C348 General Relativity Lecture notes 1-2 (UCL) by ucaptd three

General Relativity

Lecture Notes C348

c Mitchell A Berger Mathematics University College London 2004

Contents 1 Manifolds, Vectors, and Gradients 1.1 Manifolds . . . . . . . . . . . . . . . . . 1.1.1 Product Manifolds . . . . . . . . 1.2 Co-ordinate Transformations . . . . . . 1.3 Things that Live on Manifolds . . . . . 1.3.1 Scalar fields . . . . . . . . . . . . 1.3.2 Curves . . . . . . . . . . . . . . . 1.3.3 Parametrized Surfaces . . . . . . 1.3.4 Vectors . . . . . . . . . . . . . . 1.3.5 Gradients . . . . . . . . . . . . . 1.4 Transformation Laws . . . . . . . . . . . 1.4.1 Vectors . . . . . . . . . . . . . . 1.4.2 Gradients . . . . . . . . . . . . . 1.4.3 Notation . . . . . . . . . . . . . . 1.5 Duality between Vectors and Gradients 1.5.1 The directional derivative . . . . 1.5.2 Invariance . . . . . . . . . . . . . 1.5.3 Geometric Interpretation . . . .

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2 Tensors and Metrics 2.1 Tensors . . . . . . . . . . . . . . . . . . . . . 2.2 Getting the transformation correct . . . . . . 2.3 Things to do with tensors . . . . . . . . . . . 2.4 The Leviâ&#x20AC;&#x201C;Civita Tensor . . . . . . . . . . . . 2.4.1 Odd / Even Permutations . . . . . . . 2.5 3-Vector Identities . . . . . . . . . . . . . . . 2.6 Metrics . . . . . . . . . . . . . . . . . . . . . 2.7 Euclidean Metrics . . . . . . . . . . . . . . . 2.7.1 E 3 Euclidean 3-space . . . . . . . . . 2.7.2 E 3 : Cylindrical Co-ordinates . . . . . 2.8 Arc Length . . . . . . . . . . . . . . . . . . . 2.9 Scalar Product and Magnitude for Vectors . . 2.10 Raising and Lowering Operators . . . . . . . 2.11 Signature of the Metric . . . . . . . . . . . . 2.12 Riemannian and Pseudo-Riemannian metrics i

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1 1 4 5 5 5 5 6 7 8 9 9 9 10 14 14 15 15

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16 16 17 18 20 21 21 23 24 25 26 26 27 28 29 29

2.13 Map Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13.1 Cylindrical Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13.2 Mercator Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Special Relativity 3.1 Minkowski Space-time . . . . . . . . . . . . 3.2 Units . . . . . . . . . . . . . . . . . . . . . . 3.3 Einstein’s Axioms of Special Relativity . . . 3.4 Space-Time Diagrams . . . . . . . . . . . . 3.5 The Poincar´e and Lorentz Groups . . . . . 3.5.1 Group Axioms . . . . . . . . . . . . 3.6 Lorentz Boosts . . . . . . . . . . . . . . . . 3.6.1 Deriving the transformation matrix . 3.7 Simultaneity . . . . . . . . . . . . . . . . . . 3.8 Length Contraction . . . . . . . . . . . . . . 3.9 Relativistic Dynamics . . . . . . . . . . . . 3.9.1 The 4-momentum . . . . . . . . . . 3.9.2 Forces . . . . . . . . . . . . . . . . . 3.9.3 Energy-Momentum Conservation . . 3.9.4 Photons . . . . . . . . . . . . . . . .

30 30 31

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34 34 35 37 37 41 42 43 43 46 46 48 48 49 50 50

4 Maxwell’s Equations in Tensor Form 4.1 Maxwell’s Equations – Review . . . . . . . . . . 4.1.1 Internal Structure Equations . . . . . . . 4.1.2 Source Equations . . . . . . . . . . . . . . 4.1.3 Lorentz Force . . . . . . . . . . . . . . . . 4.1.4 Charge Conservation . . . . . . . . . . . . 4.2 The Faraday Tensor . . . . . . . . . . . . . . . . 4.3 Internal Structure Equations . . . . . . . . . . . 4.4 Source Equations . . . . . . . . . . . . . . . . . . 4.5 Charge Conservation . . . . . . . . . . . . . . . . 4.6 Lorentz Force . . . . . . . . . . . . . . . . . . . . 4.7 Potential Form . . . . . . . . . . . . . . . . . . . 4.7.1 Advantage – Internal Structure Equations 4.7.2 Advantage – Source Equations . . . . . . 4.8 Gauge Transformations . . . . . . . . . . . . . . 4.9 Lorentz Gauge . . . . . . . . . . . . . . . . . . . 4.10 Light Waves . . . . . . . . . . . . . . . . . . . . .

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52 52 52 53 54 54 55 55 56 57 57 58 58 58 59 59 60

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62 62 63 64 64 65 65

5 The Equivalence Principle 5.1 Inertial mass . . . . . . . . . . 5.2 Free Fall . . . . . . . . . . . . . 5.2.1 Locally Inertial Frames 5.3 Geodesics . . . . . . . . . . . . 5.3.1 Examples . . . . . . . . 5.3.2 The Geodesic Equation

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5.3.3

Covariant Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 Covariant Derivatives 6.1 Non-Euclidean Geometry . . . . . . . . . . . . . . . . . 6.2 The Covariant Derivative . . . . . . . . . . . . . . . . . 6.3 Derivatives of Other Tensors . . . . . . . . . . . . . . . 6.3.1 The gradient of the metric in General Relativity 6.4 Covariant Directional Derivatives and Acceleration . . . 6.5 Newton’s Law of motion . . . . . . . . . . . . . . . . . . 6.6 Twin Paradox . . . . . . . . . . . . . . . . . . . . . . . . 6.6.1 The rapidity . . . . . . . . . . . . . . . . . . . . 7 Orbits 7.1 Noether’s Theorem . . . . . . . . . . . . . . . 7.2 The Schwarzschild Metric . . . . . . . . . . . 7.2.1 Symmetries and Conserved Quantities 7.2.2 Orbits in the Equatorial Plane . . . . 7.3 Precession of Mercury’s Orbit . . . . . . . . . 7.3.1 Method . . . . . . . . . . . . . . . . . 7.3.2 Newtonian Solution . . . . . . . . . . 7.3.3 Relativistic Correction . . . . . . . . . 7.4 Deflection of Starlight . . . . . . . . . . . . . 7.4.1 Newtonian Theory . . . . . . . . . . . 7.4.2 Relativistic Theory . . . . . . . . . . . 7.5 Energy Conservation on Geodesics . . . . . .

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Chapter 1

Manifolds, Vectors, and Gradients ... ‘You must follow me carefully. I shall have to controvert one or two ideas that are almost universally accepted. The geometry, for instance, they taught you at school is founded on a misconception.’ ‘Is not that rather a large thing to expect us to begin upon?’ said Filby ....

The Time Machine, HG Wells 1895

1.1

Manifolds

Mathematics provides a common mathematical term for curved surfaces, curved spaces, and even curved space-times – the manifold. In essence, a manifold is an N-Dimensional surface. This means that each point of the manifold can be located by specifying N numbers or coordinates. More formally, Definition 1.1 Manifold • A manifold M is a set of points which can locally be mapped into RN for some N = 0, 1, 2, . . . . The number N will be called the dimension of the manifold. • The mapping must be one-to-one. • If two mappings overlap, one must be a differentiable function of the other. Example 1.1 Let M be a two dimensional surface. Suppose we have a point P and we wish to say where this point is. We can do this by specifying two coordinates x and y for P . In figure 1.1 P is mapped to (x, y) = (0.5, 1.5), i.e. P is given the co-ordinates (0.5, 1.5). Note that the lines of constant x never cross each other (similarly for the lines of constant y) – if they did, then the coordinates of the crossing point would not be unique. Example 1.2 Suppose we try to map the same surface M using polar coordinates r and φ. Here we run into difficulties: • The φ function is not single-valued (one-to-one) at r = 0. Any coordinate pair (r, φ) = (0, φ) refers to the origin. • The mapping is not continuous for all values of φ as angles only have a range of 2π. – we must cut the plane, say at the φ = π or − π line. Then φ will not be continuous across this line. 1

Manifolds, Vectors, and Gradients

Figure 1.1: The manifold in example 1.1.

Figure 1.2: The plane divided into two regions. Polar coordinates work in Region I, even if they do not work in all space.

1.1 Manifolds

Figure 1.3: Spherical coordinates.

Note that polars are still well-defined in regions which avoid the origin and the cut. In the figure, polar coordinates are single valued and continuous in region I, but not in the entire plane. We remedy the situation by using a different co-ordinate system in region II. (For example, we could employ Cartesian coordinates in region II.) Example 1.3 M = S 1 (circle). There is one coordinate φ which can be chosen to go from −π to π. The point mapped to φ = π also maps to φ = −π. Thus we will need at least two co-ordinate patches. To summarize: we cover a manifold M with co-ordinate patches. In each patch, the coordinates form a cross-hatch. The minimum number of patches needed depends on the topology of M . In the previous example, M = S 1 needs two patches. A manifold M = R2 consisting of the entire plane needs only one (using Cartesian coordinates rather than polars). Example 1.4 The 2-sphere M = S 2 . Spherical coordinates do not cover the entire sphere in a one-to-one continuous fashion; azimuthal angle φ is many-valued at the poles and is discontinuous at φ = ±π. On a globe of the Earth, φ corresponds to longitude, which is discontinuous at the international date line drawn near (±180o ).

Exercise 1.1 Cover the 2-sphere S 2 with just two coordinate patches. Example 1.5 S n (n-sphere)

n o S n = (X1 , X2 , . . . , Xn+1 ) | (X1 )2 + (X2 )2 + · · · + (Xn+1 )2 = 1

(1.1)

Manifolds, Vectors, and Gradients

n o S 2 = (x, y, z) | x2 + y 2 + z 2 = 1 n o S 1 = (x, y) | x2 + y 2 = 1 n o S 0 = x | x2 = 1 n o = −1, 1

e.g.

Example 1.6 B n (n-ball) n o B n = (X1 , X2 , . . . , Xn ) | (X1 )2 + (X2 )2 + · · · + (Xn )2 < 1

(1.2)

(1.3)

i.e B n is the solid volume inside S n−1 .

1.1.1

Product Manifolds

Example 1.7 M = T 2 (2-Torus). The 2-torus can be represented as the surface of a doughnut. Let θ represent angle the short way around, and φ angle the long way around. Both these coordinates are discontinuous at ±π. M =T 2

We can also represent T 2 as a rectangular box (slit and open out a torus into a rectangular shape). The edges are co-existent (θ = π is equivalent to θ = −π, similarly for φ).

Thus we can write T 2 as the set of points n T n = (θ, φ) | −π ≤ θ ≤ π,

o −π ≤ φ ≤ π .

(1.4)

1.2 Co-ordinate Transformations

Note that in the definition of T 2 , the coordinates θ and φ are completely independent. By themselves, each gives a circle S 1 . We can express this independence with the notation T 2 = S 1 × S 1 . We call T 2 a product manifold, as it can be generated by considering all combinations of θ (the first S 1 ) and φ (the second S 1 ).

1.2

Co-ordinate Transformations

Suppose we know the coordinates of points in a manifold in one coordinate system. We may need to be able to find the same points in other coordinate systems as well, for various reasons. Somebody we work with may use another system, so there will be a communication problem unless we can translate between systems. Or perhaps the equations we wish to solve are easier in a different system. Also, if we know general ways of going from one system to another, we can make sure that our solutions work independently of which particular coordinate system we use. We will only deal with coordinate systems which are differentiable functions of each other. Suppose point P has co-ordinates X = (X1 , X2 , . . . , Xn ) in one co-ordinate system, and Y = (Y1 , Y2 , . . . , Yn ) in another. Then the co-ordinates (Y1 , Y2 , . . . , Yn ) must be smooth (differentiable) functions of (X1 , X2 , . . . , Xn ). For example, Y1 = Y1 (X1 , X2 , . . . , Xn ).

(1.5)

Also, the co-ordinate (Jacobian) transformation matrix is ∂Ya , ∂Xb

(1.6)

where a labels the row, and b labels the column in the matrix.

1.3

Things that Live on Manifolds

1.3.1

Scalar fields

Definition 1.2 Scalar Fields Scalar fields are functions which assign numbers to points on the manifold. More formally, a scalar field is a function f which maps a manifold M to the set of real numbers: f : M → R. Example 1.8 M = Surface of the Earth f (P ) = Temperature at point P (→ weather map) N.B. We may wish to use complex functions ψ : M → C, for example to describe quantum wave-functions.

1.3.2

Curves

The scalar fields defined above map the manifold M to the Real line R. Suppose we now do the reverse. For each real number λ, we will obtain a point γ(λ) on M . If we string these points together, we will get a curve on M .

Manifolds, Vectors, and Gradients

Figure 1.4: A curve γ : R → M Example 1.9 Suppose M = R3 , infinite three-dimensional space. Let γ(λ) = (7 cos 3λ, 7 sin 3λ, λ).

(1.7)

This curve has the shape of a helix. The helix winds around the z axis at a radius of 7, and makes a complete turn each time λ increases by 2π/3. Definition 1.3 Curves A curve is a mapping of the real line (or part of the real line, or a circle) into the manifold M . Formally this is written γ : R → M for the whole real line, or γ : [0, 1] → M (unit interval), or γ : S 1 → M (Circle).

1.3.3

Parametrized Surfaces

Most manifolds cannot be visualized, especially if their dimension is much greater than 3. Fortunately, one and two-dimensional manifolds provide many useful visual examples. We must first be able to imbed, or place, the manifold in ordinary 3-space R3 . A one-dimensional manifold can be drawn using a curve with one parameter λ as in the previous section. For two-dimensional manifolds (surfaces) imbedded in 3-space, we can specify the surface by giving two parameters. Example 1.10 Drawing a sphere. We will use spherical polar coordinates θ and φ as parameters, accepting that there will be coordinate singularities at the poles (and a discontinuity at φ = ±π). The imbedding satisfies S :(θ, φ) → (x, y, z) x = x(θ, φ) = sin θ cos φ y = y(θ, φ) = sin θ sin φ

(1.8)

z = z(θ, φ) = cos θ

Exercise 1.2 Consider an ellipsoid E given by the equation 9x2 + 4y 2 + z 2 = 36.

(1.9)

Find a parametrization E : (θ, φ) → (x, y, z) which satisfies this equation. That is, find the functions x(θ, φ), y(θ, φ), and z(θ, φ).

1.3 Things that Live on Manifolds

Exercise 1.3 Here we find a parametrization of the Northern half of a sphere which does not use spherical coordinates: let the two parameters be t and u, where S :(t, u) → (x, y, z) x = x(t, u) = t

(1.10)

y = y(t, u) = u z = z(t, u) = ? Find the function z(t, u).

1.3.4

Vectors

A vector has: • A Magnitude • A Direction • A Base Point N.B. Forget “position vectors”. Position is given by co-ordinates, not by vectors. In flat Euclidean space, one can draw an arrow from the origin to any point, and in some elementary books this is called a vector. We will not do this, as we may wish to study highly warped manifolds where such arrows will also need to be warped (and hence their direction and magnitude will not be well defined). Vectors in differential geometry always have a base point, and give a direction and magnitude proceeding from that point. Start with a curve γ : R → M (or [0,1], or S 1 → M ). The curve provides a set of n coordinate functions of λ, i.e. γ(λ) = X1 (λ), . . . , Xn (λ) . (1.11) These n functions have derivatives which show how fast they increase with λ. Taken together, they show the direction the curve is travelling. Definition 1.4 Tangent Vectors The tangent vector to the curve γ is given by  dX1 / dλ   .. V(λ) =   . . n dX / dλ 

(1.12)

Manifolds, Vectors, and Gradients

Note that the set of coordinates of a point (e.g. (X1 , . . . , Xn )) is not a vector!

1.3.5

Gradients

Gradients are formed from scalar functions. Definition 1.5 Gradients Given a function f : M → R, ∇f =

∂f ∂f ∂f , ,..., 1 2 ∂X ∂X ∂Xn

(1.13)

This is different to a tangent vector. For one thing, it is determined everywhere on M , whereas a tangent vector is only defined on a single curve. Example 1.11 Consider an ordinance survey map, giving height h as a function of position on the Earths surface h : S 2 → R. The contours are lines of constant h. But note that these are not parametrized curves! (We have not been given a way to choose λ = 0, 1, 2, . . . etc). Thus we have no way of defining tangent vectors to the contours (at least not until we introduce metrics, in §2.6).

Exercise 1.4 Consider a torus T2 with coordinates (X1 , X2 ) = (u, v), where −π < u ≤ π, −π < v ≤ π. Suppose there is a curve γ : R → T2 , where γ(λ) = (2πλ, 6πλ) (i.e. u(λ) = 2πλ and v(λ) = 6πλ)). Find the tangent vector dγ V(λ) = . dλ Consider the function f (u, v) = sin(v). Find the gradient ∇f and the directional derivative V · ∇f .

1.4 Transformation Laws

Figure 1.5: Curve on a torus.

1.4 1.4.1

Transformation Laws Vectors

In co-ordinates X = (X1 , . . . , Xn )  dX1 / dλ   .. VX =  . . n dX / dλ 

(1.14)

In another co-ordinate system Y = (Y1 , . . . , Yn )   dY1 / dλ   .. VY =  . . n dY / dλ

(1.15)

How do we relate VX and VY ? By the chain rule N

dY1 X ∂Y1 dXa = dλ ∂Xa dλ

(1.16)

a=1

∂Y1 dXa a=1 ∂Xa dλ

 PN  ∴ VY = 

1.4.2

.. .

∂YN dXa a=1 ∂Xa dλ

   .

(1.17)

Gradients

Chain rule again. This time, we let c be the ’dummy’ index which is summed from c = 1 to c = N (we can use any letter, of course). N

X ∂f ∂Xc ∂f = ∂Y1 ∂Xc ∂Y1 c=1

(1.18)

Manifolds, Vectors, and Gradients

and so, the transformation is N N X X ∂f ∂Xc ∂f ∂Xc , . . . , c 1 ∂X ∂Y ∂Xc ∂YN

∴ ∇Y f =

c=1

1.4.3

! .

(1.19)

c=1

Notation

A) Einstein’s Summation Convention The placement of indices is important in geometry and relativity. The vectors we have looked at (like V 1 ) have been given indices on top, i.e. superscripts. Meanwhile, the gradients (like ∇1 f ) have indices lower down, i.e. subscripts. This makes it easier to distinguish between them. It also allows us to define consistent rules for calculating inner products. But first we will simplify the notation. In numerous expressions in geometry and relativity (and particle theory as well) there are places where the index a is repeated twice and summed over. If we do not bother to write down the summation sign, then the expressions will be less cluttered. Definition 1.6 Einstein Summation Given two objects, one indexed with superscripts A = (A1 , . . . , AN ), and one with subscripts B = (B1 , . . . , BN ), we define c

Bc A ≡

N X

Bc Ac

(1.20)

c=1

In a derivative like

∂f ∂Xc

the index c will be considered a subscript.

Example 1.12 In the previous section, equation (1.19) becomes ∂f ∂Xc ∂f ∂Xc ∇Y f = ,..., . ∂Xc ∂Y1 ∂Xc ∂YN

(1.21)

B) Co-ordinate system labels We will label Co-ordinate systems by either primed symbols (X0 ) and unprimed symbols (X), or by capital letters; for example Earth F rame E = E1 , . . . , EN Spaceship F rame S = S1 , . . . , SN (1.22) Lab F rame L = L1 , . . . , LN C) Differentials ∂f = ∂a f ∂Xa

∂f 0 a = ∂a f ∂X0

∂f = ∂Ea f. ∂Ea

(1.23)

D) Vector Components In, for example, the lab frame L the tangent vector to the curve

γ(λ) = (L1 (λ), . . . , LN (λ))

(1.24)

 dL1 / dλ dγ   .. V= =  . dλ N dL / dλ

(1.25)



1.4 Transformation Laws

with components

dLa . dλ

VLa =

(1.26)

We sometimes refer to the whole of the vector V by referring to a typical component V a . Similarly, we may refer to ∇f as ∂a f . E) Transforms • Vectors: For transformations between the X frame and the Y frame, equation (1.16) becomes VY1 =

∂Y1 dXa ∂Y1 a dY1 = = V . dλ ∂Xa dλ ∂Xa X

(1.27)

The formula for an arbitrary component of V is ∂Yb a V . ∂Xa X

VYb =

(1.28)

• Gradients: an arbitrary component of ∇f in equation (1.19) can now be expressed as ∂Y b f =

∂Xc ∂X c f ∂Yb

(1.29)

• For transformations between primed and unprimed co-ordinates, these expressions become 0b

∂X = V a. ∂Xa

∂b0 f =

∂Xc ∂X0

∂c f

(1.30)

(1.31)

• Note that the transformation matrices in equations (1.28) and (1.29), i.e. ∂Eb /∂Sa and ∂Sc /∂Eb , are inverses. Proof: The (a, c) component of the product matrix is N X ∂Eb ∂Sc ∂Sa ∂Eb

b=1

N X ∂Sc ∂Eb ∂Eb ∂Sa b=1 ∂Sc

∂Sa

(1.32) (1.33)

by the chain rule. But ∂Sc = ∂Sa i.e.

( 1 0

c=a c 6= a

∂Sc ∂Eb = δca ∂Eb ∂Sa

(1.34)

(1.35)

Manifolds, Vectors, and Gradients

where δca

  1 0 0 ··· 0 1 0 · · ·   = 0 0 1 · · ·   .. .. . .

(1.36)

is the identity matrix. QED This result is very important. Gradients and vectors have different (in fact, inverse) transformation laws. Thus a gradient is not a vector!

Exercise 1.5 Show that for any vector W with components W a , δcaW a = W c.

(1.37)

• How do two successive transformations work? Suppose we transform from co-ordinates X → Y → Z: ∂Za b where V , ∂Yb Y ∂Yb c ⇒ V , ∂Xc X ∂Za ∂Yb c V . ∂Yb ∂Xc X

VZa = VYb = VZa =

(1.38) (1.39) (1.40)

But by the chain rule ∂Za ∂Yb ∂Za = ∂Xc ∂Yb ∂Xc so, (as we should expect) VZa =

(1.41)

∂Za c V . ∂Xc X

(1.42)

Now try X → Y → X: ∂Xa ∂Yb c V ∂Yb ∂Xc X = δ a c VXc

VXa = =

VXa

(1.43) (1.44)

good!

(1.45)

Example 1.13 Polar Co-ordinates. C = (C1 , C2 ) = (x, y) 1

P = (P , P ) = (r, φ)

Cartesian

(1.46)

Polar

(1.47)

where C1 = x = r cos φ = P1 cos P2 2

C = y = r sin φ = P sin P

(1.48) (1.49)

1.4 Transformation Laws

or, going the other way,

p x2 + y 2 = (C1 )2 + (C2 )2 y C2 = φ = arctan = arctan 1 x C

P1 = r =

The transformation matrix (Jacobian) is: 1 ∂Ca ∂Ca ∂C /∂P1 ∂C1 /∂P2 = = ∂C2 /∂P1 ∂C2 /∂P2 ∂ Pb ∂ Pb ∂x/∂r ∂x/∂φ = ∂y/∂r ∂y/∂φ cos φ −r sin φ = sin φ r cos φ

(1.50) (1.51)

(1.52) (1.53) (1.54)

and the other way ∂Pb = ∂Cd

∂r/∂x ∂r/∂y ∂φ/∂x ∂φ/∂y x/r y/r = −y/r2 x/r2 cos φ sin φ = − sin φ cos φ .

(1.55) (1.56) (1.57)

And the two matrices are inverses. Next: Check the transformation law for tangent vectors in polar co-ordinates. Let γ be a circle of radius R, parametrized by λ = φ/(2π). Cartesian coordinates (1.58) γ(λ) = C1 , C2 = (x(λ), y(λ))

(1.59)

= (R cos 2πλ, R sin 2πλ)

(1.60)

Polar Coordinates γ(λ) =

P1 , P2

(1.61)

= (r(λ), φ(λ))

(1.62)

= (R, 2πλ)

(1.63)

Manifolds, Vectors, and Gradients

The tangent vector can be computed separately for each coordinate system: Cartesian coordinates dCa dλ dx/ dλ = dy/ dλ −2πR sin 2πλ = . 2πR cos 2πλ

VCa (λ) =

(1.64) (1.65) (1.66)

Polar coordinates dPa dλ dr/ dλ = dφ/ dλ 0 = . 2π

⇒ VPa (λ) =

(1.67) (1.68) (1.69)

Let’s check the transformation law, using equation (1.54): ∂Ca b (at r = R, φ = 2πλ) V ∂Pb P 0 cos 2πλ −r sin 2πλ = 2π sin 2πλ r cos 2πλ −2πR sin 2πλ = . 2πR cos 2πλ

VCa (λ) =

(1.70) (1.71) (1.72)

IT WORKS!

1.5 1.5.1

Duality between Vectors and Gradients The directional derivative

Two objects in maths are called ‘dual’ if they combine to give a single number (in R or C ). In elementary matrix algebra, row vectors and column vectors multiply to give a number. In quantum theory, a bra vector hψ| and a ket vector |φi combine to give the complex number hψ|φi. On a manifold, we can combine a vector, V, and a gradient, ∇f , naturally by using Einstein summation: V · ∇f = V a ∂a f =

N X

V a ∂a f.

a=1

This sum gives a number called the directional derivative. Why is this ‘natural’ ?

(1.73)

1.5 Duality between Vectors and Gradients 1.5.2

Invariance

Exercise 1.6 Prove from the transformation laws that the directional derivative is the same in all co-ordinate systems. Solution For the transformation from X → X0 , Va =

∂Xa 0b

∂X 0c ∂X 0 ∂a f = ∂f ∂Xa c ! a ∂X0 c ∂X 0 ⇒ V · ∇f = V a ∂a f = V b ∂c0 f 0 b ∂Xa ∂X 0

(1.74)

= ( δ c b ) V b ∂c f 0

= V c ∂c f Thus 0

V · ∇f = V a ∂a f = V c ∂c f.

(1.75)

The directional derivative has the same form in any two co-ordinate systems, and gives the same number.

1.5.3

Geometric Interpretation

Suppose f is the temperature as a function of position. We can measure f (λ) as we travel along γ(λ). We can find df / dλ by the chain rule: N

X ∂f dXa df = dλ ∂Xa dλ a=1

= ∂a f V a

(1.76)

= V · ∇f. In words, if the curve γ(λ) has tangent vector V, then the directional derivative V · ∇f gives the rate of change of f along the curve, df /dλ.

Chapter 2

Tensors and Metrics ‘You know of course that a mathematical line, a line of thickness nil, has no real existence. They taught you that? Neither has a mathematical plane. These things are mere abstractions.’ ‘That is all right,’ said the Psychologist. ‘Nor, having only length, breadth, and thickness, can a cube have a real existence.’ ‘There I object,’ said Filby. ‘Of course a solid body may exist. All real things’ ‘So most people think. But wait a moment. Can an instantaneous cube exist?’ ‘Dont follow you,’ said Filby. ‘Can a cube that does not last for any time at all, have a real existence?’ Filby became pensive. The Time Machine, HG Wells 1895

2.1

Tensors

Tensors are classified by their rank: Rank 0 1

Scalars (functions) Vectors and forms (inc gradients)

Two indices - appear like matrices

3 .. .

Three indices - appear like a stack of matrices .. .

000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

· ··· 000000 111111 000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111

00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

.. .

0th Rank Definition 2.1 Scalars A tensors with no subscripts or superscripts. They are functions of position on the manifold, and are completely independent of coordinates. 1st Rank Definition 2.2 Vectors Any tensor with one superscript which transforms like a tangent vector. In many books these are called contravariant vectors. We give the symbol for a vector V an overline; V has components V a . Definition 2.3 One-forms Any tensor with one subscript which transforms like a gradient. We will usually refer to one-forms simply as forms.We give forms an underline; the form W has components Wa indexed by a subscript. Two-forms will be introduced in section 4.2 (they not only have two subscripts but also must be antisymmetric). Many books refer to forms as covariant vectors or co-vectors.

2.2 Getting the transformation correct

Example 2.1 Suppose W = f ∇g, where f, g are functions. Then, going from X → X0 we have 0

f →f =f

(2.1)

g→g =g (2.2) a 0 ∂X ∂b → ∂b = ∂a (2.3) ∂X0 b a 0 ∂X ⇒ Wb → Wb = Wa (2.4) ∂X0 b NOTE: W has been constructed from a scalar function f and the gradient of another scalar g. However, it may not itself be a gradient – i.e there may not exist a third function h such that W = ∇h. Higher Rank Tensors There are 2 equivalent definitions of 2nd order (and higher) tensors. 1) A tensor with (for example) one superscript and subscript transforms as a vector on the superscript and as a form on the subscript. Example 2.2 Let M be a mixed tensor with components M a b . Then 0

M 0a b =

∂X a ∂Xd c M d ∂Xc ∂X0 b

(2.5)

2) A tensor with 1 superscript and 1 subscript is dual to the product of one vector and one form. Thus given V and W, the components M a b of M must have the following property: the quantity α = M a b V b Wa

(2.6)

is scalar, i.e. the same in all reference frames. In general, a tensor with p upper indices and q lower indices will ‘eat’ p forms and q vectors, and return a scalar. p p tensors A type q tensor has p upper indices and q lower indices. Definition 2.4 Type q The product with q vectors and p forms (summing over all indices) returns a scalar. For example, if T ab c d ef g is a type 43 tensor (7th rank) then µ = T ab c d ef g Aa Bb C c Dd E e F f Gg

(2.7)

is a scalar where A, B, D, G are forms, and C, E, F are vectors.

2.2

Getting the transformation correct

Example 2.3 0

R a b c =?Rd e f Step 1 write three sets of

∂X ∂X

(2.8)

on right: 0

R abc =

∂X ∂X ∂X d f R e ∂X ∂X ∂X

(2.9)

Step 2 Pair up indices a→d b→e c→f

(2.10) (2.11) (2.12)

Tensors and Metrics

Step 3 The a, b, and c indices correspond to primed coordinates. Fill in the a, b, and c 0 indices, together with primes, in the same positions as appear in R a b c : 0

a c b

∂X a ∂X ∂X c d f = R e . ∂X ∂X0 b ∂X

(2.13)

Step 4 Fill in the remaining indices by the pairings from (2): 0

R abc =

∂X a ∂Xe ∂X c d f R e . ∂Xd ∂X0 b ∂Xf

(2.14)

Note that the d, e, and f indices appear once on top and once on bottom, so they are summed over.

Exercise 2.1 How does the tensor Rab cd

(2.15)

transform under a coordinate transformation?

Exercise 2.2 Suppose that in some coordinate system the tensor δ a b has the form δ

( 0 = 1

a 6= b a = b.

if if

(2.16)

Show that it also has this form in any other coordinate system.

2.3

Things to do with tensors

Each of these results in a new tensor: a. Addition Ra b c + S a b c = T a b c

All terms have the same indices!

(2.17)

b. Composition M a b = V a Wb

(2.18)

Example 2.4 V 1 = 1, V 2 = 2, W1 = 20, W2 = 30 gives 30 60

(2.19)

(= 80 in the above example)

(2.20)

M ab =

20 40

c. Contraction between Tensors (Einstein summation) µ = V a Wa d. Contraction within a Tensor P ab bc = Qa c M a a = 80 = µ in the above example For 2nd rank mixed tensors, M a a = T r(M), the trace of M.

(2.21) (2.22)

2.3 Things to do with tensors

e. Symmetrizing and anti-Symmetrizing (for 2nd rank tensors) Given a tensor with components T ab , let

1 ab T + T ab 2 1 ab = T − T ab 2

S ab =

and Aab

(2.23) (2.24)

Then (1) S ab + Aab = T ab (2) S

(2.25)

(3) Aab = −Aba

(Symmetric)

(2.26)

(Antisymmetric)

(2.27)

Example 2.5 T ab = S ab =

0 2

0 3 2

1 3 3 2

0 2 = (T T )ab 1 3 0 − 21 = 1 0 2

T ba = Aab

(2.28) (2.29)

Tip: Watch out for bad tensor expressions: • W b = Ga a V a b . This equation does not make sense because there are three a’s – thus we do not know which two to sum over. • ψ = U aa . 2 a’s on top – summing would not give a scalar. • V c = X a Wa T ca Aa Bad – now there are four a’s. Perhaps this really means V c = X a Wa T cb Ab , but it could also mean V c = X a Wb T cb Aa .

Exercise 2.3 Determine which tensor equations are valid, and describe the errors in the other equations. a. Dab = Ta W a b b. Eab = F c ba Cc + Ld Sdba c. Zmn = Y a m a n d. Pc = J a a K a a Rc e. Aab = Bba + gab Dc Dc f. F c b = Gca Hda g. f = Ja Ka La M a + N b b

Tensors and Metrics

Exercise 2.4 Let space be two dimensional, with coordinates (X1 , X2 ). Suppose tensors with components V a , Wa , P ab , Qab , and M a b are measured to have the values V1 W1 P ab

= 2, V 2 = 3; = 4, W2 = 5; 2 â&#x2C6;&#x2019;1 = ; 3 6 ab

(2.30) (2.31) Qab =

0 4

2 ; 7 ab

M ab =

4 2

3 . 1 ab

(2.32)

Calculate the following tensors: Îą Tb F ac Gab

= = = =

V a Wa ; P ab Wa ; P ab Qbc ; M c b Qca .

(2.33) (2.34) (2.35) (2.36)

Theorem. Let S ab be a symmetric tensor, and Aab an anti-symmetric tensor. Then their double contraction vanishes: Aab S ab = 0. (2.37) Proof. To evaluate the double contraction of Aab and S ab , note that we can exchange the dummy labels a and b. Thus Âľ â&#x2030;Ą Aab S ab = Aba S ba . (2.38) Now use the symmetry and anti-symmetry of S ab and Aab : Âľ = Aba S ba = (â&#x2C6;&#x2019;Aab )(+S ab ) = â&#x2C6;&#x2019;Âľ.

(2.39)

This can only be true if the double contraction Âľ â&#x2030;Ą Aab S ab is zero.

2.4

The Leviâ&#x20AC;&#x201C;Civita Tensor

The Leviâ&#x20AC;&#x201C;Civita tensor is completely antisymmetric. It is actually a tensor density (see section ??). In 2-dimensions: 0 1 ab = â&#x2C6;&#x2019;1 0 In 3-dimensions: ab1

ab2

ab3

ďŁŤ ďŁś 0 0 0 = ďŁ0 0 1ďŁ¸, 0 â&#x2C6;&#x2019;1 0 ďŁŤ ďŁś 0 0 â&#x2C6;&#x2019;1 = ďŁ0 0 0 ďŁ¸, 1 0 0 ďŁŤ ďŁś 0 1 0 = ďŁâ&#x2C6;&#x2019;1 0 0ďŁ¸. 0 0 0

Note that all these matrices are anti-symmetric. In general, in N -dimensions: ďŁą ďŁ´ if any 2 of a, b, c, . . . , N are equal ďŁ˛0 ab...N = 1 if (a, b, c, . . . , N ) is an even permutation ďŁ´ ďŁł â&#x2C6;&#x2019;1 if (a, b, c, . . . , N ) is an odd permutation

(2.40)

(2.41)

(2.42)

2.5 3-Vector Identities 2.4.1

Odd / Even Permutations

A permutation is an re-ordering. e.g. (4, 2, 1, 3) is a permutation of (1, 2, 3, 4). For N = 3, there are 6 permutations: (1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1) We can get from the start (1, 2, 3) to any other permutation by swapping pairs of numbers. Thus swapping 1,2 sends (1,2,3) â&#x2020;&#x2019; (2,1,3), swapping 1,3 sends (2,1,3) â&#x2020;&#x2019; (2,3,1), etc. Similarly, all permutations on N objects can be reached from the identity (1, 2, . . . , N ) by a sequence of swaps. Sometimes two different sequences of swaps will result in the same permutation. But when this happens both sequences will consist of an even number of swaps, or an odd number of swaps. Definition 2.5 Even and Odd Permutations Starting from (1, 2, . . . , N ), an even number of swaps results in an even permutation; an odd number of swaps results in an odd permutation. Example 2.6 N = 4: 1234 = 1 1243 = â&#x2C6;&#x2019;1 2143 = 1

2.5

(2.43)

3-Vector Identities

a. â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2020;&#x2019; A Âˇ B = Ai Bi .

(2.44)

â&#x2C6;&#x2021;f = â&#x2C6;&#x201A;i f .

(2.45)

â&#x2020;&#x2019; i â&#x2C6;&#x2019; â&#x2C6;&#x2021; Ă&#x2014; A = ijk â&#x2C6;&#x201A;j Ak

(2.46)

Example 2.7 Verify equation (2.46) for the y-component of the curl. â&#x2020;&#x2019; â&#x2C6;&#x2019; Solution The y-component of â&#x2C6;&#x2021; Ă&#x2014; A is â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2021; Ă&#x2014; A = 2jk â&#x2C6;&#x201A;j Ak (2nd component = y-component) y

= 231 â&#x2C6;&#x201A;3 A1 + 213 â&#x2C6;&#x201A;1 A3 â&#x2C6;&#x201A; â&#x2C6;&#x201A; = (1) Ax + (â&#x2C6;&#x2019;1) Az â&#x2C6;&#x201A;z â&#x2C6;&#x201A;x â&#x2C6;&#x201A;Ax â&#x2C6;&#x201A;Az = â&#x2C6;&#x2019; . â&#x2C6;&#x201A;z â&#x2C6;&#x201A;x Example 2.8 Show that

Solution Translate to

â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2020;&#x2019; â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; A Ă&#x2014; B = â&#x2C6;&#x2019; B Ă&#x2014; A. â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; i A Ă&#x2014; B = ijk Aj Bk .

(2.47)

Tensors and Metrics but ijk = â&#x2C6;&#x2019; ikj â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; i â&#x2021;&#x2019; A Ă&#x2014; B = â&#x2C6;&#x2019; ikj Aj Bk = â&#x2C6;&#x2019; ikj Bk Aj

(Aj and Bk are just numbers, so we can reverse their order in a multiplication without affecting the result). Now we replace the dummy indices j â&#x2020;&#x2019; k and k â&#x2020;&#x2019; j. They are being summed over, so it does not matter which one is called which: â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; i â&#x2021;&#x2019; A Ă&#x2014; B = â&#x2C6;&#x2019; ijk Bj Ak â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; i =â&#x2C6;&#x2019; BĂ&#x2014;A

â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2020;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2020;&#x2019; â&#x2C6;´ A Ă&#x2014; B = â&#x2C6;&#x2019;B Ă&#x2014; A .

Example 2.9 Show that â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f = 0.

(2.48)

Solution Translate to i

(â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f ) = ijk â&#x2C6;&#x201A;j â&#x2C6;&#x201A;k f = ijk â&#x2C6;&#x201A;k â&#x2C6;&#x201A;j f ikj

= â&#x2C6;&#x2019;

(partial derivatives commute)

â&#x2C6;&#x201A;k â&#x2C6;&#x201A;j f

relabel j â&#x2020;&#x201D; k

â&#x2021;&#x2019; (â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f ) = â&#x2C6;&#x2019; ijk â&#x2C6;&#x201A;j â&#x2C6;&#x201A;k f i

â&#x2021;&#x2019; (â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f ) = â&#x2C6;&#x2019; (â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f ) i

â&#x2021;&#x2019; (â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f ) = 0

â&#x2C6;´ â&#x2C6;&#x2021; Ă&#x2014; â&#x2C6;&#x2021;f = 0

Exercise 2.5 Translate the following 3-vector identities into index notation, and prove them: A Âˇ (B Ă&#x2014; C) = B Âˇ (C Ă&#x2014; A) = C Âˇ (A Ă&#x2014; B); â&#x2C6;&#x2021; Âˇ (f A) = A Âˇ â&#x2C6;&#x2021;f + f â&#x2C6;&#x2021; Âˇ A; â&#x2C6;&#x2021; Âˇ (A Ă&#x2014; B) = B Âˇ â&#x2C6;&#x2021; Ă&#x2014; A â&#x2C6;&#x2019; A Âˇ â&#x2C6;&#x2021; Ă&#x2014; B.

(2.49) (2.50) (2.51)

2.6 Metrics

2.6

Metrics

Definition 2.6 Metric Given two nearby points, X1 , . . . , XN and X 1 + dX1 , . . . , X N + dXN , a distance ds can be defined by introducing a new object, the metric tensor gab . The distance satisfies ds2 = g11 dX1 dX1 + g12 dX1 dX2 + . . . + gN N dXN dXN

(2.52)

ds2 = gab dXa dXb .

(2.53)

N.B. Throughout these notes, ds2 means ( ds)2 , not d(s2 ). Example 2.10 In Euclidean geometry with N = 3 ds2 = or with

dX1

dX2

dX3

ds2 = gij dXi dXj gij = δij (i, j = 1, 2, 3)

2 (2.54)

Let’s decompose gab into symmetric and anti-symmetric parts: gab ds2

= Sab + Aab = (Sab + Aab ) dXa dXb

(2.55) (2.56)

where Sab = Sba and Aab = −Aba . Consider the anti-symmetric contributions to ds2 , Aab dXa dXb . This is the double contraction of an anti-symmetric tensor Aab with a symmetric tensor dXa dXb , so by equation (2.37) Aab dXa dXb = 0.

(2.57)

We have just shown that Aab is useless, and so we get rid of it. Thus we define gab to be symmetric gab = gba .

(2.58)

Tensors and Metrics

How many components does gab have? If the manifold is N dimensional, then an arbitrary tensor of rank r has N r components. But since the metric is symmetric, not all these components will be independent.   g11 g12 . . . g1N  g21 g22 . . . g2N    gab =  . (2.59) ..  .  .. .  gN 1 |

gN 2 N2

...

gN N

}

components

By symmetry (gab = gba ), g11  g12   =  g13  ..  .

g12 g22 g23

g13 g23 g33

... ... ...

 g1N g2N   g3N   = gba ..  . 

g1N

g2N

g3N

...

gN N



⇒ gab

so there are

2.7

N (N +1) 2

(2.60)

independent components.

Euclidean Metrics

E 2 – Euclidean Plane Cartesian: By the Pythagorean theorem, ds2

dC1

dC2

= dx2 + dy 2

(2.61) 2

= gC 11 dx + gC 12 dx dy + gC 21 dy dx + gC 22 dy . So gC 11 = gC 22 = 1 while gC 12 = gC 21 = 0, or gC ab =

Polars: Two methods to find the metric:

1 0

0 . 1

(2.62)

2.7 Euclidean Metrics

Method 1: Draw pictures: dφ a = 2(r + dr) sin 2 dφ = 2(r + dr) 2

(2.63)

= r dφ

(+O( dφ)3 )

(2.64)

(+O( dr dφ)).

(2.65)

Thus ds2 = dr2 + r2 dφ2

(2.66)

or, in terms of the metric tensor ds2 = gab dPa dPb , gP ab =

1 0 0 r2

(2.67)

Method 2: We know that gC ab = ∴ gP ab

1 0

0 1

∂Cc ∂Cd = ∂Pa ∂Pb

1 0

0 . 1 cd

(Cartesians on top (c,d), polars on bottom (a,b)).

2.7.1

E 3 Euclidean 3-space

Cartesian: 

gC ab

1 = 0 0

0 1 0

 0 0 . 1

(2.68)

Spherical: • If we just move in r: ds2r = dr2

(2.69)

• If we just move in θ, we move on a radius r: ds2θ = r2 dθ2 (2.70) • If we just move in φ, we move on a constant radius r sin θ: ds2φ = r2 sin2 θ dφ

(2.71)

Tensors and Metrics

Thus ds2 gP ab

2.7.2

(2.72) (2.73)

E 3 : Cylindrical Co-ordinates

X1 , X2 , X3 = (ρ, φ, z)

(2.74)

ds2 = dρ2 + ρ2 dφ2 + dz 2

(2.75)



⇒ gCyl

2.8

dr2 + r2 dθ2 + r2 sin2 θ dφ2   1 0 0 0 . = 0 r 2 2 0 0 r sin2 θ

1 0 = 0 ρ2 0 0

 0 0 1

(2.76)

Arc Length What is the arc length P → Q? We define Z

ds(λ)

(2.77)

Thus Z

L= P Z Q

= P

√

ds2 (2.78)

p gab dXa dXb

2.9 Scalar Product and Magnitude for Vectors

Example 2.11 E 2 with Cartesian coordinates. Here Z Qp L= dx2 + dy 2 .

(2.79)

We can divide by

√

dλ2 and multiply by dλ to make s 2 2 Z Q dy dx + dλ. L= dλ dλ P

(2.80)

If we are able to choose λ = x as our parameter, so that y = y(x) (i.e. y is a well behaved function of x. See diagram),

111111111111111 000000000000000 x

Good

Bad

then Z

L= P

dy dx

p 1 + cos2 (x) dx

2.9

+ 1 dx

y 1 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 11111111111111 00000000000000 0 1 x 0 1 0 1 0 1 0 1 0 1 0 1 0 1 -1 0 1 0 1 0 1 0 1 0 1

e.g. y = sin x Z

Scalar Product and Magnitude for Vectors

Given a vector, V, with components V a the sum mations, and so it is meaningless!

V a V a is not invariant under co-ordinate transfor-

Tensors and Metrics

But, if we have a metric, we can define V · V = gab V a V b

This is invariant!

(2.81)

The magnitude of V is

V = V · V

(2.82)

V · W = gab V a W b

(2.83)

A · B = g ab Aa Bb

(2.84)

Similarly, with two vectors V, W For two forms A, B where g ab is the inverse of the metric tensor, in the sense that g ab gbc = δ a c .

(2.85)

A = A · A.

(2.86)

Also

2.10

Raising and Lowering Operators

Note that V · W = gab V a W b is a scalar. We can write this as V · W = (gab V a ) W b

(2.87)

Then gab V a is ‘dual’ to the set of vectors – It takes a vector, W, and returns a scalar number. Therefore, given a vector V, we can define a form V where Vb = gab V a .

(2.88)

Similarly V · W = V a gab W b

(2.89)

(2.90)

⇒ V · W = V Wb = Vb W a

(2.91) (2.92)

⇒ Wa = gab W a

By contraction with the metric, we can ‘raise’ or ‘lower’ the index. Exercise 2.6 Suppose that Aab is anti-symmetric. Let Aab = gac gbd Acd where gac is the (symmetric) metric tensor. Show that Aab is antisymmetric as well.

Exercise 2.7 Let T

2 5

3 . 7

Find its symmetric and antisymmetric parts Aab and S ab . Next, let the metric be 1 2 gab = . 2 1

(2.93)

(2.94)

Find Aab . Show explicitly that Aab S ab = 0.

(2.95)

2.11 Signature of the Metric

2.11

Signature of the Metric

One can show, using linear algebra, that any metric, gab , can be diagonalized by transforming to a suitable co-ordinate system. e.g. If in coordinates X, the metric is −3 −5 gab = −5 62 Then there exists coordinates X0 where 0 gab

1 0 0 −1

In general, we can make the values on the diagonal 1 or -1 by transforming to suitable co-ordinates • Find eigenvectors and eigenvalues of gab • Transform to co-ordinates along the eigenvectors (diagonalize metric) • divide each eigenvector by it’s eigenvalue (all go to 1,-1) There may be a few ways of doing this. However, it can be shown that the sum of the diagonal elements will always be the same. This is called the signature of the metric. Example: E 3 with Cartesian coordinates

gab

 1 = 0 0

0 1 0

 0 0 1

(2.96)

Signature = 3. Example: Minkowski Metric  1 0 0 0 0 −1 0 0  = 0 0 −1 0  0 0 0 −1 

gab = ηab

(2.97)

Signature = -2.

2.12

Riemannian and Pseudo-Riemannian metrics

Is the magnitude of a vector always positive? Definition 2.7 Riemannian metric For a Riemannian metric, any vector V 6= 0 satisfies ||V||2 = V · V > 0.

(2.98)

A metric which is not Riemannian is called Pseudo-Riemannian. • Signature = N metrics are Riemannian. To see this note that we can calculate V · V (at some point P ) in any coordinate system, including the system with only ±1 diagonal elements for the metric. Then V · V = gab V a V b = ±(V 1 )2 ± (V 2 )2 ± ... ± (V N )2 . If signature = N then there will be only pluses, and the sum will be positive.

(2.99)

Tensors and Metrics

• Signature < N metrics are pseudo-Riemannian. On the other hand, if the signature is less than N , then there will be some minuses in the sum. Say for example that V · V = (V 1 )2 + (V 2 )2 + · · · + (V n−1 )2 − (V N )2 . Then we could choose a vector V whose only non-zero element is V N . This would make V · V < 0. Euclidean metrics are Riemannian, while the Minkowski metric is pseudo-Riemannian.

2.13

Map Projections

The Earth, radius r = R = constant, has the metric line element (in spherical coordinates S = (θ, φ)) ds2 = R2 dθ2 + sin2 θ dφ2 , (2.100) with 0 ≤ θ ≤ π, −π ≤ φ ≤ π. ⇒ gS ab = R

2.13.1

1 0

0 sin2 θ

(2.101)

Cylindrical Map Paper

Project out onto paper

Globe

y y

0 x

We wish to project the Earth onto a piece of paper of width w and height h. The co-ordinates on the map will be M = (x, y) with

x = y

w φ 2π h cos θ 2

(2.102) (2.103)

Rcos θ

Let us call the metric for the map projection gM ab . Using this metric we can calculate the arclength of any path drawn on the map (from Buenos Aires to Glasgow for example). The calculation will give the distance along the corresponding path on the Earth’s surface.

2.13 Map Projections

Problem: find the metric gM ab . Method 1: Use the transformation law for tensors: ∂Mc ∂Md g cd . ∂Sa ∂Sb S Method 2: Directly transform the metric line element: first,

(2.104)

gM ab =

= R2 dθ2 + sin2 θ dφ2 2π dφ = dx w 4π 2 ⇒ ds2 = R2 dθ2 + 2 sin2 θ dx2 . w ds2

(2.105) (2.106) (2.107)

Next, from the formula for y, dy ⇒ dθ2

h = − sin θ dθ 2 4 dy 2 = . h2 sin2 θ

(2.108) (2.109)

Also, sin2 θ ⇒ dθ2 = Thus

1−

2y h

4 dy 2 . − 4y 2

(2.110) (2.111)

4π 2 h2 − 4y 2 4 2 2 ds = R dx + 2 dy . (2.112) w2 h2 h − 4y 2 Changing the aspect ratio w/h will stretch or compress the map in the vertical direction. We should choose this ratio to give the least distortion. Note that at y = 0 (the equator) the y dependence disappears. The metric will be symmetric in x and y at the equator if w = πh. In this case, the line element and metric are 2 h − 4y 2 1 2 2 2 2 ds = 4R dx + 2 dy . (2.113) h4 h − 4y 2 2

gab = 4R2

2.13.2

h2 −4y 2 h4

0 1 h2 −4y 2

Mercator Projection

Idea: Align compass bearings to constant directions on the map.

! .

(2.114)

Tensors and Metrics ds2 = a2 + b2 a = R sin θ dθ b = −R dθ ⇒ ds2 = R2 w Again, let x = 2π φ. Choose y so that the slope of

dy dx

(2.115) (2.116) (2.117)

dθ2 + sin2 θ dφ2

(2.118)

of a curve is a constant for constant ψ: −R dθ 1 dθ =− . R sin θ dφ sin θ dφ

cot ψ =

(2.119)

On the map, however, cot ψ

dy dx

(2.120)

w 1 dθ 1 dθ =− sin θ dφ 2π sin θ dx w 1 = − 2π sin θ

= − ⇒

dy dθ

(2.121) (2.122)

After integration, Z

dy 0 = −

∴y=

w 2π

Z θ

dθ sin θ

w θ log cot 2π 2

We can now find gM ab . First, we derive a simple identity: let 2π θ ye ≡ y = log cot . w 2 Then sin θ =

1 = sech ye. cosh ye

Proof: 2 eye + e−ey 2 = cot θ/2 + tan θ/2 2 = cot θ/2 1 + tan2 θ/2 2 = cot θ/2 (sec2 θ/2) = 2 cos θ/2 sin θ/2.

sech ye =

Now ds2 = R2 dθ2 + sin2 θ dφ2 2 2π = R2 (− sin θ dy)2 + sin2 θ dx2 w

(2.123)

(2.124)

2.13 Map Projections

Thus we have found the Mercator line element: ds2 =

2πR w

sech 2 ye dx2 + dy 2 ,

(2.125)

and the Mercator metric: gM ab =

2πR w

sech 2 ye

1 0

0 . 1

(2.126)

Exercise 2.8 London has latitude 51◦ and longitude 0◦ . Suppose you travel on the surface of the earth, following a geodesic (great circle) which leaves London in the direction due East. Where will you first hit the equator? (This problem can be done without any equations!)