv  nk 1  0.231    0.045    ... 2  T  T  

…(41)

Variation of Cv with temperature is plotted in fig. 2. For T  T0 , the condensation temperature, we have

 Cv T T

For one mole of gas

  Cv T T  0.5134  0

15  nk 4

nk  R, so that

 Cv    Cv  

15  0.5134  R 4

Fig.2 (iii)

Entropy-Using the expressions of specific heat, we can calculate the entropy of the gas by the expression

S

Cv dT . T

Case I. For T  T0 , using equation (40), we have

S

T  1 15  0.5134  nk   T 4  T0 

T  5   0.5134  nk   2  T0 

3/ 2

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298

2  Cv T T0 3

…(42)

5  0.5134  nk. 2

Case II. For T  T0 ,

(S  S )T T0  (S )T T0 3/ 2 3  1 T 3 5  T0   T0    0.5134  nk   nk 1  0.231   0.045    ... dT , 0 2 2 T  T    T

on using equation (41), so 3/ 2    3 T 2  T    S  ( S )T T0  nk  log   0.231 1   0    ... …(43) 2  T0 3    T   

integrating upto second term only. We note that

( S )T T0 

2  Cv T T0 3

and from figure 2 we find that Cv shows a sudden drop for the temperature below T0 and consequently, entropy will also decrease suddenly. A decrease in entropy means decreases in disorder or increase in order, We have seen earlier that at T  T0 , a large number of particles condense into the ground state

(  0) which is attributed with zero entropy (entropy

k loge 1  0, since statistical weight of ground state is one). Since   p 2 / 2m,   0, implies

p  0 and therefore we can say equivalenty, that condensed particles condense in the momentum space. They acquire the same momentum

p  0 and, thereby, create an ‘order in

momentum space’. Therefore decrease in entropy implies that some sort of condensation is taking place – greater is the decrease in entropy more is the acquisition of orderly state i.e, greater is the condensation. For ground state, for which entropy is zero, it means, condensation is maximum or, in otherwords a large number of particles will rapidly fall into the ground state. As the ground state is also characterized by

p  0 , Bose Einstein condensation is sometimes

described as a ‘condensation in momentum space.’ Liquid Helium: As an application of Bose Einstein statistics, we may investigate the qualitative nature of the superfluid transition of liquid helium at 2.2K. Ordinary helium consists almost entirely of

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neutral atoms of the isotope 2H . As the total angular momentum of these atoms is zero, their discussion must fall under the jurisdiction of Bose Einstein statistics. Helium exhibits peculiar properties at low temperatures. It is found that (i)

Helium gas at atmospheric pressure condenses at 4.3K (its critical temperature being 3

5.2 K) into a liquid of very low density, about 0.124 gm/cm . (ii)

Further cooling to about 0.82 K does not freeze it and it is believed that it remains liquid all the way down to absolute zero. The solid state of helium does not form unless it is subjected to an external pressure of atleast 23 atmospheres. The phase diagram is shown in figure 3.

(iii)

For He in liquid phase, there is another phase transition, called λ – transition, which divides the liquid state into two phases He I and II.K. onnes, while liquefying helium, noted that at about 2.2K, density appeared to pass through an abrupt maximum and then decreasing slightly therefore.

Investigations also revealed that critical

temperature is at 2.186K. and that it represents a transition to a new state of matter known as liquid He II. In liquid He II state, it was found that 6

(a) Heat conductivity is very large of the order of 3.10 times greater, (b) Coefficient of viscosity gradually diminishes as the temperature is lowered, and appears to be approaching zero at absolute zero temperature, and (c) Specific heat measurements by keesom show that specific heat curve, fig. 4, is discontinuous at 2.186 K. The shape of the specific heat curve resembles the shape of letter λ and therefore this peculiar transition is called λ-transition and the discontinuity temperature 2.186K, is called ‘λ-point’. Since experimentally it was found that at ‘λ-point’ 

liquid He II state has no latent heat, Keesom concluded that transition He I He II at

T is a second order transition. The transition temperature decreases as the pressure is increased, tracking out the λ-line in figure. 3.

Fig.3

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Fig. 4 5.2 Explanation based on Bose Einstein condensation model: London’s Theory; Explanation of these pecularities of liquid He at low temperatures, based on B-E statistics, was given by London who suggested that He II is a liquid analogous to B-E gas and that λ-transition in liquid helium is the counter part of Bose Einstein condensation in the ideal gas. In Bose Einstein gas, degeneracy is

1 n  D g sV

 2 mkT    2  h 

3/ 2

(see equation 24)

London suggested that helium atoms are light enough and through the density (n/V) of the liquid is sufficiently high for the right hand side to be large and degeneracy to be well marked but is low enough for the liquid to behave as a gas. He concluded this λ-transition as a result of Bose Einstein condensation and gave an analogy between ‘λ-point’ and Bose Einstein condensation temperature T0 defined by equation (28), giving

 2 mkT0  gs   2  h  or

3/ 2

V

n F

3/ 2

(0)



n 2.612

 h2  n T0    2 mk  2.612Vg s 

2/3

In this expression, when we put V = 27.4 cm for a gram molecule of helium in liquid state, we get T0 = 3.12 K which is quite close to the observed value T  2.186K for the λ – point. This agreement in the value T0 and T favours the London explanation.

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Further for the discontinuity in specific heat curve at λ – point, London suggested that similar discontinuity occurs in the specific heat and hence the existence of two liquid components He I and He II is automatically explained. Again the decrease in entropy below T , being zero at 0.5 K, is very well explained by Bose Einstein condensation because, in the latter, we have shown that at T  T0 most of the particles rapidly fall into the ground state which is characterized by zero entropy. Though London approximated the properties of liquid helium as to resemble with those of Bose Einstein gas yet there is no reason to expect that liquid helium, with mutual interaction between the particles, should resemble with perfect Bose-Einstein gas in any important respect. Later on, Tisza introduced the two fluid hypothesis. Superfluidity: Tisza’s Two Fluid Model and Second Sound: According to this two fluid hypothesis: Liquid helium consists of two independent components, a normal fluid and super fluid. (a) The normal fluid: (i)

has energy in excess of the zero point energy,

(ii)

behaves as an ordinary liquid and possesses no unusual properties except that it is liquid at low temperatures.

(b) The super fluid: (i)

possesses only zero point energy,

(ii)

shows some queer properties; it has zero entropy, viscosity and flows without resistance through channels – the narrower the base, freer is its flow through them. Due to the latter reason it is named as super fluid.

(iii)

It consists of ‘condensed’ atoms. Its formation starts at λ-point and as the temperature is lowered, its percentage goes on increasing till at absolute zero, all the atoms go into superfluid state. In this model, the exact nature of these fluids is not specified. Both fluids have their own mass densities n ,  s , and velocity of fluids is

 n , s ,

refer to normal and superfluid. The total mass density



where subscripts n and s

and mass current



of the

system are given by

  n   s

  nn  ss .

…(1)

At the λ-point, all atoms are normal and therefore

n  1. 

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At absolute zero, whole fluid is made up of super-fluid so that

n  0.  Therefore, as the temperature rises,

n 

increases. In fig. 5 its variation is plotted. The

data for this plot are due to the experiment performed by Andronickashvili, According to these experimental results:

n   n 



(T / T )5 / 6 for T<T 1 for T>T

varies as the fourth power of T upto 0.6K beyond which the variation is more complicated.

Fig.5 The hypothesis has been successful in explaining various thermodynamic properties of liquid helium, e.g. mechano-caloric effect, the fountain effect etc. For examples, if two tanks are connected through such a thin tube that only the super-fluid component can flow through, then a transfer of mass from one tank to the other through the thin tube increases the entropy per unit mass in the former tank and lowers it in the latter. As a result, temperature of former tank will increase and that of latter will decrease. This is the mechano-cloric effect. The inverse effect that of creation of a pressure differential by heating is known as fountain effect. Tisza model has met with the following short-comings: (a) It has no molecular basis to explain the nature of two fluids. (b) It is incomplete as it does not provide hydrodynamic equations. Second sound: In two fluid model a sound wave in He II must mean a sinusoidal oscillation of

s

n

and

in phase with each other, because only when they oscillate in phase can the total mass

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density vary sinusoidally. We can also imagine a new mode of oscillation in which normal fluid o

and super fluid oscillate out of phase by 180 . Suppose by local heating of He II, a temperature gradient is established, disturbing the equilibrium of the system.

Tisza assumes that this

temperature difference will be smoothed out by a current of normal fluid in the direction of falling temperature and a current of super-fluid in the opposite direction such that the total mass density remains constant, satisfying the relation

nn  ss  0,

  0,

…(2)

that is, there is no net transport of mass across any plane in the liquid. This possible mode of motion would not be a sound wave because the total mass density is constant throughout the liquid. Instead, it would represent a sinusoidal variation of entropy per unit mass because the superfluid has not entropy. This phenomenon in which normal fluid and superfluid oscillate out of phase, giving rise to thermal waves for smoothing out the temperature difference, is called second sound. Smoothing out of temperature difference by the flow of normal fluid and superfluid suggests that some potential energy is connected with relative motion of normal and superfluid type of atoms which is periodically transferred into kinetic energy and then back again. This peculiar type of heat conduction (which is not by diffusion) will be a reversible process, taking place with a characteristic velocity u2, called the velocity of second sound. We shall now proceed to find u2. As mentioned earlier, in this phenomenon total mass density remains constant but sinusoidal variation of entropy per unit mass is considered. Since the entropy of superfluid is zero, we can express total entropy, S, per gram of the liquid interms of the entropy S n of the normal component by the relation

 n Sn   S

…(3)

For sake of mathematical convenience, let us consider the transport of entropy in one direction only, say x-axis. If we imagine a parallelepiped of unit area of cross section and of length

x

with one pair of sides vertical to the flow of liquid then amount of entropy entering the

volume per second at one face is

 ( n Sn )n

 (  S )

…(4)

and the amount of entropy leaving the volume at the other face per second is

 (  S ) 

 (  Sn ) x x

…(5)

so that the net transport of entropy per unit volume per second will be

 (  S ) 

 (  Sn ) x  (  S ) x

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 (  Sn ) x. x

…(6)

Eq. (9) in other words represents the loss of entropy within the volume in unit time. Because the original amount of entropy within the elementary volume was



 S , the loss in unit time will be

 ( S ) x

…(7)

which should be equated to eq. (6). That is

  (  Sn )   (  S ), x x

…(8)

and is called equation of continuity. If the total mass density,

 , remains constant, eq. (8) can

be written as

n Taking

n

 S S S n  . p x t

to be small, we can arrive at

n S  . x t

…(9)

Further, let the two opposite faces under consideration be at temperatures T and

Q

be the amount of heat transferred across the thickness

x

T   T . Let

per second. Then

 Q  T S , where  S is the entropy transported across

 x , which by eq. (6) is  Sn .

Therefore

 Q  T (  Sn ) The reversible work done,

…(10)

W , in transferring the heat will be

W T  Q T by the second law of thermodynamics. The work done will bring about a change in kinetic energy. Using eq. (10), we find

W  (  Sn )T

T T

= (  Sn )T Now we shall separately calculate the change of kinetic energy for thickness

…(11)

x

or volume

x

because area of cross section is unity. The kinetic energy of fluid per unit volume

U

1  nn2  ss 2  2

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Putting from equation (2),

s  

nn , s

we get

n 2n 2  1 2 U   nn   s  2 s 2  

   n  1 nn 2  s  2  s 



 nn 2 2s

…(12)

Therefore rate of change of kinetic energy per unit volume

U n n   t  s n t

…(13)

Change in kinetic energy of a volume



x

will be

n n  . x  s n t

…(14)

and is brought about the work done expressed in eq. (11). Thus equating the eqs. (11) and (14), we get



n n   x   n S T  s n t

n n T  S  s t x

Further

Q  T S

and

Q  CT  p ,

…(15)

where C is the specific heat. Thus

CT  p  T S when volume,

 , is kept constant,   0

so that

T T  S C or

T x T  x S C

T T S  x C x

…(16)

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From eqs. (16) and (15), we get

n n T S  S  s t C x or

n  2n ST  2 S   s xt C x 2

n   n  ST  2 S   s t  x  C x 2

Using eq. (9), we get

n  2 S  s t 2



S 2T  2 S C x 2

 2 S   s S 2T   2 S   t 2   n C  x 2

 U 22

2S , x 2

…(17)

which is the equation of motion of the thermal wave, called second sound. The velocity of this second sound is 1/ 2

  S 2T  U2   s   n C 

1/ 2

    n S 2T   .  C   n

which is the desired quantity. At λ-point where superfluid tends to zero,

…(18)

  -n   0, so that U 2

rapidly falls to zero. 5.3 Landau’s Theory: Landau postulated that: (a) energy levels of liquid helium II (liquid as a whole, not of individual helium atoms) consist of two sets of overlapping energy states-one representing quanta of sound or phonons and the other of quanta of vortex motion or rotons. (b) aforesaid quantas i.e. phonons and rotons, behave like gases and interchange energy with the walls. (c) the lowest phonon level is assumed to be below the lowest roton level separated form it by a finite energy gap.

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ADVANCED SOLID MECHANICS Based on these assumptions, Landau set up a quantum hydro-dynamics to explain the anomalous results of He II. We shall in brief discuss the theory: Concept of quantum liquid: In practice, helium can remain liquid down to absolute zero owing to the unusually weak inter-action between its atoms and that it has got a large zero point energy. At 2.3 K, de Broglie wavelength corresponding to the thermal motions of atoms becomes comparable with interatomic spaces i.e. quantum effect becomes very prominent, it is not necessary that solidification of helium must take place and it may be called quantum liquid. Therefore, Landau proposed the theory of quantum liquid taking its departure from known theory of solids rather than that of an ideal gas. Elementary excitation in quantum liquid: The calculation of thermodynamic quantities requires a knowledge of the energy spectrum of the given body. In calculating partition function for temperatures near absolute zero, we must take into account only weakly excited energy levels of the liquids i.e. states close to the ground state. These weakly excited states are considered as a set of separate â&#x20AC;&#x2DC;elementary excitationsâ&#x20AC;&#x2122;. As the temperature is raised above 0K, thermal agitation begins to appear and the elementary excitations arise in liquid helium. These elementary excitations behave like quasi particles possessing definite energies and momenta and move in the volume occupied by the body. For example, the excited states of a crystal in which the atoms perform small oscillations about their equilibrium position can be considered as a set of photons moving inside the crystal. Phonons are quantas of sound waves which in a crystal can be either longitudinal or transverse. We shall assume that the liquid can support only longitudinal oscillations. A molecule in a fluid is capable of much more complicated motions than merely oscillating back and forth, so the sound excitation can form only a small fraction of the total number of quantum states available to the system. Therefore, we shall assume that these waves propagating through the fluid constitute the lowest excited states. Elementary excitations obey Bose Einstein statistics: Now at low temperature, the number of elementary excitations is sufficiently small so as to neglect the interactions between them. It means the energies of these elementary excitations or quasi particles combine additively and their assembly may be regarded as an ideal gas. One of the possible types of energy spectrum of weakly excited states of a quantum liquid is characterized by the fact that the elementary excitations may appear and disappear one by one. But the angular momentum of the whole liquid can change only by a whole number. Consequently, elementary excitations which occur singly must have an integral angular momentum and will thus obey Bose-Einstein statistics. Based on these ideas, Landau postulated the energy spectrum of excitations to have the form shown in fig. 6.

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Fig. 6. Energy spectrum of elementary excitations: Basing on above discussion Landau postulated that (i) for small momenta (i.e., large wavelengths (h/p) these excitations correspond to normal sound waves in liquid and are known as phonons, (ii) the quantum states of liquid helium near the ground state can be regarded as a gas of non-interacting elementary excitations. From (i), if means that the energy

 ( p)

of elementary excitations with small p is the

linear function of the momentum (refer to fig. 6).

 ( p) For large values of momentum, the function the

 ( p)

= up,

 ( p) ceases to be linear.

For very large momenta

cannot exist at all since elementary excitation with too large momenta are unstable and decompose into several excitation with smaller momenta and energies.

The elementary excitations in the neighbourhood of p = 0 are called as phonons and those in the neighbourhood of p = p0 are called as phonons and those in the neighbourhood of p = p0 are called as rotons. There is a finite energy gap Δ between the phonons and rotons levels. The energy of rotons is given by

 

(p - p0 ) 2 , 2

where p0 is a constant and μ is the effective mass of rotons.

The empirical values of the

constants, Δ, p0 and μ are

 8.9K , p0 =2.1 10-19gm.cm./sec. and  =1.72 10-24gm. Thermodynamic quantities of liquid helium at different temperatures: On having the knowledge of the energy spectrum of elementary excitations, we can proceed to the calculation of thermodynamic quantities of liquid helium at different temperatures. At temperatures near absolute zero, elementary excitations are small and are practically phonons. The Debye’s theory can be used to calculate specific heat of quantum liquid helium

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with the modification that instead of three (one longitudinal +two transverse) directions of polarization, only one be accounted (since liquid is supposed to support only longitudinal vibrations). Refer to Debye theory art. 1(a).

d (3nkT )  3nk  3R dT

…1(a)

For this case

1 9n  3 Cl 4 V m3 or

 m3 

9nCl 3 4 V

TD3 

h3vm3 9nh3Cl 3  k3 4 Vk 3

From art. 1(a),

12 4 nk  T  Then from eq. (10) 1(a), Cv    5  TD 

Putting for TD , we get

Cv 

12 4 nkT 3 4 Vk 3 5 9nh3Cl 3

2 2 kV (kT )3  15(h / 2 Cl )3 Replacing Cl by u in case of helium, we get

Cv 

2 2 k .V (kT )3 , 3 15(hu )

which gives the specific heat of liquid helium at low temperatures, and is proportional to the cube of the absolute temperature. But above

0.7 K, CV increases much more rapidly

than predicted by T law. That means apart from photons, some other modes of excitation have set in. Thus we have to look for another mode by which energy can travel through the fluid. Classical hydrodynamics provides such a mechanism in vortex motion. Vortices in a non-viscous liquid have remarkable stability and travel in straight lines at constant velocity.

The quanta

corresponding to this form of motion are called rotons. They have a certain minimum energy of formation and can not be excited at the lowest temperatures. Since the roton energy always contains the quantity Δ, which is quite large compared to kT, Boltzmann distribution (instead of BE distribution) can be applied for description of rotons with sufficient accuracy.

The roton

contribution to specific heat is, thus, calculated to be

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 3    2  Cr  kN r       4 kT  kT   3

At low temperatures roton contribution is much lesser than phonon part and T low holds good but as the temperature increases roton contribution becomes significant. Many properties of liquid helium II have been explained on the basis of this theory. But this theory does not provide a satisfactory explanation of the phenomenon at

  po int .

Feynman has tried to relate Landau theory with London approach. He derived quantum statistically the energy level and energy gap picture of Landau while retaining the idea of BoseEinstein condensation of London. In Feynman’s work, the wave function representing an excitation in liquid helium, if of the form

   f (ri ) i

where f (ri ) is some function of position,



is the ground state wave function and the sum is

taken over each atom, i. He finds the energy value to be given by

h2 k 2  (K )  S (k ), 2m where S(k) is the structure factor of the liquid neutron scattering. For small values of k, rises linearly (phonons) and for larger k,

 (K )   

 (K )

 ( K ) , corresponding to S(k) minimum, behaves as

h 2 ( k  k0 ) 2 2

which is the form which Landau found and it agrees with the data on specific heat.

5.4 Black Body radiation and the Planck Radiation Law: The most important application of Bose-statistics to electromagnetic radiation in thermal equilibrium, called “black body radiation.” In quantum theory, radiant energy occurs in energy packets or photons or light quantum of energy hυ and momentum hυ/c where υ is the frequency and c is the velocity. Photons have zero-rest mass and a spin quantum number of 1; like all particles with spin 1, they obey Bose-Einstein statistics. Consider a black body radiation chamber of volume V containing radiation in equilibrium with the walls at temperature T. Let U d represents the energy density of radiation of frequency lying between

  d.



and

The problem is to find out U as a function of T. This was first solved by Max Planck

through the hypothesis of linear harmonic oscillator possessing discrete energy values.

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h3 In the momentum space particles with in a small volume  p  are indistinguishable. V It therefore, represents an eigen state.

At any instant, all particles, having their momenta

between p and p + dp will lie within a cell of volume 4 p dp. Therefore the total number of 2

eigen states is given by

g ( p)dp  h

For a photon

p



dp 





4 p 2 dp. h3 / V

…(1)

h c

hd c

…(2)

Substituting this value of dp in equation (1), we have

g ( )d  4V

2 c3

d.

…(3)

There is a duplication of states for the two independent directions of polarization. Therefore

2

g ( )d  8V

d.

…(4)

Equations (4) represents the total number of eigen states lying in the frequency range and

  d .



Introducing the result in Bose-Einstein distribution law, we get

dn 

g ( )d . e    1

dn  8V

2 3

…(5)

d   

c e

1

dn  2 d  8 3    . V c e 1

…(6)

The left hand side of equation (6) represents the number of photons per unit volume. Multiplying by

h , the energy of photon, gives energy density U d in the specified frequency

range. Substituting

  h ,

we have

8 h 3 d  U d  . c3 e    1 We know that formula,

  0,



…(7)

1 , moreover if we put   0, equation (7) is just Planck radiation kT

means, the conditions

 n   ni  0

should be dropped.

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Every process of emission, in nature, results in the creation of photons and similarly, every absorption proves in the absorption of photons, which may be converted into another forms of

 n

energy. Under these conditions the restriction

 0 is no longer applicable. The term  ,

then, does not involve in undetermined multiplier and equation (7) reduces to

U d 

8 h 3 d . c3 eh / kT  1

…(8)

which is Planck’s law. When





h  kT , the term eh / kT  1  h / kT and hence equation (8) can be written

U d  For

8 kT 2  d. Rayleigh Zean’s law, c3

…(9)

h  kT , equation (8) becomes U d 

8 h 3  h / kT e d. Wein’s Law c3

…(10)

The total energy density is

U 8 h   3d  3  h / kT V c 0 e 1

8 h  kT   3   c  h 





x3dx h by substituting x x e 1 kT

8 h  kT   4  3   c  h  15 4

 bT 4 where b 

8 k 4 4 15c3 h3

…(11)

Equation (11) is Stefan-Boltzmann law. Planck’s Radiation Law Statement : According to the law of radiation of wavelength lying between

 and   d 

emitted per unit volume by a perfectly black body at temperature T is given by

E d  

8 hc



d e

hc /  kT

1

…(1)

Derivation of the law Some Important Postulates : It is assumed that the photons do not interact among themselves. The photons interact only with the atoms of the cavity. It is further assumed that the photons are indistinguishable and many photons can have same energy. Photons are considered as bosons and they obey Bose-Einstein statistics.

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The most probable distribution of the particles among various energy levels for a system obeying Bose-Einstein’s distribution law is

ni 

i (   Ei )

1



 Ei / kT

e e

1

Where 

1 kT

Most important point to be noted here that photon gas differ from the ideal gas in one respect. Although the total energy of photons inside the hollow encloser at a particular temperature T remains constant, the total number of photons may be completely absorbed on striking the wall or the hot wall may emit a new photon of energy hv. i.e., the photons may be created or destroyed. In other words, for a photon gas system

  ni  0 is no longer valid. Hence  n    ni  0 i

Due to this reason, the value of

ni 

 is

equal to zero in the Bose-Einstein distribution law. Thus

gi e

Ei / kT

(   0)

1

…(2)

In this case, the cavity is large as compared to the wavelength of radiation, the energy spectrum of photons is taken to be continuous. The energy difference between successive allowed energy values is very small. Thus, replacing, g i by gi ( E )dE and ni by n( E ) dE, we get

n( E ) dE  Since

g ( E ) dE e E / kT  1

…(3)

E  h the value of g(E)dE can be written as g ( )dv . The quantity g ( )dv

corresponds to the number of oscillatory modes in the frequency range energy range E and



and

(v  dv) is the

( E  dE ).

The number of states in a black body radiation in the frequency range



and

can be obtained by calculating the spherical volume bounded to two spheres of ratio

and

(v  dv)

h(v  dv) c

hv . c The volume of the spherical shell.

4 h3 4 h3   3 (v  dv)3   3 v3 3 c 3 c 4 h3 3   3 (v  3v 2 dv  ......  v3 ) 3 c

4 h3   3 X 3v 2 dv 3 c

(higher terms neglected)

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4 h3 2   3 v dv 3 c The phase space has volume

…(4)

v  h3 and there are two states of polarization (left hand and right

hand) for the radiation. The number of states in the black body radiation in the frequency range v and

(v  dv) is given by g (v)dv  2 X 4 

V 2 v dv c3

…(5)

8 V 2 v dv c3

Substituting in equation (3), we get the number of photons in the frequency range v and

(v  dv)

given by

(8 V v 2 dv)4 1 n(v)dv  X hv / kT 3 c c 1 Since each photon has energy hv, the energy density

…(6)

E (v)dv defined as the amount of energy

per unit volume lying between the frequency range v and

E (v) dv 

hv.n (v) dv V

8 h v dv  X E (v) dv  3



(v  dv) is given by

1 e

hv / kT

…(7)

1

Equation (7) represents Planck’s radiation law for black body radiation In terms of wavelength : To express the law in terms of wavelength

d (v ) 

2



and

in equation (7), we get

E d  (or)

( ) , substituting v 

E 

8 h c3 cd  1 8 hc d . 3 . 2 . hc /  kT  5 hc /  kT 3 c   e 1  e 1

8 hc 5 ehc /  kT  1

…(8)

This equation (8) also represents Planck’s law for black radiation in terms of wavelength.

5.5 Landau spectrum of phonons Landau found the explanation of London and Tisza based on the Einstein condensation of ideal BE gas for the peculiar properties of liquid He II to be unsatisfactory. In particular, it gives wrong predictions for viscosity and second sound velocity as

T  0 . According to him all bodies

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315

T  0 should show solid-like behaviour. The observed specific heat of He II as T  0 3

follows the Debye T law which is valid for solids at low temperatures. This T dependence associated with elastic or sound waves in solids can be understood in terms of phonons, the quanta of longitudinal sound waves. For

T  1K , the potential internal motions of the liquid He II

are longitudinal sound waves and the excited state can be regarded as an aggregate of phonons. The energy of phonons is linear function of their momentum p,

 ph  u1 p ,

(1)

Where u1 is the first sound velocity This does not exhaust all the 3N degrees of freedom, if there are N atoms in the system of liquid He II. A liquid is not expected to sustain high energy transverse (shear) wave, as a solid can. Therefore, for 1K  T  T , Landau assumed the existence of excitations called rotons 7

(quanta of vortex motion) with the energy spectrum

 rot    Where

p2 . 2r

(2)

 is the energy gap (minimum energy required to excite a roton as rest), p the linear

momentum of the roton and

 r the effective mass of the roton.

Phonons, like photons, obey BE statistics. Rotons are also assumed to obey BE statistics. However, because the rot involves   kT for T  T the aggregate of rotons can be treated, to a good approximation, as a Maxwell-Boltzmann gas. Both phonons are rotons behave like quasi-particles that can move about and from the normal fluid. The superfluid has strictly zero entropy. To improve the agreement with experiments, Landau

finally proposed the energy

spectrum

( phonon), ph  u1 p,    ( p  p0 ) 2     , (roton),  rot 2r 

(3)

Where the parameters u1 , , p0 and r are adjusted to give the best fit to specific heat measurements. Landau finds

u1  22.6 X 103 cm / s,

o p   9 K , 0  2 A1 , r  0.3 m, k h

(4) 3

Where m is the mass of the He atom. The experimental value for u1 is 23.9 X 10 cm / s . The Landau spectrum is shown in Fig. 7. The active portions of the spectrum are shown by solid

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316

curves. At low temperatures (0<T<1K) only the liner part of the spectrum corresponding to phonons becomes active. At about 1K, both phonons and rotons exist.

Fig. 7 If the number of phonons and rotons per unit volume of liquid He II is not large, their aggregate can be regarded as a mixture of two ideal gases – a phonon gas and a roton gas. Then specific heat will be the sum of the contributions from the two, Cv  C ph  Crot Phonon Contribution For the longitudinal wave, vm

vm3 v2 4 3N   g (v)dv   4 V 3 dv   V 3 , 3 u1 u1 0 0

(5)

Which relates the cut-off frequency vm with u1 Then eqn gives 3

12 4 Nk  kT  16 5 k 4 3 Cv   V T ,   5  hvm  15h3u13 This is the specific heat per unit volume for phonons is

C ph  bT 3 ,

b

16 5 k 4 15h3u13

We can express it as per unit mass by using

U ph   C ph dT 

S ph   Where

C ph T

dT 

(6)

V  1/  for 1 gram. From (6)

bT 4 , 4

(7)

bT 3 , 3

(8)

 is a constant. Not that, as for photons, for phonons,   0. It is interesting to compare

and (7), remembering that the former involves two directions of polarization. In Landau’s theory phonons play an important role the

T  1K , where few rotons exist.

Unlike the theory of London and Tisza, here they contribute to the normal fluid and so lead the

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correct prediction of

4 3

nph /   U ph / u12 Let He II at

317

where U ph  bT / 4 . 4

T  1K be enclosed in a long tube along the z direction and let the walls of

the tube move with a velocity vz . The superfluid component remains at rest but phonons collide with the walls and acquire a velocity vz . For a stationary observer the energy

 of the phonon of

rest energy  0  vz pz 0 vz p cos  where cos   pz / p. Then for the phonon gas

p   p cos  d N ph 

p cos  V   2  0 0 p 2 dp sin  d d 3 0 exp (0 / kT )  1 h

By Maclaurin expansion for small v z

1 1  exp (0 / kT )  1 exp[( vz p cos  ) / kT ]  1

1 ( p cos  / kT ) exp ( / kT )  vz exp ( / kT )  1 [exp ( / kT )  1]2 The zero order term disappears when integrated over the directions,

0 sin  cos  d  0 and so

p 

exp ( / kT ) 2 V vz   0 0 p 4 dp sin  cos 2  d 3 h kT [exp ( / kT )  1]2

4 V vz  exp ( pu1 / kT ) 0 p 4 dp , 3h3 kT [exp ( pu1 / kT )  1]2

Where we have used

0 sin  cos 2  d  For uz

2 and  p(u1  vz cos) 3

pu1

u1 Integrating by parts and noting that

kT u1



  p4    exp ( pu1 / kT )  1  0

Vanishes for both limits,

p   ph vz

4 V vz 4kT  p3 dp 4 U ph   uz , 0 3 exp ( pu1 / kT )  1 3 u12 h kT u1

because

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U ph   0 d N ph 

318

pu1 4 V  p 2 dp 3 exp ( pu1 / kT )  1 h

If we write U ph per unit volume then

 ph  

4 U ph 4 bT 4 / 4  bT 4    3 u12 3 u12 3u12

In Landau’s theory for T <1K the normal fluid

(9)

 n is entirely made of  ph

Therefore, substituting (6-9) in (4) 1/ 2

    ph S 2T  u2      C ph  ph 

1  (3u12  bT 4 )1/ 2 3

 13.7 X 103 cm / s,

for T  0.

u1 31/ 2 (10)

The experiment (Fig. 8) supports this calculation, whereas Tisza wrongly predicted

u2  0 asT  0.

Fig. 8 The reason for the success of Landau’s theory is that here not particular particles but all excitations, phonons and rotons, are associated with the normal fluid.

IDEAL FERMI GAS 5.6 Thermodynamic functions of Degenerate Fermi-Dirac Gas (i)

Thermal Capacity, Cv : At such low temperatures, the gas will have a thermal capacity given by

 E  Cv     T v FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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319

 5   kT  1   12   f  

2     ...      

k  T  n f   ...    2  f  

…(1)

nT 2 1  nk  .  ... 2 f

Using equation (16),

nT 1 2  k  . 2 2/3 2 h  3n    2m  4 Vg s 

Cv 

n 2 mk 2  4 Vg s     h2  3n 

2/3

…(2)

From equation (2), we infer that heat capacity per particle would be small for large

n  and small particle mass m. If the particles are electrons then equation (2) gives V 

densities 

electronic specific heat. We note that it is proportional to the electronic specific heat. We note that it is proportional to the absolute temperature and therefore, at ordinary temperatures, the contribution to the specific heat of metals due to electrons would be negligible as compared to the 3

contribution due to the atoms since atomic specific heat is proportional to T , while at very low temperatures electronic specific heat will be significant. (ii)

Entropy: Entropy, S, can be obtained from

S



Cv dT T

n 2 mk 2  4 Vg s    h 2  3n 

n 2 mk 2  4 Vg s     h2  3n 

2/3

Using equation(2). If we use equation (1),

S

 n  f  2 

2  k     ... dT    f  

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320

n   k   f   T  ... 2   f   n f   2 (iii)

   2  kT    2     f

2     ...     

Helmholtz Free Energy:

F  E  TS 2     kT   n f 3 5  n f 1    ...  T   5  12   f    T  

2     2  kT    ...         2 f   

3 1   kT  1   kT   n f   n f     n f 5 4   f  2   f   3 1   kT  n f     5 4   f 

2     ... ,   

…(4)

which is an expression for Helmholtz free energy. Electron Gas: A metal can be considered to be composed of a system of fixed positive nuclei and a number of mobile electrons referred to as the electron gas. To study the properties of an electron gas at low temperatures in the region shall revise the earlier discussion. For electrons S 

f 

h2  3n    2m  4 V .2 

h 2  3n     8m   V 

T  0 we

1 so that g s  2s  1  2, and we get 2

2/3

…(1)

and from equation (i) =3/5nεf, we get

3 E0  n f . 5

…(2)

Further from equations (a) and (b), eqn a

gf 

4 Vg s (2m f )3/ 2 3 3h

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eqn b

321

h 2  3n  f    2m  4 Vg s 

2/3

we get

g f  n, which means that in the limit level

f

T  0 every one of the states is occupied fully up to the energy

whereas all the states above this energy level are empty. From equation (c), we can calculate the degeneracy factor for an electron gas. For

electrons

m  9.11028 gm. and g=2,

1 h2  3n   eqn. (c)   D 2mkT  4 Vg s 

2/3

we get

1 h2  3n     28 D 2  9.110 kT  8V 

2/ 3

Taking a typical metal of atomic weight 100 and density 10 so that volume of gm. atom be 10 c.c. and the number of electrons, assuming one free electron per atom, is

6.02 123.

Then,

 3  6.02 1023  1 (6.62 1027 )2    D 2  9.11028 1.38 1016  T  8  3.14 10  

2/ 3

105 1.5T 

Which means degeneracy is sufficiently high. It shows clearly that for electron gas, the classical statistics is not valid and can be applied only at temperatures of the order of

105 K (because only

then D will approach unity). Therefore at low and other ordinary working temperatures, it is necessary to use Fermi-Dirac statistics to study the electron gas in the metals.

At low

temperatures electronic contribution to the specific heat of metals is given by equation (1a). which is

Cv 

 kT  1 nk 2  .    2 f  

But from equation (1a),

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eqn (1a)

322

1 f  , D kT

we have

So that

D

Cv 

1 nk 2 D. 2

f

Using the above value of 1/D, we find that

Cv 

1 nk 2  1.5  105  T . 2

Putting nk = R, gas constant

 1.987 cal deg -1 mol-1  2cal deg-1 mol-1

 2  10, we get electronic specific heat

1 Cv  1.5 105  2 10  T 2 4

= 1.5 10  T cal / gm.atom. Pressure of the electron gas can be obtained from eqn(1b)



2 n f as 3 V

2 n f P0  . 5 V

2n h 2  3n     5V 2m  4 Vg s 

nh2  20mV

 3n     V 

2/3

using g s  2.

for a metal of atomic weight 100 and density 10 (=n/V)

P0  atoms. which means at normal temperature, the pressure of the gas is sufficiently high. Example: Calculate the Fermi energy in electron volts for Sodium assuming that it has one free electron per atom. Given density sodium  0.97 gm / cm . atomic weight of sodium = 23. 3

From eq. (1) in Electron gas, we find that Fermi energy is given by

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f 

h2  3n    8m   V 

323 2/3

Assuming one electron per sodium atom, the electron density

 Where

n is given by V

N0  , w

N0  6.1026 atoms / kg.mole, Avagadro ' s number

 =0.97 103 kg/m3 , density of sodium w=23,atomic weigth of sodium Also

h=6.62.10-34 joule  sec.

m  9.110-31kg On putting these values, we get

 f  5.032 1019 joules 5.032 1019 =  3.145eV . .6 10-19 5.7 Free Electron Model and Electronic Emission:* In free electron model of a metal, the zero of the energy is taken at the bottom. At the absolute zero of temperature, the electrons will fill n/2 lowest energy states upto the energy value

f.

The minimum amount of energy necessary to remove an electron from the metal is equal to

( Ea     f ) and is defined as the work function  of the metal. According to Fermi-Dirac distribution the number of electrons per unit volume

Ea   f  e lying between momentum space p and ( p  dp) is given by

Fig. 1.

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324

g ( p)dp , V .De / kT  1  f / kT

where

De

and

4 Vp 2 dp [See eq. (1c)] g ( p)dp  g s h3 eqn (1c)

dn( )  g s

4 mV . h3

 2m  .

 1/ 2 d De / kT  1

g s  2, so that

For electrons

8 Vp 2 dp dn  3 (  ) / kT Vh e f  1  As

8 p 2 dp h3 e

(   f ) / kT

 1

p  mv, dp  mdv, we get dn  If

 x ,  y , z

2(m / h)3 4 2 d (  ) / kT e f 1

be the velocity components of the electron in

phase space volume between the velocity rage

z  dz

x

and

x, y and z directions, then the

x  dx

and

 y  d y

and

z

and

will be

4 2 d  dx d y dz , So that

dn 

2(m / h)3 d x d y d z e

(  f ) / kT

1

The value of (   f ) is of the order of 20 kT, while kT at 300 K corresponds to an energy of about 0.03 eV. Now in (   f ) the minimum value of electrons is E x , the quantity e

(  f ) / kT



required for emission of

is much greater than one. Therefore in the denominator,

we neglect 1 and write 3

 m  (  ) / kT dn  2   e f d x d y d z h 3

 m y 2 / kT  m   / kT mx2 / kT  2   e f e 2 d x.e 2 d y h 1

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325

1  m z 2 / kT 2

d z ,

where we have substituted

1 2

  m x 2  m y 2  mz 2 . Now, supposing the surface of the metal is a plane surface perpendicular to the ġ-direction, the number of electrons with a velocity component between

x

and

x  dx

is given by

dnx  n(x )dx 3

 m   / kT  mx2 / kT    2 m( y2 z 2 ) / kT  2  e f e 2 d y d z .   e h 1

But only those electrons which are traveling in x-direction will come out in yz plane i.e. as the metal surface is in yz plane, the y-and z-components of motion of an electron would not be affected when the electron passes through the meal surface and therefore the energy required to escape must come from x component of the electron’s motion. Only those electrons will come out which have got energy greater than

E

  f  , the work function. Therefore, number of

electrons arriving per second per unit area of the metal surface will be 



nx 

n( x ) x d x

( Ea / m ) 

 m   / kT  2  e f h 

 



 



xe



1 m x 2 / kT 2

d x

(2 Ea / m )

1  m ( y 2  z 2 ) 2

d y d z .

4 mk 2  E   / kT   T 2e a f 3 h Therefore emission current density per unit area

J  enx

4 m ek 2  E   / kT   T 2e a f 3 h Putting

A and

4 m ek 2 h3

e   Ea   f  ,

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J  AT 2ee / kT .

We get

This equation is called Richardson-Dushman equation of thermionic emission. It was first derived classically upon the assumption that the electron obey Maxwell-Boltzmann statistics. Its classical form is

ne (kT )1/ 2 e e / kT 1/ 2 (2 m)

J

which differs from quantum equation in constant co-efficient and in the power of T which multiplies the exponential term. 5.8 Paul’s Theory of Paramagnetism: In metals free conduction electrons are present. As each electron is endowed with an intrinsic magnetic moment μ, it acts as an elementary magnet. Thus all free electrons, acting as elementary magnets, can orient themselves in the direction of an applied magnetic field

B and

make metals highly paramagnetic. Let us consider an assembly of N free fermions of spin

1 h each of which is described by a 2

single particle Hamiltonian,

p2  .B 2m

H

and the single particle energy levels

 i.s  where

S  1.

p2   SB 2m

Therefore for the assembly of N free fermions labeled by the occupation

numbers ni , s of the single particle levels.

 i.s , can be written as

En    i.s , ni ,s i

 p 2    p2       B  ni ,1     B  ni ,1  i  2 m   2m   where

…(1)

ni ,s  0,1

 n

i,s

N

…(2)

If we write it for particles with spin

n

i , 1

 , then

N

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n

i , 1

327

 N  N  N

then eq.(1) can be written as

En   (ni ,1  ni ,1 ) i

p2   B( N   N  ) 2m

p2   B(2 N   N ) 2m

  (ni ,1  ni ,1 ) i

…(3)

The partition function is given by



ZN 

   En

ni ,1 ;ni ,1





   B (2 N   N )

p2   ni ,1 i 2m

ni ,1 ;ni ,1



p2   ni ,1 i 2m

…(4)

where prime over summation implies that restrictions imposed by eq. (2) must be followed while carrying out the summation, i.e.,

ZN 

e

 B (2 N   N )

N 0

 e

p2   ni , 1 i 2m

ni , 1

  e

p2   ni ,1 i 2m

…(5)

ni ,1

where an arbitrary value for integer, N  , has been selected and

n

i , 1

  is subject to restriction

N 

and

  is subject to restriction  n N i , 1



 N  N

so that eq. (5) can be written as

Z N  e  BN

 e2 BN

N 0





p2 N 2m

N 0



p2 ( N  N ) 2m

…(6)

N  N

If the partition function of an ideal Fermi gas of N spinless particles of mass m be denoted by Z

0 N

then

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Z 0N 



 niN



328 p2

 2 m ni i

e  y ( N )

…(7)

then eq. (6) can be written as

Z N  e  BN

 e  2

BN 

N 0

Z 0 N Z 0 N  N

…(8)

N 1 1 log Z N    B  log  e2  BN   y ( N )  y ( N  N ) N N N 0

…(9)

Summation in eq. (9) contains (N+1) positive terms (0 to N) the logarithm of this sum is equal to the logarithm of the largest term in the sum plus a contribution of the order of log N. Therefore, we neglect the term of order

1 log N, and write N

1 log Z N   f ( N  ), N where

f ( N  )  Max.[ f ( N  )]  2N  1 f ( N  )   B    1  [ y( N  )  y( N  N  )]  N  N

…(10)

N  gives the average number of particles with spin up. We proceed to find its value. maximum

For

 

f N  we write the condition f ( N  ) 0 N 

at N   N 

  2 N  1   1  [ y( N  )  y( N  N  )]  0  B   N   N 

 N 

2 B 1 [ y( N  )] 1 [ y( N  N  )]   0 N N N  N N 

 y ( N  )   y ( N  N  )  2 B    0   N   N   N  N    N  N 

…(11)

 f   . Free energy per particle, i.e., chemical potential is given by n

Refer eqn (1d)   log 

n f    log     log   n f  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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Applying to particles with spin up, N  , we have

 N   y ( N  )   p2 N  p2     N  N   2m  2m  f 

 log 

(see eqs. 7 and 9)

and to particles with spin down, N  , we have

 N   y ( N  ) y ( N  N  ) p2   (see eqs. 7 and 9)  N  ( N  N  ) 2m  f 

 log 

Therefore putting above equations in (11), we get

 N   N  2 B   log     log  0 f f     or

 N    N   2 B 2 B log    log   f f  kT    

 N   2 B log   N    kT

N   N e2  B / kT

B  0 , N  N 

N 2

that is, when B =0, half the particles have spin up and other half spin down. At low temperature an for B>0,





e2  B / kT  1 and consequently N   N  , i.e, balance N   N  shifts in favour of

spin up and a net magnetization will result. The magnetization per unit volume will be given by

I Since



total magnetic moment . volume

is the magnetic moment of a particle, net magnetisation will be



  N  N



so that

I



 N  N



easily gives the value of magnetic susceptibility.

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UNIT QUESTIONS UNIT –I Questions 1. State and prove the laws of conservation of linear momentum, angular. Momentum and energy for a system of interacting particles. 2. What are constraints? How do they affect motion of a mechanical system? Distinguish clearly between (a) holonomic and non holonomic constraints (b) Scleronomic and rheonomic constraints. 3. What are Generalised coordinates? What is the advantage of using them? 4. What is meant by configuration space? How this concept is used to describe the motion of a system of particles? 5. Explain the term “virtual displacement and state the principle of “virtual work. 6. State and explain D’Alembert’s principle 7. Derive Lagrange’s equations of motion from D’ Alembert’s principle for a holonomic conservative system. 8. Describe the use of Rayleigh’s dissipation function. 9. State Lagrange’s equations of motion in generalized coordinates.

Apply them to the

Atwood’s machine to find the acceleration of the system. 10. Apply Lagrange’s equation to determine the motion of a bead sliding on a uniformly rotating wire in a force-free space. 11. Define generalized momentum and a cyclic co-ordinate. Shoe that generalized momentum is conserved only when the corresponding generalized coordinate is cyclic or ignorable. 12. Prove that total energy of the system where Lagrangian does not depend upon time explicitly is conserved. 13. Set up the Lagrangian and hence find the equations of motion of the (a) one-dimensional, and (b) two-dimensional oscillator. 14. A particle moves on the surface of a sphere under the influence of gravity. Use Lagrange’s equations to find the differential equations of motion in spherical polar co-ordinates, 15. (a) Prove that total energy of the system where Lagrangian does not depend upon time explicitly is conserved

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(b) Define generalised momentum and a cyclic co-ordinate. Show that generalized momentum is conserved only when the corresponding generalized co-ordinate is cyclic or ignorable. Hence prove that if a cyclic co-ordinate is rotational one, the angular momentum remains conserved. 16. What is meant by a holonomic system.

Obtain Hamilton’s cananocial equations for a

holonomic system? Show that, if the Lagrangian does not depend upto time explicitly, the Hamiltonian is equal to the total energy of the system. 17. Show that, if the Hamiltonian is not an explicit function of time, it is a constant. 18. State and prove Hamilton’s equations of motion and use them to obtain the motion of a projectile launched with speed vo at an angle α with the horizontal. 19. Derive Hamilton’s equations of motion using Legendre transformation method. What is the physical significance of the Hamiltonian? 20. Derive Hamilton’s equations or canonical equations of motion and mention in detail the physical significance of the Hamiltonian H. 21. Obtain the Hamiltonian and Hamilton’s equations for the (a) one-dimensional, (b) two dimensional harmonic oscillator. 22. Discuss the problem of scattering of charged particle by a Coulomb field and obtain Rutherford formula for scattering cross-section. 23. Derive the equation for the orbit of a particle moving under the influence of a central force consistent with the inverse square law and thereby calculate the period of motion in the case of elliptical orbits. 24. Obtain differential equation for a particle undergoing a central force and use it to verify Kepler’s laws. 25. Obtain the equation of trajectory of a particle moving under a force varying as the inverse square of distance of the particle from a fixed centre. Use this equation in analyzing the problem of α-particle scattering from nuclei. 26. State Kepler’s laws of planetary motion and deduce them from Newton’s law of gravitation. 27. Explain how the problems of two bodies, moving under the influence of a mutual central force can be reduced to a one-body problem. Discuss the Kepler problem in detail. 28. What is Lorentz transformation? Show that the four dimensional volume element is invariant under Lorentz transformation. What is the law of addition of relativistic velocities? Calculate the change in the velocity of light propagated through water flowing with velocity v, the refractive index of water being μ? 29. What are four vectors? Find the components of the momentum four vector and derive the law of variation of mass with velocity?

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30. State clearly the Lorentz transformation equations. Explain the physical significance of the parameters involved. Hence find out the laws of the addition of the two velocities u and v in relativity theory. 31. State the basic postulates of the special theory at relativity and derive Lorentz transformation from them. 32. Derive Lorentz transformation equations. State your assumptions clearly. 33. State and explain the basic postulates of the special theory of relativity. Deduce the formula for addition of velocities in relativistic mechanics. Show that when velocity of light is added to the velocity of light, we obtain the velocity of light. UNIT-II Problems 1.

Define Euler angles and derive the Euler’s equations of motion in terms of Euler’s angles.

What are Eulerian angular co-ordinates? Show them in a diagram. Find the condition that a symmetrical top may continue rotating in a vertical position for an indefinite time (sleeping top) (or) Give the theory of symmetric top moving under gravity

What do you understand by the inertia tensor of a rigid body. Discuss the force free motion of a rigid for which I1  I 2  2 I 2 the frequency of precession of the angular velocity about the axis of symmetry?

A rigid through the origin with an angular velocity

 . Obtain relations connecting the total

angular momentum L of the rigid body with angular velocity

 . Hence explain the term

‘inertia’ of a rigid body. 5.

(a) If a rigid body, with out point fixed, rotates with an angular velocity angular momentum L, show that the kinetic energy is



and has an

1 L.  . 2

(b) Derive Euler’s equations of motion for a rigid body with a fixed point 6.

A rigid body is rotating about an axis through the origin. Obtain relations connecting the components of total angular momentum with the components of the angular velocity. Explain what is meant by inertia tensor, and the ‘principle moments of intertia’

Illustrate with diagrams Euler’s angles involved in the transformation from one set of three dimensional coordinate system to another having the same origin. Obtain the complete transformation matrix for such a transformation.

(a) Obtain Euler’s equations of motion for a rotating rigid body.

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(b) Obtain the condition that a heavy symmetrical top in a gravitational field which starts spinning initially with its symmetry axis vertical may continue to spin in the same way for an indefinite period. 9. Define Euler angles and derive the Euler’s equations of motion in terms of Euler’s angles.

UNIT-III Questions 1. What are Poisson and Lagrange brackets? How are they related? Show that the Lagrange bracket is invariant under canonical transformation. 2. (a) Derive equation of motion in terms of Poisson bracket. Give the proof of Jacobi identity. Express Hamilton’s equation in Poisson bracket from (b) Prove that the Poisson bracket of two constants of motion is itself a constant of the motion even when the constants depend on time explicity. 3. Show that it the Hamiltonian and a quantity G are constants of the motion, then

G / t must

be a constant (a)

Derive equations, of motion in terms of Poisson brackets. Prove Jacobi identity

(b)

Show that the Poisson Bracket of nay two functions is invariant under canonical transformations.

4. (a)

Show that the Poisson bracket of any two function F and G does not obey the

commutative law but obeys the distributive law of algebra. (b) It H is the Hamiltonian and f is any function depending on position, momenta and time’ show that

df f   H , f  dt t When 5.



stands for Poisson bracket.

Write note on Poisson brackets.

6. What do you understand by stable and unstable equilibria? Obtain the Lagrange’s equation of motion for small oscillations of a system in the neighbourhood of stable equilibrium. 7. What are normal coordinates? 8. Set up the Lagrangian function and equation for linear triatomic molecules.

UNIT-IV

Questions

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1. State the fundamental assumptions of statistical mechanics which enables us to deduce the macroscopic properties of equilibrium states of an assembly of particles. 2. What is meant by an ensemble? Discuss microcanonical, canonical and grand canonical ensembles? Compare these three types of ensembles. 3. State and prove Liouville’s theorem. or Prove that the density of systems in the neighbourhood of some given system in phase space remains constant in time. 4. State and prove Boltzmann’s theorem connecting entropy and probability. Explain how it can be used to calculate the entropy of a monoatomic gas. 5. What do you mean by partition function? Express Helmholtz free energy and entropy in terms of the partition function. 6. Obtain partition function for Gibb’s canonical ensemble. What is the effect of (a) shifting the zero level of energy (b) decomposition of the system, on the partition function? 7. Obtain expressions of entropy S, specific heat C V, chemical potential μ and Helmholtz free energy F for a perfect monatomic gas on employing (a) microcanonical ensemble, (b) Gibb’s canonical ensemble, and (c) grand canonical ensemble. 8. Write short notes on

(i) Micro-canonical and canonical ensemble (ii) Grand canonical

ensemble (iii) Gibb’s paradox. 9. What is indistinguishability of a particles? What role it plays in quantum statistics? What are its consequences? 10. What is Bose-Einstein statistics? expression. ni 

   Ei

1

What are the basic postulates used? Derive an

for the most probable distribution of the particles of a system

obeying B.E. statistics. 11. What are Fermions? Write down the postulates of Fermi-Dirac statistics.

Derive an

expression for the probability distribution of particles governed by Fermi-Dirac statistics. 12. Applying Fermi-Dirac distribution to an electron gas show that the distribution of energy among free electrons of a conductor is

n( E )dE 

1 2

8 2 V m E dE .    E  h3 e 1 3/ 2

13. What is occupation index. Under what conditions do Bose-Einstein & Fermi-Dirac statistics yield classical statistics. 14. What is the difference between classical and quantum statistics? 15. Distinguish between Classical Statistics, Fermi-Dirac Statistics and Bose-Einstein Statistics. 16. Obtain an expression for Bose-Einstein distribution law. 17. Discuss the postulates of quantum statistics. 18. Compare the basic postulates of M-B, B-E and F-D statistics.

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UNIT-V Questions 1. Write short notes on: a. Planck’s radiation law b. Fermi-Dirac distribution law. c.

Bose-Einstein distribution law.

2. Discuss the free electron gas model for metals. Obtain the distribution law for the electron gas in a metal at absolute zero. 3. Describe the peculiar behaviour of liquid helium 4. Explain why behaviour of liquid helium cannot be explained by classical statistics. How it is overcomed by quantum statistics. 5. What is electron gas? Starting from Fermi-Dirac distribution law, drive the expression for energy distribution of free electrons in a metal. 6. Starting from Bose-Einstein energy distribution law, derive Planck’s law of black body radiation.

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