Chapter 3 Relativistic Kinematics
Chapter 3 3.1 Relativistic Transformation 3.2 Relativistic Transformation II 3.3 Four Vector, Space Time 3.4 Kinematics: Basics 3.5 Interactive Exercises
3.2 Relativistic Transformation II
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Relativistic Momentum Newton's second law, F = dp/dt, keeps its same form under a Galilean transformation, but we might not expect it to do so under a Lorentz transformation. There may be similar transformation with the conservation laws of linear momentum and energy. We need to take a careful look at our previous definition of linear momentum to see if it is still valid at high speeds. According to Newton's second law, for example, an acceleration of a particle already moving at very high speeds could lead to a velocity greater than the speed of light. That would cause a problem with the Lorentz transformation, so we expect that Newton's second law might somehow be modified at high speeds. Because we believe the conservation of linear momentum is a fundamental result, we begin by considering a collision that has no external forces and no accelerations. Frank (Fixed or stationary system) is at rest in system K holding a ball of mass m. Mary (Moving system) holds a similar ball in system K' that is moving in the x direction with velocity v with respect to system K as shown in Figure 1a. Frank throws his ball along his y axis, and Mary throws her ball with exactly the same speed along her negative y' axis.
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System K' according to Mary
System K according to Frank
(a)
(b)
Fig 1: Frank is in the fixed K system, and Mary is in the moving K' system. Frank throws his ball along his +y axis, and Mary throws her ball along her -y' axis The balls collide. The event is shown in Frank;s system in (a) and in Mary's system in (b)
The two balls collide in a perfectly elastic collision, and each twin measures the speed of his or her own ball to be u0 both before and after the collision. We show the collision according to both observers in Figure 1. Consider the conservation of momentum according to Frank as seen in system K. The velocity of the ball thrown by Frank has components in his own K of
uFx = 0
(1)
uFy = u0 Using the definition of momentum from introductory physics, p = mv, the momentum of the ball thrown by Frank is entirely in the y direction:
pFy = mu0 Dayalbagh Educational Institute
(2) 5
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Because the collision is perfectly elastic, the ball returns to Frank with speed u0 along the negative y axis. The change of momentum of his ball as observed by Frank in system K is
(3) In order to confirm the conservation of linear momentum, we need to determine the change in the momentum of Mary's ball as measured by Frank. We will let the primed speeds be measured by Frank (except that is u0 always the speed of the ball as measured by the twin in his or her own system). Mary measures the initial velocity of her own ball to be u'Mx = 0 and u'My = -u0 , because she throws it along her own negative y' axis. In order to determine the velocity of Mary's ball as measured by Frank we need to use the velocity transformation equation (4)
(4)
If we insert the appropriate values for the speeds just discussed, we obtain
(5)
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Before the collision, the momentum of Mary's ball as measured by Frank becomes Before Before
(6)
For a perfectly elastic collision, the momentum after the collision is After After
(7)
The change in momentum of Mary's ball according to Frank is (8) The conservation of linear momentum requires the total change in momentum of the collision, ΔpM = ΔpMy , to be zero. The addition of Equations (3) and (8) clearly does not give zero. Linear momentum is not conserved if we use the conventions for momentum from introductory classical physics even if we use the velocity transformations from the special theory of relativity. There is no problem with the x direction, but there is a problem with the y direction along the direction the ball is thrown in each system. Let us look for a modification of the definition of linear momentum that preserves both it and Newton's second law. Dayalbagh Educational Institute
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Similar to what we did in deriving the Lorentz transformation, we assume the simplest, most reasonable change that may preserve the conservation of momentum. We assume that the classical form of momentum mu is multiplied by a factor that may depend on velocity. Let the factor be Γ(u). Our trial definition for linear momentum now becomes
p = Γ(u) mu
(9)
In following example we show that momentum is conserved in the collision just described for the value of Γ(u) given by (10) Notice that the form of Equation (10) is the same as that found earlier for the Lorentz transformation. We even give Γ(u) the same symbol: Γ(u) = γ. However, this γ is different; it contains the speed of the particle u, whereas the Lorentz transformation contains the relative speed v between the two inertial reference frames. This distinction should be kept in mind because it can cause confusion. Because the usage is so common by physicists, we will use γ for both purposes.
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However, when there is any chance of confusion, we will write out and use for the Lorentz transformation. We will also write out often in order to avoid confusion. We can make a plausible determination for the correct form of the momentum if we use the proper time discussed previously to determine the velocity. The momentum becomes (11) We retain the velocity u = dr/dt as used classically. All observers do not agree as to the value of dr/dt, but they do agree as to the value of dr/dτ, where dτ is the proper time measures in the moving system K'. The value of dt/dτ (= γ). The definition of the relativistic momentum becomes, from Equation (11),
Relativistic momentum
(12)
where (13) This result for the relativistic momentum reduces to the classical result for small values of u/c. The classical expression is good to an accuracy of 1% as long as u< 0.14c. Dayalbagh Educational Institute
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Some physicists like to refer to the mass in Equation (12) as the rest mass m0 and call the term m = Îłm0 the relativistic mass. Tn this manner the classical form of momentum, mu, is retained. The mass is then imagined to increase at high speeds. Most research physicists prefer to keep the concept of mass as an invariant, intrinsic property of an object. We prefer this latter approach and will exclusively use the term mass to mean rest mass. Although we may use the terms mass and rest mass synonymously, we will not use the term relativistic mass. The use of relativistic mass to often leads the student into mistakenly inserting the term into classical expressions where it does not apply. Show that linear momentum is conserved for the collision just discussed and shown in Figure 2.25. Using the relativistic momentum we can modify the expressions obtained for the momentum of the balls thrown by Frank and Mary. From Equation (2), the momentum of the ball thrown by Frank becomes
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For an elastic collision, the magnitude of the momentum for this ball before and after the collision is the same. After the collision, the momentum will be the negative of this value, so the change in momentum becomes, from Equation (3), (14) Now we consider the momentum of Mary's ball as measured by Frank. Even with the addition of the γ factor for the momentum in the x direction, we still have ΔpMx= 0. We must look more carefully at ΔpMy . First, we find the speed of the ball thrown by Mary as measured by Frank. We use Equations (5) to determine (15) The relativistic factor γ for the momentum for this situation is
The value of pMy is now found by modifying Equation (6) with this value of γ.
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Inserting the value of uM from Equation (15) gives (16) The momentum after the collision will still be the negative of this value, so the change in momentum becomes (17) The change in the momentum of the two balls as measured by Frank is given by the sum of Equations (14) and (17) Thus Frank indeed finds that momentum is conserved. Mary should also determine that linear momentum is conserved.
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Relativistic Energy We now turn to the concepts of energy and force. When forming the new theories of relativity and quantum physics, physicists resisted changing the wellaccepted ideas of classical physics unless absolutely necessary. In this same spirit we also choose to keep as many definitions from classical physics as possible and let experiment dictate when we are incorrect. When we studied motion in introductory physics, we learned that the concept of force is in practice defined by its use in Newton's laws. We retain here the classical definition of force, which is best represented be Newton's second law. In the previous section we studied the concept of momentum and found a relativistic expression for momentum in Equation (12). Therefore, we modify Newtonâ&#x20AC;&#x2122;s second law to include our new definition of linear momentum, and force becomes: (18) We introduce the kinetic energy as the work done on a particle by a net (unbalanced) force. We retain the same definitions of kinetic energy and work.
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The work W12 done by a force F to move a particle from position 1 to position 2 along a path s is defined to be (19) where K1 is defined to be the kinetic energy of the particle at position 1. For simplicity, let the particle start from rest under the influence of the force and calculate the kinetic energy K after the work is done. The force is related to the dynamic quantities by Equation (18). The work W and kinetic energy K are (20) where the differential path ds is given by udt. Because the mass is invariant, it can be brought outside the integral. The relativistic factor Îł depends on u and cannot be brought outside the integral. Equation (20) becomes
The limits of integration are from an initial value of 0 to a final value of Îłu. (21)
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The integral in Equation (21) is straightforward if done by the method of integration by parts. The result, called the relativistic kinetic energy, is
(22) Equation (23) does not seem to resemble the classical result for kinetic energy, K = ½mu2. However, if it is correct, we expect it to reduce to the classical result for low speeds. Let’s see if it does. For speeds u << c, we expand γ in a binomial series as follows:
where we have neglected all terms of power ( u/c)4 and greater, because u << c. This gives the following equation for the relativistic kinetic energy at low speeds: (23) which is the expected classical result. One of the most common mistakes that students make when first studying relativity to use either ½mu2 or ½ γmu2 for the relativistic kinetic energy. It is important to use only equation (22) for the relativistic kinetic energy. Dayalbagh Educational Institute
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Although equation (22) looks much different than the classical result, it is the only correct one, and neither ½mu2 or ½ γmu2 is a correct relativistic result. Equation (22) is particularly useful when dealing with particles accelerated to high speeds. For example, the speeds produced have been in the three-kilometer-long electron accelerator at the Stanford Linear Accelerator Laboratory. This accelerator produces electrons with a kinetic energy of 8 x 10-9 J (50 GeV) or 50 x 109 eV. The electrons have velocities so close to the speed of light, that the tiny difference from c is difficult to measure directly. The speed of the electrons is inferred from the relativistic kinetic energy of Equation (22) and given by 0.99999999995c. Such calculations are difficult to do with calculators because of significant-figure limitations. As a result, we use kinetic energy or momentum to express the motion of a particle moving almost at the speed of light. The classical and relativistic speeds for both electrons and protons are shown in figure 2 as a function of their kinetic energy. Physicists have found that experimentally it does not matter how much energy we give a massive body. Its speed can never quite reach c. Let's see if this is consistent with our relativistic energy calculations.
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(a)
2
Classical
0
5 10 15 Kinetic energy (MeV)
Classical
v/c
17
(b)
2
Electrons
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v/c 1
0
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Relativistic
1
0
Electrons
0
10,000 5000 Kinetic energy (MeV)
Fig 2: the velocities (v/c) of (a) electrons and (b) protons are shown versus kinetic energy for both classical (incorrect) and relativistic calculations.
Example 1
Electrons in a television set are accelerated by a potential difference of 25,000 volts before striking the screen. Calculate the velocity of the electrons and determine the error in using the classical kinetic energy result.
Solution 1 The work done to accelerate an electron across a potential difference V is given by qV, where q is the charge of the particle. The work done to accelerate the electron from rest is the final kinetic energy K of the electron. The kinetic energy is given by
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In order to determine the correct speed of the electrons, we must use the relativistically correct kinetic energy given by Equation (22). We will first determine γ; from that, we can determine the speed u. From Equation (22) we have (24) From this equation, γ is found to be (25) The quantity mc2 for the electron is determined to be
mc2 (electron) = (9.11 x 10-31 kg) (3 x 108 m/s)2 = 8.19 x 10-14 J The relativistic factor is now found from equation (25) to be
Equation can be rearranged to determine β2 as a function of γ2, where β = u /c. (26)
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The value of β is 0.30, and the correct velocity, u = βc, is 0.90 x 108 m/s. We determine the error in using the classical result by calculating the velocity using the nonrelativistic expression is K = ½mu2, and the velocity is given by Nonrelativistic
The (incorrect) classical speed is about 4% greater than the (correct) relativistic speed. Such an error is significant enough to be important in designing electronic equipment and in making test measurements. Relativistic calculations are particularly important for electrons, because they have such a small mass, and therefore are easily accelerated to speeds very close to c.
Total energy and rest energy We rewrite Equation (2.58) in the form (27) The term mc2 is called the rest energy and is denoted by E0. Dayalbagh Educational Institute
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Rest energy
(28)
This leaves the sum of the kinetic energy and rest energy to be interpreted as the total energy of the particle. The total energy is denoted by E and is given by
(29)
Equivalence of mass and energy These last few equations suggest the equivalence of mass and energy, a concept attributed to Einstein. The result that energy = mc2 is one of the most famous equations in physics. Even when a particle has no velocity, and thus no kinetic energy, we still believe that the particle has energy through its mass, E0 = mc2. Nuclear reactions are certain proof that mass and energy are equivalent. The concept of motion as being described by kinetic energy is preserved in relativistic dynamics, but a particle with no motion still has energy through its mass. We must modify two of the conservation laws that we learned in classical physics. Mass and energy are no longer two separate conservation laws. We must combine them into one law of the conservation of mass-energy. Dayalbagh Educational Institute
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Even though we often say “ energy is turned into mass” or “mass is converted into energy” or “ mass and energy are interchangeable,” we must understand that what we mean is that mass and energy are actually equivalent. Mass is another form of energy, and we use the mass-energy and energy interchangeably. Consider two blocks of wood, each of mass m and having kinetic energy K, moving toward each other as shown in figure 3.
(a)
(b)
Fig 3: (a) Two blocks of wood, one with a spring attached and both having mass m, move with equal speeds v and kinetic energies K toward a head-on collision. (b) The two blocks collide, compressing the spring, which locks in place. The system now has increased mass, M = 2m + 2K / c2 , with the kinetic energy being converted into the potential energy of the spring.
A spring placed between them is compressed and locks in place as they collide. Let's examine the conservation of mass-energy. The energy before the collision is Mass-energy before:
E = 2mc2 + 2K
(30)
and the energy after the collision is Mass-energy after:
E = Mc2
(31)
where M is the (rest) mass of the system. Dayalbagh Educational Institute
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Because energy is conserved, we have E = 2mc2 + 2K = Mc2, and the new mass M is greater than the individual masses 2m. The kinetic energy went into compressing the spring, so the spring has increased potential energy. Kinetic energy has been conserved into mass, the result being that the potential energy of the spring has caused the system to have more mass. We find the difference in mass ∆M by setting the previous two equations for energy equal and solving for ∆M = M – 2m. (32) Linear momentum is conserved in this head-on collision. The fractional mass increase in this case is quite small and is given by f = ∆M/2m. If we use Equation (32), we have (33) For typical masses and kinetic energies of blocks of wood, this fractional increase in mass is too small to measure. For example, if we have blocks of wood of mass 0.1 kg moving rapidly at 10 m/s, Equation (33) gives
Where we have used the non-relativistic expression for kinetic energy because the speed is so low. Dayalbagh Educational Institute
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This absurdly small answer shows us that questions of mass increase are inappropriate for macroscopic objects like blocks of wood and automobiles crashing into one another. Such small increases cannot even be measured, but in the next section, we will look at the collision of two high-energy protons, in which considerable energy is available to create additional mass. Mass-energy relations will be essential in such reactions.
Relationship of energy and momentum Physicists believe that linear momentum is more fundamental concept than kinetic energy. There is no conservation of kinetic energy, whereas the conservation of linear momentum is inviolate as far as we know. A more fundamental result for the total energy in equation (29) might include momentum rather than kinetic energy. Let's proceed to find a useful result. We begin with Equation (12) for the relativistic momentum written in magnitude form only.
We square this result, multiply by c2, and rearrange the result.
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We use Equation (2.62) for β2 and find
The first term on the right-hand side is just E2, and the second term is E02. The last equation becomes We rearrange this last equation to find the result we are seeking, a relation between energy and momentum. (34) or (35) Equation (34) is a useful result to relate the total energy of a particle with its momentum. The quantities (E2 – p2c2) and m are invariant quantities. Note that when a particle’s velocity is zero and it has no momentum, Equation (34) correctly gives E0 as the particle’s total energy.
Mass less Particles Equation (34) can also be used to determine the total energy for particles having no mass.
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For example, Equation (34) predicts that the total energy of a photon is
E = pc
Photon
(36)
The energy of a photon is completely due to its motion and not at all to its rest energy. We can show that the previous relativistic equations correctly predict that the speed of a photon must be the speed of light c. We use Equations (29) and (36) for the total energy of a photon and set the two equations equal.
E = γmc2 = pc If we insert the value of the relativistic momentum from Equation (12) we have
γmc2 = γmuc The fact that u = c follows directly from this equation after careful consideration of letting m → 0 and realizing that γ → ∞.
u=c
Massless particle
(37)
The rest energy of the proton is given by
The rest energies of the elementary particles are usually quoted in MeV or GeV.
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Example 2 A 2-GeV proton hits another 2-GeV proton in a head-on collision. (a) Calculate v, β, p, K, and E for each of the protons. (b) What happens to the kinetic energy?
Solution 2 (a) By the convention just discussed, a 2-GeV proton has a kinetic energy of 2 GeV. We used Equation (29) to determine the total energy.
E = K + E0 = 2GeV + 938 MeV = 2.938 GeV where we have used 938 MeV for the proton’s rest energy. The momentum is determined from Equation (34).
The momentum is calculated to be
In order to find β we need the relativistic factor γ. There are several ways to determine γ; one is to compare the rest energy with the total energy. From Equation (29) we have ,
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We use Equation (26) to determine β.
The velocity of a 2-GeV proton is O.95c or 2.8 X 10 8 m/s. (b) When the two protons collide head-on, the situation is very similar to the case when the two blocks of wood collided head-on with one important exception. The time for the two protons to interact is less than 10-20 s. If the two protons did momentarily stop at rest, then the two-proton system would have its rest mass increased by an amount given by Equation (32), 2K/c2 or 4 GeV/c2. The result would be a highly excited system. In fact, the collision between the protons happens very quickly, and many possibilities exist. The two protons may either remain or disappear, and new additional particles may be created. Two of the possibilities are
p+p→p+p+p+p
(38)
p + p → π+ + p
(39)
where the symbols are p (proton), p (anti-proton), π (pion), and d (deuteron). We will study more about the possibilities later when we study nuclear and particle physics. Whatever happens must be consistent with the conservation laws of charge, energy, and momentum, as well as with other conservation laws to be learned. Dayalbagh Educational Institute
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Such experiments are routinely done in particle physics. In the analysis of these experiments, the equivalence of mass and energy is taken for granted.
Reconstructing Events Consider a particle decaying into two other particles Fig 4
B A C Particle A may live for a very short amount of time before decaying, and we will only observe B and C in our detectors e.g. Δ++ → p + π+ with a lifetime of 10-24s We can still infer the existence of particle A using special relativity and the conservation of 4-momentum. Since (for the 4-vectors of these particles) PA = PB + PC when we measure PB and PC we can calculate PA. Remembering that
Hence, we can calculate the mass of the decaying particle (taking into account background events). Clearly this can be generalized to N particles. Dayalbagh Educational Institute
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Reconstructing Events: Example Imagine we undertake a particle physics experiment, and are able to measure the energy and momenta of a proton (p) and pion (π+). The 4-momentum of each is (E/c, Px Py , Pz) p = (1.76, 1.48, - 0.17, 0.0) = (0.24, 0.10, 0.17, 0.0) in units of GeV/c. Summing these two 4-vectors gives P + π+ = (2.00, 1.58, 0.0, 0.0) This must be the 4-momentum of the particle which decayed to give the p and π+ and so the rest mass is √2.02 - 1.582 = 1.23 GeV/c2 The mass of the decaying particle was therefore 1.23 GeV/c2, corresponding to the Δ+ + particle.
Example 3 A particle with rest mass m1 = 5x10-20 kg collides with a particle of rest mass m2 = 7 x 10-20 kg. (fig. (a)). After the collision they go off as a single particle, which has a rest mass m3 (fig (b)). Find (I) m3 and (II) its velocity u3 after the collision. (III) Verify that momentum p' and total energy E' are conserved for an observer in reference frame S' traveling with a velocity ν = 3c/5 to the right relative to S. U1 = 12c/13
m1
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m2 29
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u3=? m3
Fig (b) 3c/5
21c/29
m2
m1
m3 Fig (c)
Solution 3 a) For an observer in reference frame S, the particle with rest-mass m2 is at rest and has no momentum of the system is possessed by the particle of rest-mass m1. Since
Before the collision = = The total energy before collision is
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=
= Since m3 has the total energy momentum of the system after the collision,
= kg Note that, b)
c) The velocities of the particles, as measured by observers in reference frame S' (fig:c), are:
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The momentum of m1 before the collision is :
x
The momentum of m2 before the collision is :
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The momentum of m3 after the collision is p'3= 0, since u'3= 0. Thus momentum, is conserved in reference frame S'. Before the collision,
After the collision
Example 4 A stationary body explodes into two fragments each of mass 1.0kg that move apart at speeds of 0.6c relative to the original body.
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Solution 4: The rest energy of the original body must equal the sum of the total energies of the fragments. Hence
and
Example 5: Consider Earth as reference frame S, and rocket 1 as reference frame S'. Rocket 2 moves with speed u'=0.60c with respect to Earth. The velocities are along the same straight line which we take to be the x (and x') axis. Calculate the speed of rocket 2 with respect to Earth.
Solution 5:
Using the equation
The speed of rocket 2 with respect to Earth is Dayalbagh Educational Institute
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