Solutions to Problems from Wald’s Book“General Relativity” Lucas Braune1 September 29, 2010 1 E-mail address: [email protected] 2 Contents 1 Introduction Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 2 Manifolds and Tensor Fields Problem 1 . . . . . . . . . . . . . Problem 2 . . . . . . . . . . . . . Problem 3 . . . . . . . . . . . . . Problem 4 . . . . . . . . . . . . . Problem 5 . . . . . . . . . . . . . Problem 6 . . . . . . . . . . . . . Problem 7 . . . . . . . . . . . . . Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 7 8 8 10 11 13 14 15 3 Curvature Problem 1 . Problem 2 . Problem 3 . Problem 4 . Problem 5 . Problem 6 . Problem 7 . Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 17 19 21 23 25 27 28 30 4 Einstein’s Equation Problem 1 . . . . . . . Problem 2 . . . . . . . Problem 3 . . . . . . . Problem 4 . . . . . . . Problem 5 . . . . . . . Problem 6 . . . . . . . Problem 7 . . . . . . . Problem 8 . . . . . . . Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 31 31 34 38 38 38 39 40 41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 CONTENTS 5 Homogeneous, Isotropic Cosmology Problem 1 . . . . . . . . . . . . . . . . . . Problem 2 . . . . . . . . . . . . . . . . . . Problem 3 . . . . . . . . . . . . . . . . . . Problem 4 . . . . . . . . . . . . . . . . . . Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 43 44 46 48 50 6 The Schwarzschild Problem 1 . . . . . . Problem 2 . . . . . . Problem 3 . . . . . . Problem 4 . . . . . . Problem 5 . . . . . . Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 53 54 54 59 59 59 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Introduction Problem 1 Car and garage paradox: The lack of a notion of absolute simultaneity in special relativity leads to many supposed paradoxes. One of the most famous of these involves a car and a garage of equal proper length. The driver speeds toward the garage, and a doorman at the garage is instructed to slam the door shut as soon as the back end of the car enters the garage. According to the doorman, “the car Lorentz contracted and easily fitted into the garage when I slammed the door.” According to the driver, “the garage Lorentz contracted and was too small for the car when I entered the garage.” Draw a spacetime diagram showing the above events and explain what really happens. Is the doorman’s statement correct? Is the driver’s statement correct? For definiteness, assume that the car crashes through the back wall of the garage without stopping or slowing down. Solution Let c = 1. The spacetime diagram can be found below. In it the primed coordinates are those assigned to events by the driver and the unprimed ones are those assigned by the doorman. At the origin is the event “the driver has just reached the doorman and is about to enter the garage”. It follows that the world lines of the driver and doorman are the t′ -axis and the t-axis, respectively. Let L denote the common proper length of the car and the garage. The world line of the back wall of the garage is thus the one parallel to the t-axis and passing through the point (x, t) = (L, 0). Similarly, the world line of the rear end of the car is the one parallel to the t′ -axis and passing through (x′ , t′ ) = (−L, 0). 5 0) where it is in the diagram given the position of (x. This is because the expression “by the time the door was closed” means different things to the driver and the doorman. To the driver. t′ ) = (L. This agrees with the doorman’s statement. 0) and (x′ . The intersection of this hyperbole’s branches and the x′ -axis are the points (x′ . So what happens here? The driver is correct to say he had crashed through the back wall of the garage by the time the doorman shuts the door. t′ ) = (−L. This event is at the intersection of the world lines of the doorman and that of the car’s rear end. This seems to be a contradiction since from both statements one would conclude that the car would and would have not crashed by the time the door was closed. Analyzing the spacetime diagram. The intersection of the t′ = 0 line with the world line of the back wall of the garage is at a smaller value of x′ than L. Let tslam be time coordinate recorded by the doorman as he slams the door shut. it refers to a line parallel to the x′ -axis. the expression refers to those events in a line parallel to the x-axis. 0) indicated on the diagram. t) = (L. It turns out this conclusion would be wrong. The doorman is also correct when he says the the back wall was intact by the time he closed the door. This agrees with the driver’s account. one concludes that both statements are correct. The line t = tslam intersects the world line of the driver at a smaller value of x than that of the back wall of the garage. 0) is that the interval between each of these points and the origin is L2 − 02 = L2 . INTRODUCTION ′ t 3′ x L x L −L t 6 A justification for the placing of the point (x′ . To the doorman. . So they lie on the hyperbole x2 − t2 = L2 (doorman coordinates).6 CHAPTER 1. t′ ) = (L. thus completing the demonstration given in section 2. but every open subset of R2 is noncompact. The functions fi± act as projections of the ±xi > 0 portion of the sphere into the xi = 0 plane. √ 1 − (x1 )2 − (x2 )2 ) . see appendix A. x3 ) = (x1 . x2 ) . x2 . It is clear that the composition f2+ ◦ (f3+ )−1 is C ∞ .) Solution of (a) This is done by finding an expression for fi± ◦ (fj± )−1 and identifying it as C ∞ . Solution of (b) A couple of stereographic projections which omit different points from the sphere will do. so that f2+ (x1 .1 that S 2 is a manifold. (It is impossible to cover S 2 with a single chart. x3 ) f3+ (x1 . as follows from the fact S 2 is compact. x2 . x2 ) = (x1 . It follows that (f3+ )−1 (x1 . (b) Show by explicit construction that two coordinate systems (as opposed to the six used in the text) suffice to cover S 2 . Take for example the case i = 2. j = 3 with both signs being ‘+’.Chapter 2 Manifolds and Tensor Fields Problem 1 (a) Show that the overlap functions fi± ◦ (fj± )−1 are C ∞ . x3 ) = (x1 . Books on Complex Analysis usually include very nice descriptions of the stereographic projection. 7 . x2 . 2]. . (c) Let Y1 .8 CHAPTER 2. thereby defining the functions C γ αβ = −C γ βα by ∑ [Yα . γ Use the Jacobi identity to derive an equation satisfied by C γ αβ . (This equation is a useful algebraic relation if the C γ αβ are constants. . X] + [Z. . (Hint: For n = 1. ∂xµ {z } Hµ (x) Problem 3 (a) Verify that the commutator. Y . The desired result follows from considering the following integrals. as will be the case if Y1 . .) Solution I didn’t follow the Hint exactly to do this. Let γ be the path t 7−→ (1 − t)a + tx. (b) Let X. X]. Y ]. 0 then prove it for general n by induction. 1]. satisfies the linearity and Leibnitz properties. . ∫ 1 d (F ◦ γ(t))dt F (x) − F (a) = dt 0 ∫ 1 = ∇F ((1 − t)a + tx) · (x − a)dt 0 = n ∑ µ=1 ∫ (x − a ) µ µ |0 1 ∂F ((1 − t)a + tx)dt . Z be smooth vector fields on a manifold M .) . . use the identity ∫ 1 F (x) − F (a) = (x − a) F ′ [t(x − a) + a]dt . . Yβ ] = C γ αβ Yγ . defined by equation (2. at each point. and hence defines a vector field. where t ∈ [0.2. Y ] = 0 . Verify that their commutator satisfies the Jacobi identity: [[X. Z]. Yβ ] in this basis. Then. .2.14).2). Yn are left [or right] invariant vector fields on a Lie group [see section 7. Z] + [[Y. MANIFOLDS AND TENSOR FIELDS Problem 2 Prove that any smooth function F : Rn −→ R can be written in the form equation (2. Yn be smooth vector fields on an n-dimensional manifold M such that at each p ∈ M they form a basis of the tangent space Vp . we may expand each commutator [Yα . X]. First note that if v and w are vector fields and f is a function. Yδ ] − Yδ (C γ αβ )[Yα . Z]. The proof of the Jacobi identity is completed by taking the formulas corresponding to each term in the cyclic sum and checking that they add to zero. The one I will derive follows from the Jacobi identity applied to the fields Yα . Y ](f ) is the same as the above. X] is obtained the same way. ∑ [[Yα . Y ]. Yδ ] = [C γ αβ Yγ . Y ].PROBLEM 3 9 Solution of (a) First we check linearity. Z −→ Y . Solution of (c) I’m not sure what equation Wald expected us to prove. Yδ ] γ = ∑ C γ αβ [Yγ . Z](f ) = [X. w](f g) =v[w(f g)] − w[v(f g)] + f v[w(g)] + v(g)w(f = v(f )w(g) ) + g v[w(f )]− − f w[v(g)] − w(g)v(f w(f − ) − g w[v(f )] )v(g) =f [v. w] = f [v. Yβ ]. Z] on a function f . then [f v.ω C γ αβ C ω γδ Yω − Yδ (C γ αβ )C ω αβ Yω . w](g) + g [v. w](αf + βg) = v[w(αf + βg)] − w[v(αf + βg)] = α v[w(f )] + β v[w(g)] − α w[v(f )] − β w[v(g)] = α[v. [[X. Next we obtain a formula for the terms involved in the Jacobi identity. Let u and v be vector fields. Yβ ] γ = ∑ γ. [v. The formula for [[Y. It is easy to check that both sides of this equation applied to any function yield the same result. but with the substitutions X −→ Z . w](f ) + β[v. w](f ) Solution of (b) This is just a boring computation. f and g be functions and a and b be real numbers. [v. Y ](f )} = X[Y {Z(f )}] − Y [X{Z(f )}] − Z[X{Y (f )}] + Z[Y {X(f )}] The formula for [[Z. w]−w(f )v. Let us compute the action of [[X. w](g) Next we check the Leibnitz property. Yβ and Yδ . Y −→ X . Y ]{Z(f )} − Z{[X. .) Solution of (a) Even though Wald does not use the Einstein summation convention in his book. . . If the C γ αβ terms are constant. the components of the commutator of two vector fields v and w are given by µ [v. Problem 4 (a) Show that in any coordinate basis. I will use it in this problem. [v. ∗ ∗ Show that the components (Y γ )µ of Y γ in any coordinate basis satisfy ∑ ∂(Y γ )µ ∂(Y γ )ν ∗ ∗ − = C γ αβ (Y α )µ (Y β )ν . w] = ∑( ν ∂wµ vν ν ∂x −w ν ∂v ∗ µ ∂xν ) . wν Xν ](f ) =v µ Xµ [wν Xν (f )] − wν Xν [v µ Xµ (f )] ( (( =v µ {Xµ (wν )Xν (f ) + ( wν( X( µ Xν (f )}− v µ X − wν {Xν (v µ )Xµ (f ) + ν Xµ (f )} ={v ν ∂ν − wν ∂ν v µ }Xµ (f ) The terms on the third line canceled because of the equality of mixed partial derivatives for smooth functions in Rn . ν µ ∂x ∂x ∗ ∗ α. I will use ∂ν as shorthand for ∂/∂xν . . w](f ) =[v µ Xµ . Y n be the dual basis.10 CHAPTER 2. MANIFOLDS AND TENSOR FIELDS Taking a cyclic sum over α. . Let Y 1 . Yn be as in problem 3(c). Moreover. . one finds that the coefficient of each Yω vanishes. the last 3 terms vanish. . . Let {Xµ } = {∂/∂xµ } be a coordinate basis. .β (Hint: Contract both sides with (Yσ )µ (Yρ )ν . β and δ and equating the result to zero. that is ∑ {C γ αβ C ω γδ + C γ δα C ω γβ + C γ βδ C ω γα γ − Yδ (C γ αβ )C ω αβ − Yα (C γ βδ )C ω βδ − Yβ (C γ δα )C ω γβ } = 0 . The last line is the desired result. ∗ (b) Let Y1 . . Prove that in a neighborhood of each p ∈ M there exist coordinates y1 . Because the vector fields Y1 . We use this relation (in the second equation below) to show that Tab (Yσ )a (Yρ )b also equals C γ σρ . . . the equations ∂f /∂xµ = Fµ with µ = 1. . . . yn such that Y1 .] Use this fact together with the results of problem 4(b) to obtain the new coordinates. . ∗ The last equation follows from Y β (Yρ ) = δ β ρ . 2) as a sum of terms (dxµ )a (dxν )b is unique. . . This of course implies ∗ ∗ ∗ 0 = ∂ν {(Y β )µ (Yρ )µ } = (Y β )µ ∂ν (Yρ )µ + (Yρ )µ ∂ν (Y β )µ . Suppose [Yα . . ∗ ∗ Sab (Yσ )a (Yρ )b = C γ αβ (Y α )µ (Y β )ν (Yσ )µ (Yρ )ν ∗ ∗ = C γ αβ Y α (Yσ ) Y β (Yρ ) = C γ σρ . . First we compute Sab (Yσ )a (Yρ )b . Yn form basis for each tangent space. . Yn are the coordinate vector fields Yµ = ∂/∂y µ . Yρ ]ν = C γ σρ Problem 5 Let Y1 . n for the unknown functions f have a solution if and only if ∂Fµ /∂xν = ∂Fν /∂xµ . ∗ ∗ Tab (Yσ )a (Yρ )b = (Yρ )ν (Yσ )µ ∂ν (Y γ )µ − (Yσ )µ (Yρ )ν ∂µ (Y γ )ν ∗ ∗ = −(Yρ )ν (Y γ )µ ∂ν (Yσ )µ + (Yσ )µ (Y γ )ν ∂µ (Yρ )ν ∗ = (Y γ )ν {(Yσ )µ ∂µ (Yρ )ν − (Yρ )µ ∂µ (Yσ )ν } ∗ = (Y γ )ν [Yσ . (Hint: In an open ball of Rn . . Yβ ] = 0 for all α.1 of appendix B for a statement of generalizations of this result. . ∗ ∗ Sab = {C γ αβ (Y α )µ (Y β )ν }(dxµ )a (dxν )b .PROBLEM 5 11 Solution of (b) In order to use the Hint. . . it follows that Tab and Sab coincide. . Yn be smooth vector fields on an n-dimensional manifold M which form a basis of Vp at each P ∈ M .β.) . Now we show that Tab (Yσ )a (Yρ )b = Sab (Yσ )a (Yρ )b for all σ and ρ. This implies the desired result since the expression of a tensor of type (0. . . [See the end of section B. we define two tensor fields Tab and Sab by ∗ ∗ Tab = {∂ν (Y γ )µ − ∂µ (Y γ )ν }(dxµ )a (dxν )b . . ∗ ∗ ∗ (Y n )n because the vectors Y 1 . Finally. . ∗ (Y n )1 . This is an assumption which we can and will make. . . It states that F will be a diffeomorphism provided that det DF (ψ(p)) ̸= 0 and that we sufficiently restrict F ’s domain to a smaller neighborhood of ψ(p) if necessary. We now use the mathematical fact described in the Hint (which goes by the name of Poincar´e Lemma). because [Yα . . Because this problem is of course solvable.ν . . . .12 CHAPTER 2. . . . xn ) = (y 1 . Now 1∗ ∗ (Y )1 . To do so we must begin with an arbitrary one ψ : O −→ U ⊂ Rn and then change variables in U so as to construct a new coordinate system which satisfies the desired condition. This was (somewhat) clear to me when I first did this problem. Restricting the domain if necessary. ̸= 0 det DF = det . . The tool to be used to show that F is diffeomorphism is the Inverse Function Theorem. Y n are linearly independent. . ∂xν ∂xµ Here (x1 . .. we get that F is a diffeomorphism. . By problem 4(b). . we show that F ◦ ψ is the desired coordinate system.. Problem 3(c) shows that. . . The claim is that F is a diffeomorphism (change of variables) and that F ◦ ψ is the desired coordinate system. . It guarantees the existence of a function y γ : U −→ R such that ∂y γ ∗ = (Y γ )σ ∂xσ provided that U is an open ball. y n ). xn ) are the coordinates associated with ψ. I just kept putting the pieces from the previous problems together kind of randomly until the following solution came to be.. . We wish to construct a coordinate system about p which satisfies a certain condition. . the coefficients C γ αβ must vanish. this implies that ∗ ∗ ∂(Y γ )µ ∂(Y γ )ν = . Yα = ∑ (Yα )µ ∂ ∂xµ (Yα )µ ∂y ν ∂ ∂xµ ∂y ν µ = ∑ µ. (Y 1 )n . Define F : U −→ Rn by F (x1 . But how I was supposed to come up with the change of variables was not. Yβ ] = 0. MANIFOLDS AND TENSOR FIELDS Solution Let p ∈ M be an arbitrary point in spacetime. . It follows that T (vi ) = αi = 0 for all i. ∗ ∗ (b) Let v1 . v n be the dual basis. α (c) Prove that the operation of contraction. α ω= ∑ ∗ ω(vα )v α . Show that w= ∑ ∗ v α (w)vα . ∗ Also. .2). the ∑ vectors {v µ } span V ∗ : Let S be any linear functional in V ∗ . . Solution of (a) ∗ ∗ ∗ The set {v µ } is linearly independent: Suppose T = α1 v 1 + · · · + αn v n = 0. It follows that they are the same linear functional. . vn be a basis of the vector space V and let v 1 . . let T be any tensor defined of V . Solution of (c) Let {vµ } and {wν } be bases for V . . Applying v α to both sides of this equation ∗ yields v α (w) = cα .3.ν = ∑ δν α ν ∂ ∂y ν ∂ ∂ = α ν ∂y ∂y Problem 6 ∗ (a) Verify that the dual vectors {v µ } defined by equation (2. We wish to show that contracting T using one basis or the other yields the same . . . vn } of V . . . This proves the first equation.PROBLEM 6 13 = ∑ ∗ (Yα )µ (Y ν )µ µ. . ∗ Then S and α S(vα )v α agree on the basis {v1 . . equation (2. “dual vector” and “element of V ∗ ” are different names for the same object. . The second follows from the fact that both sides agree on a basis of V . Let w ∈ V and let ω ∈ V ∗ . is independent of the choice of basis.1) constitute a basis of V ∗ . Solution of (b) ∗ Write w = c1 v1 + · · · + cn vn .3. Please note that “linear functional”. . . . . . v σ . (a) Show that one always can find an orthonormal basis v1 . . . then the linear functional g(vi . . . .γ = ∑ ∗ T (. dropping the requirement of nondegeneracy.γ = ∑ ∗ ∗ v γ (vσ )T (.14 CHAPTER 2. . . . . contradicting the nondegeneracy of g. . .9) and Theorem (2.. vn of V . .11). . . . . · ) would vanish on a basis of V .e. . . Instead of doing so. (Hint: Use induction. . . . . . ) = ∑ ∗ ∗ ∗ T (. . . . n ∑ ∗ T (. . .σ. .) (b) Show that the signature of g is independent of the choice of orthonormal basis. . . Let v1 .γ = ∑ ∗ δ γ σ T (. . vγ . . . we show that there can be no self-orthogonal (null) vector in a orthonormal basis of a space V with metric g. . . . .4).i. vn be a orthonormal basis. . . v σ . Artin considers symmetric bilinear forms in general in (2. vγ . . . . . A discussion of direct sums which is sufficient for our purposes can be found on the first page of section 6 in chapter 3. . wµ . . a basis such that g(vα . . . . .9) and (2. .σ. . . .11). .γ = ∑ ∗ ∗ ∗ v γ (wµ (vσ )wµ )T (. . . To apply these results to the case of a metric. . vσ . . . . . . . . ) σ. . . . ) µ. . . ) σ Problem 7 Let V be an n-dimensional vector space and let g be a metric on V .γ µ=1 = ∑ ∗ ∗ ∗ wµ (vσ )v γ (wµ )T (. . wµ . vγ . . . . . . v σ . . and also on the concept of the direct sum of vector subspaces. . vγ . though in a slightly more general form. Solution I don’t know how to do this without having to reproduce here a few definitions and propositions of Linear Algebra. The solutions to items (a) and (b) are given in Proposition (2.2) and (2. . . . . wµ (vσ )v σ . If vi were self-orthogonal. . . . v σ . . . . ) µ. v γ (wµ )vγ . . . . These two theorems build on Propositions (2. . . v σ . ) µ. . . I refer the reader to section 2 in chapter 7 of Michael Artin’s book “Algebra”. MANIFOLDS AND TENSOR FIELDS result. . . . . vβ ) = ±δαβ . respectively. . .σ. ) σ. . gµν and g µν . tan ϕ = y/x is given by ds2 = dr2 + r2 dθ2 + r2 sin2 θ dϕ2 . ϕ). . x3 ) = (x.” defined by t′ = t . z′ = z . where tan ϕ = y/x. x′ = (x2 + y 2 )1/2 cos(ϕ − ωt) . one has dxµ = ∂xµ ∂xµ 1 d˜ x + · · · + d˜ xn . θ. . . Find the components. x ˜n ) be coordinates on some manifold M . y = r sin ϕ sin θ . Show that the metric components gµν in spherical polar coordinates r. y. xn ) and (˜ x1 . . . We apply this formula in the case where (x1 . (b) The spacetime metric of special relativity is ds2 = −dt2 + dx2 + dy 2 + dz 2 . of the metric and inverse metric in “rotating coordinates. x ˜2 . . ∂x ˜1 ∂x ˜n (2. three-dimensional Euclidean space is ds2 = dx2 + dy 2 + dz 2 . cos θ = z/r . x ˜3 ) = (r. . .7). we use the formulas x = r cos ϕ sin θ . θ. According to the tensor transformation law (2. ϕ defined by r = (x2 + y 2 + z 2 )1/2 . x2 . . z = r cos θ . z) and (˜ x1 .3.1) as should be. y ′ = (x2 + y 2 )1/2 sin(ϕ − ωt) . To compute the derivatives involved.PROBLEM 8 15 Problem 8 (a) The metric of flat. Solution of (a) Let (x1 . dx = dy = sin ϕ sin θ dr + r sin ϕ cos θ dθ + r cos ϕ sin θ dϕ . dx = [−ωx′ sin ωt′ − ωy ′ cos ωt′ ] dt′ + cos ωt′ dx′ − sin ωt′ dy ′ . dy = [ωx′ cos ωt′ − ωy ′ sin ωt′ ] dt′ + sin ωt′ dx′ + cos ωt′ dy ′ . The components of the inverse metric can now be found [ ] [ ]−1 using the relation g µν = gµν . −1 2 ′2 ω (x + y ′2 ) − 1 −ωy ′ ωx′ 0 [ µν ] −ωy ′ 1 0 0 g = ′ ωx 0 1 0 0 0 0 1 ′ ′ −1 −ωy ωx 0 −ωy ′ 1 − ω 2 y ′2 ω 2 x′ y ′ 0 = ′ 2 ′ ′ 2 ′2 ωx ω xy 1−ω x 0 0 0 0 1 The matrix inversion can be easily done using Crammer’s rule. To compute the relevant derivatives. we use the equation [ ] [ ][ ] x cos ωt′ − sin ωt′ x′ = . Solution of (b) This item can be solved by proceeding as in the solution of item (a). y sin ωt′ cos ωt′ y′ One finds dt = dt′ . (This is supposed to be read as a matrix equation). From this. dz = cos θ dr − r sin θ dθ .16 CHAPTER 2. We now expand dx2 + dy 2 + dz 2 using these expressions and find dr2 + r2 dθ2 + r2 sin2 θ dϕ2 . MANIFOLDS AND TENSOR FIELDS The resulting equations are ∂x ∂x ∂x dr + dθ + dϕ ∂r ∂θ ∂ϕ = cos ϕ sin θ dr + r cos ϕ cos θ dθ − r sin ϕ sin θ dϕ . The components gµν of the metric in rotating coordinates are given in the last equation above. we compute −dt2 + dx2 + dy 2 + dz 2 = (−1 + ω 2 x′2 + ω 2 y ′2 ) dt′2 + dx′2 + dy ′2 + dz ′2 −ωy ′ (dt′ dx′ + dx′ dt′ ) + ωx′ (dt′ dy ′ + dy ′ dt′ ) . dz = dz ′ . . f . ˜ a be a torsion-free (Hint: Repeat the derivation of eq.8].1. (a) Show that there exists a tensor T c ab (called the torsion tensor ) such that for all smooth functions. Derive the analog of equation (3.8). Solution of (a) ˜ a be some torsion free derivative operator. show that there exists a unique derivative operator ∇a with torsion T c ab such that ∇c gab = 0. So we have the operators ∇a and ∇ ˜ a∇ ˜ b f − C c ab ∇c f ∇a ∇b f = ∇ for some tensor field C c ab .29). since the one in text uses nowhere that ˜ a satisfy property (5). There is no need to repeat Let ∇ here the derivation of eq. we have ∇a ∇b f − ∇b ∇a f = −T c ab ∇c f .) (b) Show that for any smooth vector fields X a .1. (3. expressing this derivative operator in terms of an ordinary derivative ∂a and T c ab .1. Y ]c . letting ∇ derivative operator. Y a we have T c ab X a Y b = X a ∇a Y c − Y a ∇a X c − [X. (c) Given a metric. This is the desired result with T c ab = C c ab − C c ba . 17 . [3. gab .1. It follows that ( ( c c (( ˜( ˜( ˜( ˜ ∇ a ∇b f − ∇ b ∇a f = ( ∇ a∇ b f(− ∇b ∇a f − (C ab − C ba )∇c f .Chapter 3 Curvature Problem 1 Let property (5) (the “torsion free” condition) be dropped from the definition of derivative operator ∇a in section 3. We now use the relation T c ab = C c ab −C c ba proved at the end of the solution of item (a).2) ∂c gab = Cbca + Cacb .1. We show that ∇a is unique. (3. 2 (3.4) . CURVATURE Solution of (b) To do this we use property (4): For all vector fields X a and functions f . This is the same as ∂a gbc = Ccab + Cbac .3). We have 0 = ∇a gbc = ∂c gbc − C d ab gdc − C d ac gdb . Y ](f ) = X a ∇a (Y b ∇b f ) − Y a ∇a (X b ∇b f ) = X a ∇a Y b ∇b f + X a Y b ∇a ∇ b f − Y a ∇a X b ∇b f − Y a X b ∇a ∇b f = (X a ∇a Y b − Y a ∇a X b )∇b f + X a Y b (∇a ∇b f − ∇b ∇a f ) = [X a ∇a Y c − Y a ∇a X c ](f ) − T c ab X a Y b (f ) This shows that both sides of the equation we wish to prove applied on an arbitrary function yield the same result.1) By index substitution. This relation implies ∂a gbc + ∂b gac − ∂c gab = Tbac + Tabc + Tcab + 2Ccba .1.14) continues to hold even if the torsion free condition is dropped from the definition of a derivative operator. [X. Then we proceed as in the proof of Theorem 3. we find ∂a gbc + ∂b gac − ∂c gab = Ccab + Cbac + Ccba + Cabc − Cbca − Cacb . Let ∂a be an ordinary derivative operator.1) and (3. (3. the above derivation is in fact a proof that the equation holds. (3.3) By adding equations (3.2) and then subtracting equation (3. one has X(f ) = X a ∇a f .18 CHAPTER 3.1. ∂b gac = Ccba + Cabc . The first step is to note that equation (3. Solution of (c) Suppose one has a derivative operator ∇a with torsion T c ab such that ∇c gab = 0. Because tangent vectors are maps of functions into numbers. or 1 Ccba = {∂a gbc + ∂b gac − ∂c gab − Tbac − Tabc − Tcab } . ∇σ ϵ12 = 0. we can use equations (3. and thus. one finds ϵab ∇c ϵab = 0.PROBLEM 2 19 Because the tensor C c ab determines ∇a . This equation reads in any coordinate system as ϵ12 ∇σ ϵ12 + ϵ21 ∇σ ϵ21 = 2ϵ12 ∇σ ϵ12 = 0 . there exists a solution. Solution of (a) In order to make progress in this problem. Consider the equation ∇a β = ϵab ∇b α. A solution of the equation ∇a ∇a α = 0 is called a harmonic function. Problem 2 Let M be a manifold with metric gab and associated derivative operator ∇a .14) to define a derivative operator with the desired property. a a where ∂a is the derivative operator associated with some coordinate system. β. ∇a ∇a β = 0.4) and (3. Because ϵ12 ̸= 0. (β is called the harmonic function conjugate to α.) (b) By choosing α and β as coordinates. show that the metric takes the form ds2 = ±Ω2 (α. . (3. β)[dα2 + (−1)s dβ 2 ] . A derivation of this case follows.5) The exterior derivative operator d and the Poincar´e Lemma are introduced and very briefly discussed (two pages) in section 1 of appendix B. The Poincar´e Lemma states that equation (dβ)a = ∇a β = ϵab ∇b α locally has a solution β if and only if d(ϵce ∇e α)ab = ∇a (ϵbc ∇c α) − ∇b (ϵac ∇c α) = 0 . On the other hand. this completes the proof of the uniqueness. (a) Show that the integrability conditions (see problem 5 of chapter 2 or appendix B) for this equations are satisfied. Show that β is also harmonic. This is the two-dimensional case of equation (B. it is necessary to know ∇c ϵab = 0. From ϵab ϵab = 2(−1)s . In the case where M is a two-dimensional manifold. follows.1. where s is the number of minuses occurring in the signature of the metric. Please note the definitions of (dβ)a as the dual of (∂/∂β)a and as the exterior derivative of the function β agree. Equation (2. so that the above derivation also settles the question of existence. let α be harmonic and let ϵab be an antisymmetric tensor field satisfying ϵab ϵab = 2(−1)s . locally.2. the desired result.1) from problem 8 of chapter 2 shows that (dβ)dual = ∂a β = ∇a β = (dβ)exterior .11). 20 CHAPTER 3. It is not hard to see that the off-diagonal coefficients g αβ = g βα vanish: g αβ = g ab (dα)a (dβ)b = ∇b α∇b β = ∇b α ϵbc ∇c α =0. Solution of (b) Instead of computing the coefficients of the metric in the coordinates (α. Writing 1/g αα as ±Ω2 is the same as saying that g αα does not change sign in M . which implies the desired result d(ϵce ∇e α) = 0. this is the case if one makes the hypothesis that M is connected. g αα since the matrix with components of the metric is the inverse of the matrix with components of the inverse metric. It remains to show that g ββ = (−1)s g αα . The harmonicity of α is expressed in coordinates as ∇1 ∇1 α + ∇2 ∇2 α = 0 . From this it follows that ds2 = 1 [dα2 + (−1)s dβ 2 ] .5) holds. β) coordinates. β is also harmonic. β) directly. CURVATURE We use the fact that α is harmonic to show that the integrability condition (3. ∇a ∇a β = ∇a (ϵab ∇b α) = ϵab ∇a ∇b α = 0 The equation ϵab ∇a ∇b α = 0 holds because the tensor ϵab is antisymmetric while ∇a ∇b α is symmetric. we first find those of the inverse metric g ab . Because g αα is continuous and never zero. This shows d(ϵce ∇e α)12 = 0. The expression in the third line vanishes because ∇b ∇c α is symmetric while ϵbc is antisymmetric. These coefficients can then be used to show that the metric takes the desired form in (α. Finally. We now compute ∇1 (ϵ2σ ∇σ α) = ϵ2σ ∇1 ∇σ α = ϵ21 ∇1 ∇1 α . . I don’t think this hypothesis can be dropped. ∇2 (ϵ1σ ∇σ α) = ϵ12 ∇2 ∇2 α = (−ϵ21 )(−∇1 ∇1 α) . (3. of course.13).ρ Multiplying this equation by g αα /2.6) These equations imply Rcbad = Rbcda = Rbdca + Rdcba = Rbdca + Rcdab . µ. (3.7) .15). µ.2. not all of these components are independent. one finds g αα (−1)s = (ϵβα g αα )(ϵβα g αα )g ββ = g ββ . one would have (dβ)α = 0 and (dβ)β = 1. even if α and β were not conjugate harmonic functions.2. Problem 3 (a) Show that Rabcd = Rcdab . The relation ϵab ϵab = 2(−1)s is expressed as ∑ 2(−1)s = g µσ g νρ ϵσρ ϵµν = g αα g ββ ϵαβ ϵαβ +g ββ g αα ϵβα ϵβα = 2g αα g ββ ϵβα ϵβα .14) and (3. on account of the symmetries (3.2. Rabcd = Racbd + Rcbad . µ.2. In these coordinates the relation (dβ)a = ϵab (dα)b is expressed as ∑ (dβ)α = ϵαµ g µν (dα)ν = 0 .14) and (3.6) yields Rabcd = Racbd + Rbdca + Rcdab .ν Now. (b) In n dimensions. Solution of (a) Equations (3. However.ν (dβ)β = ∑ ϵβµ g µν (dα)ν = ϵβα g αα .13).2. (3.2. the Riemann tensor has n4 components. Show that the number of independent components is n2 (n2 − 1)/12.PROBLEM 3 21 We show g ββ = (−1)s g αα by working with (α. (3. or Rabcd − Rcdab = Racbd − Rbdac .15) are equivalent to Rabcd = −Rbacd . Rabcd = −Racdc . β) components.ν. so the condition that they are is equivalent to ϵβα g αα = 1.σ. Plugging this in (3. I’ve had a hard time coming up with or finding a clear argument for the claim that some specific subset of equations has the nice properties above.1)*n + (b .2. b_. n = 4. index[a_. Instead of solving this problem completely.15). k <= n.14). This together with equation (3. Perhaps because of notational difficulties. i <= n. So there are 3 equations for each multi-index µνσρ.22 CHAPTER 3.6) and claiming that it contains all the information of (3. but no redundancies. The code used follows. The first two of these equations are (3. CURVATURE Writing Rabcd − Rcdab as Rbadc − Rdcba and applying equation (3. Solution of (b) The word “independence” here has the meaning given to it in the context of Linear Algebra. c_. For[l = 1.6). For[i = 1. There is a number of solutions to this problem on the internet.2. l <= n. k++. This dimension equals the number of independent components. For[k = 1.13) and (3.6). The components Rµνσρ satisfy Rµνσρ + Rνµσρ = 0 . Then the number of equations in this subset is subtracted from n4 yielding the correct number n2 (n2 − 1)/12. j <= n.2. For[j = 1. d_] := d + (c . i++.1)*n^2 + (a . l++.8) Rµνσρ − Rµσνρ + Rσµνρ − Rσνµρ + Rνσµρ − Rνµσρ = 0 .7) results in Rabcd − Rcdab = Rbadc − Rdcba = Rbdac − Racbd . n^4}]. (3. the third is (3. M = ConstantArray[0. the number of independent components is the number of components minus the number of independent equations. Most of them involve picking some subset of the 3n4 equations (3.1)*n^3. I convinced myself that the number of independent components is n2 (n2 −1)/12 for some cases of interest by asking Mathematica to compute the dimension of the null space of the matrix associated with the linear system (3. {3*n^4.7) shows that Rabcd = Rcdab . { . Rµνσρ + Rµνρσ = 0 . The equations aren’t independent of course. j++. giving in total 3n4 equations. i. index[i. index[i. k. k. 1. M[[ 1*n^4 + index[i. j. The second line associates each multi-index µνρσ with a number between 1 and n4 . -1. index[i. i. Length[NullSpace[M]] The first line sets the dimension to 4. k. j. k] ]] = 1. Solution of (a) It is easy to check that ga[c gd]b = 12 (gac gdb − gad gcb ) has the symmetries of the Riemann tensor. l]. j. j. j. for n = 3.) (b) By similar arguments. j. index[i. M[[ M[[ M[[ M[[ M[[ M[[ } ] ] ] ] 2*n^4 2*n^4 2*n^4 2*n^4 2*n^4 2*n^4 + + + + + + index[i. index[i. the Riemann tensor takes the form Rabcd = Rga[c gd]b . j. k. equation (3. j. i. show that in three dimensions the Weyl tensor vanishes identically. index[k. (Hint: Use the result of problem 3(b) to show that ga[c gd]b spans the vector space of tensors having the symmetries of the Riemann tensor. index[k. l]. M[[ 0*n^4 + index[i. Any positive integer (of reasonable size) could go here. The nested ‘for’ loops set the coefficients for the three equations each index satisfy. j. where J is the number associated with µνρσ. j. k. l]. k. k. 1. l] ]] = 1. index[j. k.. j. The equations for multi-index µνρσ are represented in lines J. Problem 4 (a) Show that in two dimensions.e. -1. l]. k. that it satisfies (3. j.28) holds with Cabcd = 0. k. j. k. l]. j. index[j. l]. -1. Because in two dimensions the vector space of . i. i. index[i. j.e. j. index[i. i. M[[ 1*n^4 + index[i. J +n4 and J + 2n4 . k. l] l] l] l] l] l] ]] ]] ]] ]] ]] ]] = = = = = = 1. l]. index[i.2. index[i. k. k. l] ]] = 1. index[i.. k. l]. The third line creates a matrix with a column for each of the n4 variables and an equation for each of the 3n4 equations. i. l]. l]. l. l] ]] = 1. j. k.PROBLEM 4 23 M[[ 0*n^4 + index[i. index[j. k.6) and the two equations preceding it. it follows that Rabcd = λga[c gd]b . First b with d. where λ is some (scalar) function.1)*n^2 + (a . R = (λ/2)δa a = λ Solution of (b) The tensor Sabcd = 2 2 (ga[c Rd]b − gb[c Rd]a ) − R ga[c gd]b n−2 (n − 1)(n − 2) has the symmetries of the Riemann tensor. we contract both sides. Now. M = ConstantArray[0.9). To show that λ = R. CURVATURE tensor having the symmetries of the Riemann tensor has 22 (22 − 1)/12 = 1 dimension. c_. this adds at most n(n+1)/2 = 6 equations. Because Sabcd and Rabcd have these properties. then the number of free components drops to zero. n^4}].24 CHAPTER 3. Rac = (λ/2)(gac δd d − gad δc d ) = (λ/2)gac Then we contract a with c. because the Weyl tensor does. as shown in the piece of code below for the case where the index in (3. For[i = 1. If these equations are independent of one another and of the ones used to reduce the number of independent components to 6. n = 3. It turns out that the independence hypotheses of the previous paragraph are satisfied.1)*n^3. .9) was raised with the flat metric. σ Because of symmetry in α and β. It follows that a unique tensor has the symmetries of the Rabcd and satisfies (3. i <= n. the components Sµνρσ satisfy ∑ (3. tensors with the symmetries of the Riemann tensor have at most n2 (n2 −1)/12 = 6 independent components.1)*n + (b . index[a_. b_. Sabcd = Rabcd . d_] := d + (c . {3*n^4 + n^2. because the Weyl tensor is trace-free. In dimension n=3. i++.9) Sασβ σ = Rαβ . -1. j. 1. index[i. -1. i. j. . j. . index[i. i. . index[i. l] l] l] l] l] l] ]] ]] ]] ]] ]] ]] = = = = = = 1. index[i. l]. index[k. k. index[i. index[i. k. M[[ 1*n^4 + index[i. l]. k. M[[ M[[ M[[ M[[ M[[ M[[ } ] ] 2*n^4 2*n^4 2*n^4 2*n^4 2*n^4 2*n^4 + + + + + + index[i. Let x1 (t). k. k++. i. j. l <= n. index[i. -1. l] ]] = 1.3. Solution of (a) To do this we work with coordinates. k] ]] = 1. confirming the uniqueness of the tensor having the symmetries of Sabcd . k <= n. l]. i. l]. . index[i. For[l = 1. } ] For[k = 1. j. l]. k. a. j.1) is satisfied. k. k. j. l]. Let s be some other parameter for this curve. a <= n. { M[[ 0*n^4 + index[i. j. } ] ] Length[NullSpace[M]] (See explanation of the [very similar] code used in problem 3[b]. j. index[i. j. k.2) can be reparameterized so that equation (3. index[j. l++. k. j++.1) n + j. For[a = 2. a++.3. index[k. j. . M[[ 0*n^4 + index[i. k. where a and b are constants. j. k.PROBLEM 5 25 For[j = 1.1) n + j. 1. Problem 5 (a) Show that any curve whose tangent satisfies equation (3. xn (t) be the components of a curve parameterized by t. l. l] ]] = 1. j. j <= n. index[j. 1] ]] = -1. index[i. (b) Let t be an affine parameter of a geodesic γ. l]. { M[[ 3*n^4 + (i . k. { M[[ 3*n^4 + (i . a] ]] = 1. k. index[j. i. l] ]] = 1. j. j. l]. k. 1. index[i. j. k. l]. Show that all other affine parameters of γ take the form at + b. M[[ 1*n^4 + index[i. j.) Mathematica returns that the length of the null space is zero. j. k. l]. k. j. index[i. ν ( )2 ∑ ( )2 2 µ dT dt dt µ σ ν µd t + + Γ σν T T =T 2 ds dt ds ds σ. Suppose T a satisfies equation (3. ∑ dS µ ∑ µ Γ σν S σ S ν + ds σ. so t = as + b for some constants a and b. CURVATURE and let T µ = dxµ /dt.3. σ σ It follows that d2 t/ds2 = 0.ν ( ) ∑ ( )2 d dt µ dt µ σ ν = T + Γ σν T T ds ds ds σ. . we compute 0= ∑ σ S ∇σ S = σ µ d2 t Tµ 2 ds ( + dt ds )2 ∑ *0 T ∇σ T µ . To do so.26 CHAPTER 3. σ ∑ S σ ∇σ S µ = 0 . Solution of (b) Let T µ and S µ be as in item (a).2).ν ( ) d2 t dt 2 = T µ 2 + αT µ ds ds S σ ∇σ S µ = σ This shows that it is sufficient to choose s such that d2 t +α ds2 ( dt ds )2 =0. We show that one can choose s so that S a satisfies (3.3. that is. Suppose this time that ∑ T σ ∇σ T µ = 0 . take a nontrivial solution t(s) of this differential equation and use it to (implicitly) define s for each value of t. σ As before. ∑ T σ ∇σ T µ = αT µ σ for some function α. S µ = dxµ /ds be the components of the tangents with respect to each parameter.1). v[[1]] = r. v[[\[Mu]]]] + D[g[[\[Mu]. \[Rho]]]. The code used in Mathematica follows. 3}]. v = ConstantArray[0. For[\[Mu] = 1. v[[\[Rho]]]]). v[[3]] = \[Phi]. Solution of (a) Plugging equation (3. g[[1. For[\[Rho] = 1. Γϕ θϕ = Γϕ ϕθ = cot θ . 2]] = r^2. (a) Calculate the Christoffel components Γσ µν in this coordinate system. we find that the nonzero components of the Christoffel symbol are Γr θθ = −r . 3. \[Mu]. g[[2. \[Rho]++. \[Mu]++. \[Nu]++. { \[CapitalGamma][[\[Rho]. . For[\[Nu] = 1. 3]] = r^2*Sin[\[Theta]]^2. \[Mu] <= 3. \[Rho]]])*(D[g[[\[Nu]. \[Rho] <= 3. (b) Write down the components of the geodesic equation in this coordinate system and verify that the solutions correspond to straight lines in Cartesian coordinates. Γθ rθ = Γθ θr = Γϕ rϕ = Γϕ ϕr = 1/r . \[CapitalGamma] = ConstantArray[0. 1]] = 1. v[[2]] = \[Theta].PROBLEM 6 27 Problem 6 The metric of Euclidean R3 in spherical coordinates is ds2 = dr2 + r2 (dθ2 + sin2 θ dϕ2 ) (see problem 8 of chapter 2). \[Nu]]]. 3}].30) into Mathematica. \[Nu]]] = 1/(2*g[[\[Rho]. g[[3.1. v[[\[Nu]]]] D[g[[\[Mu]. \[Nu] <= 3. g = ConstantArray[0. Γr ϕϕ = −r sin2 θ . {3. {3}]. \[Rho]]]. Γθ ϕϕ = − cos θ sin θ . {3. 2 ˙2 ˙2 r¨ − rθ − r sin θ ϕ = 0 θ¨ + (2/r)r˙ θ˙ − sin θ cos θ ϕ˙ 2 = 0 ¨ ϕ + (2/r)r˙ θ˙ + 2 cot θ θ˙ϕ˙ = 0 (3.30).5) yields d2 xµ /dt2 = 0. Let C be one of these curves. By construction. Writing the cartesian components of this equation using (3.4).4a and (b) by the tetrad methods of section 3.10) represent curves in Cartesian coordinates.28 CHAPTER 3. the tangent vector T a of C satisfies T a ∇a T b = 0.4.3.10) The solutions of (3. because Γµ σν = 0 in these coordinates.11) The other components can be found from this one using the symmetries Rabcd = −Rbacd = −Rabdc = Rcdab of the Riemann tensor. t)[−dt2 + dx2 ].ν The result is the following system. We compute the Christoffel components via equation (3.5) d2 xµ ∑ µ ∂xσ ∂xν Γ σν + =0. an arbitrary Lorentz metric on a two-dimensional manifold locally can always be put in the form ds2 = Ω2 (x. dt2 dt dt σ. We compute the component Rtxt x using equation (3. CURVATURE } ] ] ] Solution of (b) The components of the geodesic equation are obtained by substitution of the Christoffel components of item (a) into equation (3. Problem 7 As shown in problem 2. Γt tt = Γt xx = Γx tx = Γx xt = Ω−1 ∂t Ω Γt tx = Γt xt = Γx tt = Γx xx = Ω−1 ∂x Ω Step 2.3.1. . Solution of (a) I will use in this problem ∂x f as shorthand for ∂f /∂x. Rtxt x = Ω−2 ([∂t Ω]2 − [∂x Ω]2 ) + Ω−1 (∂xx Ω − ∂tt Ω) (3. Calculate the Riemann curvature tensor of this metric (a) by the coordinate basis methods of section 3.4b. This shows that C is a straight line in cartesian coordinates. Step 1. This equation reduces in the present case to 2 2 R1212 = Ω−1 (∂t ω212 − ∂x ω112 ) + ω112 − ω212 . e2 }. The term with ωb11 vanishes because the connection one-forms ωbσν are antisymmetric in σν. Letting σ = 1. (e2 )a = Ω−1 (∂/∂x)a . We compute R1212 using equation (3. these two coefficients determine all others.PROBLEM 7 29 Solution of (b) Step 1. ∂[a (e2 )b] = ∂t Ω(dt)[a (dx)b] + ∂x Ω(dx)[a (dx)b] = ∂t Ω(dt)[a (dx)b] . it follows ∂[a (e1 )b] = −∂t Ω(dt)[a (dt)b] − ∂x Ω(dx)[a (dt)b] = ∂x Ω(dt)[a (dx)b] .12) for ωbσν . We choose an orthonormal basis. We compute the left hand side of ∑ ∂[a (eσ )b] = η µν (eµ )[a ωb]σν . where (e1 )a = Ω−1 (∂/∂t)a . we similarly find ω221 = Ω−2 ∂t Ω. (e1 )b = gab (e1 )a = −Ω(dt)b (e2 )a = gab (e2 )a = Ω(dx)b From the definition of the ordinary derivative operator. .12) µ.21). Step 3. Letting σ = 2 and contracting (e2 )a ∂[a (e2 )b] . (3. It follows that ω112 = −Ω−2 ∂x Ω. Step 4. we have ∂[a (e1 )b] = −(e1 )[a ωb]11 + (e2 )[a ωb]12 . We solve equation (3.ν First we lower the indices of the vectors in the orthonormal basis. Step 2. we have in one hand (e1 )a ∂[a (e1 )b] = (e1 )a (e2 )[a ωb]12 = −(1/2)(e2 )b ω112 . Contracting (e1 )a ∂[a (e1 )b] . and on the other (e1 )a ∂[a (e1 )b] = (e1 )a ∂x Ω(dt)[a (dx)b] = (1/2)Ω−2 (∂x Ω)(e2 )b .4. The natural one here is {e1 . In two dimensions. 13) and expression (3.13). We compute R1212 from (3. e1 ]a − (e1 )a [e3 .11) amounts to multiplying the right hand side by Ω2 . The following three equations are (3. Solution Instead of labeling indices with Greek letters.15).13) It wouldn’t be very nice not to check that (3. (3. we have 3ω[123] = ω123 + ω231 + ω312 .16) ω231 + ω123 = (e3 )a [e2 . Because of the antisymmetry (3. e2 ]a } .15).23) with the roles of the indices interchanged. e3 ]a + (e3 )a [e2 .15) a (3.23) to solve for ωλµν in terms of commutators (or antisymmetrized derivatives) of the orthonormal basis vectors. CURVATURE Plugging here the expressions for ω112 and ω221 = −ω212 obtained in the preceding step.11). I will use numbers. we find R1212 = Ω−4 ([∂t Ω]2 − [∂x Ω]2 ) + Ω−3 (∂xx Ω − ∂tt Ω) .4. equation (3.30 CHAPTER 3. It is clear that formula ω123 = 3ω[123] − 2ω[23]1 holds.4. ω123 + ω312 = (e2 )a [e1 . Use this formula together with equation (3. e1 ] ω312 + ω231 = (e1 )a [e3 . e3 ]a (3.4.15) and subtracting equation (3.14) and (3.14) a (3. 2 . 2ω[23]1 = ω231 + ω312 . Problem 8 Using the antisymmetry of ωaµν in µ and ν. e2 ] Adding equations (3.11) found in item (a) agree. it is clear that lowering the raised index x in (3. show that ωλµν = 3ω[λµν] − 2ω[µν]λ . R1212 = Rabcd (e1 )a (e2 )b (e1 )a (e2 )b = Rabcd [Ω−1 (∂/∂t)a ][Ω−1 (∂/∂x)a ][Ω−1 (∂/∂t)a ][Ω−1 (∂/∂x)a ] = Ω−4 Rtxtx = Ω−4 {[∂t Ω]2 − [∂x Ω]2 + Ω(∂xx Ω − ∂tt Ω)} This is precisely (3. Because of the simple form of the metric.16) from the result yields 1 ω123 = {(e2 )a [e1 .4. Chapter 4 Einstein’s Equation Problem 1 Show that Maxwell’s equation (4. −8π∇b jb = −g bf Rbd F d f + g ae g bf Rbaf d Fe d = −Rbd F db + g ae Rad Fe d = −Rdb F db + Rad F ad = 0 Problem 2 (a) Let α be a p-form on an n-dimensional oriented manifold with metric gab ..ap is a totally antisymmetric tensor field (see appendix B). i. one can show that it equals zero. −8π∇b jb = 2∇b ∇a Fab = ∇b ∇a Fab − ∇b ∇a Fba = −g ae g bf (∇a ∇b − ∇b ∇a )Fef = −g ae g bf (Rabe d Fdf + Rabf d Fed ) Playing with the right hand side a little bit.12) which relates the Riemann tensor to the commutator of two derivative operators.e.3.12) implies strict charge conservation. Solution To do this I applied the derivative operator to both sides of Maxwell’s equation and looked at the right hand side from different points of view until I could show it vanishes.2.. αa1 . ∇a j a = 0.. 31 . The idea that worked was using equation (3. bn−p ϵb1 .ap b1 .e. dF = 0 ..cp = In order to use equation (B. the totally antisymmetric tensor field determined up to sign by equation (B..3. . . . of α by ∗αb1 .bn−p c1 . . Gauss’s law of electromagnetism continues to hold in curved spacetime.bn−p ϵb1 .an is the natural volume element on M ..... . . Thus. we obtain Σ ∗j = (1/4π) S ∗F . Maxwell’s equations (4. ap b2 . .32 CHAPTER 4. Tab . .. p!(n − p)! 1 p ∗ ∗ αc1 .. then so does F˜ab .ap α ϵa1 . ap . .cp .. EINSTEIN’S EQUATION We define the dual. bn−p a1 . ...13). ∗α..a ϵa1 . . ap b1 ....bn−p .. bn−p b1 ..cp (n − p)! 1 = αa . where s is the number of minuses occurring the signature of the gab ..13) can be written as d ∗ F = 4π ∗ j . Show that the stress-energy. .2... ap b1 .. Solution of (a) By the definition of the ∗ operation. i. We call F˜ab a duality rotation of Fab by “angle” β. and − S ∗F = S E na dA is just the integral of the normal component of Ea = Fab tb on S. bn−p has sign (−1)p . of the solution F˜ab is the same as that of Fab . . But − Σ ∗j = Σ j ta dΣ a is just the total electric charge ∫ ∫ eain the volume Σ..bn−p c1 . p! where ϵa1 . It follows immediately from part (b) that if Fab satisfies the source-free Maxwell’s equations (j a = 0).. (b) Show that in differential forms notation (see appendix B). and from this it isn’t hard to see that the sign of a1 . where Σ is a three-dimensional ∫ ∫ a hypersurface with two-dimensional boundary S.12) and (4.2. 2π] the tensor field F˜ab = Fab cos β + ∗Fab sin β. Note that if we apply Stokes’s theorem (see appendix B) to the first equa∫ ∫ tion. we permute the raised indices. Show that ∗ ∗ α = (−1)s+p(n−p) α. ... where t is the unit normal to Σ. (c) Define for each β ∈ [0. .bn−p = 1 a1 . 1 ∗ αb1 . bn−p b1 a1 .3. The permutation a1 .....9). .ap b1 . bn−p c1 .1.1) = {∇e F ab ϵabcd + ∇c F ab ϵabde + ∇d F ab ϵabec } . It follows that ∗ ∗ αc1 .ap δ [a1 c1 .bn−p a1 . (dF )abc = 3∇[a Fbc] ...bn = 0..2) are the same.12) and again the definition of ∗.2. so it is clear that (4..1) and (4. we fix some coordinate system and adhere to the Einstein summation convention... 4π(∗j)ecd = (∗∇a Fba )ecd = ∇a F ba ϵbecd . (4..ap ] δ a1 c1 . . 4π(∗j)123 = ∇µ F νµ ϵν123 = ∇µ F 4µ ϵ4123 = (∇1 F 14 + ∇2 F 24 + ∇3 F 34 )ϵ1234 1 (d ∗ F )123 = {∇1 F µν ϵµν23 + ∇2 F µν ϵµν31 + ∇3 F µν ϵµν12 } 2 = ∇1 F 14 ϵ1423 + ∇2 F 24 ϵ2431 + ∇3 F 34 ϵ3412 = (∇1 F 14 + ∇2 F 24 + ∇3 F 34 )ϵ1234 . Below is a (somewhat sketchy) proof of the fact that equation (4.PROBLEM 2 33 is (−1)p(n−p) . The second equality follows from (B.13). (d ∗ F )ecd = ∇e (∗F )cd + ∇c (∗F )de + ∇d (∗F )ec 1 = {∇e (F ab ϵabcd ) + ∇c (F ab ϵabde ) + ∇d (F ec ϵabcd )} 2 1 (4. d ∗ F = 4π ∗ j..3. .2.11).3.cp = (−1)p(n−p) αa .12) implies its version in differential forms notation. This is what is necessary to establish d ∗ F = 4π ∗ j.... the λµνcomponent of the right hand sides of equations (4. Here we make the case λµν = 123.4) of the exterior derivative operator. ∇a ϵb1 . though this would be necessary for one to conclude that both equations are indeed equivalent.. Solution of (b) By the definition (B.. the fourth is the desired result. Using Maxwell’s equation (4... . By the definition of the ∗ operation.2) For what follows..cp . .cp p!(n − p)! 1 p = (−1)s+p(n−p) αa1 .a ϵb1 . 2 Equation (B.. δ ap cp = (−1)s+p(n−p) αc1 . was used in the last line. I did not show that the converse holds.13) and dF = 0 are equivalent... δ ap ] cp = (−1)s+p(n−p) α[a1 .. It isn’t hard to show that for each particular multi-index λµν.3.ap ϵb1 . 4 (4. ⃗ = 2Mω B ⃗ . Thus it only remains to prove that (4. By definition. 3R .34 CHAPTER 4. This amounts to showing 1 Fac Fb c − gab Fde F de − ∗Fac ∗ Fb c = 0 .google. so I went to the oracle1 for help. I then found the following (symmetric in Fab and ∗Fab ) expression for the electromagnetic stress-energy tensor Tab : Tab = 1 {Fac Fb c + ∗Fac ∗ Fb c } . ∗Fab = (1/2)ϵabcd F cd .3) It is easy to check using (4. ⃗ and B ⃗ (b) Show that the “gravitational electric and magnetic fields” E inside a spherical shell of mass M and radius R (with M ≪ R) slowly rotating with angular velocity ω ⃗ are ⃗ =0 E 1 http://www. so 1 ∗Fac ∗ Fb c = ϵacij ϵb c kl F ij F kl . EINSTEIN’S EQUATION Solution of (c) I had some trouble with the messy algebra here.4) yields after some algebra 1 ∗Fac ∗ Fb c = Fac Fb c − gab Fde F de . Problem 3 (a) Derive equation (4.4) We disentangle the ϵ terms using equation (B. 2 This is the desired result.2. ϵacij ϵb c kl = gbd gke glf ϵcaij ϵcdef = −3!gbd gke glf δ [d a δ e i δ f ] j = gba gkj gli − gba gki glj + gbi gka glj − − gbj gka gli + gbj gki gla − gbi gkj gla Plugging this into equation (4.com/ .3) agrees with the expression for the electromagnetic stress-energy tensor given in Wald’s book.24).13).3) and ∗ ∗ Fab = −Fab (see item (a)) that Fab and F˜ab yield the same Tab . 2 We compute ∗Fac ∗ Fb c . 8π (4.4. they are reproduced here for reference. is changed from what it would be without the shell.1. 2. 3) γ00 = 12 γ 00 . . σ. Aµ = − 14 γ 0µ Fµν = ∂µ Aν − ∂ν Aµ ∑ ϵµ0 σν Fσν Bµ = − 21 Eµ = Fµ0 σ.ν The relevant Christoffel components are as usual computed using equation (3. ⃗ where Ω ⃗ = 2B ⃗ = 4 (M/R)⃗ S. γ0σ = γ 0σ . The following formulas will be used in this problem. 3) γ µν = 0 . Show that the inertial components.” defined by parallel propagation alon a geodesic. = Ω ω.24) is done in an analogous manner. At the center of the shell the local standard of “nonrotating. Solution of (a) Below is a proof that a1 = −E 1 − 4(v 2 B 3 − v 3 B 2 ) . where ua is the tangent to his world line. first analyzed by Thirring and Lense (1918) and discussed further by Brill and Cohen (1966).ν We start with geodesic equation.PROBLEM 3 35 (c) An observer at rest at the center of the shell of part (b) parallelly propagates along his (geodesic) world line a vector S a with S a ua = 0. 3 This effect. we use the equalities (σ = 1. may be interpreted as a “dragging of inertial frames” caused by the rotating shell.5) The verification of the other components of equation (4. ⃗ precess according to dS/dt ⃗ ⃗ × S. which follow from 1 1 γµν = γ µν − γ = γ µν − γ 00 .30). in a manner in accord with Mach’s principle. ν = 1. To relate expressions found with to Aµ . (4. 2 2 We also use below the relation (µ. ∑ a1 = − Γ1 σν v σ v ν . 2.4. σ=1 This completes the proof of (4. then one may take γ µν = 0. Now. Bµ = ϵ0µ 12 F12 + ϵ0µ 13 F13 + ϵ0µ 23 F23 . Γ1 00 = 21 (−∂1 γ00 ) = − 14 ∂1 γ 00 = ∂1 A0 = ∂1 A0 − ∂0 A1 = F10 = E1 . EINSTEIN’S EQUATION This follows from fact that in Linearized Theory. In the present case. 2. Γ1 0σ = 12 (∂σ γ01 − ∂1 γ0σ ) = 2(∂1 Aσ − ∂σ A1 ) = 2F1σ . where ρ is the mass-energy density and.6) The Christoffel components then read ∑ Γ1 ρσ = 12 α η 1α (∂ρ γσα + ∂σ γρα − ∂α γρσ ) = 0 . by definition. 3. so that v 2 B 3 − v 3 B 2 = v 2 ϵ03 12 F12 − v 3 ϵ02 13 F13 = v 2 ϵ0312 F12 − v 3 ϵ0213 F13 = v 2 F12 + v 3 F13 = 3 ∑ F1σ v σ . Solution of (b) The stress-energy tensor for this configuration has the form (4. ν = 1. σ=1 Here it was used that v 0 = 1 (approximately) and Γ1 0σ = Γ1 σ0 .4. if Tµν = 0 for some multiindex µν and ∂0 Tµν = 0. for µ = 1. −J2 0 −J3 (4. Jµ is the xµ -component of . where ρ and σ are nonzero indices. 3 because Tµν is assumed to have the form −J1 −J2 −J3 ρ −J1 .6).36 CHAPTER 4.14).5). 2. The geodesic equation now looks like a1 = −E1 v 0 v 0 − 3 ∑ Γ1 0σ v 0 v σ − σ=1 = −E 1 − 4 3 ∑ 3 ∑ Γ1 σ0 v 0 v σ σ=1 F1σ v σ . we take γ µν = 0 for µ. This fact is discussed and used in the text after equation (4. Using the approximation ∂0 γµν = 0. The other components of dS/dt = 2B By computations made in items (a) and (b).11) in Griffiths’ book. Γ1 0σ = 2F1σ .7). we find that Aµ satisfies ∇2 Aµ = −4πJµ . (4. Note that the vector J⃗ = (J1 . J2 . Plugging this into (4. 0. 1 From the Linearized Einstein’s ∑ ν Equation. we find ∑ dS 1 = −2 F1ν S µ dt 3 ν=1 2 3 = −2(S B − S 3 B 2 ) . We use this equation to show dS 1 = 2(B 2 S 3 − B 3 S 2 ) . essentially. where in the second line a computation made in item (b) was used. and the answer agrees with B ω.1. where J0 := −ρ. This is. J3 ) also represents the current density of a charged sphere of radius R and charge2 M rotating with angular velocity ω ⃗.ν ∑ dS µ =− Γµ 0ν S ν . 0).PROBLEM 3 37 the momentum density. The problem of finding B ⃗ is solved in example and conclude that E ⃗ = (2M/3R)⃗ (5. equations (5. This completes the proof of (4.24) from Griffiths’ book “Intro⃗ = (A1 .8) dt ⃗ ⃗ ×S ⃗ are verified by proceding similarly. So −A0 and A static and vector potentials determined by the charge and current densities ⃗ This analogy with Electromagnetism allows one to use Gauss’ Law ρ and J. A2 . 2 Am I missing something like 4π or ϵ0 in front of M ? . ⃗ = 0. Γ1 00 = E1 = 0 .19). The components of the propagated vector S µ satisfy dS µ ∑ σ µ + u Γ σν S ν = 0. (4. A3 ) are the electroduction to Electrodynamics”. 0. Solution of (c) The components of the tangent ua to the observer’s world line relative to the global inertial coordinates read (1.62) and (2.7) dt ν This is just equation (3. it follows that Aµ = 4 γ 0µ satisfies Maxwell’s equation ν ∂ ∂ν Aµ = −4πJµ .8). or dt σ. ∂ a Tab = 0) in Minkowsky spacetime. there exists a tensor field sab = −sba such that v a = ∂b sab . [This follows from applying the converse of the Poincar´e lemma (see the end of section B. .] Use this fact to show that Tab = ∂ c Wcab where Wcab = W[ca]b . Solution This is a straightforward computation. (This is established most easily from the spinor decomposition of the Weyl tensor given in chapter 13. show that in vacuum. (2) for Rab by substituting ηab + γab for gab and keeping precisely the terms quadratic in γab .2.) Problem 6 As discussed in the text.38 CHAPTER 4. Then use the fact that ∂ c Wc[ab] = 0 to derive the desired result. We define the Bel-Robinson tensor in terms of the Weyl tensor by 1 f Tabcd = Caecf Cb e d f + ϵae hi ϵb ej k Chicf Cj k d 4 3 f e f = Caecf Cb d − ga[b Cjk]cf C jk d . converved tensor field (i.1 in appendix B) to the 3-form ϵabcd v d .4. 2 where ϵabcd is defined in appendix B and equation (B. a four-index tensor Tabcd can be constructed out of the curvature in a manner closely analogous to the way in which the stress tensor of the electromagnetic field is constructed out of Fab (eq.16). we have ∇a Tabcd = 0. please let me know via email.2. Should anyone really need the solution to this problem.4.51). Tab = Tba .13) was used. [4. Rab = 0. It follows that Tabcd = T(abcd) .27]).e.) (a) Show that T a acd = 0. in general relativity no meaningful expression is known for the local stress-energy of the gravitational field. Show that there existsa tensor field Uacbd with the symmetries Uacbd = U[ac]bd = Uac[bd] = Ubdac such that Tab = ∂ c ∂ d Uacbd . Problem 5 Let Tab be a symmetric. (b) Using the Bianchi identity (3.2. equation (4.5) for Rab . (Hint: For any vector field v a in Minkowsky spacetime satisfying ∂a v a = 0. derive the formula.. EINSTEIN’S EQUATION Problem 4 Starting with equation (3. However. 3 a 1 f f g [a Cjk]cf C jk d = [g a a Cjkcf + g a j Ckacf + g a k Cajcf ]C jk d 2 2 1 f = [4Cjkcf + Ckjcf + Ckjcf ]C jk d 2 1 f = 2Cjkcf C jk d 2 This implies f T a acd = Caecf C ae d f − Cjkcf C jk d = 0 . t30 ). x3 )dx1 dx2 dx3 . (b) Let ξa be a gauge transformation which vanishes outside a bounded region of space.PROBLEM 7 39 Solution of (a) We begin expanding the anti-commutator in the second expression for Tabcd and contracting a with b. x1 . . equation (4. t20 . of Σ. The integrand in the second line is the divergence of the vector field (t10 . x1 .2 imply that this field goes to zero sufficiently rapidly. Problem 7 (a) Show that the total energy E.e. is time independent. the value of E is unchanged if the integral is performed over a time translate. x3 ) in spacetime such that x0 = t. Show that E[γab ] = E[γab + 2∂(a ξb) ] by comparing them ′ ] where ξ ′ is a new gauge transformation which with E[γab + 2∂(a ξb) a agrees with ξa in a neighborhood of the hyperplane Σ but vanishes in a neighborhood of another hyperplane Σ′ . so that the divergence theorem gives the third equality. x2 . Then the total energy associated with γab computed at Σt is ∫ E(t) = t00 d3 x ∫Σt = t00 (t. R3 We compute the derivative dE/dt.. x2 .55). The decay hypothesis on γab described in the text after figure 4. Σ′ . ∫ ∂t00 3 dE = d x dt ∂t 3 R ∫ ∂t10 ∂t20 ∂t30 3 = + + d x=0 1 ∂x2 ∂x3 3 R ∂x In the second equality it was used that ∂ a tab = 0.4. Solution of (a) Let Σt be the collection of points (x0 . i. 4. It is easy to check that qxx = 32 M L2 and that the other components qµν vanish.4.58) and (4. The spring is set into oscillation. √ where L0 is the equilibrium length of the spring and ω = 2K/M .40 CHAPTER 4. we compute qµν . the integral in formula (4.4. equation (4. Here the origin of time is chosen so that L = L0 at t = 0.59). what fraction of the energy of oscillation of the spring is radiated away during one cycle of oscillation? Solution Let L denote the distance between the two masses.48) becomes a sum. L − L0 = A sin ωt . Because mass is concentrated. By elementary mechanics. Qxx = qxx − 31 qxx = M L2 Qyy = Qzz = − 13 qxx = − 21 M L2 other components = 0 . Letting the x-axis coincide with the line through the centers of the masses and the origin of space be at the midpoint between them.4. EINSTEIN’S EQUATION Solution of (b) ∫ E[γab ] = = ′ ∫Σ Σ′ ∫ = ∫Σ t00 [γab ]d3 x ′ t00 [γab + 2∂(a ξb) ]d3 x ′ t00 [γab + 2∂(a ξb) ]d3 x t00 [γab + 2∂(a ξb) ]d3 x = Σ = E[γab + 2∂(a ξb) ] Problem 8 Two point masses of mass M are attached to the ends of a spring of spring constant K. In the quadrupole approximation. From this we compute Qµν and P using equations (4.58). PROBLEM 9 41 . )2 ( ) d3 Qxx . . 1 1 1+ + dt3 . ret 4 4 . ) ( . 1 3 d3 2. = M 3 (L0 + A sin ωt) . where E = 21 KA2 . Solution To do this we proceed in three steps. Let the system execute its motion in the xy-plane with the origin of the coordinates coinciding with its center of mass. 4R These expressions allow us to write √ −8E 3 ω(E) = . so the energy of the system is M E=− 3 . dω/dt. 30 2 dt 1 P = 45 ( ret 16 A2 K 3 cos2 ωtret (4A sin ωtret + L0 )2 = 15 M The energy radiated in one period is ∫ 2π/ω Erad = P dt . M5 . Assuming the validity of equation (4.4. 15 Problem 9 A binary star system consists of two star of mass M and of negligible size in a nearly Newtonian circular orbit of radius R around each other. calculate the rate of increase of the orbital frequency due to emission of gravitational radiation. we find that the ratio Erad /E. is 16 √ 2πK 3/2 M −1/2 (4A2 + L20 ) . and the potential energy is −M 2 /2R. We relate the time derivative of the orbital frequency. The kinetic energy is M 2 /4R. The elementary relation M Rω 2 = M2 4R2 (left hand side is mass times acceleration and √ right hand side is force) gives the orbital frequency of the system: ω = M/4R3 . Step 1. 0 Computing the integral (using Maple). to the power P radiated in form of gravitational radiation.58). We compute P . √ dω 3 2 7/2 15/2 = M R dt 5 . We put the results of the two previous steps together. Qxy = Qyx = 6M sin ωt cos ωt − 2M .4. Using equation (4.58). sin ωt) with the origin of time suitably chosen. The position vector of the second mass is x⃗2 = −x⃗1 . Using Maple to compute P via (4. we compute the components of the quadrupole moment tensor. Qxy = 6M sin2 ωt − 2M .42 CHAPTER 4. EINSTEIN’S EQUATION so that dω dω dE = = 2−1/2 3(M R)−3/2 P .4. dt dE dt Step 2. other components = −2M . we find 2 P = M 5 R9 . qxx = 6M cos2 ωt qxy = qyx = 6M sin ωt cos ωt qxy = 6M sin2 ωt other components = 0 The trace-free quadrupole moment tensor is then given by Qxx = 6M cos2 ωt − 2M . 5 Step 3. The position vector of one of the masses is x⃗1 = R(cos ωt.8). k = 0. where r = sin ψ. can be put in the form above by use of spherical coordinates. ∑ ∂xµ ∂xν . Isotropic Cosmology Problem 1 Show that the Robertson-Walker metric. Consider the case of 3-sphere geometry. To write ds2 in the desired form. = (ds )ψψ dr 1 = . change the coordinates from (ψ.1.3. ϕ). (∂r/∂ψ)2 1 1 = = . (5. can be expressed in the form [ ] dr2 2 2 2 2 ds2 = −dτ 2 + a2 (τ ) + r (dθ + sin θ dϕ ) . By the tensor transformation law (2.ν ( )2 dψ 2 . θ.11).1) 1 − kr2 What portion of the 3-sphere (k=+1) can be covered by these coordinates? Solution The metric of Euclidean space.Chapter 5 Homogeneous. ∂r ∂r µ. equation (5. The metric is ds2 = dψ 2 + sin2 ψ(dθ2 + sin2 dϕ2 ) .8). θ. 2 cos ψ 1 − r2 (ds2 )rr = (ds2 )µν 43 . ϕ) to (r. 2. θ. 2π) × (0. (5.2. We compute the Ricci tensor components using the coordinate component methods of section 3. by letting (ψ. Now. ϕ). it is clear that z can take any possible value. ϕ) coordinates by this method will yield formula (5. π). half of the 3-sphere is (essentially) covered. θ. x and y can also take take any values which would be possible for points on a 3-sphere.1). we essentially cover the 3-sphere in the same way spherical coordinates essentially cover the 2-sphere in 3-space. θ. we note that the points on the 3-sphere satisfy w2 ≤ 1 . So it is clear that with the above range.44 CHAPTER 5. Step 1. Similarly. 1) while ψ ranges in (π/2. for the 3-sphere (k = +1) and hyperboloid (k = −1) cases.4a.15). θ. To see how much of the 3-sphere the coordinates (r. If we fix some choice of w. Solution We do this in two steps. To see this. w can take any possible value. Letting r range in (−1. x = cos ϕ cos θ cos ψ y = cos ϕ cos θ sin ψ z = cos ϕ sin θ w = sin ϕ Using the tensor transformation law. w2 + z 2 ≤ 1 . ISOTROPIC COSMOLOGY It is easy to see that computing the other coefficients of ds2 in (r. w2 + z 2 + y 2 + x2 = 1 .14) and (5. π) × (0. w2 + z 2 + y 2 ≤ 1 . we can verify expressions such as dx = − sin ϕ cos θ cos ψ dψ − cos ϕ sin θ cos ψ dθ − cos ϕ cos θ sin ψ dϕ and use them to show that dx2 + dy 2 + dz 2 + dw2 indeed equals dψ 2 + sin2 ψ(dθ2 + sin2 θ dϕ2 ) . ϕ) range in (0. . This justifies the “definition” given above. ϕ) cover. we recall the definition of (ψ. HOMOGENEOUS. 3π/2). Problem 2 Derive Einstein’s equations. 5) . gab X a X b gϕϕ Using Mathematica or Maple to plug the results of step 1 into the left hand side of (5. R∗∗ = Rϕϕ Rab X a X b = . Then we compute the Christoffel symbol components using equation (3. The best coordinate system to use is the one of the previous problem because it allows us to handle all three geometries at the same time.30). where u is some spacelike unit vector. R∗∗ = Rab ua ub . By definition.2) and (5. First we choose a coordinate system.2.2) R∗∗ − (5.2. we have.1. a2 a k a˙ 2 2¨ a − 2− 2− = 8πP .3) and simplifying the resulting expressions yields 3k 3a˙ 2 + 2 = 8πρ .1.5). Finally.PROBLEM 2 45 We review the method here.4. Notice that Rτ τ was found in step 1 and R = −Rτ τ + R∗∗ . in terms of ρ. k and a and its derivatives. Γτ rr = aa′ 1 − kr2 Γτ θθ = r2 aa˙ Γτ ϕϕ = r2 aa˙ sin2 θ Γr rr = Γr τ r = Γr rτ = Γθ τ θ = Γθ θτ = Γϕ τ ϕ = Γϕ ϕτ = kr 1 − kr2 Γr θθ = −r(1 − kr2 ) Γr ϕϕ = −r(1 − kr2 ) sin2 θ Γθ rθ = Γθ θr = Γϕ rϕ = Γθ ϕr = Γθ ϕϕ = − sin θ cos θ Γϕ θϕ = Γϕ ϕθ = 3a˙ ′ a 2k + 2a˙ 2 + a¨ ˙a = 2 1 − kr a˙ a 1 r cos θ sin θ Rτ τ = − Rθθ = −1 + 2kr2 − cot2 θ + csc2 θ + 2r2 a˙ 2 + r2 a¨ a Rrr Rϕϕ = r2 sin2 θ(2k + 2a˙ 2 + a¨ a) Step 2.5) and (3. so all that is needed is to find a formula for R∗∗ . The nonvanishing Christoffel and Ricci components follow.4. The computations (3.14) and (5.15) by writing Rτ τ + 12 R = 8πρ . we compute the Ricci tensor components using (3. P . for instance.4) (5. (5. We derive equations (5.3) 1 2R = 8πP . a a a (5.30) are straightforward and can easily be done using Mathematica or Maple. Letting X a = (∂/∂ϕ)a . 4) and (5.2.2. (We show the implication (5.15).7) . (The resulting space-time actually is spacetime homogeneous and isotropic and is known as the de Sitter spacetime. See Hawking and Ellis 1973 for further details.15) in the general case of possibly nonzero cosmological constant in the next problem.2. ISOTROPIC COSMOLOGY It is easy to check that the pair of equations (5. these equations become a˙ 2 3k = 8πρ + Λ − 2 . HOMOGENEOUS.12) are Gτ τ − Λ = 8πρ .5) is equivalent to the pair (5.17) with cosmological constant Λ. The solutions with k = ±1 correspond to different choices of spacelike hypersurfaces in this spacetime. a2 a a ¨ k a˙ 2 −2 = 8πP − Λ + 2 + 2 .14) and (5.) For a dust filled Einstein universe (P = 0). So the analogs of equations (5. a a a 3 (5.2.2.14) and (5.) Problem 3 (a) Consider the modified Einstein’s equation (5. Using the expressions for Gτ τ and G∗∗ found in the solution of the preceding problem.4) and (5.) Solution of (a) The Einstein equation with cosmological constant reads Gab + Λgab = 8πTab .2.11) and (5.2. (b) Consider the modified Einstein equation with Λ > 0 and Tab = 0. (These solutions are called Einstein static universes. Show that static solutions of these equations are possible if and only if k = +1 (3-sphere) and Λ > 0. relate the “radius of the universe” a to the density ρ. Examine small perturbations from the “equilibrium” value of a. G∗∗ + Λ = 8πP .6) (5.14) and (5. Obtain the spatially homogeneous and isotropic solution in the case k = 0.15) with the Λ-terms.46 CHAPTER 5.2. and show that the Einstein static universe is unstable. Write out the analogs of equations (5.5) ⇒ (5.2. 8) becomes 2 a˙ 2 3 3 = − 2 +Λ 2 2 a a a (5.8) admits a static (a˙ = 0) solution.8) admits a static solution. Λ > 0. it follows that Λ = 4πρ.PROBLEM 3 47 This system is equivalent to 2 a˙ 3k 3 = 8πρ + Λ − a2 a2 a ¨ 3 = −4π(ρ + 3P ) + Λ .8). We now write a in terms of ρ in the case of a static. Any deviation from C2 = 0 results in a exponential evolution of the scale factor. 4πρ Finally.e. we find a= √ 1 . it follows from the first equation that k > 0. Consider equations (5.14) and (5. System (5. a a Thus a ¨ = a˙ 2 /a.8) constitutes the analog of equations (5.9). a (5.9) 1 ¨ 3a =− 2 +Λ . Conversely. the conditions k = 0 and Tab = 0 imply that the scale factor a satisfies the equations a˙ 2 =Λ a2 a ¨ 3 =Λ. the metric has the form ds2 = −dτ 2 + a2 (τ )(dx2 + dy 2 + dz 2 ) . dust filled universe.2. P = 0.8).2. The solutions of this differential equation are a = C1 eC2 τ . Plugging this into the first equation (and using using k = 1). k = 1.. the static solution is unstable. From the second equation in (5. i..e.8) together with the conditions ρ = 1/4πa.1 Solution of (b) When k = 0. The system (5.8). Any suggestions? . Suppose now that (5. k = +1. From this (and the assumption ρ > 0).15). a 3 1 Problem with the stability analysis: The function a = C1 eC2 τ does not exactly satisfy (5. i. then it is possible to choose physically plausible ρ and P such that (5. if k = 1 and Λ > 0.6) into (5. we study the stability of the static solution for the dust filled universe. According to (5. The second equation implies (assuming ρ + 3P > 0) that Λ > 0.8) where the second equation is obtained by plugging (5. I would be very thankful if somebody sent me an email with a correct solution.10) and a˙ = da/dτ . where hab is defined by equation (5.1. any metric of this form with H = 2 Λ/3 satisfies Einstein’s equation. So the metric takes the form ds2 = −dτ 2 + eHτ (dx2 + dy 2 + dz 2 ) . To show uc ∇b hca = (a/a)h ˙ ab we work with the coordinates appearing on the Robertson-Walker metric as it is on the description of problem 1. The term hca ∇b uc expands to c c ua uc ∇ b u + gca ∇b u = ∇b ua . it was noted that a = C1 eC2 τ solves these equations. we find uc hca = uc gca + uc uc ua = ua − ua = 0 . (b) Show that along any null geodesic we have dω/dλ = −k a k b ∇a ub = 2 . Conversely. ISOTROPIC COSMOLOGY In item (a). Using the defining equation for hab . ∑ uσ ∇ν hσµ = −∇ν hτ µ σ = −∂ν hτ µ+ ∑ Γσ ντ hσµ + σ ∑ Γσ νµ hτ σ σ = Γν ντ hνµ = (a/a)h ˙ νµ The last two equalities are established by using the expressions for the Christoffel symbol components computed in the solution of problem 2. where λ is the affine parameter along the geodesic. Solution of (a) This isn’t really a “solution” since I got a sign wrong in the end.48 CHAPTER 5.6). (5. −(a/a)ω ˙ (c) Show that the result of (b) yields equation (5. It follows that ∇b (uc hca ) = 0 = uc ∇b hca + hca ∇b uc . The . Problem 4 Derive the cosmological redshift formula (5.3.6) by the following argument: (a) Show that ∇a ub = (a/a)h ˙ ab .3. HOMOGENEOUS.10) The equality on the right translates (except for a minus sign) into the equation we are trying to prove. where H = 2C2 and the coordinates were rescaled in such a way that the constant C1 does not√ appear in this expression. So the formula for ω gives dτ /dλ. because the geodesic is null. 0) and therefore uc = (−1. except for a minus sign. The minus sign is there to account for the fact that uc = (1. This implies da da =ω dλ dτ and consequently d(aω)/dλ = 0. 0). We get this component by contracting −uc k c . Plugging the results of the preceding paragraph in equation (5. Solution of (b) One of the equalities is established by noticing that ω = −ka ua and.PROBLEM 4 49 components for the plane geometry (k = 0) can also be found on page 97 of the book. k a ka = 0. a b −k a k b ∇a ub = −(a/a)k ˙ k hab a b = −(a/a)(k ˙ k ua ub + k a k b gab ) 2 k a k = −(a/a)(ω ˙ + a) The other equality follows from the very clever observation that a geodesic satisfies the geodesic equation: k a ∇a k b = 0. d(aω) da dω =ω +a dλ dλ dλ da da =ω − ω2 dλ dτ Now. To this effect we show that the derivative of aω along the null geodesic vanishes. 0.10). dω/dλ = −k a ∇a (kb ub ) = −(k a ∇a kb )ub − (k a ∇a ub )kb = −k a k b ∇a ub Solution of (c) We wish to show ω2 /ω1 = a(τ2 )/a(τ1 ) or. 0. which is the equation we wished to prove. 0. . the τ component of k a is dτ /dλ. a1 ω1 = a2 ω2 . equivalently. 0. we get (a/a)h ˙ ba + ∇b ua = 0 . 50 CHAPTER 5. 1 − cos η 0 . (a) Show that for all three spatial geometries the change ∫ τ2 in the coordinate ψ of the ray between times τ1 and τ2 is ∆ψ = τ1 dτ /a(τ ).1. k) = − +a dλ dλ holds for any k. Solution of (b) We assume the light ray’s world line to be a radial.” (c) Show that in the radiation filled spherical model. ISOTROPIC COSMOLOGY Problem 5 Consider a radial (dθ/dλ = dϕ/dλ = 0) null geodesic propagating in a Robertson-Walker cosmology. HOMOGENEOUS. null geodesic. ψ = (x2 + y 2 + z 2 )1/2 . [Here. dλ dλ We use this to compute ∆ψ. null geodesic. It follows that dτ dψ = ±a .” Solution of (a) Let k a be the tangent to a radial. According to item (a). τ = 12 C(η − sin η) . a light ray emitted at the big bang travels precisely halfway around the universe by the time of the “big crunch. the change in the angular coordinate ψ between two consecutive values of η such that a(η) = 0 is ∫ 2π 1 − cos η ∆ψ = dη = 2π.] (b) Show that in the dust filled spherical model. Equation ( )2 ( )2 dτ 2 2 dψ 0 = ds (k.1: a = 12 C(1 − cos η) . The solution to the spherical dust filled model is given in table 5. a light ray emitted at the big bang travels precisely all the way around the universe by the time of the “big crunch. in the flat case (k = 0). equation (5.11). ∫ λ2 ∫ λ2 ∫ τ2 dψ 1 dτ dτ ∆ψ = dλ = ± dλ = ± λ1 dλ λ1 a dλ τ1 a(τ ) Here λi is the parameter value corresponding to τi . ψ is defined to be the ordinary radial coordinate. The solution to the spherical radiation filled model is given in table 5.1: √ a(τ ) = B 1 − (1 − τ /B)2 . √ Here B corresponds to C ′ in the formula given in the table.PROBLEM 5 51 Solution of (c) We assume the light ray’s world line to be a radial. The integral is easily computed by making the substitution sin t = 1 − τ /B. . According to item (a). null geodesic. the change in the angular coordinate ψ between two consecutive values of η such that a(η) = 0 is 1 ∆ψ = B ∫ 0 2B √ dτ 1 − (1 − τ /B)2 = −π. ISOTROPIC COSMOLOGY .52 CHAPTER 5. HOMOGENEOUS. Using the first equation to write H(˜ r) = r2 /˜ r2 and plugging this in the second equation. We wish to find r˜ so that H(˜ r)˜ r2 = r2 .3). as usual.1.1) .) (b) Show that in isotropic coordinates the Schwarzschild metric is ) ( (1 − M/2˜ r )2 2 M 4 2 2 ds = − dt + 1 + [d˜ r + r˜2 dΩ2 ].Chapter 6 The Schwarzschild Solution Problem 1 Let M be a three-dimensional manifold possessing a spherically symmetric Riemannian metric with ∇a r ̸= 0. (This shows that every spherically symmetric three-dimensional space is conformally flat. (1 + M/2˜ r )2 2˜ r Solution to (a) By spherical symmetry and the condition ∇a r ̸= 0. dΩ2 = dθ2 + sin2 θ dϕ2 . where r is defined by equation (6. r˜ d˜ r or d˜ r r˜ = . we find r dr = h(r) . the metric on M can be written as ds2 = h(r)2 dr2 + r2 dΩ2 . √ dr H(˜ r) = h(r) d˜ r for some H. dr h(r)r 53 (6. (a) Show that a new “isotropic” radial coordinate r˜ can be introduced so that the metric takes the form ds2 = H(˜ r)[d˜ r2 + r˜2 dΩ2 ]. where. Solution It took me 15 minutes to do this with Mathematica. spherically symmetric spacetime. we know this has a solution.1. where the (eµ )a are defined by equation (6. In one hand we want H(˜ r) to have the form (1 + M/2˜ r)4 . Feel free to send me an email if you’d like a copy of the code. The computer didn’t get stuck trying to do any “dumb algebra” and everything went fine.5). one gets ( M r = r˜ 1 + 2˜ r )2 . . using the coordinate component method of section 3. Problem 2 Calculate the Ricci tensor.5). We use this solution as the new coordinate.3.1.54 CHAPTER 6.1) with h(r) = (1 − 2M/r)−1 hold. equation (6.13) in a static.3. equation (6. Thus the Schwarzschild metric takes the desired form in these coordinates. Compare the amount of labor involved with that of the tetrad method approach given in the text.1. spherically symmetric spacetime. THE SCHWARZSCHILD SOLUTION From the theory of ordinary differential equations. Putting these two equalities together. On the other 2 2 we saw in the solution to item (a) that H(˜ r) = r /˜ r . Solution to (b) We do this by defining r˜ in a way that seems natural and checking that it yields the correct form for the metric.6).12) and (4. Problem 3 Consider the source-free (j a = 0) Maxwell’s equations (4. It is straightforward to check that with this relation between r and r˜. (a) Argue that the general form of a Maxwell tensor which shares the static and spherical symmetries of the spacetime is Fab = 2A(r)(e0 )[a (e1 )b] + 2B(r)(e2 )[a (e3 )b] .4a. equations ( ) (1 − M/2˜ r)2 2M = 1− (1 + M/2˜ r)2 r and (6. Rab . for a static. Solution to (b) In differential forms notation. . f ∗ dr = dr . dF = 0 . B. C. f ∗ dθ = dθ . Also. F = A(r)e0 ∧ e1 . Gab = 8πTab . It is intuitively clear that under the action of f (a rotation by π). r. Because Fab is assumed to be static. [The solution obtained with B(r) ̸= 0 is a “duality rotation” of this solution. the functions A. they do not depend on θ of ϕ because of spherical symmetry. Similarly. It follows that f ∗ Fab = A(e0 )[a (e1 )b] + B(e0 )[a (e2 )b] − C(e0 )[a (e3 )b] + + D(e1 )[a (e2 )b] − E(e1 )[a (e3 )b] + G(e2 )[a (e3 )b] . where A. ϕ).D and E vanish. . Let f be a spacetime symmetry (a diffeomorphism) which rotates θ by π. Show that the general solution is the Reissner-Nordstrom metric. ds2 = − 1 − + 2 dt2 + 1 − + 2 r r r r Solution to (a) The most general form of a 2-form Fab in Schwarzschild spacetime is Fab = A(e0 )[a (e1 )b] + B(e0 )[a (e2 )b] + C(e0 )[a (e3 )b] + + D(e1 )[a (e2 )b] + E(e1 )[a (e3 )b] + G(e2 )[a (e3 )b] . the general solution of Maxwell’s equations with the form of part (a) is A(r) = −q/r2 . So it only remains to show that B. since B(r) = 0. E and G are functions of (t. the differential dϕ becomes f ∗ dϕ = −dϕ. representing the field of a magnetic monopole. . C. ) )−1 ( ( 2M q2 2M q2 dr2 + r2 dΩ2 . . where q may be interpreted as the total charge. . This can also be checked using the definitions found in Appendix C of the book.] (c) Write down and solve Einstein’s equation. Maxwell’s equations for F read d∗F =0 . with the electromagnetic stress-energy tensor corresponding to the solution of part (b). one can show that B = D = 0. The other differentials remain unchanged: f ∗ dt = dt . it follows that C = E = 0. D. θ.PROBLEM 3 55 (b) Show that if B(r) = 0. Because f ∗ Fab = Fab . G do not depend on t. . Let M be an n-dimensional manifold. (2) and (3). which is also called the “Hodge star operator”. It is a nice exercise to prove (6. The ∗ operator defined in the problem set for chapter 4. THE SCHWARZSCHILD SOLUTION because there are no sources (see problem 2(b) in chapter 4). e1 . ∗{A(r)e0 ∧ e1 } = A(r)e2 ∧ e3 . if it is a function.2).56 CHAPTER 6.4) (6. ∗(e0 ∧ e1 ∧ e2 ) = e3 . (6. e2 . . then df is the 1-form df = n ∑ ∂f i dx . .3) (6. Now we discuss (6. has a complicated formula. . but instead the following characterization of it: Theorem. This is the equality we wished to prove. (3) For every form ω. . (2) If ω and η are forms of orders k and l. ∂xi i=1 where {x1 . then d(ω ∧ η) = dω ∧ η + (−1)k ω ∧ dη . d(dω) = 0 . ∗(e0 ∧ e1 ) = e2 ∧ e3 .5) From this we conclude that A(r)r2 equals some constant −q. defined for all k. respectively. or A(r) = −q/r2 .5) using only properties (1). such that: (1) If f is a 0-form. that is. I wont prove this theorem here. There exists a unique linear map d : Λk −→ Λk+1 . we find 0 = d ∗ F = d{A(r)r2 sin θ dθ ∧ dϕ} = d{A(r)r sin θ} dθ ∧ dϕ ∂ = (A(r)r2 ) sin θ dr ∧ dθ ∧ dϕ . e3 } is a positive basis of the space of 1-forms.2) Writing e2 = r dθ and e3 = r sin θ dϕ and using a few properties of the d operator. Denote by Λk the vector space of all k-forms on M . If {e0 . I didn’t have in mind the formula for the d operation given in Appendix B of the book. ∂r 2 (6. xn } can be any set of coordinates on M . When doing these computations.3)-(6. then ∗(e0 ) = e1 ∧ e2 ∧ e3 . but what it really does is simple. As will be discussed below. (e0 ∧ e1 ∧ e2 )abc = 3(e0 ∧ e1 )[ab (e2 )c] . .17).1. one finds that (e0 ∧ e1 ∧ e2 )abc = 6(e0 )[a (e1 )b (e2 )c] . 1 ∗(A(r) e0 ∧ e1 )cd = (A(r) e0 ∧ e1 )ab ϵabcd . (e0 ∧ e1 )ab = 2(e0 )[a (e1 )b] .2).PROBLEM 3 57 In other contexts one defines the Hodge star operation by requiring it to be linear and that these equations hold for every even permutation of the indices {1. Plugging the first equation into the second and expanding. space components don’t.8) since using an orthonormal basis indices are raised and lowered as in Minkowsky spacetime: time components change sign. √ ϵ = |g| dt ∧ dr ∧ dθ ∧ dϕ . This is all really nice. (6. Putting (6. but we now turn to prove (6. . According to (B. one has A(r)(e0 ∧ e1 )ab = A(r)[(e0 )a (e1 )b − (e0 )b (e1 )a ] .6a-d) 2 {e0 . it isn’t hard to believe that ϵ = e0 ∧ e1 ∧ e2 ∧ e3 = 24(e0 )[a (e1 )b (e2 )c (e3 )d] .7) and (6. and therefore A(r)(e0 ∧ e1 )ab = −A(r)[(e0 )a (e1 )b − (e0 )b (e1 )a . . . 3. (6.2.7) Now. Having proved this.2) using the regular abstract index notation. Now. According to the definition given in chapter 4. 2. Looking at these formulas again. e3 }) that |g| = f gr sin θ. It is easy to see from (which define the orthonormal basis √formulas √ (6. 4}. from definition. This is (6. we find ϵ = e0 ∧ e1 ∧ e2 ∧ e3 .6) We must find what each term on the right hand side means. .6) and contracting (this isn’t hard thanks to orthonormality) results in ∗(A(r) e0 ∧ e1 )cd = (1/2)A(r)4(e2 )[c (e3 )d] = A(r)(e2 ∧ e3 )cd .8) into (6. 2 (6. 8π Einstein’s equations then read ) ( q2 h′ 1 1 . It follows that f h = K for some constant K. Similarly. [8πT00 = G00 ] = 2 + 2 1− r4 rh r h ) ( q2 1 f′ 1 − 4 = − . one gets the correct formula for h (which is the factor multiplying dr2 in the metric). f h This can be rewritten as f ′ h + f h′ = 0 = (f h)′ .2. one can assume f = 1/h. 4 Remember that raising or lowering the index of e0 produces a minus sign. This is the correct formula for f (which is the factor multiplying −dt2 in the metric). Adding the equations [8πTii = Gii ] for i = 1. Solving this equation. Changing the scale with which time is measured if necessary. So if Ba = 3(e0 )a +11(e1 )a . r2 dr h Multiplying both sides of [8πT00 = G00 ] by r2 and integrating.) The right hand side of the first equation can be seen to equal ( ( )) 1 d 1 r 1− . one gets f ′ h′ + =0.2. [8πT22 = G22 ] r4 2rf h 2rh2 2 f h dr fh Tab = (See [6. [8πT11 = G11 ] 1− r rf h r2 h ) ( q2 f′ 1 d f′ h′ √ √ + = − . Fac Fb c = A2 [(e0 )a (e1 )c − (e0 )c (e1 )a ][−(e1 )b (e0 )c + (e0 )b (e1 )c ] = A2 [(e0 )a (e0 )b − (e1 )a (e1 )b ] . THE SCHWARZSCHILD SOLUTION Solution of (c) First we compute Tab 1 = 4π { } 1 c de Fac Fb − gab Fde F . one obtains ) ( q2 1 r 1− = − + 2M . . h r where 2M is the constant of integration. then B a = −3(e0 )a +11(e1 )a .3]-[6. Putting these results together yields A2 [(e0 )a (e0 )b − (e1 )a (e1 )b + (e2 )a (e2 )b + (e3 )a (e3 )b ] .58 CHAPTER 6. This completes the proof of the fact that the general solution of Einstein’s equation in these circumstances is the Reissner-Nordstrom metric. and that raising or lowering the index of the other basis elements has no effect on the components. Fdc F cd = −2A2 .5] in book. 2 and multiplying the result by rh. with the other end of the string held by a stationary observer at large r.e. Suppose a particle of mass m is held stationary by a (massless) string. Use conservation of energy arguments to show that the force exerted by the observer at infinity on the other end of the string is F∞ = V F . z [with ξ a = (∂/∂t)a ] such that the components of gab approach diag(−1.3. Hence. . show that the maximum lifetime of any observer in region II is τ = πM [∼ 10−5 (M/M⊙ ) s]. 1.gab ) is asymptotically flat.e.e. Figure 6.gab ) be a stationary spacetime with timelike Killing field ξ a .PROBLEM 4 59 Problem 4 Let (M . equation (6. Problem 6 Show that any particle (not necessarily in geodesic motion) in region II (r < 2M ) of the extended Schwarzschild spacetime. Thus the magnitude of the force exerted at infinity differs from the force exerted locally by the redshift factor.. i. y. 1) as r → ∞.45). any observer in region II will be pulled into the singularity at r = 0 within this proper time. for the general relativistic time delay. x. Let V 2 = −ξ a ξa . Let F denote the force exerted by the string on the particle.. the “energy as measured at infinity” of a particle of mass m and 4-velocity ua is E = −mξ a ua . (b) Suppose in addition that (M . must decrease its radial coordinate at a rate given by |dr/dτ | ≥ [2M/r − 1]1/2 . Problem 5 Derive the formula. (See chapter 11 for further discussion of asymptotic flatness. (a) Show that the acceleration ab = ua ∇a ub of a stationary observer is given by ab = ∇b ln V . where r = (x2 + y 2 + z 2 )1/2 .9. i. geodesic) motion from r = 2M with E → 0. According to part (a) we have F = mV −1 [∇a V ∇a V ]1/2 . that there exist coordinates t. Show that this maximum time is approached by freely falling (i. 1.) As in the case of the Schwarzschild metric..