Solutions: The Geometry of Spacetime

Transcription

1 Solutions: The Geometry of Spacetime James Callahan

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3 iii Preface This book consists of detailed solutions to the exercises in my text The Geometry of Spacetime (Springer, New York, 2000). My Web site, callahan/ contains additional material, including an errata file for the text, spacetime/errata.pdf This version of the solutions manual takes into account all corrections to the text found in the errata file up to 3 December 200. Given the nature of the material, it is likely that the solutions themselves still have some errors. I invite readers of this book to contact me at callahan@math.smith.edu to let me know about any errors or questions concerning the solutions found here. I will post a solutions errata file on my Web site. This solutions manual is available only directly from Springer, via its Web site J. Callahan January 20

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5 Contents Relativity before Einstein. Spacetime Galilean Transformations The Michelson Morley Experiment Maxwell s Equations Special Relativity Kinematics 5 2. Einstein s Solution Hyperbolic Functions Minkowski Geometry Physical Consequences Special Relativity Kinetics Newton s Laws of Motion Curves and Curvature Accelerated Motion Arbitrary Frames Uniform Rotation Linear Acceleration Newtonian Gravity Gravity in Special Relativity Surfaces and Curvature The Metric Intrinsic Geometry on the Sphere De Sitter Spacetime Curvature of a Surface Intrinsic Geometry Theorema Egregium Geodesics Curved Spacetime Mappings Tensors General Relativity The Equations of Motion The Vacuum Field Equations The Matter Field Equations v

6 vi CONTENTS 8 Consequences The Newtonian Approximation Spherically Symmetric Fields The Bending of Light Perihelion Drift

7 Solutions: Chapter Relativity before Einstein. Spacetime. a) 0. z 0.5. b) The maximum z occurs where 0 = z = 2t 3t 2, or t = 2/3. At that time, z = (2/3) 2 (2/3) 3 = 4/27. c) The velocity at departure is z (0) = 0. The velocity at return is z () =. d) Replacing the factor ( t) by ( t) 2 in z(t) will make z () = 0. Then, for example, any graph of the form z = kt 2 ( t) 2 has the correct velocities; when k = 4, the maximum value of z is. 2. a) Because z (t) = g and z (0) = v 0, integration gives z (t) = gt + v 0. Because z(0) = h, another integration gives z(t) = gt 2 /2+v 0 t + h. b) z From left to right, the three graphs have v 0 = 2,0,+0. c) An object hits the ground when gt 2 /2+v 0 t + h = 0, or t = v 0 v gh = 3.30, 3.499, g t t respectively, for v 0 = 2,0,+0. The order is what we would expect. 3. The worldline of M is z = vt + λ. This meets the worldline of the signal (i.e., z = t) when t = z = vt + λ. Thus ( v)t = λ, or t = λ v. At the event E, z = z = t. The worldline of the reflected signal is therefore z z = (t t ) or z = t + 2t. This meets the worldline of G (i.e. z = vt) when vt 2 = z = t 2 + 2t, or t 2 = 2t +v = 2λ..2 Galilean Transformations. According to R, the spatial distance between E and E 2 is 0. According to G, the coordinates of E are (, v) and the coordinates of E 2 are (2, 2v), and so the spatial distance between them is v. 2. Let M be the matrix whose columns are the vectors X and Y, in that order. It is a standard result of linear algebra that detm = A(X,Y); see Exercise 4, 5. (Solutions page 40). In particular, the parallelogram determined by X and Y is oriented: it has positive orientation if detm > 0 and negative orientation if detm < 0. If detm = 0, the parallelogram collapses to a line segment (and has area 0). Now use the given matrix L to construct the product matrix LM. By the definition of matrix multiplication, this is a matrix whose columns are L(X) and L(Y), in that order. Therefore, A(L(X),L(Y)) = detlm = detl detm = detl A(X,Y). Thus L reverses orientation if detl < 0 and collapses the image to a line segment if detl = A linear transformation L preserves areas if detl = ±. Because dets v =, S v preserves areas.

8 2 SOLUTIONS: CHAPTER. RELATIVITY BEFORE EINSTEIN 4. a) We have ( )( ) 0 0 S v S w = = v w ( ) 0 = S v+w v+w, and similarly, S w S v = S w+v = S v+w. Therefore S v S v = S v S v = S 0 = I, implying S v = S v. b) We have S w S v = S w+v = S v+w = S v S w. The previous work shows that the map v S v : R G 2 is an onto group homomorphism. Since the kernel of this map is just v = 0, the homomorphism is an isomorphism. 5. a) The given information implies that G s position at time t = τ is x = 0, y = v y t, z = v z t. Thus the event with coordinates (τ, ξ, η, ζ) in G s frame has coordinates t = τ, x = ξ, y = η + v y τ, z = ζ + v z τ in R s frame. If we set aside the x = ξ coordinate, this can be written in matrix form as t y = 0 0 v y 0 τ η, so S v = 0 0 v y 0, z v z 0 ζ v z 0 with v = (v y,v z ). b) Direct calculation gives S v S w = 0 0 v y w y 0 v z 0 w z 0 = 0 0 v y + w y 0 = S v+w = S w+v. v z + w z 0 In particular, S v S v = S 0 = I = S v S v, implying S v = S v. c) The groupg 3 is commutative because w+v = v+w. 6. In Exercise 5a, above, change the x-coordinate of G s position at time t = τ to x = v x τ, and change the x- coordinate of the event to x = ξ + v x τ. This makes the matrix form of the relation between R s coordinates of the event and G s coordinates equal to S v = v x 0 0 v y 0 0, v z 0 0 with v = (v x,v y,v z ). An immediate calculation gives S v S w = S v+w = S w S v, so multiplication in G 4 is commutative. Furthermore, as in Exercise 5, S v S v = S 0 = I, so G 4 is a group..3 The Michelson Morley Experiment. a) The figure below is similar to the SPACE diagram on page 8 of the text, except that it shows the movement of G (the source of light) and the mirror that lies parallel to the track of G over time. Note that, at every instant, the line from G to the mirror remains perpendicular to the direction of motion as given by the vector v. After time T /2, G and the mirror have travelled vt /2 lightseconds to the right. By the Pythagorean theorem, the light ray from G to the mirror has therefore travelled λ 2 + v 2 T 2 /4 light-seconds. G at time 0 M at time T /2 G v λ vt /2 vt /2 G at time T That light ray is reflected back to G (after G has travelled to the right a total of vt light-seconds), travelling an equal distance and time on its return. But the distance travelled, in light-seconds, is numerically equal to the travel time T, in seconds: T = 2 λ 2 + v 2 T 2 /4 seconds. Hence T 2 4 = λ 2 + v 2 T 2 4 from which it follows that T = or 2λ ( v 2 ) T 2 4 = λ 2, Finally, the distance from G to M and back is D = 2λ, so the average speed of light is D /T =. b) The travel time T is the time t 2 that was determined in the solution to Exercise of.. Because D = 2λ, the average speed of light is D /T =, not.

9 .4. MAXWELL S EQUATIONS 3 c) By Taylor s theorem, = +v2 + O(v 4 ), = + 2 v2 + O(v 4 ), as v 0. Therefore, T T = ( ) 2λ = λ v 2 + O(v 4 ) = v 2 (λ + O(v 2 )) as v 0. We use the fact that v 2 O(v 2 ) = O(v 4 ) as v a) z Each vertical line is mapped to itself by both F v and S v. But S v simply translates points on the line t = a by the amount av, while F v also compresses them by the factor. Both maps turn horizontal lines into lines with slope v. 4. a) The vectors (,±) t (where V t is the transpose of V) lie on the two worldlines. The image worldlines therefore contain the vectors ( )( 0 v ±) ( = v ± ). The slopes of these worldlines are The map C v compresses the image vertically by the factor. The image grid thus consists of rectangles of size parallel to the coordinate axes. b) Matrix multiplication gives ( )( ) ( ) S v C v = v 0 = = F v v ; however, when v C v S v = ( 0 v z ) t F v m ± = v ± = v ±. b) The graph of w = is the unit circle centered at (0,0) in the (v,w)-plane; therefore it lies above the line w = v on the interval 0 < v <. In other words, > v there. Hence m + = v+ > v+ v =. Because > and v > 0 on the same domain, m = v > 0 =. The maximum value of m + (v) occurs when 0 = (m + ) (v) = v/. Hence v = / 2 and m + (/ 2) = If F v is to be valid, it must be consistent with the Michelson Morley experiment. That is, the image worldlines of photons must have slopes ±. Exercise 4b demonstrates that this is not so if v > 0..4 Maxwell s Equations The map F v first compresses the image vertically by the factor (this is the action of C v ) and then carries out a vertical shear with slope v (the action of S v ). t. a) Let A = (a,a 2,a 3 ), B = (b,b 2,b 3 ), C = (c,c 2,c 3 ); then ( ) b B C = 2 b 3 c 2 c 3, b 3 b c 3 c, b b 2 c c 2. The first component of A (B C) is therefore b a b 2 2 c c 2 a 3 b 3 b c 3 c = b (a 2 c 2 + a 3 c 3 ) c (a 2 b 2 + a 3 c 3 ) = b (a c + a 2 c 2 + a 3 c 3 ) c (a b + a 2 b 2 + a 3 b 3 ) = (A C)b (A B)c.

10 4 SOLUTIONS: CHAPTER. RELATIVITY BEFORE EINSTEIN (The underlined terms that have been added to the expression offset each other.) In a similar way we can show that the second and third components of A (B C) are respectively. (A C)b 2 (A B)c 2 and (A C)b 3 (A B)c 3, b) Let = A = B and F = C. Using the fact that can be treated like a vector in vector equations, together with the identity A (B C) = B(A C) (B C)C (note the reordered factors in the first term), we have ( F) = ( F) ( )F = ( F) 2 F. 2. Let F = (P,Q,R), where P(x,y,z), Q(x,y,z) and R(x, y, z) are smooth functions. By definition, F = (R y Q z,p z R x,q x P y ); subscripts denote partial derivatives. Again, by definition, ( F) = (R y Q z ) x +(P z R x ) y +(Q x P y ) z = R yx Q zx + P zy R xy + Q xz P yz = 0, by the commutativity of partial differentiation (for example, R yx = R xy, etc.). 3. From Maxwell s equation H = E + J we obtain t ( ) E 0 = ( H) = + J = t t E+ J, again using the commutativity of partial differentiation. This equation, together with Maxwell s equation E = ρ, implies ρ t = J. 4. These equalities follow from the chain rule. Because ζ = z = vt, we have ζ z = and ζ t = v. Because τ = t and E(t,z) =E(τ,ζ vτ), we find E z = E z = E ζ ζ z =E ζ, E zz = E ζ z = E ζ ζ ζ z =E ζζ, E t = E = E τ t τ t + E ζ =E τ ve ζ t ζ E tt = E τ v E ζ =E ττ ve t t τζ v ( ) E ζτ ve ζζ =E ττ 2vE τζ + v 2 E ζζ. 5. The relation between the Greek and the Roman variables is therefore τ = t vz, τ t = τ t =, ζ t = ζ t ζ = z vt ; τ z = τ z = v, = v, ζ z = ζ z =. Using E(t,z) =E(τ,ζ), we find E t =E τ τ t +E ζ ζ t = E τ ve ζ, E tt = ( ) Eττ τ t +E τζ ζ t ve ζτ τ t ve ζζ ζ t = E ττ 2vE τζ + v 2 E ζζ, E z =E τ τ z +E ζ ζ z = ve τ +E ζ ( ) E zz = veττ τ z ve τζ ζ z +E ζτ τ z +E ζζ ζ z = v2 E ττ 2vE τζ +E ζζ. 6. a) Let u = z ± ct, so u t = ±c, u z = ; then t h(z ± ct) = h (u) u t = ±ch (u), 2 t 2 h(z ± ct) = ±ch (u) u t = c 2 h (u), z h(z ± ct) = h (u) u z = h (u), 2 t 2 h(z ± ct) = h (u) u t = h (u), for any (sufficiently differentiable) function h(u). Hence, if E(t,z) = f(z ct)+g(z+ct), then E tt = c 2 ( f(z ct)+g(z+ct)) = c 2 E zz. b) It is a standard fact that the graph of w = f(z ct 0 ) is the graph of w = f(z) translated by the amount ct 0. Thus the spike moves to the point z = ct at time t, so it travels with velocity c. c) The graph of w = f(z+ct 0 ) is the graph of w = f(z) translated by ct 0. The spike moves to z = ct at time t, so it travels with velocity c.

11 Solutions: Chapter 2 Special Relativity Kinematics 2. Einstein s Solution Note: we use 0 to denote the zero vector in any R k.. We can write the straight line l passing through the point A in the direction D 0 parametrically in the form X = Dt + A, < t <. If L is linear, then L(X) = L(Dt + A) = L(D)t + L(A); this is the straight line through L(A) in the direction L(D). If L(D) = 0 for some D 0, then L collapses all of R 2 to a line (or a point), so there is nothing further to prove. To continue, we assume L is invertible. If l is parallel to l, then its direction vector D is a nonzero multiple of D: D = kd, k 0, and we can parametrize l as Y = D s+b. The image of l under L is the line L(Y) = L(D )s+l(b) = kl(d)s+l(b). The direction vector here, kl(d), is a nonzero multiple of L(D), so l and l have parallel images. If the points P j are equally spaced on l, then P j = Dt j + A, where the parameter values t j = t 0 + n t are equally spaced on the t-axis, and conversely. The image points, L(P j ) = L(D)t j + L(A), have (the same!) equally-spaced parameter values, so they are equally spaced. 2. a) Suppose α and β are distinct parallel lines whose images M(α) and M(β) intersect. Let A on α have the same image as B A on β : M(A) = M(B) Thus M fails to be, contradicting the assumption that M is invertible. It follows that the images of distinct parallel lines never intersect. By hypothesis, the images are straight lines, so they must parallel and distinct. b) Let X and Y be arbitrary nonzero vectors in R 2. Let l be the unique line containing 0 and Y, and let l 2 be the unique line containing X and X +Y. These two lines are parallel, so their images are parallel. We can take the direction of l to be Y, and the direction of its image M(l ) to be M(Y) M(0) = M(Y) because M(0) = 0, by hypothesis. Moreover, M(Y) 0 because M is invertible, so M(Y) is a valid direction vector. We can also take Y to be the direction vector for the parallel line l 2, and thus we can take M(Y) for the direction of its image M(l 2 ). Because M(X) and M(X +Y) are two distinct points on M(l 2 ), we can write M(X +Y) M(X) = k Y M(Y), k Y 0. By reversing the roles of X and Y (and constructing two more parallel lines), we can also write M(X +Y) M(Y) = k X M(X), k X 0. Combining these, we find ( k X )M(X) = ( k Y )M(Y). If the coefficients are different from 0, M(X) and M(Y) are linearly dependent. But X and Y are arbitrary nonzero vectors and M is invertible, so we are forced to conclude the coefficients are zero, so k Y = k Y =. This implies M(X +Y) = M(X)+M(Y). c) Suppose Y j X; then continuity of M implies M(Y j ) M(X). Using the additively of M established in part b, we then find M(X +Y j ) = M(X)+M(Y j ) M(X)+M(X) = 2M(X). We also have X +Y j X + X = 2X, so continuity of M implies M(X + Y j ) M(2X). Hence, because a limit is unique, M(2X) = 2M(X). d) Additivity of M allows us to write M((k )X +Y j ) = M(X +(k 2)X +Y j ) = M(X)+M((k 2)X +Y j ). 5