The Geometry of Spacetime: An Introduction to Special and General Relativity (Instructor Solution Manual, Solutions) [1, 1st ed. 2000. Corr. 2nd printing 2001] 0387986413, 9780387986418

Citation preview

Solutions: The Geometry of Spacetime James Callahan

DVI file created at 16:06, 20 January 2011

DVI file created at 16:06, 20 January 2011

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, http://maven.smith.edu/˜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 31 December 2010. 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 [email protected] 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 http://www.springer.com/instructors?SGWID=0-115-12-333200-0 J. Callahan January 2011

DVI file created at 16:06, 20 January 2011

iv

DVI file created at 16:06, 20 January 2011

Contents 1

Relativity before Einstein 1.1 Spacetime . . . . . . . . . . . . . . 1.2 Galilean Transformations . . . . . . 1.3 The Michelson–Morley Experiment 1.4 Maxwell’s Equations . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

1 1 1 2 3

Special Relativity—Kinematics 2.1 Einstein’s Solution . . . . 2.2 Hyperbolic Functions . . . 2.3 Minkowski Geometry . . . 2.4 Physical Consequences . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

5 5 9 13 17

3

Special Relativity—Kinetics 3.1 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Curves and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Accelerated Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20 20 22 27

4

Arbitrary Frames 4.1 Uniform Rotation . . . . . . 4.2 Linear Acceleration . . . . . 4.3 Newtonian Gravity . . . . . 4.4 Gravity in Special Relativity

2

5

6

7

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

30 30 31 34 37

Surfaces and Curvature 5.1 The Metric . . . . . . . . . . . 5.2 Intrinsic Geometry on the Sphere 5.3 De Sitter Spacetime . . . . . . . 5.4 Curvature of a Surface . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

39 39 43 45 47

Intrinsic Geometry 6.1 Theorema Egregium 6.2 Geodesics . . . . . . 6.3 Curved Spacetime . . 6.4 Mappings . . . . . . 6.5 Tensors . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

53 53 60 64 67 73

General Relativity 7.1 The Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Vacuum Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Matter Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80 80 86 92

. . . . .

. . . . .

. . . . .

. . . . .

. . . .

. . . . .

. . . . .

v

DVI file created at 16:06, 20 January 2011

CONTENTS

vi 8

Consequences 8.1 The Newtonian Approximation 8.2 Spherically Symmetric Fields . 8.3 The Bending of Light . . . . . 8.4 Perihelion Drift . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

DVI file created at 16:06, 20 January 2011

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

98 . 98 . 100 . 107 . 107

Solutions: Chapter 1

Relativity before Einstein 1.1 Spacetime

respectively, for v0 = −2, 0, +10. 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 t1 = z = vt1 + λ . Thus λ . (1 − v)t1 = λ , or t1 = 1−v At the event E1 , z = z1 = t1 . The worldline of the reflected signal is therefore z − z1 = −(t − t1 ) or z = −t + 2t1 . This meets the worldline of G (i.e. z = vt) when

z

1. a) 0.1

0.5

t

1.

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′ (1) = −1.

vt2 = z = −t2 + 2t1,

(1 − t)2

d) Replacing the factor (1 − t) by in z(t) will make z′ (1) = 0. Then, for example, any graph of the form z = kt 2 (1 − t)2 has the correct velocities; when k = 4, the maximum value of z is 1.

or t2 =

2t1 2λ . = 1 + v 1 − v2

1.2 Galilean Transformations

1. According to R, the spatial distance between E1 and 2. a) Because z′′ (t) = −g and z′ (0) = v0 , integration E2 is 0. According to G, the coordinates of E1 are (1, −v) gives z′ (t) = −gt + v0 . Because z(0) = h, another inte- and the coordinates of E2 are (2, −2v), and so the spatial gration gives z(t) = −gt 2 /2 + v0t + h. distance between them is v. b)

z

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, § 5.1 (Solutions page 40). In particular, the parallelogram determined by X and Y is oriented: it has positive orientation if det M > 0 and negative orientation if det M < 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,

70 60 50 40 30 20 10

0

1

2

3

4

5

t

From left to right, the three graphs have v0 = −2, 0, +10.

A(L(X), L(Y )) = det LM = det L · detM = detL · A(X,Y ).

c) An object hits the ground when −gt 2 /2 + v0t + h = 0, or q −v0 − v20 + 2gh t= = 3.301, 3.499, 4.665 −g

Thus L reverses orientation if detL < 0 and collapses the image to a line segment if det L = 0. 3. A linear transformation L preserves areas if det L = ±1. Because det Sv = 1, Sv preserves areas. 1

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 1. RELATIVITY BEFORE EINSTEIN

2

4. a) We have      1 0 1 0 1 0 Sv Sw = = = Sv+w , v 1 w 1 v+w 1

1.3 The Michelson–Morley Experiment

1. a) The figure below is similar to the SPACE diagram on page 18 of the text, except that it shows the movement of G (the source of light) and the mirror that lies parallel b) We have Sw Sv = Sw+v = Sv+w = Sv Sw . The previous to the track of G over time. Note that, at every instant, work shows that the map v 7→ Sv : R → G2 is an onto group the line from G to the mirror remains perpendicular to homomorphism. Since the kernel of this map is just v = 0, the direction of motion as given by the vector v. After the homomorphism is an isomorphism. time T⊥ /2, G and the mirror have travelled vT⊥ /2 lightseconds to the right. By the Pythagorean theorem, the 5. a) The given information implies that G’s position at light ray from G to the mirror has therefore travelled time t = τ is x = 0, y = vyt, z = vzt. Thus the event with q coordinates (τ , ξ , η , ζ ) in G’s frame has coordinates λ 2 + v2 T⊥2 /4 light-seconds. and similarly, Sw Sv = Sw+v = Sv+w . Therefore S−v Sv = Sv S−v = S0 = I, implying S−v = Sv−1 .

t = τ,

x = ξ,

y = η + vy τ ,

z = ζ + vz τ

M at time T⊥/2

in R’s frame. If we set aside the x = ξ coordinate, this can be written in matrix form as        τ 1 0 0 t 1 0 0 y = vy 1 0 η  , so Sv = vy 1 0 , ζ vz 0 1 z vz 0 1

λ

with v = (vy , vz ).

b) Direct calculation gives    1 0 0 1 0 0 Sv Sw = vy 1 0 wy 1 0 vz 0 1 wz 0 1   1 0 0 = vy + wy 1 0 = Sv+w = Sw+v . vz + wz 0 1

In particular, Sv S−v = S0 = I = S−v Sv , implying S−v = Sv−1 . c) The group G3 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 = vx τ , and change the xcoordinate of the event to x = ξ + vx τ . This makes the matrix form of the relation between R’s coordinates of the event and G’s coordinates equal to   1 0 0 0 vx 1 0 0  Sv =  vy 0 1 0 , vz 0 0 1

vT⊥/2 G at time 0

vT⊥/2 G v

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: q T⊥ = 2 λ 2 + v2 T⊥2 /4 seconds.

Hence

T⊥2 T2 = λ 2 + v2 ⊥ 4 4

or

1 − v2

from which it follows that T⊥ = √


T⊥2 = λ 2, 4

2λ 1 − v2

Finally, the distance from G to M and back √ is D⊥ = 2λ , so the average speed of light is D⊥ /T⊥ = 1 − v2.

with v = (vx , vy , vz ). An immediate calculation gives Sv Sw = Sv+w = Sw Sv , so multiplication in G4 is commu- b) The travel time Tk is the time t2 that was determined tative. Furthermore, as in Exercise 5, Sv S−v = S0 = I, so in the solution to Exercise 1 of § 1.1. Because Dk = 2λ , G4 is a group. the average speed of light is Dk /Tk = 1 − v2, not 1.

DVI file created at 16:06, 20 January 2011

1.4. MAXWELL’S EQUATIONS

3

c) By Taylor’s theorem, 1 = 1 + v2 + O(v4 ), 1 − v2

√

1 = 1 + 12 v2 + O(v4 ), 1 − v2

Each vertical line is mapped to itself by both Fv and Sv . But Sv simply translates points on the line t = a by the amount av, while Fv also compresses them by the factor √ 1 − v2. Both maps turn horizontal lines into lines with slope v.

as v → 0. Therefore, Tk − T⊥ =   1 1 t t √ − 2λ = λ v2 + O(v4 ) = v2 (λ + O(v2 )) 4. a) The vectors (1, ±1) (where V is the transpose of 1 − v2 1 − v2 V ) lie on the two worldlines. The image worldlines there2 2 4 as v → 0. We use the fact that v O(v ) = O(v ) as v → 0. fore contain the vectors      1 √ 0 1 1 √ z . = 2. a) ±1 1 − v2 v v ± 1 − v2

t

The slopes of these worldlines are √ p v ± 1 − v2 m± = = v ± 1 − v2 . 1

√ b) The graph of w = 1 − v2 is the unit circle centered at (0, 0) in the (v, w)-plane; therefore it lies above the line w = 1 − v on the interval 0 < v < 1. In other words, √ 1 − v2 > 1 − v there. Hence p The √ map Cv compresses the image vertically by the factor m+ = v + 1 − v2 > v + 1 − v = 1. 1 − v2.√The image grid thus consists of rectangles of √ size 1 × 1 − v2 parallel to the coordinate axes. Because − 1 − v2 > −1 and v > 0 on the same domain, p b) Matrix multiplication gives m− = v − 1 − v2 > 0 − 1 = −1. ! ! ! 1 0 1 0 1 0 √ √ SvCv = = = Fv ; The maximum value of m+ (v) occurs when 0 = 2 √ √ v 1 0 v 1−v 1 − v2 2 (m+ )′ (v) Hence v = 1/ 2 and √ = 1√− v/ 1 − v . m+ (1/ 2) = 2. however, ! 5. If Fv is to be valid, it must be consistent with the 1 0 Cv Sv = √ 6= Fv √ Michelson–Morley experiment. That is, the image worldv 1 − v2 1 − v2 lines of photons must have slopes ±1. Exercise 4b demonstrates that this is not so if v > 0. when v 6= 0. 3.

1.4 Maxwell’s Equations

z

t

The map √ Fv first compresses the image vertically by the factor 1 − v2 (this is the action of Cv ) and then carries out a vertical shear with slope v (the action of Sv ).

DVI file created at 16:06, 20 January 2011

1. a) Let A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ), C = (c1 , c2 , c3 ); then   b b3 b3 b1 b1 b2 . , , B × C = 2 c2 c3 c3 c1 c1 c2 The first component of A × (B × C) is therefore b b 2 − a 3 b 3 b 1 a2 1 c1 c2 c3 c1 = b1 (a2 c2 + a3c3 ) − c1 (a2 b2 + a3 c3 )

= b1 (a1 c1 + a2c2 + a3c3 ) − c1(a1 b1 + a2b2 + a3 b3 ) = (A · C)b1 − (A · B)c1 .

SOLUTIONS: CHAPTER 1. RELATIVITY BEFORE EINSTEIN

4

(The underlined terms that have been added to the expres- 5. The relation between the Greek and the Roman varision offset each other.) In a similar way we can show that ables is the second and third components of A × (B × C) are t − vz z − vt τ=√ ζ=√ , ; 2 (A · C)b2 − (A · B)c2 1−v 1 − v2 and (A · C)b3 − (A · B)c3, therefore respectively. 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) − ∇2F. 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 = (Ry − Qz , Pz − Rx , Qx − Py ); subscripts denote partial derivatives. Again, by definition, ∇ · (∇ × F) = (Ry − Qz )x + (Pz − Rx )y + (Qx − Py)z = Ryx − Qzx + Pzy − Rxy + Qxz − Pyz = 0,

by the commutativity of partial differentiation (for example, Ryx = Rxy , etc.).

∂E 3. From Maxwell’s equation ∇ × H = + J we obtain ∂t   ∂E ∂ 0 = ∇ · (∇ × H) = ∇ · + ∇ · J = ∇ · E + ∇ · J, ∂t ∂t again using the commutativity of partial differentiation. This equation, together with Maxwell’s equation ∇ · E = ρ , implies

∂ρ = −∇ · J. ∂t

4. These equalities follow from the chain rule. Because ζ = z = vt, we have ζz = 1 and ζt = −v. Because τ = t and E(t, z) = E(τ , ζ − vτ ), we find

∂E ∂E ∂ζ = = Eζ , ∂z ∂ζ ∂z ∂ Eζ ∂ Eζ ∂ ζ = = Eζ ζ , Ezz = ∂z ∂ζ ∂z ∂E ∂E ∂τ ∂E ∂ζ Et = = + = Eτ − vEζ ∂t ∂τ ∂t ∂ζ ∂t
∂ Eζ ∂ Eτ Ett = −v = Eττ − vEτζ − v Eζ τ − vEζ ζ ∂t ∂t = Eττ − 2vEτζ + v2Eζ ζ . Ez =

DVI file created at 16:06, 20 January 2011

1 ∂τ =√ , τz = ∂t 1 − v2 −v ∂ζ ζt = =√ , ζz = ∂t 1 − v2

τt =

Using E(t, z) = E(τ , ζ ), we find

−v ∂τ =√ , ∂z 1 − v2 1 ∂ζ =√ . ∂z 1 − v2

Eτ − vEζ Et = Eτ τt + Eζ ζt = √ , 1 − v2
1 Ett = √ Eττ τt + Eτζ ζt − vEζ τ τt − vEζ ζ ζt 2 1−v Eττ − 2vEτζ + v2Eζ ζ = , 1 − v2 −vEτ + Eζ Ez = Eτ τz + Eζ ζz = √ 1 − v2
1 Ezz = √ −vEττ τz − vEτζ ζz + Eζ τ τz + Eζ ζ ζz 2 1−v 2 v Eττ − 2vEτζ + Eζ ζ = . 1 − v2 6. a) Let u = z ± ct, so ut = ±c, uz = 1; then

∂ h(z ± ct) = h′ (u) · ut = ±ch′(u), ∂t

∂2 h(z ± ct) = ±ch′′ (u) · ut = c2 h′′ (u), ∂ t2 ∂ h(z ± ct) = h′ (u) · uz = h′ (u), ∂z ∂2 h(z ± ct) = h′′ (u) · ut = h′′ (u), ∂ t2 for any (sufficiently differentiable) function h(u). Hence, if E(t, z) = f (z − ct) + g(z + ct), then Ett = c2 ( f (z − ct) + g(z + ct)) = c2 Ezz . b) It is a standard fact that the graph of w = f (z − ct0 ) is the graph of w = f (z) translated by the amount ct0 . 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 + ct0 ) is the graph of w = f (z) translated by −ct0 . The spike moves to z = −ct at time t, so it travels with velocity −c.

Solutions: Chapter 2

Special Relativity—Kinematics 2.1 Einstein’s Solution

unique line containing X and X + Y . These two lines are parallel, so their images are parallel. We can take the diNote: we use 0 to denote the zero vector in any Rk . rection of ℓ1 to be Y , and the direction of its image M(ℓ1 ) 1. • We can write the straight line ℓ passing through the to be M(Y ) − M(0) = M(Y ) because M(0) = 0, by hypoint A in the direction D 6= 0 parametrically in the form pothesis. Moreover, M(Y ) 6= 0 because M is invertible, so M(Y ) is a valid direction vector. X = Dt + A, −∞ < t < ∞. If L is linear, then We can also take Y to be the direction vector for the parallel line ℓ2 , and thus we can take M(Y ) for the direcL(X) = L(Dt + A) = L(D)t + L(A); tion of its image M(ℓ2 ). Because M(X) and M(X +Y ) are this is the straight line through L(A) in the direction L(D). two distinct points on M(ℓ2 ), we can write If L(D) = 0 for some D 6= 0, then L collapses all of R2 to M(X + Y ) − M(X) = kY M(Y ), kY 6= 0. a line (or a point), so there is nothing further to prove. To continue, we assume L is invertible. By reversing the roles of X and Y (and constructing two • If ℓ∗ is parallel to ℓ, then its direction vector D∗ is more parallel lines), we can also write a nonzero multiple of D: D∗ = kD, k 6= 0, and we can parametrize ℓ∗ as Y = D∗ s + B. The image of ℓ∗ under L M(X + Y ) − M(Y ) = kX M(X), kX 6= 0. is the line Combining these, we find L(Y ) = L(D∗ )s + L(B) = kL(D)s + L(B). (1 − kX )M(X) = (1 − kY )M(Y ). The direction vector here, kL(D), is a nonzero multiple of If the coefficients are different from 0, M(X) and M(Y ) L(D), so ℓ and ℓ∗ have parallel images. are linearly dependent. But X and Y are arbitrary nonzero • If the points Pj are equally spaced on ℓ, then Pj = vectors and M is invertible, so we are forced to conclude Dt j + A, where the parameter values t j = t0 + n∆t are the coefficients are zero, so k = k = 1. This implies Y Y equally spaced on the t-axis, and conversely. The image points, M(X + Y ) = M(X) + M(Y ). L(Pj ) = L(D)t j + L(A), have (the same!) equally-spaced parameter values, so they c) Suppose Y j → X; then continuity of M implies M(Y j ) → M(X). Using the additively of M established are equally spaced. in part b, we then find 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 6= A on β : M(A) = M(B) Thus M fails to be 1–1, 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.

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).

b) Let X and Y be arbitrary nonzero vectors in R2 . Let ℓ1 be the unique line containing 0 and Y , and let ℓ2 be the

M((k − 1)X + Y j ) = M(X + (k − 2)X + Y j ) = M(X) + M((k − 2)X + Y j ).

M(X + Y j ) = M(X) + M(Y j ) → M(X) + M(X) = 2M(X).

d) Additivity of M allows us to write

5

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

6

Assume first that M((h − 1)X +Y j ) → M(hX) for all pos- The quadratic polynomial is called the characteristic itive integers h < k. In the right-hand side of the previous polynomial of M. It has two (possibly complex) roots that equation we have M((k − 2)X + Y j ) → M((k − 1)X), so are the eigenvalues of M, because then (M − λ I)X = 0 has nonzero solutions. M((k − 1)X + Y j ) → M(X) + M((k − 1)X) = M(kX). b) If λ1 , λ2 are the roots of the quadratic λ 2 − Bλ + C, Now assume that M((h − 1)X +Y j ) → hM(X) for all posi- then tive integers h < k. In the same location we can now write λ 2 − Bλ + C = (λ − λ1 )(λ − λ2) M((k − 2)X + Y j ) → (k − 1)M(X), so M((k − 1)X + Y j ) → M(X) + (k − 1)M(X) = kM(X). Because the hypotheses have been established for h = 2, the induction holds and we conclude that M(kX) = kM(X) for all positive integers k.

= λ 2 − (λ1 + λ2 )λ + λ1 λ2 ,

implying λ1 + λ2 = B, λ1 λ2 = C. For the eigenvalues (i.e., the roots of the characteristic polynomial), we have

λ1 + λ2 = tr M, λ1 λ2 = det(M). e) Because rX + −rX = 0 for any real scalar r, and M(0) = 0, additivity of M implies c) Suppose the nonzero vectors X1 and X2 were linearly dependent; then X2 = kX1 for some k 6= 0. Then 0 = M(rX + −rX) = M(rX) + M(−rX)

and hence that M(−rX) = −M(rX). In particular, if −k is a negative integer, then k is positive and M(−kX) = −M(kX) = −kM(X) by part d. Hence M(pX) = pM(X) for any integer p. f) As suggested, set Z = 1n X so pZ = np X = qX. Then nM(qX) = nM(pZ) = pM(nZ) = pM(X), so M(qX) = np M(X) = qM(X).

λ2 X2 = MX2 = kMX1 = kλ1 X1 = λ1 X2 ,

implying (λ2 − λ1 )X2 = 0. But X2 6= 0, so λ2 − λ1 = 0, contradicting the hypothesis that λ1 6= λ2 . d) The roots of the characteristic polynomial are p tr(M) ± tr(M)2 − 4 det(M) λ± = p 2 a + d ± (a + d)2 − 4(ad − bc) . = 2

g) Because q j → r, we have q j X → rX. Continuity of M therefore implies M(q j X) → M(rX). But we also have When c = b, some algebra reduces this to q j M(X) → rM(X), so by the uniqueness of the limit, p M(rX) = rM(X). Hence M is linear. a + d ± (a − d)2 + 4b2 . λ± = 3. We have Bv F = 2      a(v) b(v) 1 0 a(v) −b(v) Since (a − d)2 + 4b2 ≥ 0, the roots λ± are real. = b(v) a(v) 0 −1 b(v) −a(v) Write the dot product of two vectors as a matrix multiplication that involves the transpose operator: X1 · X2 = and FBv F = X1t X2 . The crucial step in he following sequence uses the      symmetry of M, Mt = M: 1 0 a(v) −b(v) a(v) −b(v) = = B−v . 0 −1 b(v) −a(v) −b(v) a(v) λ1 X1 · X2 = (λ1 X1 )t X2 = (MX1 )t X2 = X1t Mt X2 Therefore a(−v) = a(v) and b(−v) = −b(v). = X1t MX2 = X1t (λ2 X2 ) = X1 · (λ2 X2 ) = λ2 X1 · X2 . 4. a) Write MX = λ X as MX − λ X = MX − λ IX. A This implies (λ1 − λ2 )X1 ·X2 = 0 and because λ1 − λ2 6= 0, nonzero solution of MX = λ X is a nonzero solution of we conclude X1 · X2 = 0. That is, X1 ⊥ X2 . AX = 0, where A = M − λ I. The nonzero solution implies The eigenvalues are given by the equation A is noninvertible, or det A = 0. That is to say, p a + d ± (a − d)2 + 4b2 a − λ λ± = ; b 2 = (a − λ )(d − λ ) − bc 0 = det(M − λ I) = c d − λ if λ+ = λ− , the discriminant (a − d)2 + 4b2 must be 0, = λ 2 − (a + d)λ + ad − bc implying a = d, b = c = 0. Thus λ = λ± = (a + a)/2 = a = λ 2 − tr(M)λ + det(M) = p(λ ). and M = aI = λ I.

DVI file created at 16:06, 20 January 2011

2.1. EINSTEIN’S SOLUTION

7

  4 4 5. Because tr F = 0 and detF = −1, the eigenvalues are • Let M = . The characteristic polynomial is 2 −3 the roots of λ 2 − 1 = 0, or λ± = ±1. To find an eigenvector, solve each of the equations (F = λ± I)X = 0. We λ 2 − λ − 20 = (λ − 5)(λ + 4). have      0 0 x x The eigenvalues are therefore 5 and −4 and we have , so X = , 0 = (F − λ+ I)X = 0 −2 y 0          −1 4 8 4 2 0 x 0 M − 5I = , M + 4I = , 0 = (F − λ− I)X = , so X = . 2 −8 2 1 0 0 y y so we can take 6. See the solution to Exercise 4d.

λ =5:X =

  4 1

and λ = −4 : X =



 1 . −2

7. If M is invertible, then detM = λ1 λ2 6= 0, so the eigenvalues of M are nonzero. Suppose X is an eigenvector of M with eigenvalue λ ; then MX = λ X and this equation The matrix M is not symmetric. can be rewritten as   2 1 • Let M = . The characteristic polynomial is 1 0 1 X = M −1 (λ X) = λ M −1 X or M −1 X = X. λ λ 2 − 3λ + 2 = (λ − 2)(λ − 1), Thus X is an eigenvector for M −1 with eigenvalue 1/λ .   6 2 8. a) • Let M = . Then 2 3

λ 2 − 9λ + 14 = (λ − 7)(λ − 2); the eigenvalues are 7 and 2. For these eigenvalues, we have, respectively,     −1 2 4 2 M − 7I = , M − 2I = , 2 −4 2 1 so we can take for the corresponding eigenvector     2 −1 λ =7:X = and λ = 2 : X = . 1 2

and the eigenvalues are 2 and 1. We have     0 1 1 1 M − 2I = , M − 1I = , 0 −1 0 0 so we can take   1 λ =2:X = 0



 1 and λ = 1 : X = . −1

The matrix M is not symmetric.   2 1 • Let M = . The characteristic polynomial is 1 1

λ 2 − 3λ + 1,

Note that M is symmetric and the eigenvectors are orthog- so the eigenvalues are onal.   √ √ 6 −2 3± 9−4 3± 5 • Let M = . The characteristic polynomial = λ± = −2 3 2 2 here is the same as for the previous matrix, so the eigenvalues are again 7 and 2. For these eigenvalues, we have, We obtain the eigenvectors by respectively, ! √   1− 5 2 1     2 √ √ M − λ+ I = , X= , −1 −2 4 −2 −1− 5 5−1 1 M − 7I = , M − 2I = , 2 −2 −4 −2 1 so we can take for the corresponding eigenvector     2 1 λ =7:X = and λ = 2 : X = . −1 2

and

M − λ− I =

√ 1+ 5 2

1

1√

−1+ 5 2

!

,

 √  1− 5 X= . 2

Note that M is symmetric and the eigenvectors are orthog- Note that M is symmetric and the eigenvectors are orthogonal. onal.

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

8

b) • M =

  6 2 2 3

z

•M= (2, 3)

  2 1 1 1

z

(1, 1) (2, 1)

(6, 2)

t

•M=



 6 −2 −2 3

t

c) The object here is to build a grid for M in each case using eigenvectors B1 and B2 of M as the basis vectors. This grid is shown in gray and its basic element is shaded. Then the image of the basic grid element under M is shown in solid black outline. The sides of the image grid element are parallel to the sides of the gray grid.       6 2 2 −1 •M= ; grid basis B1 = , B2 = 2 3 1 2

z

(−2, 3)

t

(6, −2)

z

  4 4 •M= 2 −3

z

t (4, 2)

t (4, −3)

The basic grid element is stretched by the factor 7 in one direction and by the factor 2 in the other. 

     6 −2 2 1 •M= ; grid basis B1 = , B2 = −2 3 −1 2

z   2 1 •M= 0 1

z t (1, 1) (2, 0)

t The basic grid element is stretched by the factor 7 in one direction and by the factor 2 in the other.

DVI file created at 16:06, 20 January 2011

2.2. HYPERBOLIC FUNCTIONS

•M=

9

      2 4 4 1 ; grid basis B1 = 1 , B2 = 2 −3 −2 2 z

t

9. Rθ : Because tr(Rθ ) = 2 cos θ and det(Rθ ) = 1, the characteristic polynomial is λ 2 − 2 cos θ λ + 1 and the eigenvalues are √ p 2 cos θ ± 4 cos2 θ − 4 λ± = = cos θ ± − sin2 θ 2 = cos θ ± i sin θ , √ where i = −1. Therefore     ∓i sin θ − sin θ ∓i −1 = sin θ . Rθ − λ ± I = 1 ∓i sin θ ∓i sin θ

The last expression makes it clear that the eigenvectors do not depend on θ ; the eigenvectors are   ±i The basic grid element is stretched by the factor 5 in one X± = . 1 direction and by the factor -4 in the other. The negative factor is manifested as a flip, or orientation reversal. Mu : Because tr(Hu ) = 2 cosh u and det(Hu ) = 1, the       2 1 1 1 characteristic polynomial is λ 2 − 2 cosh uλ + 1. Whereas •M= ; grid basis B1 = , B2 = 2 0 1 0 −1 cos θ − 1 = − sin2 θ , the hyperbolic equivalent has z cosh2 u − 1 = + sinh2 u. Therefore, the eigenvalues of Hu are real: p 2 cosh u ± 4 cosh2 u − 4 λ± = = cosh u ± sinhu. 2 t Therefore     ∓ sinh u sinh u ∓1 1 Hu − λ±I = = sinh u , sinh u ∓ sinh u 1 ∓1 The basic grid element is stretched by the factor 2 hori- and the eigenvectors are   zontally and left unchanged in the other direction. ±1       X± = . 1 2 1 −β 2 •M= , B2 = ; grid basis B1 = , 1 1 β 2 √ where β = 5 − 1. 2.2 Hyperbolic Functions z

t

1. From the definition,   u e2u + 2 + e−2u e + e−u 2 = , cosh u = 2 4   u e2u − 2 + e−2u e − e−u = sinh2 u = ; 2 4 with subtraction, the exponential terms cancel, leaving

The basic grid element is stretched by the factor λ1 = √ (3 + 5)/2 ≈ 2.62 in √ the B1 direction and compressed by the factor λ2 = (3 − 5)/2 ≈ 0.38 in the B2 direction. For each of the five matrices, the image of the gray grid (i.e., the grid based on eigenvectors) is parallel to the original gray grid; in this it behaves like the image of the eigenvector grid for the map Bv as sketched in the text (on page 36).

DVI file created at 16:06, 20 January 2011

cosh2 u − sinh2 u =

2 −2 − = 1. 4 4

The other two identities are 1 cosh2 u − sinh2 u = = sech2 u, cosh2 u cosh2 u cosh2 u − sinh2 u 1 coth2 u − 1 = = = csch2 u. sinh2 u sinh2 u

1 − tanh2 u =

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

10

Thus

2. From the definitions we obtain the product  u  v  e − e−u e + e−v sinh u cosh v = 2 2

cosh(u ± v) = cosh u cosh v ± sinhu sinh v;

eu+v + eu−v − e−u+v − e−(u+v) . = 4

this differs from the analogous circular identities: the “±” on the right is “∓” in the circular case. Next we have

Reversing the positions of u and v, we have cosh u sinh v =

eu+v + e−u+v − eu−v − e−(u+v) 4

.

sinh(u ± v) cosh(u ± v) sinh u cosh v ± coshu sinh v . = cosh u cosh v ± sinhu sinh v

tanh(u ± v) =

In the sum of these two expressions, the middle two terms Dividing numerator and denominator by cosh u cosh v, we cancel, yielding find sinh u cosh v + coshu sinh v sinh u sinh v ± eu+v − e−(u+v) eu+v − e−(u+v) cosh u cosh v = tanh u ± tanhv . tanh(u ± v) = + = sinh(u + v). = sinh u sinh v 1 ± tanhu tanh v 4 4 1± cosh u cosh v By contrast, in the difference, the outer two terms cancel, This differs from the analogous circular identities: the yielding “±” in the denominator is “∓” in the circular case. sinh u cosh v − coshu sinh v For the final identities, we can write =

e(u−v) − e−(u−v) e−(u−v) − e(u−v) − = sinh(u − v). 4 4

Thus sinh(u ± v) = sinh u cosh v ± coshu sinh v, agreeing with the analogous circular identities.

coth(u ± v) =

1 1 ± tanhu tanh v = . tanh(u ± v) tanh u ± tanhv

Multiplying numerator and denominator by ± coth u coth v yields coth(u ± v) =

± coth u coth v + 1 1 ± cothu coth v = . ± coth v + cothu coth u ± cothv

For the next pair of identities, we begin with the prodOne way to write the analogous circular identities is ucts  u    −1 ± cotu cot v e + e−u ev + e−v ; cot(u ± v) = cosh u cosh v = cot u ± cotv 2 2 in this case, the difference is solely in the sign of the “1”. eu+v + eu−v + e−u+v + e−(u+v) , = 3. From the formulas,  u  4 v  −u −v e −e e −e eu + e−u eu − e−u sinh u sinh v = cosh u + sinhu = + = eu , 2 2 2 2 eu+v − eu−v − e−u+v + e−(u+v) eu + e−u eu − e−u . = − = e−u . cosh u − sinhu = 4 2 2 Once again the middle two terms cancel in a sum, yielding 4. To save space, let C = cosh u , S = sinh u , i = 1, 2. i i i i Then the addition formulas for cosh and sinh give cosh u cosh v + sinhu sinh v    C S C S u+v −(u+v) u+v −(u+v) 2 2 1 1 e +e e +e Hu1 · Hu2 = = + = cosh(u + v). S2 C2 S1 C1 4 4   C1C2 + S1S2 C1 S2 + S1C2 The outer two terms cancel in the difference, yielding = S1C2 + C1 S2 S1 S2 + C1C2   cosh(u1 + u2) sinh(u1 + u2) cosh u cosh v − sinhu sinh v = sinh(u1 + u2) cosh(u1 + u2 ) e(u−v) + e−(u−v) −e(u−v) − e−(u−v) − = cosh(u − v). = = Hu1 +u2 = Hu2 +u1 = Hu2 · Hu1 . 4 4

DVI file created at 16:06, 20 January 2011

2.2. HYPERBOLIC FUNCTIONS 5. a) The identity involving tanh2 u gives 1 = sech2 u = 1 − tanh2 u = 1 − v2. cosh2 u √ Thus cosh u = 1/ 1 − v2 and v . sinh u = cosh u · tanhu = √ 1 − v2

b) Making the substitutions from part a, we find  1  v   √ √ 1 2 1 v  1 − v2  √ Hu =  1 v− v = Bv . = 1  1 − v2 v 1 √ √ 1 − v2 1 − v2 c) We make use of the equation Bv = Hu , in which v = tanh u. We claim the function v = tanh u is invertible on |v| < 1. To see this, it is sufficient to note that v′ = sech2 u > 0 and tanh u → ±1 as u → ±∞. If we denote the inverse as arctanh v, then, for any |v| < 1 we can write Bv = Harctanh v . Now suppose −1 < v1 , v2 < 1; then we have Bv1 ·Bv2 = Harctanh v1 ·Harctanh v2 = Harctanh v1 +arctanh v2 . = Bv3 ,

11 This is done on page 96 of the text; the algebra is intricate.

w

6. a)

1

w = sech u u w = csch u

w = tanh u

w = coth u

The graphs of w = tanh u and w = coth u are asymptotic to w = ±1 as u → ±∞. The graphs of w = coth u and w = csch u are asymptotic to the w-axis. b) Because 1 ≤ cosh u < ∞, it follows that 1≥

1 = sech u > 0. cosh u

where

The inequalities −1 < tanh u < 1 were established in the v3 = tanh(arctanhv1 + arctanhv2 ) and thus |v3 | < 1. previous solution by using calculus. Here uis a separate proof that uses only inequalities. Because e and e−u are The addition formula for tanh allows us to write the equa- both positive, tion for v3 in a different (and simpler!) form eu − e−u u −u u −u < 1. e − e < e + e ⇒ tanh u = v1 + v2 tanh(arctanh v1 ) + tanh(arctanh v2 ) eu + e−u = . v3 = 1 + tanh(arctanhv1 ) tanh(arctanh v2 ) 1 + v1v2 For the other inequality, That is, Bv1 · Bv2 = B(v1 +v2 )/(1+v1 v2 ) . e−u + eu > e−u − eu = −(eu − e−u), Comment: The intervention of the hyperbolic rotation Hu is quite helpful here. Without it, we could still write −(eu − e−u) so 1 > = − tanh u, or −1 < tanh u.     eu + e−u 1 1 1 v1 1 v2 √ Bv1 · Bv2 = √ 1 − v21 v1 1 1 − v22 v2 1 7. a) The following parametrizations work.   1 1 + v1 v2 v2 + v1 =√ 2 2 I : x − y = 1, x < 0, x = − cosh u, y = sinh u; (1 − v21)(1 − v22) v1 + v2 1 + v1v2 2 2   II : y − x = 1, y > 0, x = sinh u, y = cosh u; v1 + v2 1 1 + v1 v2 2 2   1 + v v 1 2 III : y − x = 1, y < 0, x = sinh u, y = − cosh u. =√  v1 + v2 (1 − v21)(1 − v22) 1 y II 1 + v1 v2 −1 1 I b) 1 0 Thus it appears that v3 = (v1 + v2 )/(1 + v1v2 ), but to prove that the final matrix has the form Bv3 , we need to show the matrix coefficient has the correct form, that is, that 0 x 1 1 + v1 v2 √ =√ . (1 − v21)(1 − v22 ) 1 − v23 −1 0

−1

DVI file created at 16:06, 20 January 2011

1

III

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

12 8.

y

The segment of the circle is the graph of x = 0 ≤ y ≤ sin u. Therefore,

2

area(A + B) =

Z sin u p 0

p 1 − y2 ,

1 − y2 dy.

The substitutions y = sin s, dy = coss ds, 1 − y2 = cos2 s, s = u when y = sin u transforms this into Z u sin(2s) s u sin(2u) u 2 cos s ds = + = + ; 4 2 0 4 2 0

3/2

1 1/2 0

x

−1/2

the integral of cos2 s is a standard form. With the identity sin(2u) = 2 sin u cos u, we can write area(A + B) =

−1

sin u cos u u + . 2 2

The right triangle B has sides cos u and sin u, so its area is (sin u cosu)/2, and area(A) = u/2.

The hyperbola above was created with a PostScript program that calculated the location of each point labelled 12. a) The kth term of the Taylor series for f (u) has the with its u-value. The four lines do indeed appear to be form f ( k)(0) xk /k! (at u = 0); therefore parallel. u2 u3 u4 9. The arc subtends the fraction u/2π of the entire cireu : 1 + u + + + + · · · , 2! 3! 4! cumference, and the circular sector A occupies the same 2 u4 u6 u fraction of the entire area. The entire area is π , so the area cosh u : 1 + + + + · · · , of A must be π · u/2π = u/2. 2! 4! 6! u3 u5 u7 p sinh u : u + + + + · · · . 10. a) Solve x2 − y2 = 1 for x to get x = ± 1 + y2 and 3! 5! 7! then choose the plus sign because x > 0 on this portion of the hyperbola. The portion of the hyperbola we want is b) For f (u) = tanh u, we can use the derivatives given on given by 0 ≤ y ≤ sinh u, so the area of the region A + B is page 41: the integral Z sinh u p f (u) = tanh u f (0) = 0 1 + y2 dy. ′ 2 0 f (u) = sech u f ′ (0) = 1 2 ′′ The substitutions y = sinh s, dy = cosh s ds, 1 + y = f (u) = 2 sech u · − sechu tanh u cosh2 s, and s = u when y = sinh u, transform this integral = −2 tanh u · sech2 u f ′′ (0) = 0 into f ′′′ (u) = −2 sech2 u · sech2 u Z u sinh(2s) s u sinh(2u) u 2 − 2 tanhu · −2 tanhu sech2 u cosh s ds = + = + , 4 2 4 2 0 = −2 sech4 u + 4 tanh2 u sech2 u f ′′′ (0) = −2 0

according to the integral identity in the text, page 41. Using the identity sinh(2u) = 2 sinh u cosh u, we can write area(A + B) =

Z sinh u p

1 + y2

0

sinh u cosh u u dy = + . 2 2

b) The region B is a right triangle whose horizontal side has length cosh u and whose vertical side has length sinh u. The area of B is thus (sinh u cosh u)/2, so area(A) = u/2. 11. The region A is a circular sector and the region B is the triangle from the origin to the point (cos u, sin u) to the point (0, sin u) and back to the origin.

DVI file created at 16:06, 20 January 2011

The third-order Taylor polynomial is therefore u − u3/3.

13. If u = g(w) is the inverse of w = f (u), then du/dw is the reciprocal of dw/du, expressed as a function of w. • u = arcsinh w is the inverse of w = sinh u, and dw/du = cosh u. Furthermore, p p cosh u = sinh2 u + 1 = w2 + 1,

so we can write

1 1 du = =√ . dw cosh u w2 + 1

2.3. MINKOWSKI GEOMETRY

13

• u = arccosh w is the inverse of w = cosh u, and respectively. The first of these makes an angle of θ with dw/du = sinh u. Furthermore, respect to the positive x-axis, i.e., with respect to the first basis vector. Because p p sinh u = cosh2 u − 1 = w2 − 1 cos(θ + π /2) = cos(θ ) cos(π /2) − sin(θ ) sin(π /2) so we can write = − sin(θ ), du 1 1 sin(θ + π /2) = sin(θ ) cos(π /2) + cos(θ ) sin(π /2) = =√ . dw sinh u w2 − 1 = cos(θ ), • u = arctanh w is the inverse of w = tanh u, and we can write dw/du = sech2 u. Furthermore,     cos(θ + π /2) − sin θ sech2 u = 1 − tanh2 u = 1 − w2, = . cos θ sin(θ + π /2) so we can write Thus the image of the second basis vector makes an angle 1 1 du of θ + π /2 with the positive x-axis, and hence an angle of . = = dw sech2 u 1 − w2 θ with respect to the second basis vector itself. Hence Rθ rotates the basis vectors by θ and thus, by the suggestion given in the question, rotates the entire plane 14. a) Let θ = arcsin(tanh u); then sin θ = tanh u and by θ . p p cos θ = 1 − sin2 θ = 1 − tanh2 u = sech u. 2. Because the inverse of any action A “undoes” A, and because rotation by −θ undoes rotation by θ , we have Hence Rθ−1 = R−θ . Trigonometric identities imply sin θ tanh u sinh u cosh u tan θ = = = · = sinh u,     cos θ sech u cosh u 1 cos θ sin θ cos(−θ ) − sin(−θ ) R −θ = , = sin(−θ ) cos(−θ ) − sin θ cos θ and arctan(sinh u) = θ = arcsin(tanh u). so

b) Set u = arctanhw; then eu − e−u e2u − 1 w = tanh u = u = , e + e−u e2u + 1 and

R θ R −θ

we2u + w = e2u − 1. Hence 2u

0 = (1 − w)e − (1 + w) or so

 1+w , 2u = ln 1−w 

15. a) cosh(r)

e

2u

1+w , = 1−w

  1 1+w u = ln . 2 1−w

(b) sech(1)

2.3 Minkowski Geometry 1. The images of the standard basis vectors under any matrix multiplication are the columns of the matrix; thus for Rθ the images are     cos θ − sin θ and , sin θ cos θ

DVI file created at 16:06, 20 January 2011

   cos θ − sin θ 1 cos θ sin θ = 0 − sin θ cos θ sin θ cos θ     cos θ − sin θ cos θ sin θ 1 = = sin θ cos θ − sin θ cos θ 0

R −θ R θ =



 0 , 1  0 . 1

3. The map Rθ Rϕ is rotation by ϕ followed by rotation by θ . But this is also the single rotation by ϕ + θ = θ + ϕ , or Rϕ +θ = Rθ +ϕ . The product Rθ Rϕ is the matrix   cos θ cos ϕ − sin θ sin ϕ − cos θ sin ϕ − sin θ cos ϕ sin θ cos ϕ + cos θ sin ϕ − sin θ sin ϕ + cos θ cos ϕ   cos(θ + ϕ ) − sin(θ + ϕ ) = R θ +ϕ , = sin(θ + ϕ ) cos(θ + ϕ ) by the addition formulas for the sine and cosine. 4. Note the remark in the proof of Theorem 2.3 that, because T and L are each preserved as a whole, it is sufficient to show that future and past sets are preserved individually—that is, t always has the same sign as τ , where t = τ cosh u + ζ sinh u.

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

14

In all cases, we can adapt the proof of Theorem 2.3. • T − : On T − , τ < 0 and τ < ζ < −τ , so t < τ cosh u − τ sinh u = τ e−u < 0. • L + , L − On all of L , ζ = ±τ , so t = τ cosh u ± τ sinh u = τ e±u . Because e±u > 0, t has the same sign as τ , so individually preserved.

L ± are implying that X2 is an eigenvector R2θ F0 with eigen-

5. Let Ei be (the transpose of) (τi , ζi ), i = 1, 2. Because these are future vectors, kEi k2 = Q(Ei ) = τi2 − ζi2 > 0,

implying that X1 is an eigenvector R2θ F0 with eigenvalue +1. We also have   − cos 2θ sin θ + sin 2θ cos θ R2θ F0 (X2 ) = − sin 2θ sin θ − cos2θ cos θ     sin(2θ − θ ) sin θ = = −X2 , = − cos θ − cos(2θ − θ )

i = 1, 2.

Therefore kE1 k2 kE2 k2 = (τ12 − ζ12 )(τ22 − ζ22 )

= (τ1 τ2 )2 − (τ1 ζ2 )2 − (ζ1 τ2 )2 + (ζ1 ζ2 )2 .

Furthermore, (E1 · E2 )2 = (τ1 τ2 − ζ1 ζ2 )2

= (τ1 τ2 )2 − 2τ1 τ2 ζ1 ζ2 + (ζ1 ζ2 )2 ,

so (E1 · E2 )2 − kE1k2 kE2 k2

= −2τ1 τ2 ζ1 ζ2 + (τ1 ζ2 )2 + (ζ1 τ2 )2

= (τ1 ζ2 − ζ1 τ2 )2 ≥ 0.

Thus (E1 · E2 )2 ≥ kE1 k2 kE2 k2 ; taking the square root and using the fact that kE1 kkE2k is real and nonnegative, we get the reverse Cauchy–Schwarz inequality. 6. a) The reflection Fθ fixes the given vector X1 , so X1 is an eigenvector of Fθ with eigenvalue λ1 = +1. At the same time, Fθ maps the given X2 to its opposite, so X2 is also an eigenvector, this time with eigenvalue λ2 = −1.

value −1. The second coordinate used the cosine identity in the form cos(2θ − θ ) = cos2θ cos(−θ ) − sin2θ sin(−θ ) = cos2θ cos θ + sin 2θ sin θ .

The vectors X1 and X2 form a basis for R2 ; because Fθ and R2θ F0 agree on that basis, they agree on all of R2 , so   cos 2θ sin 2θ . Fθ = R2θ F0 = sin 2θ − cos 2θ c) Note that Fθt = Fθ and that performing a reflection twice is the identity: Fθt Fθ = (Fθ )2 = I. 7. We compute the product Fα Fβ =      cos 2β sin 2β A −B cos 2α sin 2α = , B A sin 2α − cos 2α sin 2β − cos 2β where A = cos 2α cos 2β + sin 2α sin 2β = cos(2α − 2β ) = cos 2(α − β ),

B = sin 2α cos 2β − cos2α sin 2β = sin(2α − 2β ) = sin 2(α − β ). Thus Fα Fβ = R2(α −β ).

8. a) Let M1 and M2 be the columns of M. Then Mt M is the 2 × 2 matrix whose entries are   M1 · M1 M1 · M2 b) The map F0 is reflection across the x-axis, so it leaves M2 · M1 M2 · M2 the x-coordinate of a vector unchanged but changes the sign of the y-coordinate. Thus If this is equal to the identity matrix I, then     1 0 cos2θ sin 2θ kM1 k2 = M1 · M1 = kM2 k2 = M2 · M2 = 1, M1 · M2 = 0, F0 = and R2θ F0 = . 0 −1 sin 2θ − cos2θ as asserted. We have b) The points (cos α , sin α ) parametrize the unit circle   in a 1–1 fashion when A − π ≤ α < A + π (for any A). cos 2θ cos θ + sin 2θ sin θ R2θ F0 (X1 ) = Because M1 is on the unit circle, there is a unique α = θ sin 2θ sin θ − cos2θ sin θ     for which     cos θ cos(2θ − θ ) a cos θ = X1 , = = . M1 = = sin(2θ − θ ) sin θ sin θ c

DVI file created at 16:06, 20 January 2011

2.3. MINKOWSKI GEOMETRY For the vectors perpendicular to M1 on the unit circle, we can take α = θ ± π /2, implying       b cos(θ ± π /2) ∓ sin θ M2 = = = d sin(θ ± π /2) ± cos θ   − sin θ . =± cos θ If the plus sign holds, then   cos θ − sin θ M= = Rθ ; sin θ cos θ this is rotation by θ . If the minus sign holds, then   cos θ sin θ M= = Fθ /2 ; sin θ − cos θ

15 • Kv−u = H−u Kv Hu . We have    cosh 2v − sinh 2v cosh u sinh u Kv Hu = sinh 2v − cosh 2v sinh u cosh u   cosh(2v − u) − sinh(2v − u) = , sinh(2v − u) − cosh(2v − u) and H−u Kv Hu =    cosh u − sinh u cosh(2v − u) − sinh(2v − u) − sinh u cosh u sinh(2v − u) − cosh(2v − u)   cosh((2v − u) − u) − sinh((2v − u) − u) = sinh((2v − u) − u) − cosh((2v − u) − u)   cosh(2v − 2u) − sinh(2v − 2u) = = Kv−u . sinh(2v − 2u) − cosh(2v − 2u)

10. a) With the same crucial addition formula, we find    this is reflection across the line through the origin that cosh 2v2 − sinh 2v2 cosh 2v1 − sinh 2v1 makes the angle θ /2 with the x-axis. Kv1 Kv2 = sinh 2v2 − cosh 2v2 sinh 2v1 − cosh 2v1   9. • H−u = K0 Hu K0 This is an immediate calculation: cosh(2v1 − 2v2) sinh(2v1 − 2v2 ) = = Hu ,     sinh(2v1 − 2v2) cosh(2v1 − 2v2) 1 0 cosh u sinh u 1 0 K0 Hu K0 = 0 −1 sinh u cosh u 0 −1 where u = 2(v1 − v2 ).    1 0 cosh u − sinh u b) We have = 0 −1 sinh u − cosh u      cosh 2u2 − sinh 2u2 cosh u1 sinh u1 cosh u − sinh u Hu1 Ku2 = = = H−u , sinh 2u2 − cosh 2u2 sinh u1 cosh u1 − sinh u cosh u   cosh(u1 + 2u2) − sinh(u1 + 2u2) = = Kv , sinh(u1 + 2u2) − cosh(u1 + 2u2) because cosh u = cosh(−u), − sinh u = sinh(−u). • H−u = Kv Hu Kv According to Theorem 2.10, where v = 12 u1 + u2.   cosh 2v − sinh 2v 11. Consider the linear map t = −τ , z = ζ that simply Kv = ; sinh 2v − cosh 2v reverses time; it given by the matrix   therefore, by using the addition formulas, we obtain −1 0 M= . 0 1    cosh u sinh u cosh 2v − sinh 2v Hu Kv = This will preserve the Minkowski norm if Mt J1,1 M = J1,1 . sinh u cosh u sinh 2v − cosh 2v   We find cosh(u + 2v) − sinh(u + 2v)      = . sinh(u + 2v) − cosh(u + 2v) −1 0 1 0 −1 0 Mt J1,1 = = 0 1 0 −1 0 −1      For the next product, the crucial addition formula is −1 0 −1 0 1 0 sinh A cosh B − cosh A sinh B = sinh(A − B), and it gives (Mt J1,1 )M = = = J1,1 , 0 −1 0 1 0 −1 Kv Hu Kv =    so M preserves the Minkowski norm. By construction, M cosh 2v − sinh 2v cosh(u + 2v) − sinh(u + 2v) maps the future timelike set T + to the past T − , so M is a sinh 2v − cosh 2v sinh(u + 2v) − cosh(u + 2v) nonorthochronous Lorentz map.   cosh(2v − (u + 2v)) sinh(2v − (u + 2v)) More generally, let Au = MHu , Bu = Hu M = Atu : = sinh(2v − (u + 2v)) cosh(2v − (u + 2v))     − cosh u − sinh u − cosh u sinh u Au = , Bu = . = H−u . sinh u cosh u − sinh u cosh u

DVI file created at 16:06, 20 January 2011

16

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

Because M and Hu both preserve the Minkowski norm, respectively. If u < 0, then e−u > eu and the semi-major so do Au and Bu . Because Hu preserves T + , whereas M and semi-minor axes are simply reversed. maps it to T − , Au also maps T + to T − . Because Hu also 14. We take as our starting point the fact that the general preserves T − , Bu maps T + to T − . ellipse centered at the origin is the locus 12. a) Because cosh(−u) = cosh(+u), the points P−u 0 < k2 = Ax2 + 2Bxy + Cy2, with B2 < AC. and P+u have the same t-coordinate, for every u. The line connecting them is thus parallel to the z-axis. Thus (If B2 > AC, then the locus is a hyperbola.) the first three lines, being parallel to the z-axis, are parSuppose the inverse of X = LU is given by allel to each other. The tangent vector at the point Pu is      Pu′ = (sinh u, cosh u) = (0, 1) at P0 . This vector is parallel u a b x U = = = L−1 X, to the z-axis and is thus parallel to the other three lines. v c d y b) The hyperbolic rotation H1/2 maps the point Pu to the 2 2 2 point Pu+1/2. Therefore the three given lines are the im- and let C be the circle u + v = r > 0 in the (u, v)-plane. ages, under H1/2 , of the three lines in part (a). These lines We show that L(C) is an ellipse centered at the origin in are thus parallel, because any linear map maps parallel the (x, y)-plane. We have lines to parallel lines. The tangent to the curve at u = 0 is r2 = (ax + by)2 + (cx + dy)2 likewise mapped to the tangent to the curve at u = 1/2, so this image line is parallel to the other three image lines. = a2 x2 + 2abxy + b2y2 + c2 x2 + 2cdxy + d 2y2 = Ax2 + 2Bxy + Cy2 13. a) A hyperbolic rotation Hu preserves the lines z = ±t; because it is linear, it preserves at least the direction where of any parallel line z = ±t + k. Thus the image of the given rectangle is another rectangle with sides parallel to A = a2 + c2 , B = ab + cd, C = b2 + d 2 . the given one. The lengths of those sides may, however, be altered, to new lengths, a′ and b′ , let us say. It remains to show that B2 < AC. We have The side of length a in√ the given rectangle A is deter√ mined by the vector a(1/ 2, 1/ 2)t . Its image in A′ is B2 = a2 b2 + 2abcd + c2 d 2 , thus determined by the vector AC = a2 b2 + a2 d 2 + b2c2 + c2d 2 ; √    √   cosh u sinh u a(cosh u + sinhu)/√2 a/√2 = therefore sinh u cosh u a/ 2 a(sinh u + coshu)/ 2  √  1/√2 AC − B2 = a2 d 2 − 2abcd + b2 c2 = (ad − bc)2 ≥ 0. = aeu , 1/ 2 But ad −bc = detL−1 6= 0 because L−1 is invertible; hence from which it follows that a′ = aeu . AC − B2 > 0, as required. The other side of√the given rectangle A is determined √ Now suppose Cb is a circle centered at a point P = (p, q) by the vector b(1/ 2, −1/ 2)t , and its image in A′ is other than the origin. We must show L(C) is again an determined by the vector ellipse, but one whose center may not be at the origin. √   √   Let T : R2 → R2 be the map that translates the P to the 1/ √2 b(cosh u − sinhu)/√2 = be−u , b is a circle centered origin: T (X) = X − P. Then C = T (C) b(sinh u − coshu)/ 2 −1/ 2 b is an ellipse centered at at the origin, so L(C) = L(T (C)) from which it follows that b′ = be−u . the origin. But b) The map Hu is a symmetric linear map; part (a) shows L(T (X)) = L(X − P) = LX − LP = Tb(L(X)), us that we can take (1, ±1)t as eigenvectors and e±u as corresponding eigenvalues. By Exercise 14, Hu will map where Tb is the translation that moves the point L(P) to the a circle of radius r into an ellipse whose axes are in the origin. Finally, then, directions (1, ±1)t . If u > 0, then eu > e−u and the semi-major and semib = Tb−1 L(C) = L(C) + L(P) L(C) minor axes are     is an ellipse whose center has been translated from the reu 1 re−u 1 √ and √ , origin to the point L(P). This completes the proof that the 2 1 2 −1

DVI file created at 16:06, 20 January 2011

2.4. PHYSICAL CONSEQUENCES

17

image of an arbitrary circle under an invertible linear map The points W = Gt X therefore satisfy the equation is an ellipse.   

1/λ12 0 w −1 2 2 t W= w z r =W D Now let L be symmetric. We must show how the imz 0 1/λ22 age ellipse is related to the eigenvalues and eigenvectors 2 2 w z of L. The discussion just ended implies we can restrict = 2 + 2. λ1 λ2 ourselves to circles and ellipses centered at the origin. By the solution to Exercise 4d in §2.1 (Solutions Because |λ1 | ≤ |λ2 |, we see that W lies on the ellipse E page 6), L has real eigenvalues λ1 and λ2 with associated whose semi-minor axis has length |λ1 |r and lies along the mutually orthogonal eigenvectors G1 and G2 . Note: the w-axis; its semi-major axis has length |λ2 |r and lies along stated hypothesis, λ1 ≤ λ2 , is modified to read |λ1 | ≤ |λ2 |. the z-axis. (If |λ1 | = |λ2 |, the ellipse is in fact a circle.) Let G1 and G2 be unit vectors, so that Now Gt = R−θ and G = Rθ ; therefore W = Gt X implies X = Rθ W , so points in the X-plane are the image under G1 · G1 = G2 · G2 = 1, G1 · G2 = 0. rotation by θ of points in the W -plane. The set L(C) is Then if G is the matrix whose columns are G1 and G2 (we the ellipse E after it has been rotated by Rθ . Its semi-axis lengths are unchanged. Its minor and major axes are the write G = (G1 G2 )), we have coordinate axes after rotation by θ ; these are lines deter t   t tG
mined by the eigenvectors G G G G 1 2 1 1 1 G1 G2 = Gt G = = I, Gt2 Gt2 G1 gt2 G2     cos θ − sin θ , G2 = , G1 = sin θ cos θ the identity matrix. It follows from exercise 8 of this section that G is either a rotation or a reflection, depending on the sign of G2 . Choose G2 so that G is a rotation. (When respectively. we want to emphasize this, we shall write G = Rθ for an appropriate θ ). 2.4 Physical Consequences Now consider how G is related to L. By construction, L(G1 G2 ) = (LG1 LG2 ) = (λ1 G1 λ2 G2 ),

1. a)

1.0

and  t G1 G LG = (λ1 G1 , λ2 G2 ) Gt  2 t    λ1 G1 G1 λ2 Gt1 G2 λ1 0 = = = D, 0 λ2 λ1 Gt2 G1 λ2 GT2 G2

0.8

a diagonal matrix. Taking the inverse of both sides, we get

0.2

t

D

−1

−1 t

t

= G LG −1

GD G = L .


−1

0.6

0.4

0.2

−1 −1

t −1

= G L (G ) ,

We can now consider once again the circle C of radius r in the (u, v)-plane. The point U is on C if r2 = U t U; by letting L−1 X = U, we convert this to a condition on X = L(U):
r2 = (L−1 X)t (L−1 X) = X t (L−1 )t L−1 X.

Substituting L−1 = GD−1 Gt (and the transpose of this), we find
2
t r2 = X t GD−1 Gt GD−1 Gt X = Gt X D−1 Gt X.

DVI file created at 16:06, 20 January 2011

0.4

0.6

0.8

1.0

The graph makes it plausible that f (v) ≈ 2v when v ≈ 0 and f (v) ≈ 1 when v ≈ 1. To check, note that 1 = 1 − v2 + v4 − · · · = 1 + O(v2) 1 + v2 so

as v → 0,

2v = 2v + O(v3) as v → 0. 1 + v2 (Taylor’s theorem gives the same result directly.) Because |O(v3 )| ≪ |v| when v ≈ 0, it follows that f (v) ≈ 2v when v ≈ 0.When v ≈ 1, f (v) =

f (v) =

2 2v ≈ = 1. 1 + v2 1 + 1

SOLUTIONS: CHAPTER 2. SPECIAL RELATIVITY—KINEMATICS

18

the ends of a different object P moving with R. (For clarity, the events of P are not shown.) To make the relation 2v 1 between the figures symmetric, we take ∆z = 1 for them, f (v) = = , 1 + v2 2 too. By Fitzgerald contraction, R considers the events markso 4v = 1 + v2, or v2 − 4v + 1 = 0. The quadratic formula ing the ends of Q at time t = 0 in R’s frame to be √ O and gives √ Q′ , and these have the spatial separation ∆z = 1 − v2. √ 4 ± 16 − 4 The symmetry of the figures makes it clear that the events = 2 ± 3. v= 2 marking the ends of P at time τ = 0 in √ G’s frame are O √ ′ 2 Because v > 1 is not possible, v = 2 − 3 ≈ 0.268. In and P ; their spatial separation is ∆ζ = 1 − v . The symmetry of the figures resolves the paradox. fact, on the graph of f we see f (0.268) ≈ 0.5. √ 4. To construct the first-order Taylor polynomial, we just 2. The relevant equation is ∆t = ∆τ / 1 − v2, where ∆t is need the values of f (0) and f ′ (0). We have f (0) = 1 and is lifetime lifetime in the fixed (laboratory) frame and ∆τ √ √ √ √ √ in the particle’s own frame. Because v = 0.99, 1 − v2 ≈ 1 + v · −1/2 1 − v − 1 − v · 1/2 1 + v ′ , f (v) = 0.141, so the particle’s lifetime in its own frame is 1+v −1/2 − 1/2 p f ′ (0) = = −1. ∆τ = 1 − v2 ∆t ≈ 0.141 × 10−10 sec = 1.41 × 10−11 sec. 1 The polynomial is therefore P(v) = 1 − v. A different argument uses τ z 3. a) ζ √ 1 = 1 − A + O(A2) 1 ± v = 1 ± 12 v + O(v2 ) and B′ A 1+A R to write B′ A √ B τ t G R 1 − v 1 − 12 v + O(v2) O A′ O √ = G 1 + v 1 + 12 v + O(v2)

B A′ = 1 − 1 v + O(v2) 1 − 1 v + O(v2 ) b) The velocity v we seek is a solution of the equation

2

In the two frames above, the events A and A′ are simultaneous for G; assume τ = 1 for them. Similarly, the events B and B′ are simultaneous for R; t = 1 for them. clock runs slow by the factor √ Because R says G’s 1 − v2,√the event A′ on R’s worldline happens at R’s time t = 1 − v2. The symmetry we see in the two frames ′ guarantees that the corresponding √ event B on G’s world2 line happens at G’s time τ = 1√ − v . Thus G says that R’s clock runs slow by the factor 1 − v2. The symmetry of the figures resolves the paradox. b) z

ζ P

R G

O

ζ

z Q

P′

2

2

t

Q

P

Q′

Q τ

= 1 − v + O(v ).

5. [Note: in the Times-Roman fonts used here, Greek nu and Roman math vee are virtually indistinguishable: nu, ν ; vee, v. To clarify, we put a tilde over the nu, νe.] The text produces a single formula that abandons the distinction between nadv and nrec by letting v represent the time rate of change of distance between source G and observer R. We adopt this approach here; it means we need only analyze the receding source. We assume, as in the text, that there are events along G’s worldline that are spaced ∆τ = 1/νe seconds apart. In the Galilean formulation, photon worldlines of slope −1 + v emanate from these events in R’s frame, and we seek to determine the spacing ∆t between these worldlines where they intersect R’s worldline.

Q τ

R

O

t

t G In the two frames above, the shaded region represents an object Q that moves with G. The events O and Q mark the two ends of Q at the time τ = 0; assume their spatial separation is ∆ζ = 1. Similarly, the events O and P mark

DVI file created at 16:06, 20 January 2011

slope +1 + v G ∆τ

∆t

v ∆τ a

R slope −1 + v

2.4. PHYSICAL CONSEQUENCES

19

Recall that, in a Galilean transformation, all time intervals are measured horizontally. Thus ∆t = ∆τ + a, and a can be determined from the slope −1 + v =

−v∆τ , a

so

a=

−v∆τ . −1 + v

ζ

7. a)

V

G

z

EV EV V

−a O

τ

R

tV tT

O

τ t

Thus −1 + v − v −v∆τ ∆τ = ∆τ = , ∆t = ∆τ + −1 + v −1 + v 1−v and

1 1−v n= = νe(1 − v). = ∆t ∆τ

G T

ET T

ET

In the two frames, the flashes from T and V are the events ET and EV , respectively. Given the stated circumstances, 6. The positions of the t- and z-axes make it clear that we assume v > 0. a Lorentz transformation is involved, not a Galilean. We b) According to G, the flashes are simultaneous.. They continue to denote source frequency as νe, and we con- occurred a seconds before the event O when the flashes tinue to let v represent the time rate of change of distance are seen. between source G and observer R. Thus again we concenc) According to R, the flashes are not simultaneous. Betrate on the receding observer. cause v > 0, the one from the rear of the train (ET ) hapa pens first, the one from the front (EV ), second. To deterav ∆τ mine the times in R’s frame, we can use the boost Bv to G transform G’s coordinates for ET and EV . We have av       ∆t 1 1 v −a −a R . ET = ,E = , Bv = √ −a G V a G 1 − v2 v 1 By Minkowski invariance, ∆t 2 = a2 − (av)2 , so We need only the t-coordinates; they are ∆t a= √ . −a(1 − v) −a(1 + v) 1 − v2 , tV = √ . tT = √ 2 1−v 1 − v2 Furthermore, ∆τ + av = a, so ∆τ = a(1 − v). Therefore, Their difference is the time interval between the two ∆τ ∆t flashes as seen by R: =√ 1−v 1 − v2 2av tV − tT = √ . and thus 1 − v2 p r (1 − v)(1 + v) 1+v ∆t = ∆τ = ∆τ . d) According to G, both of the flashes were emitted at 1−v 1−v time τ = −a. According to R, the flash from the rear of The frequencies therefore satisfy the relation the train was emitted before time t = −a; the one from the front, after that time: r r 1−v 1−v 1 1 r r = = νe . n= 1+v 1−v ∆t ∆τ 1 + v 1+v < −a < −a = tV . tT = −a 1−v 1+v This agrees with the Doppler calculation made in the text from the source (i.e., G’s) frame. (The two expressions for tT are equal; likewise, the two expressions for tV are equal.)

DVI file created at 16:06, 20 January 2011

Solutions: Chapter 3

Special Relativity—Kinetics 3.1 Newton’s Laws of Motion

3. The “trick” that says to balance dimensions by inserting factors of c as necessary suggests that we should have

1. The result follows from Taylor’s theorem or from the fact that the geometric series 1 + A + A2 + A3 + · · · + An + · · ·

v → v/c,

V /c =

cosh2 u − 1 1 sinh2 u = = 1− , 2 cosh u cosh2 u cosh2 u

V → V /c,

v/c + w/c 1 + (v/c)(w/c)

or V =

v+w . 1 + vw/c2

(Of course it is possible to go through all the steps of Proof 2 with the replacements made.)

or cosh2 u = 1/(1 − (v/c)2). It follows that

4. Let us denote (t, x/c, y/c, z/c) as Xsec to distinguish it from Xm = (ct, x, y, z) as used in the text. Then, of course, Xsec = (1/c)Xm . The same relation holds for 4-velocity, proper-4-velocity, and 4-momentum:

1 cosh u = p . 1 − (v/c)2

sinh2 u = cosh2 u − 1 =

w → w/c,

that are indicated by the new form that the boost Bv takes. Then the formula V = (v + w)/(1 + vw) that appears in the proof is itself replaced by

2. a) If v = c tanh u, then

b) We have

v+w . 1 + vw/c2

To confirm, we can adapt Proof 2 of Proposition 3.1 by making the replacements

has the sum 1/(1 − A) when |A| < 1. Thus, with A = 23 v2 we find
v = v 1 + 32 v2 + · · · . p= 2 2 1− 3v Finally, if v ≈ 0, then 23 v2 + · · · ≪ |v| and so p ≈ v. (v/c)2 =

V=

1 Vsec = Vm = (1, v/c), c 1 1 (1, v/c), Usec = Um = p c 1 − (v/c)2

1 (v/c)2 −1 = , 2 1 − (v/c) 1 − (v/c)2

p 1 Psec = Pm = (m, p/c). so sinh u = (v/c)/ 1 − (v/c)2. c As the text states, the boost Bv corresponds to the hyperbolic rotation Hu when v = c tanh u in conventional The first two are dimensionless; the third has the dimension of mass. It is also true that units. Because   µ cosh u sinh u Psec = m(1, v/c) = mVsec = p Vsec = µ Usec . Hu = , 1 − (v/c)2 sinh u cosh u The Minkowski norm is s

immediate substitution gives 1

Bv = p 1 − (v/c)2



 1 v/c . v/c 1

kXsec k =

20

DVI file created at 16:06, 20 January 2011

t2 −

x2 y2 z2 − − sec, c2 c2 c2

3.1. NEWTON’S LAWS OF MOTION and thus the 4-speed is kVsec k =

q 1 1 − (v/c)2 = kVm k. c

5. a) Consider the formula t = τ cosh u + ζ sinh u as written in geometric units. In conventional units, each term must have the dimensions of seconds. The first term already does, but the second has the dimension of meters. The remedy is to attach the factor 1/c (sec/m) to the second term: 1 t = τ cosh u + ζ sinh u. c For the formula z = τ sinh u + ζ cosh u

21 6. a) For X = (t, x, y, z), define Q(X) = t 2 −

x2 y2 z2 − − sec2 . c2 c2 c2

b) In J1,3 , below, the first diagonal entry is dimensionless and the second, third, and fourth have the dimensions of sec2 /m2 :   1 0 0 0 0 −1/c2 0 0  . J1,3 =  2 0 0 −1/c 0  0 0 0 −1/c2 c) With hindsight, we find it is helpful to write Hut J1,1 Hu = J1,1 also in the form J1,1 Hu = (Hut )−1 J1,1 . Suppose     1 p q p −r Hu = , then (Hut )−1 = , r s D −q s

in geometric units, its conventional form will have dimen- where D = det Hu = ps − qr. Note: J1,1 = H t J1,1 Hu imu sions of meters. But the first term has the dimensions of plies seconds, so we need to attach the factor c (m/sec): −1/c2 = det J1,1 = det Hut · detJ1,1 · detHu = (−1/c2 )D2 , z = τ c sinh u + ζ cosh u. so D = ±1 and Hu is indeed invertible. Direct computaThus, in conventional units, tion gives   ! 1 1 p r/c2 sinh u cosh u = (HUt )−1 J1,1 . Hu = c D −q −s/c2 c sinh u cosh u   p q = J1,1 Hu = . −r/c2 −s/c2 b) The point (τ , ζ ) = (1, 0) is on G’s worldline; its image (t, z) under Hu will give G’s velocity as v = z/t. We have Because we know D = ±1 already, by comparing the diagonal entries we find D = +1. The off-diagonal entries !     1 imply q = r/c2 . The sinh function is 1–1 and maps R onto 1 cosh u t cosh u sinh u = = , c itself; hence there is a unique u for which r = c sinh u and 0 c sinh u z c sinh u cosh u q = (1/c) sinh u. Now consider the original identity; once again direct so v = c sinh u/ coshu = c tanh u. computation gives c) Because v = c tanh u exactly as in Exercise 2, above,  2  p − r2 /c2 pq − rs/c2 we know = Hut J1,1 Hu pq − rs/c2 q2 − s2 /c2   1 v/c 1 0 cosh u = p , sinh u = p . = J1,1 = . 1 − (v/c)2 1 − (v/c)2 0 −1/c2

As we did in that exercise, we obtain the boost Bv from The new information here is the hyperbolic rotation Hu we now have from part (a) by p2 = 1 + r2/c2 and s2 = c2 q2 + 1; replacing sinh u and cosh u by the corresponding expressions in v: given the values of q and r, we can express these equations   1 1 v/c2 as Hu → p = Bv . 2 v 1 p = ± cosh u and s = ± cosh u. 1 − (v/c)

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 3. SPECIAL RELATIVITY—KINETICS

22

Because 1 = D = ps − qr = ps − sinh2 u, it follows that the curvature vector is ps = 1 + sinh2 u ≥ 1 and the signs of p and s must be the k(s) = u′ (s) = (0, 0) for all s. same. Therefore we have determined that ! 1 Because the curvature κ = 0, the radius of curvature func± cosh u sinh u Hu = . c tion is undefined. c sinh u ± cosh u • The curve is the straight line traversed from (−2, 3) to (13, 23). Its length is L = 25. This is, in fact, as far as we can go. Only by specifying, in addition, that Hu is orthochronous can we eliminate the b) • Here x′ = (−6 sin 2q, 6 cos 2q) and kx′ k = 6, so the possibility of the minus sign in the formula for Hu . arc-length function is 7. a) When v = 30 m/sec and c = 3 × 108 m/sec, then v/c2 = 3.3 × 10−16 and 1 p = 1.000 000 000 000 005; 1 − (v/c)2

therefore the boost is



1 3.3 × 10−16 B30 m/sec = 1.000 000 000 000 005 × 30 1   1 0 = to 13 decimal places. 30 1



b) When v = 30 m/sec, the Galilean shear is   1 0 S30 m/sec = . 30 1

s(q) =

Z q

6 dq = 6q.

0

Thus q = s/6 and L = s(π ) = 6π . • y(s) = x(s/6) = (3 cos(s/3), 3 sin(s/3)), 0 ≤ s ≤ 6π . • The unit tangent is u(s) = (− sin(s/3), cos(s/3)), and 1 k(s) = (− cos(s/3), − sin(s/3)). 3 Therefore κ (s) = 1/3, so the radius of curvature function is ρ (s) = 3 for all s. • The curve is the circle of radius 3 traversed counterclockwise once around the origin. Its length is L = 6π .

c) • Here x′ = (− sin q, cosq), kx′ k = 1, and s(q) = q. • Because s = q, the curve already has its arc-length parametrization: y(s) = x(s), 0 ≤ s ≤ 2π . To 13 decimal places, B30 m/sec = S30 m/sec . More gener• u(s) = (− sin s, cos s) and k(s) = −(coss, sin s). ally, as c → ∞, we have κ (s) = 1 = ρ (s) for all s. Therefore q 2 • The curve is the circle of radius 1 traversed once v/c → 0 and 1/ 1 − (v/c)2 → 1; counterclockwise around the point (3, 2). Its length is therefore, L = 2π .     q d) • Here x′ (q) = (−3 cos2 q sin q, 3 sin2 q cos q) 1 0 1 v/c2 Bv = 1/ 1 − (v/c)2 → = Sv . v 1 v 1 = 3 sin q cos q (− cos q, sin q). When c ≫ |v|, a Lorentz boost is indistinguishable from a Because k(− cosq, sin q)k = 1 and sin q cos q ≥ 0 on the Galilean shear. interval 0 ≤ q ≤ π /2, we have kx′ (q)k = 3 sin q cos q and

3.2 Curves and Curvature

s(q) =

Z q 0

3 sin q cosq dq = 32 sin2 q.

p 1. a) • Here x′ = (3, 4) and kx′ k = 5, so the arc-length Therefore q(s) = arcsin 2s/3 and L = s(π /2) = 3/2. • To determine y(s), we shall need the facts that function is Z q sin(arcsin v) = v and 5 dq = 5q. s(q) = p 0 p cos(arcsin v) = 1 − sin2 (arcsin v) = 1 − v2 Furthermore, q = s/5 and arc length is L = s(5) = 25. • The arc-length parametrization is Thus y(s) = x(q(s)) y(s) = x(s/5) = (3s/5 − 2, 4s/5 + 3), 0 ≤ s ≤ 25.  3  3  p p = , sin(arcsin cos(arcsin 2s/3) 2s/3) • First get the unit tangent vector   u(s) = y′ (s) = (3/5, 4/5); = (1 − 2s/3)3/2 , (2s/3)3/2 , 0 ≤ s ≤ 3/2.

DVI file created at 16:06, 20 January 2011

3.2. CURVES AND CURVATURE

23

• The tangent vector is   u(s) = y′ (s) = − (1 − 2s/3)1/2 , (2s/3)1/2 ;

Note that ku(s)k2 = 1 − 2s/3 + 2s/3 = 1, confirming u(s) is a unit vector for all s. The curvature vector is ! 1 1 1 p k(s) = u′ (s) = , ,p 3 1 − 2s/3 2s/3

and

1 κ (s) = 3

s

1 1 1 + = 1 − 2s/3 2s/3 3

s

1 − 2s/3 + 2s/3 (1 − 2s/3)(2s/3)

• The arc-length parametrization is   3/2 2/3 2/3 (3s + 1) − 1 (3s + 1) − 1   , y(s) = x(q(s)) =  . 3 2 • The unit tangent vector is  1/2  1 ′ 2/3 u(s) = y (s) = ,1 , (3s + 1) − 1 (3s + 1)1/3

and the various aspects of curvature are 1 k(s) = u′ (s) = (3s + 1)4/3

1 (3s + 1)2/3 − 1

1 1 =p . κ (s) = kk(s)k = , 2s(3 − 2s) (3s + 1)2/3 − 1 1 p ρ (s) = = (3s + 1)2/3 − 1. Therefore ρ (s) = 2s(3 − 2s), 0 ≤ s ≤ 3/2. κ (s) • Over the larger domain 0 ≤ q ≤ 2π , the curve x(q) is the astroid pictured below. Our curve, with 0 ≤ q ≤ π /2, (There are a lot of intermediate calculations.) • is the portion in the first quadrant. 2.0

!


1/2 , −1 ,

1.0

1.5

0.5 1.0

0.5 -1.0

0.5

-0.5

1.0

0.5

1.0

1.5

2.0

2.5

-0.5

-1.0

p e)p • Here x′ (q) = (q2 , q) and kx′ (q)k = q4 + q2 = q q2 + 1 when 0 ≤ q ≤ 2. The arc-length function is Z q p

q q2 + 1 dq q (q2 + 1)3/2 − 1 3/2 1 2 = 3 (q + 1) = . 3 0

s(q) =

0

Its inverse is

q(s) =

q (3s + 1)2/3 − 1,

and L = s(2) = (53/2 − 1)/3 ≈ 3.39.

DVI file created at 16:06, 20 January 2011

f) • Here x′ (q) = eq (cos q − sinq, cosq + sin q) and p √ kx′ (q)k = eq 2 cos2 q + 2 sin2 q = eq 2; therefore

s(q) =

Z q 0

√ q √ q √ q e 2 dq = e 2 = 2 (e − 1). q

0

√ The is q(s) = ln(1 + s/ 2), and L = s(π /2) = √ inverse 2 (eπ /2 − 1). √ √ • We have eq = 1 + s/ 2; if we set A = ln(1 + s/ 2), we can write the arc-length parametrization as √
y(s) = (1 + s/ 2) cos A, sin A . • In the same terms, the unit tangent vector is


1 u(s) = √ cos A − sinA, cos A + sin A , 2

SOLUTIONS: CHAPTER 3. SPECIAL RELATIVITY—KINETICS

24

and the curvature vector is
1 k(s) = √ √ − sin A − cosA, − sin A + cosA . 2( 2 + s)

Thus

√ 1 1 κ (s) = √ √ · 2= √ 2( 2 + s) 2+s √ and ρ (s) = 2 + s. • The curve is a portion of a spiral, called an equiangular spiral because the radius from the center (the origin in this case) to any point on the spiral makes a constant angle with the tangent at that point. In fact, the cosine of the angle in this case is

e2q (cos2 q + sin2 q) 1 x · x′ √ = =√ , ′ kxkkx k eq · eq 2 2 ◦ so the angle itself is 45 . The angle at which the curve crosses a coordinate axis (each of which is a radius) does indeed appear to be 45◦.

4

In the figure below, each point X(q) is labelled with its q-parameter value, tangent vectors X′ (q) are white with black outline, and acceleration vectors X′′ (q) are plotted at half their actual length. 0

X′ X X′′ −1

1

−2

2

The speed is kX′ (q)k = 2/(1 + q2); this increases (i.e., the parameter point “speeds up”) while q < 0, and it decreases (the parameter point “slows down”) while q > 0. The tangent vectors increase in length until q = 0, and then they decrease. The acceleration vector always points inward. While the parameter point is speeding up, acceleration X′′ makes an acute angle with velocity X′ ; while the point is slowing down, the angle is obtuse.

3

b) The alternative curvature formula is s (X′ · X′ )(X′′ · X′′ ) − (X′ · X′′ )2 κ= . (X′ · X′ )3

2

We have already seen that X′ · X′ = kX′ k2 = 4/(1 + q2)2 . In addition, 1

0.5

1.0

1.5

2. a) The tangent (velocity) and acceleration vectors are   −4q 2 − 2q2 , , X′ (q) = (1 + q2)2 (1 + q2)2   −12q + 4q3 12q2 − 4 ; , X′′ (q) = (1 + q2)3 (1 + q2)3 the values to plot are as follows. q 0 ±1 ±2 X (1, 0) (±1, 0) (±4/5, −3/5) X′ (2, 0) (0, ∓1) (−6/25, ∓8/25) X′′ (0, −4) (∓1, 1) (±8/125, 44/125)

DVI file created at 16:06, 20 January 2011

X′′ · X′′ =

16 16 + 48q2 + 48q4 + 16q6 = , 2 6 (1 + q ) (1 + q2)3

X′ · X′′ =

−8q −8q − 16q3 − 8q5 = ; (1 + q2)5 (1 + q2)3

therefore (X′ · X′ )(X′′ · X′′ ) − (X′ · X′′ )2 = =

64q2 64 − 2 5 (1 + q ) (1 + q2)6

64(1 + q2) − 64q2 64 = = (X′ · X′ )3 . (1 + q2)6 (1 + q2)6

From this it follows that κ ≡ 1.

c) Because the domain of the parameter q is (−∞, ∞), the arc-length formula has the form q Z q Z q 2 dq ′ kX (q)k dq = s(q) = = 2 arctan q 2 1 + q −∞ −∞ −∞

= 2(arctan(q) − arctan(−∞)) = 2 arctan(q) + π .

3.2. CURVES AND CURVATURE

25

d) The total length of X is s(∞) = 2 arctan(∞) + π = 2π . 4. Page 116 of the text establishes that the length of the portion of the ellipse e) The inverse of s(q) = 2 arctan(q) + π is q(s) = tan



s−π 2



,

x2 y2 + = 1, a2 b2

0 < s < 2π .

0 < b < a,

Set (s− π )/2 = A. Because the arc-length parametrization in the first quadrant is given by Y(s) is obtained from X(q), we first calculate Z π /2 q L=a 1 − k2 sin2 q dq, 2 2 2 2 1 + q = 1 + tan A = sec A = 1/ cos A. 0 √ Hence where k = a2 − b2 /a. The length of the complete ellipse is then 4L. The integral itself can be determined by 2q 2 sin A/ cosA = = 2 sin A cos A = sin 2A, 2 2 numerical integration. Its value is also available in tables; 1+q 1/ cos A it is known as the complete elliptic integral of the second 2 2 1 − q2 1 − sin A/ cos A 2 2 kind, denoted E(k). = = cos A − sin A = cos 2A, √ 1 + q2 1/ cos2 A For the first ellipse, a = 4, b = 3, and k = 7/4; therefore E(k) = 1.38147, L = 4 × E(k) = 5.52587, and the and it now follows that total circumference is 22.1035. √ Y(s) = X(q) = (sin(s − π ), cos(s − π )), 0 < s < 2π . For the second ellipse, a = 12, b = 5, and k = 119/5; The domain of s excludes the endpoints 0 and 2π be- therefore E(k) = 1.16033, L = 12 × E(k) = 13.924, and cause the image curve must exclude the point (0, −1) = the total circumference is 55.6959. (sin(−π ), cos(−π )) = (sin(2π − π ), cos(2π − π )).

f) The unit tangent vector and the curvature vector are

-1.0 1.0

5. a)

-0.5 0.5 0.0

U(s) = Y′ (s) = (cos(s − π ), − sin(s − π ),

K(s) = U′ (s) = (− sin(s − π ), − cos(s − π )) = −Y(s). Thus K is the unit vector that, everywhere on the circle, points back to its center. Unlike X′′ , it has constant length and is everywhere perpendicular to the curve. 3. Two modifications are needed. First, because ϕ (q) is now a decreasing function, we have ϕ ′ (q) < 0. Hence, in the first displayed equation on page 112 of the text, where |ϕ ′ (q)| is replaced by ϕ ′ (q), it must now be replaced by −ϕ ′ (q), transforming the displayed equation into

0.0 0.5 1.0

-0.5 -1.0

2

1

kx′ (q)k dq = · · · = −kX′ (Q)k dQ.

The second modification involves the second displayed equation. We can clarify the second equality in the original form by writing Z b

kx′ (q)k dq =

Z ϕ −1 (b)

kX′ (Q)k dQ =

Z B

kX′ (Q)k dQ.

0

The illustration is drawn with a = 0.2. The curve is a helix; more particularly, a circular helix with radius 1 But if we incorporate the first modification with the now- whose axis is the z-axis. The parameter a determines the changed values of ϕ −1 (a) and ϕ −1 (b), the new version of pitch of the helix; this is the amount of vertical space the second equality becomes between vertically adjacent points xa (q) and xa (q + 2π ). Z ϕ −1 (b) Z b The pitch is thus 2π a; in the figure it is thus about 1.26.
−kX′ (Q)k dQ kx′ (q)k dq = Lx = When a → 0, the helix becomes a circle (in the plane ϕ −1 (a) a z = 0). When a → ∞, the pitch increases without limit and Z B Z A
′ ′ the helix becomes a straight line (parallel to the axis of the kX (Q)k dQ = LX . −kX (Q)k dQ = = A B helix). a

ϕ −1 (a)

A

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 3. SPECIAL RELATIVITY—KINETICS

26

′ ′ b) √ We have xa (q) = (− sin q, cos q, a) and kxa (q)k = 2 1 + a . If we take q = 0 for the starting point of the helix, then

sa (q) =

Z qp

1 + a2 dq =

0

p qa (s) = q/ 1 + a2,

6. a) Note that the curve here is a generalization of the one analyzed in Exercise 1(f); our analysis here will generalize that analysis. Write xa (q) as eaq (cos q, sin q); then

p 1 + a2 q,

x′a (q) = aeaq (cos q, sin q) + eaq(− sin q, cosq) = eaq (a cosq − sin q, a sin q + cosq). We shall also need kx′a (q)k2 = e2aq a2 cos2 q − 2a cosq sin q + sin2 q

and therefore       s as s ya (s) = cos √ , sin √ ,√ . 1 + a2 1 + a2 1 + a2

+ a2 sin2 q + 2a sinq cos q + cos2 q

= e2aq (a2 + 1).




′ To get the curvature vector, we first need the unit tangent The cosine of the angle between xa and xa is ′ ua (s) = ya (s) = xa · x′a e2aq (a cos2 q + a sin2 q) a √ = =√ .       ′ aq · eaq a2 + 1 2+1 kx kkx k a e a a s s 1 √ − sin √ , cos √ ,a . This gives us the measure of the angle and shows that it is 1 + a2 1 + a2 1 + a2 constant. √ Then ka (s) = u′a (s) = b) Because kx′a (q)k = eaq a2 + 1, the arc-length we       seek is 1 s s √ √ − cos , − sin , 0 . 0 √ √ Z 0 1 + a2 1 + a2 1 + a2 p eaq a2 + 1 a2 + 1 aq 2 a + 1 dq = . e L= = a a −∞ −∞ c) • Using the curvature vector itself, we have

κa = kka (s)k =

1 . 1 + a2

7. a) Because

kx(q)k2 = cos4 q + cos2 q sin2 q + sin2 q
= cos2 q cos2 q + sin2 q + sin2 q

• To use the alternate formula, we need x′a (from above) and x′′a = (− cos q, − sin q, 0). We have x′a · x′a = 1 + a2,

x′a · x′′a = 0,

every point of x(q) is at unit distance from the origin, and hence is on the unit sphere.

x′a · x′a = 1,

and hence we find

κa (s) =

s

b) 1 (1 + a2) · 1 − 02 = , (1 + a2)3 1 + a2

1.0

in agreement with the first method. Note that the curvature is constant and depends only on the “pitch” parameter a. d)

= cos2 q + sin2 q = 1,

0.5

1.0 0.8 0.6

0.0

0.4 0.2

-0.5 0.5

1.0

1.5

2.0

2.5

1.0

3.0

Sketched is the graph of κa as a function of a. According to the formula, κa → 0 as a → ∞. According to the our geometric understanding of the helix itself, the helix becomes “straighter” as a increases; this is consistent with its curvature κa decreasing to 0.

DVI file created at 16:06, 20 January 2011

0.5 0.0

-1.0 1.0

-0.5

0.5 0.0 -0.5

-1.0 -1.0

3.3. ACCELERATED MOTION

27

The curve starts at x(−π /2) = (0, 0, −1), the south pole b) The proper time calculation is similar to the arc-length of the unit sphere. It ends at x(π /2) = (0, 0, 1), the north calculation in Exercise 5 already referred to. We have pole. The curve crosses the equator when z = sin q = 0, X′ (t) = (1, −rω sin ω t, rω cos ω t), that is, when q = 0. c) At the point x(q), the curve’s direction is given by so it follows (using the Minkowski norm in its simple, x′ (q) = (−2 cos q sin q, − sin2 q + cos2 q, cos q). The direc- geometric form) that tions at the starting and ending points are therefore kX′ (t)k2 = 1 − r2ω 2 sin2 ω t − r2 ω 2 cos2 ω t = 1 − r2ω 2 . ′ ′ x (−π /2) = (0, −1, 0) and x (π /2) = (0, 1, 0), Note that, for X(t) to be a timelike worldcurve, we must have respectively. The illustration reflects these facts. 1 0 < 1 − r2ω 2 or r < ; ′ ω d) First write x (q) as (− sin 2q, cos 2q, cos q); then in other words, the period of the motion (and hence the kx′ (q)k2 = sin2 2q + cos2 2q + cos2 q = 1 + cos2 q, frequency) puts a bound on the size of the radius of the motion. and the arc length is therefore If we take t = 0 as the initial time in R’s frame, then the proper time function is Z π /2 p Z tp L= 1 + cos2 q dq.  p −π /2 τ (t) = 1 − r2ω 2 De = 1 − r2ω 2 t. 0

To cast this in the form of an elliptic integral of the second kind (cf. E(k) in the solution to Exercise 4, above), first Proper time for X is thus proportional to t; the proportionality constant is less than 1, so R says that X’s clock runs use the symmetry of the curve itself to write slow. Z π /2 p 2. Note the first sentence of the proof of Proposition 3.3: L=2 1 + cos2 q dq 0 “Since X′ (q) . . . is a future timelike vector, t ′ (q) > · · · 0.” It follows that the function t = t(q) is invertible for all q Then 1 + cos2 q = 1 + 1 − sin2 q = 2(1 − k2 sin2 q) where for which X(q) is defined. Let q = q(t) denote the inverse, k2 = 1/2, so and let q Z π /2 √ xb(t) = x(q(t)), yb(t) = y(q(t)), b z(t) = z(q(t)). L=2 2 1 − k2 sin2 q dq. 0

√ The integral is E(1/ 2) = 1.35064, and L = 3.8202.

Then

b = X(q(t)) = (t, xb(t), yb(t),b X(t) z(t))

gives the new parametrization.

3.3 Accelerated Motion 1. a) This curve is similar to the helix in Exercise 5 of the previous section 3.2. In R’s (t, y, z)-spacetime, X(t) is a circular helix whose axis is the t-axis. Its shape has two aspects: its radius (r, in this case), and its pitch: the amount by which t changes in one complete cycle. In one cycle ω t increases by 2π ; therefore the increase in t (i.e., the pitch) is 2π /ω . That is, ω determines the pitch of the helix. (In the spatial (y, z)-plane, the helix defines motion in a circle of radius r with a period of 2π /ω . Thus the pitch of the helix in spacetime gives the period of motion in space. Its reciprocal, ω /2π , is the frequency of the periodic motion.)

DVI file created at 16:06, 20 January 2011

3. By analogy with the proof of Theorem 3.1, we seek to construct a smooth function Q = ϕ (q) for which X(q) = b is 1–1 on its image (i.e., G’s worldb ϕ (q)). Because X X( b cuve), the inverse X−1 is defined there, and we take b −1 (X(q)), ϕ (q) = X

b ϕ (q)) = X(q), implying X(

as desired. If we can prove ϕ is differentiable, then we can apply the chain rule to get b ′ (ϕ (q))ϕ ′ (q). X′ (q) = X

b ′ (q) and X′ (q) are both future vectors, their Because X norms are positive, implying that ϕ ′ (q) is also positive, for all q. Therefore, b ′ (ϕ (q))k · |ϕ ′ (q)| = kX b ′ (ϕ (q))kϕ ′ (q), kX′ (q)k = kX

SOLUTIONS: CHAPTER 3. SPECIAL RELATIVITY—KINETICS

28 and thus ∆τq =

Z q2 q1

kX′ (q)k dq = =

Z ϕ (q2 )

ϕ (q1 ) Z Q2 Q1

b ′ (ϕ (q))kϕ ′ (q) dq kX

b ′ (Q)k D′ s = ∆τQ . kX

The endpoints of the integral transform as shown because b ϕ (qi )) = X(Q b i ), Ei = X(qi ) = X(

i = 1, 2,

the first two terms on the right cancel each other, leaving us with just r(∆P) = −

1 b ′ (Q)k2 kX

Dividing by k∆Pk, we find

b ′ (Q) · R(∆Q). X

r(∆P) r(∆P) ∆Q = k∆Pk ∆Q k∆Pk   1 ∆Q b ′ (Q) · R(∆Q) =− . X ′ 2 b ∆Q k∆Pk kX (Q)k

implying ϕ (qi ) = Qi , as claimed. To complete the proof we must show ϕ is differentiable. As noted in the proof of Theorem 3.1, the key As ∆Q → 0, the expression in parentheses goes to 0 and b −1 is differentiable at every ∆Q/k∆Pk remains bounded, because is to show that the inverse X b point on the curve X(q). The proof then indicates that R(∆Q) ∆P b ′ the needed derivative is the vector DP whose value at the b ′ (Q). = X (Q) + →X b ∆Q ∆Q point P = X(Q) is given by the formula DX(Q) = b

1

b ′ (Q)k2 kX

b ′ (Q). X

Therefore, because ∆Q → 0 if and only if ∆P → 0, we have r(∆P)/k∆Pk → 0 as ∆P → 0.

4. By the fundamental theorem of calculus, (The proof identifies DP as a certain gradient vector
b −1 (P), but we do not need this fact for our argument dτ p ∇ X = 1 − k2 sin2 t. here.) dt The proof that DP is indeed the derivative involves tak- By the inverse function theorem, b ing a nearby point P + ∆P on the curveX(Q) and showing dt 1 1 1 that ϕ′ = = =√ . =p r(∆P) 2 d τ d τ /dt 1 − k sin2 ϕ (τ ) 1 − k2 sin2 t → 0 as ∆P → 0 k∆Pk If we write ϕ ′ = (1 − k2 sin2 ϕ )−1/2 , then where b −1 (P + ∆P) − X b −1(P) − DP · ∆P. r(∆P) = X

To begin, note that P + ∆P is on the curve and near P; b + ∆Q) for some ∆Q, and then therefore, P + ∆P = X(Q we have r(∆P) = ∆Q −

1 b ′ (Q) · ∆P. X ′ b kX (Q)k2

b is itself differentiable at Q, we can write Because X

ϕ ′′ = − 21 (1 − k2 sin2 ϕ )−3/2 (−2k2 sin ϕ cos ϕ · ϕ ′ )

= (1 − k2 sin2 ϕ )−3/2 (k2 sin ϕ cos ϕ )(1 − k2 sin2 ϕ )−1/2 =

k2 sin ϕ cos ϕ . (1 − k2 sin2 ϕ )2

We have (ϕ ′ )2 cos ϕ + ϕ ′′ sin ϕ =

cos ϕ − k2 sin2 ϕ cos ϕ + k2 sin2 ϕ cos ϕ (1 − k2 sin2 ϕ )2 cos ϕ = ; (1 − k2 sin2 ϕ )2

b + ∆Q) = X(Q) b b ′ (Q)∆Q + R(∆Q), X(Q +X

=

where R(∆Q)/∆Q → 0 as ∆Q → 0. If we rewrite the last equation as b + ∆Q) − X(Q) b b ′ (Q)∆Q + R(∆Q), ∆P = X(Q =X

then the last expression for r(∆P) takes the form   1 b ′ (Q) · X b ′ (Q) ∆Q r(∆P) = ∆Q − X b ′ (Q)k2 kX 1 b ′ (Q) · R(∆Q). X − ′ 2 b kX (Q)k

DVI file created at 16:06, 20 January 2011

cos ϕ k2 sin2 ϕ cos ϕ + 2 (1 − k2 sin ϕ ) (1 − k2 sin2 ϕ )2

therefore
A(τ ) = ϕ ′′ , k(ϕ ′′ sin ϕ + (ϕ ′ )2 cos ϕ )   2 k sin ϕ cos ϕ k cos ϕ = , (1 − k2 sin2 ϕ )2 (1 − k2 sin2 ϕ )2
k cos ϕ k sin ϕ , 1 . = 2 2 2 (1 − k sin ϕ )

3.3. ACCELERATED MOTION 5. a) Let (t, z) = X(q) = (cosh(q), sinh q). Then, for all q, t = cosh q > 0 and t 2 − z2 = cosh2 q − sinh2 q = 1; thus X(q) satifies all the conditions to parametrize the hyperbola. Moreover, X′ (q) = (sinh q, cosh q), Q(X(q)) = sinh2 q − cosh2 q = −1 < 0 for all q, so the hyperbola is spacelike and we write (cf. p page 53 of the text) kX(q)k = −Q(X(q)) = 1.

b) The given formula for the Minkowski length is l=

Z q 0

1 · dq = q.

6. By definition, p = mv, so p = CV and hence pc = me v. If we write E = mc2 in the form E/c = mc, then a substitution gives pc = Ev/c. 7. When v = c, the equation established in the previous exercise takes the form pc = E. Now apply this result to the displayed formula in the proof of Theorem 3.9:

µ 2 c2 = kPk2 =

E2 p 2 c2 − p2 = 2 − p2 = 0. 2 c c

This forces µ = 0.

DVI file created at 16:06, 20 January 2011

29

Solutions: Chapter 4

Arbitrary Frames in the (τ , ζ )-plane. (Note the substitution Z = ζ in this equation.) The graph of a function of this type √ is shown immediately above, in the solution to Exer1. a) Solve τ = T0 / 1 − r2ω 2 for r. To begin, we have cise 1(c). Hence F maps a rectangular grid in the (T, Z)1 − r2ω 2 = T02 /τ 2 , so plane as shown in the text. Evidence for this can also q be obtained by using Mathematica’s ParametricPlot 1 − T02 /τ 2 1 − T02 /τ 2 command. 2 , or r = r = . ω2 ω b) The image of the horizontal grid line ϕ = b is the 2 2 (same) horizontal line in the (θ , ϕ )-plane. The image of b) For r to be real, we must have 1 − T0 /τ ≥ 0 and 2 2 the vertical line d = a is the graph of hence τ ≥ T . As τ → ∞, we have T /τ → 0, implying

4.1 Uniform Rotation

r → 1/ω .

0

0

θ = k sec ϕ ,

c) An immediate calculation gives T2 dr √ 0 . = dτ ωτ 3 1 − T02 /τ 2 √ As τ → T0 , we have √ 1 − T02 /τ 2 → 0 and hence r′ → 0. As τ → ∞, however, 1 − T02 /τ 2 → 1 but the factor τ 3 in the denominator of r′ will force r′ → 0. r′ =

where k = a/R, a constant. The graph of a multiple of the secant function has the form shown in the text, so F maps a rectangular grid in the (d, ϕ )-plane as shown in the text. 3. According to Taylor’s theorem (see the solution to Exercise 1(c), § 1.3, Solutions page 3) √

0.4

1 = 1 + 12 v2 + O(v4 ) as v → 0. 2 1−v

The substitution v = rω gives

0.3

π πω 2 2 √ r + O(r4 ) as r → 0. =π+ 2 1 − r2 ω 2

0.2

4. The starting point that we need is Taylor’s formula

0.1

0

2

4

6

8

cos x = 1 − x2/2 + x4/24 + O(x6) as x → 0.

10

The graph of r = r(τ ) in the illustration above uses T0 = 2 and ω = 3. It agrees with the graph in the text. 2. a) The image of the horizontal grid line Z = b is the (same) horizontal line ζ = b in the (τ , ζ )-plane. The image of the vertical grid line T = a is the graph of a τ=p 1 − ζ 2ω 2

30

DVI file created at 16:06, 20 January 2011

Hence   d 1 − cos R   2 2R d4 d2 6 = 2 1−1+ 2 − + O(d ) d 2R 24 R4

A 2 R2 = 2 2 πd d

= 1−

d2 + O(d 4 ) as d → 0. 12 R2

4.2. LINEAR ACCELERATION

31

5. a) Because the radius r is represented by d in the fig- where A = −x2 /2 + O(x4 ) and so O(A2 ) = O(x4 ). The ure in the margin on page 153, it is best for us to work next step is with d. The circle C in that figure has radius d when mea2 2 sured on the surface, but its circumference is shown to be ln(sech x) = ln(1 + A + O(A )) = ln(1 + B) = B + O(B ) 2π R sin ψ . Here the measure of ψ is d/R radians, so where B = A + O(A2 ) and so O(B2 ) = O(A2 ) = O(x4 ). Thus circumference 2π R sin(d/R) sin(d/R) = = →1 (ατ )2 2π d 2π d d/R ln(sech(ατ )) = − + O(τ 4 ); 2 as d → 0. (We have used the fact that sin(x)/x → 1 as this confirms the formula for ζ once again. x → 0, with x = d/R.) We obtain the second ratio from the previous Exercise: 2. a) First we check that M −1 ◦ M is the identity. We have area = 1 + O(d 2) → 1 as d → 0. t sinh ατ π d2 = tanh ατ ; = z cosh ατ b) From Exercise 3, above, we have hence   circumference 1 1 1 −1 t = π + O(r2 ) → π as r → 0. tanh = tanh−1 (tanh ατ ) = ατ = τ , diameter α z α α

4.2 Linear Acceleration

as required. Next, we have

α 2 (z2 − t 2) = α 2 ·

e2αζ (cosh2 ατ − sinh2 ατ ) = e2αζ ; α2

1. Let f (τ ) = ln(sech(ατ )). One way to confirm the formula is to construct a Taylor polynomial for f from its hence derivatives, as follows. 1 1 1 ln(α 2 (z2 − t 2)) = ln(e2αζ ) = 2αζ = ζ , 1 2α 2α 2α · − sech(ατ ) tanh(ατ ) · α f ′ (τ ) = sech(ατ ) as required. = −α tanh(ατ ), Now check that M ◦ M −1 is the identity. We have 2 2 2 ′′ f (τ ) = −α sech (ατ ) · α = −α sech (ατ ),   t −1 t 2 ′′′ , so tanh ατ = . ατ = tanh f (τ ) = −α · 2 sech(ατ ) · − sech(ατ ) tanh(ατ ) · α z z = 2α 3 sech2 (ατ ) tanh(ατ ). This equation allows us to establish the domain of M −1 , 2 ′′′ ′ ′′ Therefore, f (0) = 0, f (0) = 0, f (0) = −α , f (0) = 0, as follows. The equation implies |t/z| < 1, because so | tanh x| < 1 for all x. Given that α > 0, the formula for z implies z > 0. Therefore |t/z| < 1 can be written in the α f (τ ) ζ= = 0 + 0 · τ − τ 2 + 0 · τ 3 + O(τ 4 ) form |t| < z; this is the domain of M −1 . α 2 To apply M, we need to express sinh ατ and cosh ατ α = − τ 2 + O(τ 4 ) as τ → 0. in terms of the expression for tanh ατ that M −1 gives us. 2 Start with the standard identity 1 − tanh2 x = sech2 x; it Another way to confirm the formula is to use some gives more-or-less well known Taylor expansions: 1 1 . cosh2 x = or cosh x = p cosh x = 1 + x2/2 + O(x4), 1 − tanh2 x 1 − tanh2 x 1/(1 − A) = 1 + A + O(A2), From sinh2 x = cosh2 x − 1 we get ln(1 + B) = B + O(B2). tanh x sinh x = p . Then 1 − tanh2 x 1 1 = sech x = Therefore cosh x 1 + x2/2 + O(x4) 1 t/z 1 = = 1 + A + O(A2) sinh ατ = p , cosh ατ = p . 1−A 2 1 − (t/z) 1 − (t/z)2

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 4. ARBITRARY FRAMES

32 The second equation in M −1 gives us 2αζ = ln(α 2 (z2 − t 2 )) or

eαζ

p = α z2 − t 2 .

Now we can evaluate M; the first equation is

p t/z eατ sinh ατ = z2 − t 2 · p α 1 − (t/z)2 q t = t, = 1 − (t/z)2 · p 1 − (t/z)2

as required. The second equation is similar:

p 1 eατ cosh ατ = z2 − t 2 · p α 1 − (t/z)2 q z = z, = 1 − (t/z)2 · p 1 − (t/z)2

again, as required.

b) The figure below shows the image in the (t, z)-plane of the square −1 ≤ τ , ζ ≤ 1 under the map M with α = 1. As the domain in the (τ , ζ )-plane is enlarged, more of the quadrant z > |t| is filled. 5

a straight line with nonzero slope there into a horizontal line. As we shall then see, M −1 transforms the hyperbolic rotation in R’s frame into a translation in G’s frame. The figure indicates that the correct hyperbolic rotation is H−α k , where k is chosen so that (b sinh α k, b cosh α k) is the point where the sloping line is tangent to the family of hyperbolas Q(t, z) = z2 − t 2 = b2 . The tangent point occurs where the function q(t) = Q(t, vt + z0 ) has an extreme. Thus, vz0 . 0 = q′ (t) = 2(vt + z0 )v − 2t, or t = 1 − v2 The z-coordinate of the tangent point is therefore z0 vz0 + z0 = , z=v 1 − v2 1 − v2 so   vz0 z0 (b sinh α k, b cosh α k) = , 1 − v2 1 − v2 We can use Q to determine the relation between b, k and v, z0 , as follows. First,   z0 vz0 , b2 = Q(b sinh α k, b cosh α k) = Q 1 − v2 1 − v2 z20 z20 2 (1 − v ) = , (1 − v2)2 (1 − v2) √ so b = z0 / 1 − v2. Next, =

4

1 cosh α k = √ , 1 − v2

3

v sinh α k = √ , 1 − v2

implying tanh α k = v, α k = tanh−1 v. It follows that the hyperbolic rotation Hα k maps (or “boosts”) the horizontal line z = b to the line z = vt + z0 , and hence that H−α k 1 reverses the process. Let 0     -3 -2 -1 0 1 2 3 t t cosh α k − z sinh α k tˆ = = H−α k −t sinh α k + z cosh α k z zˆ c) The text itself deals with the cases v = 0, ±1. Therefore, we assume, first of all, that |v| < 1. This question is By construction, zˆ = b when z = vt + z If we substitute 0 addressed later in the text (page 337), where it is shown into the equation for zˆ the original equations for t and z that the image curves in the (τ , ζ )-plane are just horizon- provided by M, we obtain tal translates of the curves for v = 0. eαζ z b = zˆ = (− sinh ατ sinh α k + cosh ατ cosh α k) α ζ eαζ z = vt + z 0 = cosh(α (τ − k)), α τ G z=b by the addition formula for the hyperbolic cosine. Solving αζ (bsinh α k, bcosh α k) this equation for ζ gives e = α b sech(α (τ − k)), or 2

R

t

ζ=

ln α b 1 + ln(sech(α (τ − k))). α α

To derive this result afresh, note that a suitable hyper- This is the one of the original curves in the (τ , ζ )-plane bolic rotation in the (t, z)-plane (R’s frame) will convert translated horizontally by τ = k = tanh−1 (v)/α .

DVI file created at 16:06, 20 January 2011

4.2. LINEAR ACCELERATION

33

If |v| > 1, no hyperbolic rotation will make z = vt + z0 horizontal. However, a hyperbolic rotation can make it vertical. Analogy with the case |v| < 1 then suggests that we first analyze a vertical line t = a (in the way that the text first analyzed a horizontal line). The image of the vertical t = 0 is the ζ -axis, so we just consider t = a with a 6= 0. The first equation for M gives a=

eαζ sinh ατ . α

This point has the alternate form (a cosh α k, a sinh α k), in which −z0 a= √ , v2 − 1

cosh α k = √

v v2 − 1

1 sinh α k = √ , 2 v −1

,

and hence α k = tanh−1 (1/v). We claim the hyperbolic rotation H−α k transforms z = vt + z0 into the vertical line tˆ = a. To see this, we take (from the previous page)

tˆ = t cosh α k − z sinh α k Solving this for ζ (and keeping in mind that sinh ατ must tv vt + z0 −z0 have the same sign as a) gives =√ −√ =√ = a. 2 2 v −1 v −1 v2 − 1 1 α a ln(α |a|) ln(± sinh ατ ) ζ (τ ) = ln − . = Now substitute into the same formula for tˆ the original α sinh ατ α α equations for t and z provided by M: When a < 0, we must use the minus sign in the last term and thus take τ < 0; when a > 0, we use the plus sign
eαζ a = tˆ = sinh ατ cosh α k − cosh ατ sinh α k and τ > 0. The graph of ζ (τ ) is a vertical translate (by α ln(α |a|)/α ) of one of the two graphs eαζ = sinh(α (τ − k)). ln(− sinh ατ ) ln(sinh ατ ) α ζ− = − , τ < 0, ζ+ = − , τ > 0. α α This is one of the original curves in the (τ , ζ )-plane transThese appear on the left and right, below, respectively.

lated horizontally by τ = k = tanh−1 (1/v)/α .

d) The images of the vertical lines t = t0 were found in the previous part of this solution, and they match the roughly vertical curves in the figure provided. The text (page 160) shows that images of the hyperbolas are the graphs of equations of the form   q 1 2 2 2 ζ = ln α z0 cosh ατ + α z0 sinh ατ + 1 α

as z0 varies. To obtain the curves below the τ -axis, it is necessary to allow z0 to be negative, as long as the argument of the logarithm remains positive; this forces z0 > −1/α . The Mathematica command Plot[Table[Log[(Exp[2 a] - 4) Cosh[t/4]/4 + Sqrt[(Exp[2 a] - 4)ˆ2 Sinh[t/4]ˆ2/16 + 1]] 4, {a, -4, 4}], {t, -12, 12}]

The positive ζ -axis is a vertical asymptote for both produces the following output, which matches the set of graphs. Because sinh x ≈ ex /2 when x is large, the line roughly horizontal curves in the figure provided. 40 ζ = ∓τ + ln(2)/α is also an asymptote for ζ± , respectively. The vertical translates of these curves, together with the 20 image of t = 0, fill the entire (τ , ζ )-plane. Now suppose z = vt + z0 with z0 6= 0 (and with |v| > 1 continuing to hold). An analysis similar to the one performed above shows that this line is tangent to one of the hyperbolas t 2 − z2 = a2 at the point t=

vz0 , v2 − 1

z=

−z0 . v2 − 1

DVI file created at 16:06, 20 January 2011

-10

5

-5

-20

-40

10

SOLUTIONS: CHAPTER 4. ARBITRARY FRAMES

34

3. a) The signal’s worldline is the graph of the equation 4.3 Newtonian Gravity zsignal = t + 1/α . This intersects the worldline of C when s 1. We take 4 × 107 meters as the circumference of the 1 eα h α2 2 earth (assumed to be a sphere), so its radius is (2/π )× 107 th + = zsignal = zC = th + 1, 2 α h α α e meters, and or 9.8 m/sec2 × (2/π )2 × 1014 m2  2  2α h 2α h M = = 5.95 × 1024 kg. e e α 1 2t h −11 m3 /kg sec2 2 2 6.67 × 10 t + 1 = t + . + 2 = 2 th2 + h α α α e2 α h h α2 A bit of algebra gives th =

αh  αh

e2 α h − 1 e = 2α α

−e 2

−α h 

=

eα h sinh α h. α

Substituting this value of th into the formula for zC = zh then gives eα h eα h p sinh2 α h + 1 = cosh α h. zh = α α b) According to R, C’s velocity at any time t is v=

eα h

α 2t

t dzC p = ; = · dt α e2α h α 2t 2 /e2α h + 1 zC

at time th the velocity is therefore vh =

sinh α h th = = tanh α h. zh cosh α h

2. Write R = r2 = (x − x0 )2 + (y − y0 )2 + (z − z0 )2 and then Φ = R−1/2 for ease of calculation. We have

∂Φ ∂R = − 21 R−3/2 = −R−3/2(x − x0 ), ∂x ∂x and then 3(x − x0)2 − R ∂ 2Φ −5/2 2 −3/2 . = 3R (x − x ) − R = 0 ∂ x2 R5/2 Because the three variables appear symmetrically in R, we have

∂ 2 Φ 3(y − y0)2 − R , = ∂ y2 R5/2

∂ 2 Φ 3(z − z0 )2 − R . = ∂ z2 R5/2

Therefore,

For F(h) we now have s r 1 − tanh α h (1 − tanh α h)2 = F(h) = 1 + tanh α h 1 − tanh2 α h 1 − tanh α h = cosh α h − sinh α h = e−α h . = sech α h

∂ 2Φ ∂ 2Φ ∂ 2Φ + 2 + 2 ∂ x2 ∂y ∂z 3(x − x0)2 + 3(y − y0)2 + 3(z − z0)2 − 3R = 0. = R5/2

∇2 Φ =

The appearance of R in the denominator requires R 6= 0, which happens if (x, y, z) 6= (x0 , y0 , z0 ).

The map (T, h) → (τ , ζ ) is realized by the Mathematica 3. a) We can take the potential function to be command ParametricPlot[{T Exp[-h/3], h}, 1 1 Φ(x, y) = − p −p . {T, -1.5, 1.5}, {h, -3, 3}, 2 2 (x − 1) + y (x + 1)2 + y2 PlotRange -> {{-2, 2}, 4{-1, 1}}, MeshShading -> {{White, White}, {White, White}}]

2

Its output, shown below, demonstrates that the image grid has the form shown in the text.

1

1.0

0

0.5

0.0

-1 -0.5

-2 -1.0 -2

-1

0

1

2

DVI file created at 16:06, 20 January 2011

-3

-2

-1

0

1

2

3

4.3. NEWTONIAN GRAVITY

35

b) The work done is the negative of the difference in po- d θ d θ are automatically zero. Here are the remaining six tential: terms: W = −Φ(0, 0) − (−Φ(0, ∞)) =

1 1 + − 0 = 2. 1 1

4. The work done is −Φ(−1, 0, 0) − (−Φ(∞, 0, 0)) = 5 1 √ +√ − 0 = 5 12 . 1+0+0 4+0+0

r sin ϕ cos ϕ cos θ sin θ dr d ϕ , r sin2 ϕ cos2 θ dr d θ , r cos ϕ sin ϕ cos θ sin θ d ϕ dr, r2 cos ϕ sin ϕ cos2 θ d ϕ d θ , −r sin2 ϕ sin2 θ d θ dr,

5. Let ϕ (t) = f (t α ,t β ,t γ ). Taylor’s theorem says that

−r2 sin ϕ cos ϕ sin2 θ d θ d ϕ .

Because d ϕ dr = −dr d ϕ , the first and third terms cancel each other. The second and fifth terms combine as

ϕ (ε ) = ϕ (0) + εϕ ′ (0) + O(ε 2 ).

Because O(ε 2 ) ≈ 0 when ε is small, we can rewrite this r sin2 ϕ cos2 θ dr d θ + r sin2 ϕ sin2 θ dr d θ = r sin2 ϕ dr d θ as ϕ (ε ) − ϕ (0) ≈ εϕ ′ (0). The fourth and sixth terms combine in a similar way to By the chain rule, give ′ r2 cos ϕ sin ϕ d ϕ d θ . ϕ (t) = fx (t α ,t β ,t γ )α + fy (t α ,t β ,t γ )β + fz (t α ,t β ,t γ )γ = ∇ f (t α ,t β ,t γ ) · (α , β , γ ),

Therefore, dU dV reduces to

ϕ (0) = ∇ f (0, 0, 0) · (α , β , γ ); ′

therefore, the equation ϕ (ε ) − ϕ (0) ≈ into

εϕ ′ (0)

r sin2 ϕ dr d θ + r2 cos ϕ sin ϕ d ϕ d θ . translates

f (εα , εβ , εγ ) − f (0, 0, 0) ≈ ε ∇ f (0, 0, 0) · (α , β , γ ). The condition that (α , β , γ ) be a unit vector is not needed. 6. a) The basic fact we use is that if w = f (x, y, z), then dw = fx dx + fy dy + fz dz. Given the expressions for U, V , and W , in terms of r, ϕ , and θ , we have dU = sin ϕ cos θ dr + r cos ϕ cos θ d ϕ − r sin ϕ sin θ d θ ,

dV = sin ϕ sin θ dr + r cos ϕ sin θ d ϕ + r sin ϕ cos θ d θ , dW = cos ϕ dr − r sin ϕ d ϕ . b) We have dr dq d p = dr(dq d p) = dr(−d p dq) = −dr d p dq,

dr dq d p = (dr dq)d p = (−dq dr)d p = −dq dr d p, but

dr dq d p = −dq dr d p = −dq(dr d p) = −dq(−d p dr) = +dq d p dr. c) The exterior product dU dV apparently has nine terms, but the three terms involving dr dr, d ϕ d ϕ , and

DVI file created at 16:06, 20 January 2011

The exterior product dV dW has only four terms that are not automatically zero; they are −r sin2 ϕ sin θ dr d ϕ ,

r cos2 ϕ sin θ d ϕ dr, r sin ϕ cos ϕ cos θ d θ dr,

−r2 sin2 ϕ cos θ d θ d ϕ . The first two terms combine to −r sin θ dr d ϕ , giving dV dW = −r sin θ dr d ϕ + r sin ϕ cos ϕ cos θ d θ dr + r2 sin2 ϕ cos θ d ϕ d θ .

There are also just four nonzero terms in dW dU: r cos2 ϕ cos θ dr d ϕ , −r cos ϕ sin ϕ sin θ dr d θ ,

−r sin2 ϕ cos θ d ϕ dr,

+r2 sin2 ϕ sin θ d ϕ d θ . The first and third terms combine to give r cos θ dr d ϕ , yielding dW dU = r cos θ dr d ϕ + r cos ϕ sin ϕ sin θ d θ dr + r2 sin2 ϕ sin θ d ϕ d θ .

SOLUTIONS: CHAPTER 4. ARBITRARY FRAMES

36

d) To check the correctness of the results of the previous b) div A = 1/5 + 1/5 = 2/5. part, we compute dU dV dW three ways: as (dU dV )dW , (dW dU)dV , and dU(dV dW ). The first is −r2 sin2 ϕ · sin ϕ dr d θ d ϕ + r2 cos2 ϕ sin ϕ d ϕ d θ dr = r2 sin ϕ dr d ϕ d θ .

2

1

The second is r2 sin ϕ cos2 θ dr d ϕ d θ + r2 cos2 ϕ sin ϕ sin2 θ d θ dr d ϕ + r2 sin2 ϕ · sin ϕ sin2 θ d ϕ d θ dr

0

-1

= r2 sin ϕ cos2 θ dr d ϕ d θ + r2 sin ϕ sin2 θ dr d ϕ d θ = r2 sin ϕ dr d ϕ d θ .

-2

The third is -2

r2 sin2 ϕ cos2 θ sin ϕ dr d ϕ d θ

-1

0

1

2

c) div A = 0 + 0.

+ r2 cos2 ϕ sin ϕ cos2 θ d ϕ d θ dr + r2 sin ϕ sin2 θ d θ dr d ϕ = r2 cos2 θ sin ϕ dr d ϕ d θ + r2 sin ϕ sin2 θ dr d ϕ d θ

1.0

= r2 sin ϕ dr d ϕ d θ . 0.5

e) We have 0.0

dx = cos θ dr − r sin θ d θ , dy = sin θ dr + r cos θ d θ .

-0.5

Therefore, dx dy = r cos2 θ dr d θ − r sin2 θ d θ dr

= r cos2 θ dr d θ + r sin2 θ drd θ = r dr d θ .

-1.0 -1.0

1.0

1.0

0.5

0.5

0.0

0.0

-0.5

-0.5

-1.0

-1.0 -0.5

0.0

0.0

0.5

1.0

0.0

0.5

1.0

d) div A = 0 + 0 = 0.

7. a) div A = 1/5 + 1/5 = 2/5.

-1.0

-0.5

0.5

1.0

DVI file created at 16:06, 20 January 2011

-1.0

-0.5

4.4. GRAVITY IN SPECIAL RELATIVITY

4.4 Gravity in Special Relativity 1. a) The magnitude of the force on the test mass m at the surface of the sun is GM mα = F = m 2 = mα , X where α is the corresponding acceleration, M is the mass of the sun, and X is its radius. Thus α = GM/X 2 . In conventional units, the value of α is (6.67 × 10−11 m3 /kg sec2 )(2 × 1030 kg) = 272 m/sec2 . (7 × 108 m)2 To convert the value to geometric units (in which 3 × 108 meters = 1 second), multiply by 1/(3 × 108 m/sec) to get

α = 272

1 sec m × = 9.07 × 10−7 sec−1 . sec2 3 × 108 m

37 To recalculate the angle in conventional units, we must use the equation ζ = −αη 2 /2c2 . Then ζ ′ = −αη /c2 , so

θ≈

−α X (−272 m/sec2 )(7 × 108 m) = 2 c (3 × 108 m/sec)2 ≈ −2 × 10−6 radians.

d) The photon’s path has two parts: (1) from infinity to a point tangent to the surface of the sun and, (2) from the point of tangency on to infinity. The two parts are symmetric, so the total deflection is twice the deflection along one part. Part (c) of this exercise shows that the latter is about 2 × 10−6 radians, so the total deflection is about 4 × 10−6 radians. This agrees with Einstein’s 1911 calculation. 2. a) The numbers N and νe are frequencies with the dimensions of Hertz (1 Hz = 1 sec−1 ) while Φ has the dimensions (m/sec)2 in conventional units and hence is dimensionless in geometric units. Therefore the equation

b) The formula ζ = −αη 2 /2 for the distance a photon N = νe(1 − ∆Φ) drops is expressed in geometric units. The corresponding formula in conventional units is ζ = −αη 2 /2c2 (page 195 balances dimensionally when geometric units are used, of the text). In conventional units, but must be converted to   ∆Φ −(272 m/sec2 )(7 × 108 m)2 e 1 − ν N = ζ= = −740 meters. c2 2 × (3 × 108 m/sec)2

To determine ζ ’s value in geometric units, we need the value of η = X in geometric units. As above, we have the value X = 7 × 108 m ×

7 1 sec = seconds. 3 × 108 m 3

The value of ζ is therefore

ζ = −9.07 × 10

−7

sec

−1

 2 7 1 × sec2 × 3 2

= −2.47 × 10−6 seconds.

c) At its initial position (where η = 0), the photon’s initial direction is horizontal. Hence its change in direction is the angle θ made by the tangent line to the graph ζ = −αη 2 /2 at the point η = X. (Note this is the formula for ζ in geometric units.) The tangent of θ is the slope of that tangent line, which is ζ ′ = −αη = −α X. Because θ is so small, tan θ ≈ θ ; therefore,

θ ≈ −9.07 × 10−7 sec−1 ×

7 sec ≈ −2 × 10−6 radians. 3

DVI file created at 16:06, 20 January 2011

in order to balance when conventional units are used. The general form of this adjustment is described in the text on page 101. (For typographic reasons, we continue to put a tilde over the Greek letter nu to distinguish it from the roman letter vee.)

b) The Newtonian gravitational potential (page 171) is Φ(r) = −GM/r, where r is the distance from the center of the sun, M is its mass, and G is the gravitational constant. If X is the radius of the sun and E is the distance from the center of the earth to the center of the sun, then the potential difference is   1 1 − . ∆Φ = Φ(E) − Φ(X) = GM X E The value of E is about 150 million km, or 1.5 × 1013 meters; the other values appear in the previous exercise. Thus, 1/X − 1/E = 1.42 × 10−9 m−1 , GM = 1.334 × 1020 m3 /sec2 , and ∆Φ = (1.334 × 1020 m3 /sec2 )(1.42 × 10−9 m−1 ) = 1.9 × 1011 (m/sec)2 .

SOLUTIONS: CHAPTER 4. ARBITRARY FRAMES

38 The value of the potential difference in geometric units is ∆Φ 1.9 × 1011 (m/sec)2 = = 2.1 × 10−6, c2 (3 × 108 m/sec)2 or about 2 × 10−6, in agreement with the text.

c) As a function of the emission frequency νe, the red shift is N − νe = −∆Φ · νe = −(2 × 10−6) νe Hz.

DVI file created at 16:06, 20 January 2011

Solutions: Chapter 5

Surfaces and Curvature 5.1 The Metric

d) This surface is the “northern” hemisphere of radius 1 centered at the origin. It is shown above, on the left.

1. a) There is no surface to be drawn for this part of the question. The figures for the following parts are all made using the Mathematica command ParametricPlot3D (even if Plot3D would have been sufficient). In parts (b), (c), and (d), the lines q1 = constant run roughly from the front left to the back right; the lines q2 = constant run roughly from the back left side to the front right. The Mathematica 7 command includes the options

e) This surface, shown above on the right, is also the “northern” hemisphere of radius 1 centered at the origin. Here, though, the coordinate lines q1 = constant are the vertical lines of longitude; the coordinate lines q2 = constant are the horizontal lines of latitude. f) This surface is the entire sphere of radius 1 centered at the origin (below left). The coordinate lines are still the longitude and latitude lines.

BoundaryStyle -> Directive[Thin], MeshShading -> {{None, None}, {None, None}}, Axes -> False

b) The surface, shown below on the left, is saddleshaped. Furthermore, the figure shows it is made up of two families of straight lines.

g) This surface has the same image as the previous one. It arises from the previous one by making the replacement cos q2 → sin q2 , together with the change in domain of q2 from [−π /2, π /2] to [0, π ]. The coordinate lines are still longitude and latitude lines, respectively. c) This surface, shown above on the right, is also saddle- h) This surface is shown above on the right. The circular 1 shaped, but its orientation in space is different from that functions of q still appear, so the surface still has rotational symmetry around the vertical (i.e., z-) axis. The coof the previous surface. ordinate lines are still what we can call the longitude and latitude lines, respectively. The longitude lines are now hyperbolas, though, because they are parametrized by the hyperbolic cosine and hyperbolic sine functions. The surface is called a hyperboloid of one sheet; it has the shape of a cooling tower at a power plant. 2. a) (Nothing to do.) 39

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

40 x1 = (1, 0, q2 ),

b)

g)

x2 = (0, 1, q1 ), x1 × x2 = (−q2 , −q1 , 1),     1 + (q2)2 q1 q2 g11 g12 , = g21 g22 q1 q2 1 + (q1)2 p √ g = 1 + (q1)2 + (q2 )2 .

The coordinate lines in the figure do not appear to be orthogonal except along the coordinate lines q1 = 0 and q2 = 0. This is confirmed by the fact that g12 = q1 q2 6= 0 except when q1 = 0 or q2 = 0. 1

x1 = (1, 0, 2q ),

c)

2

x2 = (0, 1, −2q ),

x1 × x2 = (−2q1, 2q2 , 1),     1 + 4(q1)2 −4q1q2 g11 g12 = , g21 g22 −4q1q2 1 + 4(q2)2 p √ g = 1 + 4(q1)2 + 4(q2 )2 .

Again, the coordinate lines do not appear to be orthogonal except along the coordinate lines q1 = 0 and q2 = 0, and this is confirmed by g12 = −4q1q2 . p d) Set h(q1 , q2 ) = 1 − (q1)2 − (q2)2 to help simplify expressions. Then x1 = (1, 0, −q1 /h), x2 = (0, 1, −q2 /h), x1 × x2 = (q1 /h, q2 /h, 1),     1 + (q1/h)2 q1 q2 /h2 g11 g12 , = g21 g22 q1 q2 /h2 1 + (q2/h)2 p √ g = 1 + (q1/h)2 + (q2 /h)2 .

x1 = (− sin q2 sin q1 , sin q2 cos q1 , 0), x2 = (cos q2 cos q1 , cos q2 sin q1 , − sin q2 ),

x1 × x2 = −(sin2 q2 cos q1 , sin2 q2 sin q1 , sin q2 cos q2 ),   2 2   g11 g12 sin q 0 , = g21 g22 0 1 √ g = sin q2 . The coordinate lines (i.e., the longitude and latitude lines) are clearly orthogonal everywhere (except at the poles, q2 = 0, π ). This is confirmed by g12 = 0. h)

x1 = (− cosh q2 sin q1 , cosh q2 cos q1 , 0), x2 = (sinh q2 cos q1 , sinh q2 sin q1 , cosh q2 ), x1 × x2 =

(cosh2 q2 cos q1 , cosh2 q2 sin q1 , − cosh q2 sinh q2 ),     cosh2 q2 0 g11 g12 = , g21 g22 0 2 sinh2 q2 + 1 p √ g = cosh q2 2 sinh2 q2 + 1.

The coordinate lines (i.e., the longitude and latitude lines) are clearly orthogonal everywhere; this is confirmed by g12 = 0. 3. a) We have
y1 = 1, 0, (q1 + 0.9q2)2 + 2(q1 − 0.9q2)(q1 + 0.9q2)
= 1, 0, (q1 + 0.9q2)(3q1 − 0.9q2) ,
y2 = 0.9 1, 0, (q1 + 0.9q2)(q1 − 2.7q2) .

Thus, when q1 = q2 = 0, we see that y1 = (1, 0, 0) and Again, the coordinate lines do not appear to be orthogonal y2 = (0.9, 0, 0), so y1 and y2 are collinear. except along the coordinate lines q1 = 0 and q2 = 0, and b) Let (x, 0, z) be any point for which x 6= 0. We must this is confirmed by g12 = q1 q2 /h2 . show that there is a point (q1 , q2 ) for which y(q1 , q2 ) = 2 1 2 1 (x, 0, z). By definition of y, x = (− cos q sin q , cos q cos q , 0), e) 1

x2 = (− sin q2 cos q1 , − sin q2 sin q1 , cos q2 ),

x1 × x2 = (cos2 q2 cos q1 , cos2 q2 sin q1 , cos q2 sin q2 ),   2 2   cos q 0 g11 g12 = , g21 g22 0 1 √ g = cos q2 . Note: on the interval −π /2 ≤ q2 ≤ π /2, cos q2 ≥ 0, so p cos2 q2 = | cos q2 | = cos q2 . The coordinate lines (i.e., the longitude and latitude lines) are clearly orthogonal everywhere (except at the poles, q2 = ±π /2). This is confirmed by g12 = 0. f) Same as part (e).

DVI file created at 16:06, 20 January 2011

x = q1 + 0.9q2 = x,

z = q1 − 0.9q2)x2 ,

or q1 − 0.9q2 = z/x2 . Therefore, q1 =

1 z x+ 2 , 2 x

q2 =

z 1  x− 2 . 1.8 x

This determines (q1 , q2 ) for any (x, z) with x 6= 0. Because y(0, 0) = (0, 0, 0), we have shown y(R2 ) is the entire plane y = 0 excluding the nonzero points on the z-axis. 4. We have M = v ∧ w, where     w1 v , v= 1 , w= w2 v2

5.1. THE METRIC and we know from the text that area M = kvkkwk sin θ . Therefore, (area M)2 = kvk2kwk2 sin2 θ = kvk2 kwk2(1 − cos2 θ ) = kvk2kwk2 − (v · w)2

= (v21 + v22 )(w21 + w22 ) − (v1 w1 + v2 w2 )2 = v21 w22 + v22 w21 − 2(v1w1 v2 w2 ) = (v1 w2 − v2 w1 )2 .

From this it follows that areaM = ±(v1 w2 − v2 w1 ). To determine the sign, take v = e1 , w = e2 , the standard basis vectors. By construction, the area of e1 ∧ e2 is +1, whereas v1 w2 − v2 w1 = 1 · 1 − 0 · 0 = +1. Therefore the + sign should be taken, and areav ∧ w = v1 w2 − v2 w1 . 5. a)

x1 = (1, 0, a), x2 = (0, 1, b), x1 × x2 = (−a, −b, 1),     1 + a2 ab g11 g12 , = g21 g22 ab 1 + b2 p √ g = 1 + a2 + b2 .

The coordinate lines are not orthogonal unless a = 0 or b = 0, because g12 = ab. b) If we let f1 = ∂ f /∂ q1 , f2 = ∂ f /∂ q2 , then x1 = (1, 0, f1 ), x2 = (0, 1, f2 ), x1 × x2 = (− f1 , − f2 , 1),     1 + ( f1 )2 f1 f2 g11 g12 = , g21 g22 f1 f2 1 + ( f2 )2 p √ g = 1 + ( f1 )2 + ( f2 )2 .

41 d) This sphere is the particular surface of revolution for which r = R cos q2 and z = R sin q2 , where r and z are as given in the previous part of this exercise. Therefore r′ = −R sin q2 , z′ = R cos q2 , and by substituting into the previous answer, we get x1 = R cos q2 (− sin q1 , cos q1 , 0), x2 = R(− sin q2 cos q1 , − sin q2 sin q1 , cos q2 ),

x1 × x2 = R2 cos q2 (cos q1 cos q2 , sin q1 cos q2 , sin q2 ),   2 2 2   R cos q 0 g11 g12 = , g21 g22 0 R2 √ g = R2 cos q2 . As a surface of revolution, the sphere’s coordinate lines are orthogonal. e) This torus is the particular surface of revolution for which r = a + r cosq2

and z = r sin q2 .

Therefore r′ = −r sin q2 , z′ = r cos q2 , and x1 = (a + r cosq2 )(− sin q1 , cos q1 , 0), x2 = r(− sin q2 cos q1 , − sin q2 sin q1 , cos q2 ),

x1 × x2 =

r(a+r cos q2 )(cos q1 cos q2 , sin q1 cos q2 , sin q2 ),     (a + r cosq2 )2 0 g11 g12 , = g21 g22 0 r2 √ g = r(a + r cosq2 ). The coordinate lines are orthogonal.

6. The parametrized curve (r, z) = (r(q), z(q)) is called the generator of the surface of revolution. Think of the (r, z) plane as a vertical plane that passes through In general, the coordinate lines are not orthogonal unless the z-axis and rotates around that axis. Then the curve f1 = 0 or f2 = 0, that is, unless f is independent of either (r(q), z(q)) “sweeps out” the surface, forming the longitude lines of the surface. (Because the curve is q1 or q2 . parametrized, it can have a complicated shape and interc) x1 = r(− sin q1 , cos q1 , 0), sect itself, for example. The surface of revolution will then reflect this complexity.) In exercises 1d, 1f, and 1g, x2 = (r′ cos q1 , r′ sin q1 , z′ ), the curve is a semicircle of radius 1 centered at the origin; x1 × x2 = r(z′ cos q1 , z′ sin q1 , −r′ ), on the sphere, it is a meridian that runs from the south pole   2   r 0 g11 g12 to the north. In 5d, the same is true but the semicircle now = , g21 g22 0 (r′ )2 + (z′ )2 has radius R. In 5e, the generator is a full circle of radius p √ r whose center is at the point (r, z) = (a, 0) away from the ′ 2 ′ 2 g = r (r ) + (z ) . origin. The coordinate lines are longitude and latitude lines on the surface of revolution, and they are orthogonal.

DVI file created at 16:06, 20 January 2011

7. The circle (a + r cosq2 , r sin q2 ) is the generator of the torus Ta,r . The outer half of the torus corresponds to the

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

42

part of the circle where cos q2 ≥ 0, or −π /2 ≤ q2 ≤ π /2. The inner half is where cosq2 ≤ 0; by periodicity of the circular functions, we can take π /2 ≤ q2 ≤ 3π /2. By Theorem 5.1, the area of a region is the integral of √ g over the parameter domain that determines the region. √ From Exercise 5e we have g = r(a + r cosq2 ), so area(outer half) =

Z 2π

Z π /2

dq1

−π /2

0

c) The curve wraps three times around the torus longitudinally while advancing twice around the central axis. The two figures below are analogous to those for the previous curve.

r(a + r cosq2 ) dq2

π /2 = 2π (raq + r sin q ) 2

2

2

= 2π (raπ + 2r ), 2

−π /2

and area(inner half) =

Z 2π 0

dq1

Z 3π /2 π /2

r(a + r cosq2 ) dq2

9. a) The curve is just one-quarter turn of a rather elongated helix that is shown in the figure on the left, below. 4

3π /2 = 2π (raq2 + r2 sin q2 )

4

π /2

= 2π (raπ − 2r2).

The total area of Ta,r is therefore 4π 2ra. Incidentally, Opus’s centroid theorem for the area of a surface of revolution says that the area equals the product of the length of the generating curve ( 2π r in this case) and the distance its geometric centroid travels as the surface is generated. Here the centroid is just just the center of the generating circle; it lies a units from the axis, so it travels a distance of 2π a. Hence the area of the torus is 2π r · 2π a = 4π 2ra. By determining the centroids for the inside and the outside of the torus, we could even get their separate areas using Pepper’s theorem.

2 2

1.5 1.5

1.0 1.0

-1.0 0.5

-0.5

-1.0 0.5

-0.5

0.0 0.00

0.0

8. a) This can be done without integrals. Each meridian 0.5 0.5 1.0 1.0 of longitude is a complete circle of radius r, so its length -0.5 -0.5 2 is just 2π r. The parallel of latitude with a fixed value of q is a circle centered on the z-axis with radius a + r cos q2 . Its length is therefore 2π (a + r cosq2 ). Meridians all have b) The figure on the right, above, shows the curve (in the same length, but the parallels of latitude have varying gray) with the five unit tangent vectors (in black). We length, depending on q2 . have b) The curve wraps once around the torus, as shown in 1 s s √  ′ z (s) = − sin , cos , 3 , the figures below. On the left, the torus is represented by a 2 2 2   number of its longitude circles, allowing the whole curve 1 2 s 2 s ′ 2 sin + cos + 3 = 1. kz (s)k = to be seen. On the right, the torus is opaque. 4 2 2 0.00

c) The figure with the long tangent line segments is shown on the left, below (next page). d) The tangent developable surface is shown on the right, below (next page). The image is not especially clear; one way to see the cusp edge better is to produce the figure using Mathematica and then manipulate the output image with a mouse.

DVI file created at 16:06, 20 January 2011

5.2. INTRINSIC GEOMETRY ON THE SPHERE

43 e) The segment in question has length

4

kx(c1 , b) − x(c1, a)k = kz(c1 ) + bz′(c1 ) − z(c1 ) − az′(a)k

4

= (b − a)kz′(c1 )k = b − a

because z′ (c1 ) is a unit vector, by construction. 2

2

5.2 Intrinsic Sphere

1.5

Geometry

on

the

1 -1

1.0 0

-1.0

0

0.5

-0.5

1 0.0 0.00

0.5

-1 1.0

-0.5

1. If x(q1 , q2 ) is the parametrization of the sphere of radius R that is the adaptation of the parametrization x(q1 , q2 ) in the text, then x(q1 , q2 ) = Rx(q1, q2 ). Hence xi (q1 , q2 ) = Rxi (q1 , q2 ) and gi j = xi · x j = R2 xi · x j = R2 gi j .

10. a) When z(s) is a space curve parametrized by arc length, then z′ · z′ = 1,

z′ · z′′ = 0,

z′′ · z′′ = κ 2 ;

Therefore


g = g11 g22 − g212 = R4 g11 g22 − g212 = R4 g, √ √ and the area magnification factor is g = R2 g.

2. a) By analogy with the circle of latitude analyzed in κ (s) is the curvature function for the curve (text Chap. 3.2, the text (pages 222–223), the meridian of longitude for page 121). Therefore, if x(q1 , q2 ) = z(q1 )+ q2 z′ (q1 ), then which q1 = k can be parametrized as (q1 (t), q2 (t)) = (k,t) with π /2 ≤ t ≤ π /2. Therefore (dq1 /dt, dq2 /dt) = (0, 1) x1 = z′ (q1 ) + q2z′′ (q1 ), and s x2 = z′ (q1 ),
dqi dq j p x1 × x2 = q2 z′′ (q1 ) × z′ (q1 ), gi j q1 (t), q2 (t) = g22 (k,t) · 1 = 1. dt dt     1 2 2 g11 g12 1 + (q κ (q )) 1 , = In other words, the parametrization of the meridian is a g21 g22 1 1 unit-speed parametrization, so the point that lies d units √ g = q2 κ (q1 ). from the north pole along this meridian has the parameter value q2 = t = π /2 − d. Therefore, the circle of radius 2 √ b) The parametrization is singular where g = 0. This d centered at the north pole is given by q = π /2 − d. happens at two kinds of places. The first is where q2 = 0; Finally, the length of the entire meridian is the length of these are all the points on the original curve z(q1 ). The the parameter interval, namely π . second is where κ (q1 ) = 0; these are all the points along b) By part (a), we see that the spherical cap of radius d any line tangent to a point where the original curve has corresponds to the parameter domain  curvature equal to zero. −π ≤ q1 ≤ π , D: 1 ′′ 1 ′ 1 π /2 − d ≤ q2 ≤ π /2. c) When q is fixed, the vector z (q ) × z (q ) is fixed, as well, so Therefore, the area of the spherical cap is x1 × x2 = q2 z′′ (q1 ) × z′(q1 ) varies only in length, not direction. Therefore, the unit normal vector n(q1 , q2 ) = x1 × x2 /kx1 × x2 k is constant when q1 is fixed.

ZZ D

√ g dq1 dq2 =

Z π

−π

π /2 = 2π · sin(q )

dq1

2

π /2−d

Z π /2

π /2−d

cos(q2 ) dq2

= 2π (1 − sin(π /2 − d))

= 2π (1 − cos(d)). d) Because g12 = 1 6= 0, the coordinate lines are never orthogonal. We have used the identity sin(π /2 − θ ) = cos(θ ).

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

44

c) By the solution to Exercise 1, the metric tensor for Then dq1 /dt = dq2 /dt = 1, so the sphere of radius R is gi j = R2 gi j , where gi j is the s
dqi dq j p metric tensor of the unit sphere. Therefore, if we let gi j q1 (t), q2 (t) = g11 (t,t) · 1 + g22(t,t) · 1 1 2 (q (t), q (t)) = (k,t), −π /2 ≤ t ≤ π /2 parametrize the dt dt p meridian of longitude q1 = k, as in part (a), then = cos2 t + 1. s The length of the line is dqi dq j Z π /2 p dt = R. gi j dt dt cos2 t + 1 dt = 1.91. 0

In other words, the parametrization has speed R (rather than speed 1), so the meridian at distance d from the north pole has parameter value q2 = t = π /2 − d/R. Hence the parameter domain for the spherical cap is  −π ≤ q1 ≤ π , D: π /2 − d/R ≤ q2 ≤ π /2. √ √ Recalling that g = R2 g = R2 cos(q2 ), we see that the area of that spherical cap is ZZ p

g dq1 dq2 =

D

Z π

−π

dq1

Z π /2

π /2−d/R

R2 cos(q2 ) dq2

This is an elliptic integral whose value was found using Mathematica. It can also be found using numerical integration, another computer algebra system, or tables of elliptic integrals. 5. a) We modify the analysis found on page 228 of the text. Keep x1 = (1, 0), but let q(t) = (t, arcsin(tanht)),

q′ (t) = (1, secht).

Because q2 is now arcsin(tanht), we must also modify the term g11 = cos2 (q2 ) in the the metric tensor. We have cos2 (q2 ) = cos2 (arcsin(tanht)) = 1 − sin2 (arcsin(tanht)) = 1 − tanh2 t = sech2 t

π /2 = 2π · R2 sin(q2 )

so the components we use to compute the cosine of the angle θ between q and x1 are π /2−d/R   
sech2 t 0 1 = 2π R2 (1 − sin(π /2 − d/R)) ′ = sech2 t, q · x1 = 1 secht 0 0 1 2 = 2π R (1 − cos(d/R)).   
sech2 t 0 1 ′ ′ = 2 sech2 t q · q = 1 secht secht 0 1 By Taylor’s theorem, cos θ = 1 − θ 2 /2 + O(θ 4) when   
sech2 t 0 θ ≈ 0. If d is small in relation to R, then θ = d/R ≈ 0, so 1 = sech2 t. x1 · x1 = 1 0 we can express the area of the spherical cap as 0 0 1  4  Therefore, 
d d2 2π R2 1 − cos(d/R) = 2π R2 1 − 1 + 2 + O 1 q′ · x1 sech2 t 2R R4 cos θ = √ ′ ′ √ = √ = constant =√  4 2 · q q x · x 2 sech t 2 1 1 d . = π d2 + O so the curve is a loxodrome. R2 b) The previous computation shows that θ = 45◦.

Π

3.

2

Π 4

0

Π

Π

4

2

The curve lies entirely in the northern hemisphere, and 4. Parametrize the line as q1 (t) = q2 (t) = t, 0 ≤ t ≤ π /2. begins where the positive x-axis meets the equator.

DVI file created at 16:06, 20 January 2011

5.3. DE SITTER SPACETIME

45

c) Here is one approach: Modify the curve q(t) in the previous part to q′ (t) = (k, secht),

q(t) = (kt, arcsin(tanht)), Then

q′ · x1 = k sech2 t,

q′ · q′ = (k2 + 1) sech2 t,

x1 · x1 = sech2 t,

and the angle θ between q and the parallels of latitude satisfies k cos θ = √ . 2 k +1 This implies (k2 + 1) cos2 θ = k2 , or cos θ k= √ = cot θ . 1 − cos2 θ In other words, to construct the loxodrome that makes the angle θ with parallels of latitude, where 0 < θ < π , set k = cot θ . d) The north pole is at the center of the figure below; the circles represent parallels of latitude that are 1◦ apart. The loxodrome does appear to be an equiangular spiral, making an angle of 45◦ with latitude circles. The illustrated portion of the loxodrome begins (where t = π ) at about 85◦ north latitude.

5.3 De Sitter Spacetime 1. a) We compute the three components of X1 × X2 . Because of the relation between the Euclidean norm and the Minkowski norm, the signs of the second and third terms in the Minkowski cross product must be the negatives of their signs in the Euclidean cross product. The first component is sinh(q1 ) cos(q2 ) sinh(q1 ) sin(q2 ) 1 1 − cosh(q1 ) sin(q2 ) cosh(q1 ) cos(q2 ) = sinh(q ) cosh(q ).

The second (with the required change of sign) is sinh(q1 ) sin(q2 ) cosh(q1 ) = cosh2 (q1 ) cos(q2 ). − cosh(q1 ) cos(q2 ) 0

The third (also with the sign change) is cosh(q1 ) sinh(q1 ) cos(q2 ) = cosh2 (q1 ) sin(q2 ). − 0 − cosh(q1 ) sin(q2 )

These are the three components of X, each multiplied by cosh(q1 ). b) Given the result in part (a), it is sufficient to check that X · X1 = X · X2 = 0. We have X · X1 = sinh(q1 ) cosh(q1 ) − cosh(q1 ) sinh(q1 ) cos2 (q2 ) − cosh(q1 ) sinh(q1 ) sin2 (q2 ) = 0

X · X2 = + cosh2 (q1 ) cos(q2 ) sin(q2 )

− cosh2 (q1 ) sin(q2 ) cos(q2 ) = 0.

c) Note that X and hence X1 × X2 are spacelike vectors; therefore √ kXk = −X · X We have kX1 × X2 k = cosh(q1 )kXk and X · X =

sinh2 (q1 )−cosh2 (q1 ) cos2 (q2 )−cosh2 (q1 ) sin2 (q2 ) = −1. Thus kXk = +1 and kX1 × X2 k = cosh(q1 ). Because −g = cosh2 (q1 ) is given in the text, we finally have √ kX1 × X2 k = −g. 2. a) The expressions for x1 and x2 are the same as for X1 and X2 , respectively, but the signs of the second and third components of x1 × x2 are the negatives of those for X1 × X2 . Thus

e) The total length of q(t) is Z ∞p −∞

q′ · q′

dt =

Z ∞√ −∞

√ 2 secht dt = π 2.

DVI file created at 16:06, 20 January 2011

x1 × x2 = cosh(q1 ) sinh(q1 ), − cosh2 (q1 ) cos(q2 ),
− cosh2 (q1 ) sin(q2 ) .

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

46

Furthermore,     cosh2 (q1 ) + sinh2 (q1 ) 0 g11 g12 = , g21 g22 0 cosh2 (q1 ) q √ g = cosh(q1 ) cosh2 (q1 ) + sinh2 (q1 ) p = cosh(q1 ) cosh(2q1 ).

At this parameter point, the two normal vectors are
1 sinh(t), 0, − cosh(t) , n= p 2 2 sinh t + 1
1 N= p sinh(t), 0, cosh(t) . 2 2 sinh t + 1

They differ only in the sign of the last coordinate. MoreIt is clear in the figure on page 231 of the text that the co- over, N is always collinear with X, but n is collinear with ordinate lines are orthogonal in the Euclidean sense, and x only when t = 0. The two normals are shown in the this is confirmed here by g12 = g21 = 0. figure below. As |t| gets large, N becomes tangent to the b) Let us write x = (x, y, z), so that the curve q(t) has the surface on the Euclidean sense, so it becomes orthogonal to n in the Euclidean sense. components z x = sinh(t), x y = cosh(t) cos(2 arctan(et ) − π /2), z = cosh(t) sin(2 arctan(et ) − π /2).

Then all aspects of the analysis on pages 237–238 of the text go through, mutatis mutandis, to establish y = 1,

n

z = x.

Hence q(t) is a straight line in (x, y, z)-space. Because the earlier analysis found that 1 , cosh(t) sin(2 arctan(et ) − π /2) = tanh(t),

cos(2 arctan(et ) − π /2) =

along q(t), the normal vector x1 × x2 there has the form
cosh(t) sinh(t), −1, − sinh(t)

x 3. We know cos(−θ ) = cos(θ ) = sin(π /2 − θ ) and sin(−θ ) = − sin(θ ) = − cos(π /2 − θ ). Therefore cos(±A ∓ π /2) = cos(∓(π /2 − A)) = cos(π /2 − A) = sin(A),

sin(±A ∓ π /2) = sin(∓(π /2 − A)) = ∓ sin(π /2 − A) = ∓ cos(A).

and therefore


1 n(t) = √ sinh(t), −1, − sinh(t) . 2 sinh2 (t) + 1

Because sinh(t) is 1–1, n(t1 ) 6= n(t2 ), if t1 6= t2 .

4. a) Complementary angles have tangents that are reciprocals of each other: x = tan(θ ) = 1/ tan(π /2 − θ ).

c) Let us take advantage of the rotational symmetry of Therefore, if θ = arctan(x), then π /2 − θ = arctan(1/x), x(q1 , q2 ) = (x, y, z) = X(q1 , q2 ) to compute vectors only or on the plane y = 0 and only on the half where z > 0. This arctan(1/x) = π /2 − arctan(x). means we can take q2 = π /2 and then Hence, the substitution τ → −τ yields
x1 × x2 = cosh(q1 ) sinh(q1 ), 0, − cosh(q1 ) , q2 = 2 arctan(eτ ) − π /2

X1 × X2 = cosh(q1 ) sinh(q1 ), 0, cosh(q1 ) . → 2 arctan(1/eτ ) − π /2 = 2 π /2 − arctan(eτ ) − π /2 In other words, both these normal vectors also lie in the = −2 arctan(eτ ) + π /2 = −q2 ; plane y = 0, so it is sufficient to make a 2-dimensional drawing. that is, q2 → −q2 . The intersection of the surface x = X with the halfb) Let ϕC (τ ) = 2 arctan(eτ ) + C. By construction, each plane y = 0, z > 0, is the curve given parametrically as parametrized curve (q1 , q2 ) = (τ , ϕC (τ )) is tangent to the
L+ vector field. Because τ = q1 in this parametrization, (x, 0, z) = sinh(t), 0, cosh(t) .

DVI file created at 16:06, 20 January 2011

5.4. CURVATURE OF A SURFACE

47

reflection across the q2 -axis, which is given by the substi- 5.4 Curvature of a Surface tution q1 → −q1, is the same as τ → −τ . As the previous part shows, this substitution leads to the same result as 1. a) Nothing to do. q2 → −q2 (reflection across the q1 -axis). 1 But reflection across the q -axis clearly transforms the b) Given that A = (x′ · x′ )−1/2 , we can write L+ -vector field into the L− -vector field. Therefore, beA′ = − 21 (x′ · x′ )−3/2 (x′ · x′ )′ cause q2 = ϕC (τ ) is tangent to L+ , the reflected q2 = −ϕC (τ ) must be tangent to the reflected L+ , that is, to = − 21 (x′ · x′ )−3/2 (x′′ · x′ + x′ · x′′ ) L− = −(x′ · x′ )−3/2 (x′ · x′′ ) = −B, 5. Let P α be the past timelike set that is visible to the where B is as defined in 1b. The product rule applied to observer G at time τ = α . By analogy with future sets, t = Ax′ then gives the past light cone for α = 0 is given by the two graphs t′ = Ax′′ + A′ x′ = Ax′′ − Bx′ . 2 τ q = −2 arctan(e ) + π /2, τ < 0. c) We have q2 = 2 arctan(eτ ) − π /2, Note the domain of τ . As τ → −∞, range of values of q2 approaches π .

q2

→ ±π /2, so the Π

t′ · t′ = A2 (x′′ · x′′ ) − 2AB (x′′ · x′ ) + B2 (x′ · x′ ) =

2

= = Π

2

x′′ · x′′ (x′ · x′′ )2 (x′ · x′′ )2 (x′ · x′ ) −2 ′ ′ 2 + ′ ′ x ·x (x · x ) (x′ · x′ )3 x′′ · x′′ (x′ · x′′ )2 − ′ ′ 2 x′ · x′ (x · x )

(x′′ · x′′ )(x′ · x′ ) − (x′ · x′′ )2 , (x′ · x′ )2

The past light cone for α 6= 0 involves graphs of the form q2 = ±(2 arctan(eτ ) + C),

and therefore

where C must be chosen so that both graphs fall on the τ -axis when τ = α . Thus

Finally, then,

α

α

0 = ±(2 arctan(e ) + C),

or C = −2 arctan(e ).

The figure below shows the graphs of
q = ± 2 arctan(e ) − 2 arctan(e ) 2

τ

α

with α = 3.

Π

3

-Π

As τ → −∞ with α fixed, arctan(eτ ) → 0 so q2 → ∓(2 arctan(eα ); thus the range of values of q2 approaches 4 arctan(eα ). Therefore, if we now let α → ∞, the range of values of q2 approaches 4 · π /2 = 2π , covering the entire circle.

DVI file created at 16:06, 20 January 2011

t′ · t′ (x′′ · x′′ )(x′ · x′ ) − (x′ · x′′ )2 = . x′ · x′ (x′ · x′ )3 kt′ k κ= ′ = kx k

r

t′ · t′ = x′ · x′

s

(x′′ · x′′ )(x′ · x′ ) − (x′ · x′′ )2 , (x′ · x′ )2

3 ′ 2 2. a) Given p that x = (q, q ), we have x = (1, 3q ) and ′ 4 kx k = 1 + 9q . Therefore ! 3q2 1 x′ . ,p t= ′ = p kx k 1 + 9q4 1 + 9q4

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

48

The curve x(q) itself is shown on the left, above; it has Therefore, an inflection where q = 0. Its unit tangent map—drawn x′ (q) X′ (r) x′ (q) ϕ ′ (q) = = = ±t(q), to a different scale—is on the right. To make it clearer T(r) = kX′ (r)k kx′ (q)k |ϕ ′ (q)| kx′ (q)k that the unit tangent reverses direction (at q = 0, where it touches the x-axis), its length has been altered by the vari- where t is the unit tangent for x, and the sign is chosen deable factor eq/10. That is, the plot is actually of eq/10t(q). pending on whether ϕ preserves or reverses orientation, that is, on whether ϕ ′ is positive or negative. Thus two ′ (q) = (1, cos q) and kx′ k = b) Here x(q) = (q, sin q), so x p parametrizations of a curve have Gauss maps that are ei1 + cos2 q; therefore ther the same or antipodal. ! 4. We use Theorem 5.3 to determine K as the quotient 1 cos q t(q) = p . ,p kn1 × n2 k/kx1 × x2 k. We have 1 + cos2 q 1 + cos2 q x(q1 , q2 ) = ( f (q1 ), g(q1 ), q2 ), The curve x(q) is shown immediately below; it has x1 (q1 , q2 ) = ( f ′ (q1 ), g′ (q1 ), 0), three inflections, at the points where q = π , 2π , and 3π . x2 (q1 , q2 ) = (0, 0, 1), x1 × x2(q1 , q2 ) = (g′ (q1 ), − f ′ (q1 ), 0). The unit tangent curve t(q), shown at the left, reverses direction three times, at q = π , 2π , and 3π . Again, to make this fact more visible, its length has been altered by eq/100 . The curve also makes it evident that the tangents vary in direction between +45◦ and −45◦ .

c) Here x

= (q2 , q − q3),

x′

= (2q, 1 − 3q2)

so and p p kx′ k = 4q2 + (1 − 3q2)2 = 1 − 2q2 + 9q4, ! 2q 1 − 3q2 t= p . ,p 1 − 2q2 + 9q4 1 − 2q2 + 9q4

The curve x(q) is shown on the left, below. It has no inflections. Its unit tangent map, shown on the right, Points to the southwest initially, then turns clockwise through north and ends pointing to the southeast.

Note that x1 ×x2 depends only on q1 . The same will therefore be true for n, so we will have n2 = 0 at all points. Therefore n1 × n2 = 0, and hence K = 0, at all points. The Gaussian curvature on a cylinder is identically zero. 5. We use Theorem 5.4 to determine K(P). x(q1 , q2 ) = (q1 , q2 , q1 q2 ), we find

Given

x1 = (1, 0, q2 ), x2 = (0, 1, q1 ), x1 × x2 = (−q2 , −q1 , 1),

n = (−q2 A−1/2 , −q1 A−1/2, A−1/2 ),

where A = 1 + (q1)2 + (q2)2 . Thus,

∂ (−q2 A−1/2 ) +q2 −3/2 = A · 2q1 = q1 q2 A−3/2 , ∂ q1 2
∂ (−q1 A−1/2 ) = (q1 )2 A−3/2 − A−1/2 = − 1 + (q2)2 A−3/2, 1 ∂q ∂ (A−1/2 ) = −q1 A−3/2 . ∂ q1

The derivatives with respect to q2 are analogous. Hence n1 = n2 =

1 A3/2 1 A3/2

(q1 q2 , −1 − (q2)2 , −q1 ), (−1 − (q1)2 , q1 q2 , −q2 ),

3. The Gauss map is the map of the unit tangent vector of a curve. Suppose x(q) is one parametrization of the curve. By direct inspection, we see that Introduce a second parametrization as X(r) = x(ϕ (r)), q1 q2 −1 − (q2)2 where q = ϕ (r) and either ϕ ′ (r) > 0 for all r in the don1 = 3/2 x1 + x2 , ′ A A3/2 main, or ϕ (r) < 0 for all r in the domain. −1 − (q1)2 q1 q2 To find the unit tangent T for X, we compute n2 = x + x2 . 1 A3/2 A3/2 X′ (r) = x′ (q)ϕ ′ (r) and kX′ (r)k = kx′ (q)k|ϕ ′ (r)|

DVI file created at 16:06, 20 January 2011

5.4. CURVATURE OF A SURFACE

49

Thus, e = B(q)

1

q1 q2

A3/2

−1 − (q1)2

−1 − (q2)2 q1 q2

!

,

and e K(P) = det B(q)

(q1 )2 (q2 )2 − 1 − (q1)2 − (q2 )2 − (q1)2 (q2 )2 A3 1 2 2 2 −1 − (q ) − (q ) −1 =
3 =
2 . 1 2 2 2 1 1 + (q ) + (q ) ) 1 + (q )2 + (q2 )2 ) =

At the origin, K = −1. Moreover, r2 = (q1 )2 + (q2 )2 , so the calculation of K(P) here agrees completely with the calculation using polar coordinates in the text.

Gaussian curvature of S is the limit ratio of the area of n to the area of x. Because the image of n has zero area, Gaussian curvature of S must be zero everywhere.

c) Because n depends only on q1 , n2 = 0. The other derivative is
1 n1 = 3 4q1 (1 + 3(q1)2 ), −2(1 − 9(q1)4 ), 0 , 6. a) The curve C, shown on the left, below, has a douA ble point at (x, y) = (1, 0); this is the image of the two so we can write parameter values u = ±1. 2(1 + 3(q1)2 ) n1 = x1 + 0 · x2 , A3 n2 = 0 · x1 + 0 · x2. Hence

  1 2 e = 2(1 + 3(q ) ) 1 0 , B(q) 0 0 A3 e from which it follows that det B(q) = 0 = K(P).

The surface S, above right, is the vertical cylinder over C. The double point on C becomes a vertical line of double points on the surface. b) We have x(q1 , q2 ) = ((q1 )2 , (q1 )3 − q1, q2 ); therefore

7. a) Given that S is parametrized by x(q1 , q2 ) = ( f (q1 ), g(q1 ), q2 ), we have x1 = ( f ′ , g′ , 0), x2 = (0, 0, 1), x1 × x2 = (g′ , − f ′ , 0).

By hypothesis, ( f ′ , g′ ) is never zero; therefore, x1 × x2 = (g′ , − f ′ , 0) is never zero, either. By definition, S is therex1 = (2q1 , 3(q1 )2 − 1, 0), fore nonsingular. The unit normal is x2 = (0, 0, 1),  ′  g −f′ x1 × x2 = (3(q1 )2 − 1, −2q1, 0), n= , ,0 , A A   3(q1 )2 − 1 −2q1 p n= , ,0 , where A = ( f ′ )2 + (g′ )2 . As in the previous exercise, A A and for the same reason, n2 = 0. The other derivative is p
1 where A = 1 − 2(q1)2 + 9(q1)4 . Because n depends n1 = 3 f ′ ( f ′ g′′ − f ′′ g′ ), g′ ( f ′ g′′ − f ′′ g′ ), 0 1 ∗ A only on the single parameter q , the image S of n on the f ′ g′′ − f ′′ g′ ′ ′ unit sphere (see below) is only a curve, not a surface. Be= ( f , g , 0) A3 cause the third component of n is zero, this curve lies in f ′ g′′ − f ′′ g′ the plane z = 0; that is, it lies on the equator of the unit = x1 . sphere. A3

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

50

Therefore (following the pattern of the previous exercise), we have  ′ ′′ ′′ ′  e = f g −f g 1 0 , B(q) 0 0 A3 e = 0 = K(P). from which it follows that det B(q)

8. • The Gauss map of an ellipsoid is a 1–1 cover of the unit sphere. The image does not cover the unit sphere uniformly, though. On a region of the ellipsoid were the curvature is least, image points are “bunched up” on the unit sphere; where the curvature is largest, image points are spread the farthest. The next two figure illustrate this. On the left is the upper half of an ellipsoid whose axes are in the ratios 4 : 2 : 1. The ellipsoid is least curved where the shortest axis (the vertical z-axis) meets the ellipsoid, and most curved where the longest axis (the x-axis) meets it.

The Gauss map of this surface (on the right) covers the upper half of the unit sphere. The coordinate lines are most “bunched up” near the vertical axis and most spread out near the x-axis. • Suppose the axis of rotation of the bell-shaped surface (shown on the left, below, with a segment removed) is the z-axis.

The two boundary points on that meridian (one on the outer equator, one on the inner) therefore have antipodal images on the Gauss sphere. The image of the outer equator covers the entire equator of the sphere, and the same is true of the inner equator. To make the double cover of the equator easier to see, the radius of image points has been varied.

Each circle of latitude is covered the same way. The points on the “bottom” meridian of the torus all map to the south pole of the unit sphere. 9. Given x(q1 , q2 ) = (q1 , q2 , aq1 + bq2 + c), we find x1 = (1, 0, a), x2 = (0, 1, b), x1 × x2 = (−a, −b, 1),   −b 1 −a . , , n= 1 + a2 + b2 1 + a2 + b2 1 + a2 + b2 The image of the Gauss map of a plane is a single point on the unit sphere.

Then its Gauss map, shown on the right with the same cutaway, covers a portion of the upper hemisphere of the unit sphere (a “polar” region) and “folds back” on itself at the points that are the images of the inflection points on the bell. The radius of the image has been varied to make the fold more readily visible. • The Gauss map of the torus is a double-cover of the unit sphere. To see this, consider just the bottom half of a torus. The points on a single half-meridian on the torus have Gaussian images on the same half-meridian on the unit sphere.

DVI file created at 16:06, 20 January 2011

10. a) The solution to Exercise 5e in §5.1 (cf. Solutions page 41) provides x1 ×x2 = r(a+r cos q2 )(cos q1 cosq2 , sin q1 cos q2 , sin q2 ), from which we obtain the unit normal n(q1 , q2 ) = (cos q1 cos q2 , sin q1 cosq2 , sin q2 ) that defines the Gauss map G : Ta,r → S2 as q → n(q). Some representative images of the Gauss map are shown in the solution to Exercise 8.

5.4. CURVATURE OF A SURFACE b) We find K(P) by expressing n1 and n2 in terms of x1 and x2 . We have n1 = (− sin q1 cos q2 , cos q1 cos q2 , 0), n2 = (− cosq1 sin q2 , − sin q1 sin q2 , cos q2 ), and because x1 = (a + r cosq2 )(− sin q1 , cos q1 , 0), x2 = r(− cos q1 sin q2 , − sin q1 sin q2 , cos q2 ), we can write cos q2 x1 , a + r cosq2 ; 1 n2 = x2 . r

n1 =

 cos q2 0 2  e = B(q)   a + r cosq 1 0 r

e = and hence K(P) = det B(q)



cos q2 . r(a + r cosq2 )

51 d) The points P where K(P) = 0 are the top and bottom circles of latitude on the torus. The entire top circle is mapped to the north pole of the Gauss sphere by G ; the entire bottom circle is mapped to the south pole. 11. a) The text (pages 246–247) establishes that the curvature of the sphere of radius R is 1/R2 at each point. The solution to Exercise 5d in §5.1 (Solutions page 41) provides √ g = R2 cosq2 . If D is the upper hemisphere, then 0 ≤ q2 ≤ π /2 while we still keep 0 ≤ q1 ≤ 2π . Therefore, total curvature over D is Z 2π

dq

1

0

Z π /2 1 0

R2

·R cosq dq = 2π 2

2

2

Z π /2 0

cosq2 dq2 = 2π .

If D is the whole sphere, then −π /2 ≤ q2 ≤ π /2 and the integral with respect to q2 becomes Z π /2

cos q2 dq2 = 2, c) If Ta,r is an ordinary torus that avoids the z axis and −π /2 hence avoids self-intersections, the radius a of the “core” of the torus is larger than the radius r of the meridian cir- implying that the total curvature is 4π . Neither of these total curvature values depends on R. cle. Thus, 2 0 < a − r ≤ a + r cosq , b) When D is either the upper hemisphere or the whole so the denominator in K(P) is always positive. Hence the sphere, its total curvature is exactly equal to the area of sign of K(P) is the sign of its numerator cosq2 . Thus, its Gaussian image. The Gaussian image of any region K(P) > 0 when |q2 | < π /2. This parametrizes the outside on the sphere of radius R is independent of R; it depends half of each meridian circle and hence parametrizes the only on the shape of the region itself. The independence outer half of the torus. Likewise, K(P) < 0 when either of the Gaussian image therefore does explain why the total π < q2 < −π /2 or π /2 < q2 < π ; these values parametrize curvature is independent of R. the inner half of the meridian circles and hence the inner 12. a) For the torus Ta,r , we have half of the torus. K=

cos q2 r(a + r cosq2 )

√ g = r(a + r cosq2 );

and

see the solutions to Exercise 10d, above, and Exercise 5e in §5.1 (Solutions page 41). Therefore, the total curvature of the region given by the parameter domain D is just ZZ D

√ K g dq1 dq2 =

ZZ D

cos q2 dq1 dq2 .

The figure on the left, above, is a small portion I of the inner half of the torus; the figure on the right is its Gaussian For all three regions in question, 0 ≤ q1 ≤ 2π . For the image G (I). The top (resp. bottom) half of I maps to the outer half of the torus, −π /2 ≤ q2 ≤ π /2. Using the peritop (resp. bottom) half of G (I). However, the right verti- odicity of cos q2 , we can take: cal edge of I maps to the left vertical edge of G (I). As we outer : − π /2 ≤ q2 ≤ π /2, take vertical meridian lines from right to left along I, we find their images are meridian lines on the Gauss sphere, inner : π /2 ≤ q2 ≤ 3π /2, but taken from left to right. This implies that G reverses entire : − π ≤ q2 ≤ π . the orientation of I.

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 5. SURFACES AND CURVATURE

52 Thus, the total curvature of each region is outer =

Z 2π

dq1

0

inner =

Z 2π

dq1

0

entire =

Z 2π 0

dq1

Z π /2

−π /2

cos q2 dq2 = 4π ;

Z 3π /2 π /2 Z π −π

cos q2 dq2 = −4π ;

cos q2 dq2 = 0.

No value depends on either a or r. b) The image of the outer half of the torus is the entire unit sphere, positively oriented; its area is +4π . The image of the inner half is again the entice unit sphere, but negatively oriented; the area of that image is therefore −4π . These images are independent of a and r, thus explaining why the total curvatures are independent of a and r. 13. Partition D into two regions: D+ , on which the curvature function K is nonnegative, and D− , on which K is negative. The oriented area of G (D+ ) is nonnegative and is the total curvature of D+ . The oriented area of G (D− ) is negative, and is the total curvature of D− . The total curvature of D is the algebraic sum of these two numbers. But the algebraic sum of these numbers is also the net oriented area of G (D).

DVI file created at 16:06, 20 January 2011

Solutions: Chapter 6

Intrinsic Geometry 6.1 Theorema Egregium

2. We compute b jk as x jk ·n and obtain bij by determining the linear combinations n j = −bij xi .

1. • The plane Here x1 = (1, 0, a), x2 = (0, 1, b), so     1 + a2 0 g11 g12 = g21 g22 0 1 + b2

• The plane All x jk = 0, so b jk = 0. Furthermore, n is constant, so ni = 0, implying all bij = 0. It is therefore immediately true that b jk = gik bij

and 

g11 g12 g21 g22



  1/(1 + a2) 0 = . 0 1/(1 + b2)

• The cylinder Here x1 × x2 = (g′ , − f ′ , 0), so

• The cylinder Here x1 = ( f ′ , g′ , 0), x2 = (0, 0, 1), so   ′ 2   g11 g12 ( f ) + (g′ )2 0 = g21 g22 0 1 and

n=

  11 12   1/(( f ′ )2 + (g′)2 ) 0 g g = . 0 1 g21 g22

−f′

!

p ,p ,0 . (g′ )2 + ( f ′ )2 (g′ )2 + ( f ′ )2

while bi j = 0 otherwise. We also have  ′ ′ ′′  f ( f g − g′ f ′′ ) g′ ( f ′ g′′ − g′ f ′′ ) n1 = , , 0 ((g′ )2 + ( f ′ )2 )3/2 ((g′ )2 + ( f ′ )2 )3/2 f ′ g′′ − g′ f ′′ x1 + 0 · x2 = ((g′ )2 + ( f ′ )2 )3/2

x1 = R(− sin q1 sin q2 , cos q1 sin q2 , 0), x2 = R(cos q1 cos q2 , sin q1 cos q2 , − sin q2 ),

and

g′

Moreover, x11 = ( f ′′ , g′′ , 0) while all other xi j = 0. Therefore, g′ f ′′ − f ′ g′′ , b11 = x11 · n = p ( f ′ )2 + (g′ )2

• The sphere Here

so

and bij = gik b jk

   2 2 2  g11 g12 R sin q 0 = g21 g22 0 R2

while n2 = 0. Hence   g′ f ′′ − f ′ g′′ f ′ g′′ − g′ f ′′ = b11 = − ′ 2 ′ 2 3/2 ((g ) + ( f ) ) ((g′ )2 + ( f ′ )2 )3/2

  11 12   g g 1/(R2 sin2 q2 ) 0 . = g21 g22 0 1/R2

• The torus Ta,r The metric tensor     (a + r cosq2 )2 0 g11 g12 = g21 g22 0 r2

while bij = 0 otherwise. Thus, using gi j from the solution to Exercise 1, gi1 bi1 = g11 b11 + g21b21

was found in the solution to Exercise 5e, § 5.1 (Solutions page 41); hence  11 12    g g 1/(a + r cosq2 )2 0 = . g21 g22 0 1/r2 53

DVI file created at 16:06, 20 January 2011

= (( f ′ )2 + (g′ )2 ) · =

g′ f ′′ − f ′ g′′ +0·0 ((g′ )2 + ( f ′ )2 )3/2

g′ f ′′ − f ′ g′′ = b11 , ((g′ )2 + ( f ′ )2 )1/2

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

54 as required. In addition,

In the other direction, we find

gi1 bi2 = g11 b12 + g21b22 = 0 + 0 = b12 = b21 , gi2 bi2

= g12 b12 + g22b22

= 0 + 0 = b22 ,

also as required. In the other direction, we have 1k

11

12

g b1k = g b11 + g b12 g′ f ′′ − f ′ g′′ p +0·0 ( f ′ )2 + (g′ )2 g′ f ′′ − f ′ g′′ = = b11 , (( f ′ )2 + (g′ )2 )3/2 =

1

( f ′ )2 + (g′ )2

g1k b1k = g11 b11 + g12b12
1 1 = 2 2 2 · − R sin2 q2 + 0 · 0 = − = b11 , R R sin q g2k b2k = g21 b21 + g22b22   1 1 = 0·0+ · (−R) = − = b22 . 2 R R

In the other two sums, each term has at least on factor that is zero. Thus we confirm gik b jk = bij in every case.

• The torus Ta,r . From the solution to Exercise 5e, § 5.1 (Solutions page 41), we have as required. In all other combinations jk , at least one ik i factor is zero in each term, so g b jk = b j is satisfied vacx1 = (a + r cosq2 )(− sin q1 , cos q1 , 0), uously. x2 = r(− cos q1 sin q2 , − sin q1 sin q2 , cos q2 ), • The sphere The unit normal n is just the radius vector gik b

x divided by its length, R: 1

and hence 2

1

2

2

n = (cos q sin q , sin q sin q , cos q ). Furthermore, x11 = R(− cosq1 sin q2 , − sin q1 sin q2 , 0),

x12 = x21 = R(− sin q1 cos q2 , cos q1 cos q2 , 0),

x22 = R(− cosq1 sin q2 , − sin q1 sin q2 , − cos q2 ),

x11 = (a + r cosq2 )(− cos q1 , − sin q1 , 0),

x12 = x21 = −r sin q2 (− sin q1 , cos q1 , 0),

x22 = r(− cos q1 cos q2 , − sin q1 cos q2 , − sin q2 ).

Because n = (cos q1 cosq2 , sin q1 cos q2 , sin q2 ) (cf. the solution to Exercise 10a, § 5.4, Solutions page 50), we have

so b11 = −R sin2 q2 , Because n =

b12 = b21 = 0,

1 x1 , R

Moreover, n2 =

1 x2 , R

implying 1 b11 = − , R

b12 = b21 = 0, b22 = −r.

b22 = −R.

1 x, we also have R n1 =

b11 = −(a + r cosq2 ) cos q2 ,

b12 = b21 = 0,

1 b22 = − . R

Using the gi j from the solution to Exercise 1, we find  
1 +0·0 gi1 bi1 = g11 b11 + g21b21 = R2 sin2 q2 · − R = −R sin2 q2 = b11 ,

  1 gi2 bi2 = g12 b12 + g22b22 = 0 · 0 + R2 · − R = −R = b22 . In the other two sums, gi1 bi2 = b12 and gi2 bi1 = b21 , at least one factor in each term is zero. We have thus confirmed gi j bik = b jk in every case.

DVI file created at 16:06, 20 January 2011

cos q2 x1 , a + r cosq2 1 n2 = (− cos q1 sin q2 , − sin q1 sin q2 , cos q2 ) = x2 , r n1 = cos q2 (− sin q1 , cos q1 , 0) =

implying b11 = −

cos q2 , a + r cosq2

b12 = b21 = 0,

1 b22 = − . r

Using the gi j from the solution to Exercise 1, we find gi1 bi1 = g11 b11 + g21b21 = (a + r cosq2 )2 ·

− cos q2 +0·0 a + r cosq2

= − cosq2 (a + r cosq2 ) = b11 , −1 = −r = b22 . gi2 bi2 = g12 b12 + g22b22 = 0 · 0 + r2 · r

6.1. THEOREMA EGREGIUM

55

In the other two sums, at least one factor in each term is zero. This confirms gi j bik = b jk in every case. In the other direction, we find

• The sphere Referring to Exercises 1 and 2, we find Γ11,1 = 0, Γ11,2 = −R2 sin q2 cos q2 ,

g1k b1k = g11 b11 + g12b12 1 = · (a + r cosq2 ) cos q2 + 0 · 0 (a + r cosq2 )2 =−

cos q2 = b11 , a + r cosq2



1 g b2k = g b21 + g b22 = 0 · 0 + 2 r 1 = − = b22 . r 2k

21

22



Γ12,1 = Γ21,1 = R2 sin q2 cos q2 , Γ12,2 = Γ21,2 = 0, Γ22,1 = 0, Γ22,2 = 0.

· (−r)

To compute Γ jk,l as a sum of partial derivatives, note that the only nonconstant g jk is g11 , and it depends only on q2 :

∂ g11 = 2R2 sin q2 cos q2 , ∂ q2 ∂ g jk = 0 otherwise. ∂ ql

In the other two sums, at least one factor in each term is zero. This confirms gik b jk = bij in every case. 3. We refer to the solutions to Exercises 1 and 2 for xl and x jk . • The plane All x jk = 0, so Γ jk,l = 0

for all j, k, l.

Alternatively, to compute Γ jk,l as a sum of partial derivatives, note that all g jk are constant, so

∂ g jk = 0 and hence Γ jk,l = 0 ∂ ql for all j, k, l. The two ways of obtaining Γ jk,l agree. • The cylinder We have Γ11,1 = ( f ′′ , g′′ , 0) · ( f ′ , g′ , 0) = f ′ f ′′ + g′ g′′ , Γ11,2 = ( f ′′ , g′′ , 0) · (0, 0, 1) = 0, Γi j,k = 0,

for all other i, j, k because xi j = 0 unless i = j = 1. To compute Γ jk,l as a sum of partial derivatives, note that the only nonconstant gi j is g11 , and it depends only on q1 . Hence,

∂ g11 = 2 f ′ f ′′ + 2g′g′′ , ∂ q1 ∂ g jk = 0 otherwise. ∂ ql

It follows that the only nonzero Γ jk,l are those that have two indices equal to 1 and one equal to 2. Thus Γ11,2 = = Γ12,1 = =

  1 ∂ g12 ∂ g12 ∂ g11 + − 2 ∂ q1 ∂ q1 ∂ q2 1 (0 + 0 − 2R2 sin q2 cos q2 ) = −R2 sin q2 cos q2 , 2  1 ∂ g11 ∂ g21 ∂ g12 + − 2 ∂ q2 ∂ q1 ∂ q1 1 (2R2 sin q2 cos q2 + 0 − 0) = R2 sin q2 cos q2 . 2

By symmetry, Γ21,1 = Γ12,1 = R2 sin q2 cos q2 . The two ways of obtaining Γ jk,l agree. • The torus Ta,r Referring to the solution to Exercise 2, we find Γ11,1 = 0, Γ11,2 = r sin q2 (a + r cosq2 ), Γ12,1 = Γ21,1 = −r sin q2 (a + r cosq2 ), Γ12,2 = Γ21,2 = 0, Γ22,1 = 0,

Γ22,2 = 0.

As was true with the sphere, the only nonconstant g jk is Therefore, the only nonzero Γ jk,l is the one whose three g11 , and it depends only on q2 : indices are all 1: ∂ g11   = −2r sin q2 (a + r cosq2 ), 1 ∂ g11 ∂ g11 ∂ g11 ′ ′′ ′ ′′ ∂ q2 Γ11,1 = = f f + g g . + − 2 ∂ q1 ∂ q1 ∂ q1 ∂ g jk = 0 otherwise. ∂ ql The two ways of obtaining Γ jk,l agree.

DVI file created at 16:06, 20 January 2011

56

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

Hence, again as with the sphere, the only nonzero Γ jk,l has two indices equal to 1 and one equal to 2. Thus   1 ∂ g12 ∂ g12 ∂ g11 Γ11,2 = + − 2 ∂ q1 ∂ q1 ∂ q2
1 = 0 + 0 − (−2r sinq2 (a + r cosq2 )) 2 = r sin q2 (a + r cosq2 ),   1 ∂ g11 ∂ g21 ∂ g12 + − Γ12,1 = 2 ∂ q2 ∂ q1 ∂ q1
1 − 2r sin q2 (a + r cosq2 ) + 0 − 0 = 2 = −r sin q2 (a + r cosq2 ).

as required. For any ( j, k) 6= (1, 1), the equation

By symmetry, Γ21,1 = Γ12,1 = −r sin q (a + r cos q ). The two ways of obtaining Γ jk,l agree. 2

2

4. • The plane From the solution to Exercise 3, we know all Γ jk.l = 0; hence, Γijk = gil Γ jk,l = 0

for all i, j, k.

Moreover, b jk = 0 and x jk = 0 for all j, k. Therefore, it is immediately true that x jk = Γijk xi + b jk n. • The cylinder We have Γ111 = g1l Γ11,l = g11 Γ11,1 + g12 Γ11,2 1 = ′ 2 · ( f ′ f ′′ + g′ g′′ ) + 0 · 0 ( f ) + (g′ )2 f ′ f ′′ + g′ g′′ = ′ 2 , ( f ) + (g′ )2 Γ211 = g2l Γ11,l = g21 Γ11,1 + g22 Γ11,2 = 0 · ( f ′ f ′′ + g′ g′′ ) + 0 · 0 = 0. Suppose ( j, k) 6= (1, 1); then Γ jk,l = 0 from Exercise 3. Hence, for the same j, k, we have Γijk = gil Γ jk,l = gi1 Γ jk, 1 + gi2Γ jk,2 = 0 + 0 = 0. Next, we verify that (noting Γ211 = 0) x11 = Γi11 xi + b11n = Γ111 x1 + b11 n. p For simplicity, let A = ( f ′ )2 + (g′)2 ; then

f ′ f ′′ + g′g′′ ′ ′ ( f , g , 0), A2   g′ f ′′ − f ′ g′′ g′ − f ′ b11 n = , ,0 , A A A

Γ111 x1 =

( f ′ )2 + (g′ )2 ′′ ′′ ( f , g , 0) A2 ′′ ′′ = ( f , g , 0) = x11 ,

Γ111 x1 + b11n =

DVI file created at 16:06, 20 January 2011

x jk = Γijk xi + b jk n

is true because b jk = 0, Γijk = 0, and x jk = 0. • The sphere We have

Γ111 = g11 Γ11,1 + g12Γ11,2 1 = 2 2 2 · 0 + 0 · (−R2 sin q2 cos q2 ) = 0, R sin q

Γ211 = g21 Γ11,1 + g22Γ11,2   1 = 0·0+ · (−R2 sin q2 cos q2 ) R2 = − sin q2 cos q2 ,

Γ112 = Γ121 = g11 Γ12,1 + g12Γ12,2 1 = 2 2 2 · (R2 sin q2 cos q2 ) + 0 · 0 R sin q cosq2 = cot q2 , = sin q2 Γ212 = Γ221 = g21 Γ12,1 + g22Γ12,2 = 0 · (R2 sin q2 cos q2 ) +



1 R2



· 0 = 0.

Because Γ22,1 = Γ22,2 = 0, it follows immediately that Γ122 = Γ222 = 0. It remains to show x jk = Γijk xi + b jk n. Consider first ( j, k) = (1, 1); because Γ111 = 0, we have Γi11 xi = Γ211 x2 = −R sin q2 cos q2 (cos q1 cos q2 , sin q1 cos q2 , − sin q2 ),

b11 n = −R sin2 q2 (cos q1 sin q2 , sin q1 sin q2 , cos q2 ),

Γi11 xi + b11 n

= −R sin q2 (cos q1 , sin q2 , 0) = x11 . Next, let ( j, k) = (1, 2); because Γ212 = 0, we have Γi12 xi = Γ112 x1 =

R cos q2 (− sin q1 sin q2 , cos q1 sin q2 , 0) sin q2

= R cos q2 (− sin q1 , cos q2 , 0), b12 n i Γ12 xi + b12n

= 0 · n = 0,

= R cos q2 (− sin q1 , cos q2 , 0) = x12 .

By symmetry, we get the same result with ( j, k) = (2, 1). When ( j, k) = (2, 2), we have Γi22 = 0 so Γi22 xi + b22n = b22 n = −R(cosq1 sin q2 , sin q1 sin q2 , cos q2 ) = x22 .

6.1. THEOREMA EGREGIUM

57

• The torus Ta,r Because Γ11,1 , Γ12,2 and g12 = g21 are all zero, we have Γ111 = g11 Γ11,1 + g12Γ11,2 = 0, Γ211 = g21 Γ11,1 + g22Γ11,2   1 = 0 + 2 · (r sin q2 (a + r cosq2 )) r

Therefore, ∂ Γ111 /∂ q1 is the only nonzero partial deriva′ )2 + (g′ )2 . Then we have tive. To determine it, let A = ( f √
′
′ 1 ′ 1 1 Γ11 = 2 A /A = 2 ln A = ln A and hence the compact form √
′′ ∂ Γ111 = ln A . 1 ∂q

Now consider the mixed Riemann tensor. If even one index in Rijkl is different from 1, then sin q2 (a + r cosq2 ) = , r ∂ Γijl ∂ Γijk Γ112 = Γ121 = g11 Γ12,1 + g12Γ12,2 = = 0. ∂ qk ∂ ql
1 2 2 = · − r sin q (a + r cosq ) + 0 (a + r cosq2 )2 Moreover, if a term in Γ pjl Γipk or Γ pjl Γipk is to be nonzero, it 2 −r sin q must have the form Γ111 Γ111 , but this is impossible. There, = a + r cosq2 fore, Rijkl = 0 unless i = j = k = l = 1. For the remaining 2 2 21 22 Γ12 = Γ21 = g Γ12,1 + g Γ12,2 = 0 + 0 = 0. component, In addition, Γ22,1 = Γ22,2 = 0, so Γ122 = Γ222 = 0 as well. To show x jk = Γijk xi + b jk n, first take ( j, k) = (1, 1). then sin q2 (a + r cosq2 ) × r r(− cos q1 sin q2 , − sin q1 sin q2 , cos q2 )

∂ Γ111 ∂ Γ111 p 1 p 1 − + Γ11 Γ p1 − Γ11 Γ p1 ∂ q1 ∂ q1 = ϕ (q1 ) − ϕ (q1) + Γ111 Γ111 + Γ211 Γ121

R1111 =

− Γ111Γ111 − Γ211Γ121 = 0.

Γi11 xi = Γ211 x2 =

b11 n = − cos q2 (a + r cosq2 )× 1

2

1

2

2

(cos q cos q , sin q cos q , sin q ),

Γi11 xi + b11n = = (a + r cosq2 )(− cos q1 , − sin q1 , 0) = x11 .

Hence Rijkl = 0 for all i, j, k, l. Up to this point, it has been easy to calculate the 16 components of the mixed Riemann tensor RIjkl because all, or nearly all, have been obviously zero. For the sphere and the torus this is no longer true; therefore, we reduce the computational labor by exploiting a symmetry, namely

Next, let ( j, k) = (1, 2); because Γ212 and b12 are zero, we have Γi12 xi + b12n = Γ112 x1 =

−r sin q2 · (a + r cosq2 )(− sin q1 , cos q1 , 0) a + r cosq2 2

1

1

= −r sin q (− sin q , cos q , 0) = x12 . By symmetry, Γi21 xi + b21 n = x21 . When ( j, k) = (2, 2), we have Γi22 = 0, so Γi22 xi + b22n = b22 n = (−r) · (cosq1 cosq2 , sin q1 cos q2 , sin q2 ) = x22 . 5. • The plane Because all Γijk = 0, it follows that all ∂ Γijk /∂ ql and all Rijkl are zero. • The cylinder From the solution to Exercise 4, Γ111 =

f ′ f ′′ + g′ g′′ , ( f ′ )2 + (g′ )2

Γijk = 0

otherwise.

DVI file created at 16:06, 20 January 2011

Rijlk = −Rijkl . (See Exercise 8, below.) Think of the symmetry this way: for each fixed i and j, the 2 × 2 matrix whose kth row and lth column is Rijkl is skew-symmetric: M ij

 i    R j11 Rij12 0 Aij = = , Rij21 Rij22 −Aij 0

Aij = Rij12 .

Thus, there is just one component AiJ = Rij12 in each of the four matrices M ij , 1 ≤ i, j ≤ 2, to compute.

• The sphere The nonzero Christoffel symbols are those in which the index 1 appears exactly twice. Furthermore, they depend only on q2 , so the only nonzero partial derivatives are

∂ Γ211 = sin2 q2 − cos2 q2 , ∂ q2 −1 ∂ Γ112 ∂ Γ121 = = 2 2. ∂ q2 ∂ q2 sin q

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

58 Now take i = j = 1; then

∂ Γ112 ∂ Γ111 − + Γ112Γ111 + Γ212Γ121 ∂ q1 ∂ q2 − Γ111 Γ112 − Γ211 Γ122 = 0 − 0 + 0 + 0 − 0 − 0 = 0.

R1112 =

For i = 1, j = 2, we have R1212 =

Finally, when i = 2, j = 1, we have

∂ Γ122 ∂ Γ121 − + Γ122Γ111 + Γ222Γ121 ∂ q1 ∂ q2 − Γ121 Γ112 − Γ221 Γ122

−1 cos2 q2 +0+0− 2 2 −0 2 2 sin q sin q 2 2 1 − cos q = 1. = sin2 q2

= 0−

For i = 2, j = 1, we have R2112 =

∂ Γ212 ∂ Γ211 − + Γ112 Γ211 + Γ212Γ221 ∂ q1 ∂ q2 − Γ111 Γ212 − Γ211 Γ222

= 0 − (sin2 q2 − cos2 q2 ) +

cos q2 (− sin q2 cos q2 ) + 0 − 0 − 0 sin q2

= − sin2 q2 . Finally, for i = j = 2 we have R2212 =

∂ Γ222 ∂ q1

= 0.

−

∂ Γ221 ∂ q2

∂ Γ121 − Γ121 Γ112 ∂ q2 r2 sin2 q2 r2 + ar cosq2 − = 2 2 (a + r cosq ) (a + r cosq2 )2 r2 cos2 q2 + ar cosq2 r cos q2 = = . (a + r cosq2 )2 a + r cosq2

R1212 = −

∂ Γ211 + Γ112 Γ211 ∂ q2  a cos q2 + cos2 q2 − sin2 q2 =− r −r sin q2 sin q2 (a + r cosq2 ) + · a + r cosq2 r a 2 2 2 = − cos q − cos q + sin2 q2 − sin2 q2 r cosq2 (a + r cosq2 ). =− r

R2112 = −

6. The symmetry Rh jlk = gih Rijlk = −gih Rijkl = −Rh jkl means that there are, once again, at most four nonzero components to determine: Rh j12 , 1 ≤ h, j ≤ 2. • The plane Because all Rijkl = 0, it is also true that Rh jkl = 0 and hence that K = 0 at all points. • The cylinder Again all Rijkl = 0 = Rh jkl , so K = 0 at all points. • The sphere We have R1112 = g11 R1112 + g21R2112 = 0 + 0 = 0,

+ Γ122Γ211 + Γ222Γ221

− Γ121 Γ212 − Γ221 Γ222

R1212 = g11 R1212 + g21R2212 = R2 sin2 q2 · 1 + 0 = R2 sin2 q2 ,

R2112 = g12 R1112 + g22R2112

= 0 + R2 · (− sin2 q2 ) = −R2 sin2 q2 ,

• The torus Ta,r . As with the sphere, the only nonzero R2212 = g12 R1212 + g22R2212 = 0 + 0 = 0. Christoffel symbols are those in which the index 1 appears exactly twice, and they depend only on q2 . The only Because g = R4 sin2 q2 , Gaussian curvature has the value nonzero partial derivatives are therefore R1212 R2 sin2 q2 1 K= = 4 2 2 = 2. 2 g R ∂ Γ11 a R sin q = cos q2 + cos2 q2 − sin2 q2 , ∂ q2 r • The torus Ta,r Referring to the expressions for the varr + a cosq2 ∂ Γ112 ∂ Γ121 ious Rh jkl in the solution for the sphere, we find R1112 = = = −r . ∂ q2 ∂ q2 (a + r cosq2 )2 R2212 = 0 and Because the same Christoffel symbols and their derivatives are nonzero for the torus and the sphere, we see already that R1112 = R2212 = 0, because that is true for the sphere. The only nonzero terms in the remaining two components will be those that were nonzero for the sphere. Thus, when i = 1, j = 2, we have

DVI file created at 16:06, 20 January 2011

R1212 = g11 R1212 = (a + r cosq2 )2 ·

r cos q2 a + r cosq2

= r cos q2 (a + r cosq2 ),   cosq2 2 2 2 (a + r cosq ) R2112 = g22 R112 = r · − r = −r cos q2 (a + r cosq2 ).

6.1. THEOREMA EGREGIUM

59

Because g = r2 (a + r cos q2 )2 , Gaussian curvature has the value K=

Next, in computing the sum Γijk = gil Γ jk,l , we exclude the term containing g12 = g21 = 0:

R1212 r cos q2 (a + r cosq2 ) cos q2 = = . g r2 (a + r cosq2 )2 r(a + r cosq2 )

Γ111 = g11 Γ11,1 = 0, Γ211 = g22 Γ11,2 =

This agrees with the value found in the solution to Exercise 10b in § 5.4 (Solutions page 51).

γ pq = gkl mkp mlq .

rr′ r′ = , 2 r r Γ212 = Γ221 = g22 Γ12,2 = 0, Γ122 = g11 Γ22,1 = 0, Γ222 = g22 Γ22,2 =

∂ Γijk ∂ ql

∂ Γ112 ∂ Γ111 − + Γ112 Γ111 + Γ212 Γ121 ∂ q1 ∂ q2 − Γ111Γ112 − Γ211 Γ122 = 0 − 0 + 0 + 0 − 0 − 0 = 0,

=−

R1112 =

∂ Γijl

+ Γ pjk Γipl − Γ pjl Γipk ∂ qk ! ∂ Γijl ∂ Γijk p i p i − + Γ jl Γ pk − Γ jk Γ pl ∂ qk ∂ ql −

− Rijkl . Rijkk

−Rijkk ,

b) Setting l = k in the identity, we have = i i implying R j11 = R j22 = 0 for 1 ≤ i, j, ≤ 2. These make up 8 of the 16 components of the mixed Riemann tensor. 9. a) Let r = r(q2 ), z = z(q2 ); we have   11 12   2 g g 1/r 0
. = g21 g22 0 1/ (r′ )2 + (z′ )2

Each Γi j,k is obtained as a sum of partial derivatives of the gi j . The only nonzero derivatives are

∂ g11 = 2rr′ , ∂ q2

r′ r′′ + z′ z′′ . (r′ )2 + (z′ )2

As explained in the solution to Exercise 5, the only components of the mixed Riemann tensor we need to find are Rij12 , 1 ≤ i, j ≤ 2. We now work these out in detail. For i = j = 1, we have

8. a) For every i, j, k, l, we have Rijlk =

,

Γ112 = Γ121 = g11 Γ12,1 =

7. a) Transposition interchanges the rows and columns of a matrix, so aij = mij . b) The element Bkq in the kth row and the qth column of the matrix B = GM is the sum Bkq = gkl mlq . The element γ pq in the pth row and qth column of the matrix Γ = Mt B = Mt GM is the sum γ pq = mkp Bkq = mkp gkl mlq . We can rewrite this as

−rr′

(r′ )2 + (z′ )2

∂ g22 = 2r′ r′′ + 2z′ z′′ . ∂ q2

Thus either all three indices of Γi j,k have the value 2, or else precisely one does; hence Γ11,1 = Γ12,2 = Γ21,2 = Γ22,1 = 0. In the following, only nonzero terms are written.   ∂ g11 1 Γ11,2 = − 2 = −rr′ , 2 ∂q   1 ∂ g11 = rr′ , Γ12,1 = Γ21,1 = 2 ∂ q2   1 ∂ g22 = r′ r′′ + z′ z′′ . Γ22,2 = 2 ∂ q2

DVI file created at 16:06, 20 January 2011

For i = 1, j = 2, first set A = (r′ )2 + (z′ )2 ; then we have

∂ Γ122 ∂ Γ121 − + Γ122Γ111 + Γ222Γ121 ∂ q1 ∂ q2 − Γ121 Γ112 − Γ221 Γ122  ′ ′  ′ 2 r′ r′′ + z′ z′′ r′ r r +0+ −0 = 0− · − r A r r rr′′ − (r′ )2 (r′ )2 r′′ + r′ z′ z′′ (r′ )2 + − 2 =− r2 rA r −rr′′ (r′ )2 r′′ + r′ z′ z′′ −r′′ A (r′ )2 r′′ + r′ z′ z′′ = 2 + = + r rA rA rA −r′′ (r′ )2 − r′′ (z′ )2 + (r′ )2 r′′ + r′ z′ z′′ rA z′ (r′ z′′ − r′′ z′ ) z′ (r′ z′′ − r′′ z′ ) = = rA r((r′ )2 + (z′ )2 )

R1212 =

For i = 2, j = 1, we have

∂ Γ212 ∂ Γ211 − + Γ112 Γ211 + Γ212 Γ221 ∂ q1 ∂ q2 − Γ111Γ212 − Γ211Γ222   −rr′ ′ r′ −rr′ + · +0 = 0− A r A −rr′ r′ r′′ + z′ z′′ −0− · A A rz′ (r′′ z′ − r′ z′′ ) rz′ (r′′ z′ − r′ z′′ ) = . = A2 ((r′ )2 + (z′ )2 )2

R2112 =

60

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

In the final step, a number of calculations have been carried out. For the last component, we have i = j = 2 and

see the connection between concavity and the expression r′ z′′ − r′′ z′ . To do this we express the curve as the graph of a function r = R(z). This can be done near any point where the curve is not horizontal, i.e., at any point q where z′ (q) 6= 0. By the implicit function theorem, we can then solve the equation z = z(q) for q near q. Write the solution as q = Q(z) and then set

∂ Γ222 ∂ Γ221 − + Γ122Γ211 + Γ222Γ221 ∂ q1 ∂ q2 − Γ121 Γ212 − Γ221 Γ222 = 0.

R2212 =

For R1212 we find R1212 = g11 R1212 = r2 ·

r = R(z) = r(Q(z)). z′ (r′ z′′ − r′′ z′ ) rA

=

rz′ (r′ z′′ − r′′ z′ ) A

.

Because g = r2 [(r′ )2 + z′ )2 = r2 A, rz′ (r′ z′′ − r′′ z′ ) 1 R1212 = · 2 g A r A ′ ′ ′′ ′′ ′ ′ ′ ′′ z (r z − r z ) z (r z − r′′ z′ ) = = . 2 rA r((r′ )2 + (z′ )2 )2

K=

b) In the formula for K, we see K = 0 requires either z′ = 0 or r′ z′′ = r′′ z′ .

Concavity of the curve near q is then determined by the sign of R′′ (z). We have R′ (z) = r′ (Q(z)) · Q′ (z) =

r′ (Q(z)) . z′ (Q(z))

Here we use the fact that the inverse function z(q) and Q(z) have derivatives that are reciprocals of each other: Q′ (z) = 1/z′ (q) = 1/z′ (Q(z)). Therefore, R′′ (z) =

z′ r′′ − r′ z′′ z′ · r′′ Q′ − r′ · z′′ Q′ = . ′ 2 (z ) (z′ )3

c) Because r is assumed positive, we see K < 0 precisely For comparison with K = z′ (r′ z′′ − r′′ z′ )/rA2 , we write R′′ in the form when ′ ′ ′′ ′′ ′ z (r z − r z ) < 0. rA2 z′ (r′ z′′ − r′′ z′ ) = −K · . R′′ (z) = − We can relate this expression more closely to the shape of (z′ )4 (z′ )4 the generating curve c for the surface of revolution. The Because the factor rA2 /(z′ )4 is always positive, we see R′′ surface is given parametrically as is a negative multiple of K.
x(q1 , q2 ) = r(q2 ) cos q1 , r(q2 ) sin q1 , z(q2 ) . We can now confirm the our earlier observations. At those points on the generating curve (r(q), z(q)) where The parametric curve c(q) = (r(q), z(q)) in the vertical z′ (r′ z′′ − r′′ z′ ) < 0, K < 0 but R′′ > 0, so the curve is (r, z)-plane generates this surface when the plane is rotated concave away from the z-axis. At those points where around the z-axis. z′ (r′ z′′ − r′′ z′ ) > 0, K > 0 but R′′ < 0, so the curve is concave towards the z-axis. z z′ = 0

K0 R′′ = 0

K 0 where the curve is concave toward the axis (shown in black), and K < 0 where the curve is concave away from the axis (shown in gray). Concavity switches if either z has an extreme value (i.e., at a point where z′ = 0) or the curve has a regular inflection point. We need to

DVI file created at 16:06, 20 January 2011

6.2 Geodesics 1. According to the solution to Exercise 4, § 6.1 (Solutions page 56), all Γijk = 0. Therefore, the geodesic equations are d 2 q1 d 2 q2 = 0, = 0. 2 dt dt 2 The solutions are q1 (t) = α 1t + β 1 ,

q2 (t) = α 2t + β 2 ,

where α 1 , α 2 , β 1 , β 2 are arbitrary constants of integration. The solutions are are all constant-speed parametrized straight lines in the (q1 , q2 )-plane.

6.2. GEODESICS

61

2. a) This surface of revolution is a special case of the or dq2 1 surface analyzed in the solution to Exercise 9, § 6.1 (Solu=p . 2 2 ′ ′′ ′ ds tions page 59) for which z(q ) = q (so z = 1 and z = 0). (r )2 + 1 Therefore, the Christoffel symbols reduce to For the second derivative of q2 (s), we find (keeping in 1 mind that r′ = r′ (q2 (s))) Γ11 = 0, −rr′ , (r′ )2 + 1 r′ Γ112 = Γ121 = , r Γ212 = Γ221 = 0, Γ211 =

Γ122 = 0, Γ222 =

r′ r′′ . (r′ )2 + 1

In their turn, the geodesic equations reduce to r′ dq1 dq2 d 2 q1 + 2 = 0, dt 2 r dt dt  2  2 2 dq1 dq rr′ r′ r′′ d 2 q2 − ′ 2 + ′ 2 = 0. 2 dt (r ) + 1 dt (r ) + 1 dt b) Suppose q1 = k on the meridian; then, in the (q1 , q2 )plane, we can parametrize this meridian as

2
−3/2 −r′ r′′ d 2 q2 ′ 2 ′ ′′ dq 1 = − = . (r ) + 1 · 2r r 2 ds2 ds ((r′ )2 + 1)2

The geodesic equation for q2 now reads r′ r′′ + ′ 2 2 (r ) + 1 ((r′ )2 + 1) −r′ r′′

as desired.

= 0,

The only nonzero first or second derivative is dq1 /dt = 1. Therefore, the first geodesic equation is satisfied, and the second becomes 0− or just

dq1 d 2 q1 = 0, = dt dt 2 d 2 q2 dq2 = 0. = 1, dt dt 2

p (r′ )2 + 1

!2

c) Parametrize the parallel of latitude as (q1 , q2 ) = (t, k). We first note that this is a constant-speed parametrization, because  1 2 dqi dq j dq gi j = r2 (k) · 1 = constant. = g11 dt dt dt

(q1 , q2 ) = (k,t). Therefore,

1

rr′ (r′ )2 + 1

· 1 + 0 = 0,

r(k) r′ (k) = 0. (r′ (k))2 + 1

Because r depends only on q2 , and q2 = k, the left-hand side is a constant. Moreover, the only way this constant can equal zero is if r′ (k) = 0. Geometrically, this means These expressions do indeed satisfy the first geodesic that the generating curve r = r(z) (recall that z = q2 for the equation, but the second geodesic equation becomes surface of revolution) has either a maximum, a minimum, or an inflection with a vertical tangent. In other words, r′ r′′ 0−0+ ′ 2 = 0, a parallel of latitude is a geodesic if and only if the gen(r ) + 1 erating curve for the surface has an extreme or a vertical which is not true, in general. When the meridian is given inflection. a simple linear parametrization, it is not a geodesic. In fact, this is not even a constant-speed parametrization, as 3. a) According to the solution to Exercise 4 in § 6.1 (Soa geodesic must be (Theorem 6.2, text page 270). The lutions page 56), square of the parameter speed is f ′ f ′′ + g′g′′ Γ111 = ′ 2 , Γijk = 0 otherwise.   ′ )2 2 ( f ) + (g i j 2
dq dq dq gi j = g22 = (r′ )2 + 1 · 1 = (r′ )2 + 1. dt dt dt The geodesic equations are therefore 2  Suppose instead that the meridian is parametrized by arc f ′ f ′′ + g′g′′ dq1 d 2 q1 1 2 2 + = 0, length s, so that (q , q ) = (k, q (s)). Then dt 2 ( f ′ )2 + (g′)2 dt    2 2 d 2 q2
dq2 2 dq = 0. = (r′ )2 + 1 , 1 = g22 dt 2 ds ds

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

62

b) By assumption, q1 is arc length on the plane curve Because the geodesic equation for q2 is still c(q1 ) = ( f (q1 ), g(q1 )), so d 2 q2 ′ 2 ′ 2 ′ 2 = 0, kc k = ( f ) + (g ) = 1, ds2 and consequently the geodesics are the parametrized curves 2 f ′ f ′′ + 2g′g′′ = 0.

(q1 , q2 ) = (ϕ (α 1 s + β 1 ), α 2 s + β 2 )

This implies that the geodesic equations reduce to

that depend on the four arbitrary constants α 1 , α 2 , β 1 , β 2 .

d 2 q1 d 2 q2 = = 0, ds2 ds2 (where q1 = s) and the geodesics are therefore all constant-speed parametrized straight lines in the (q1 , q2 )plane. c) The arc-length parameter on c is now s(q1 ) =

Z q1

kc′ k dq1 =

Z q1 p

( f ′ )2 + (g′ )2 dq1 .

1 1 dq1 = ϕ ′ (s) = ′ 1 = √ . ′ 2 ds s (q ) ( f ) + (g′ )2 f ′ (ϕ (s))

Keeping in mind that = find the second derivative is

and

g′

= g′ (ϕ (s)),

we



−3/2 d 2 q1 = − 21 ( f ′ )2 + (g′)2 · 2 f ′ f ′′ + 2g′g′′ · ϕ ′ (s) ds2 f ′ f ′′ + g′ g′′ =− . (( f ′ )2 + (g′ )2 )2 Thus, when q1 = ϕ (s), the geodesic equation for q1 is satisfied, because

ϕ ′′ + =

f ′ f ′′ + g′ g′′ ′ 2 (ϕ ) ( f ′ )2 + (g′ )2

d 2 q1 f ′ f ′′ + g′g′′ + ′ 2 2 ds ( f ) + (g′)2

=−

f ′ f ′′ + g′ g′′ (( f ′ )2 + (g′ )2 )2

+



dq1 ds

2

f ′ f ′′ + g′g′′ ( f ′ )2 + (g′ )2

r + a cos q2 dq1 dq2 d 2 q1 − 2r = 0, 2 dt (a + r cosq2 )2 dt dt   dq1 2 d 2 q2  a 2 2 + cos q + cos2q = 0. dt 2 r dt To analyze the meridian q1 = k, note first that the parametrization of this meridian as (q1 , q2 ) = (k,t) is a constant-speed parametrization:

Let q1 = ϕ (s) be the inverse; then

f′

4. The solution to Exercise 4, § 6.1 (Solution page 58) provides the Christoffel symbols for the geodesic equations of the torus:

dqi dq j gi j = g22 dt dt



dq2 dt

2

= r2 · 1 = constant.

With this parametrization, d 2 q2 d 2 q1 dq1 = = 0; = 2 dt dt dt 2 therefore every term in each geodesic equation is zero. Hence, every meridian is a geodesic when parametrized by q2 . 5. a) The text, on pages 274–275, describes the steps of the solution.



1 √ ′ 2 ( f ) + (g′)2

2

b) Taking the plots below from the left side and from the top down, the value of α is, successively, –2, 0, 1, 2, 25. Π 2

= 0.

Π 4

If, more generally, q1 = ϕ (α 1 s + β 1 ), we have Π

-

dq1 = α 1ϕ ′, ds and thus we still have

d 2 q1 = (α 1 )2 ϕ ′′ , ds2

2  d 2 q1 f ′ f ′′ + g′g′′ dq1 + ′ 2 ds2 ( f ) + (g′)2 ds   f ′ f ′′ + g′g′′ ′ 2 1 2 ′′ (ϕ ) = 0. = (α ) ϕ + ′ 2 ( f ) + (g′ )2

DVI file created at 16:06, 20 January 2011

Π

Π

Π

4

4

2

2

Π

4

Π

2

c) Let q1 = q1 + β , q2 = q2 , where q1 and q2 are as given in part (a). Then qi has the same derivatives as qi , so q1 and q2 satisfies the same geodesic equations as q1 and q2 .

6.2. GEODESICS

63

Thus each Cα ,β is a geodesic. On the sphere, Cα is This implies a great circle that crosses the equator at q1 = 0, π . On α 2 + sec2 δ = α 2 + 1 + tan2 δ the sphere, Cα ,β is Cα after it has been rotated β radians = α 2 + 1 + (1 + α 2) tan2 (a1 − β ) around the polar axis so that it crosses the equator at q1 = β , π + β . Therefore, each Cα ,β is a great circle, and hence = (1 + α 2)(1 + tan2 (a1 − β )) is a geodesic. = (1 + α 2) sec2 (a1 − β ), d) If we set qβ (t) = (q1 (t), q2 (t)) = (β + q1 (t), q2 (t)), then the argument used in the previous part can just be repeated to show that qβ is a geodesic whenever q is.

and also sec2 δ = sec2 (a1 − β ) + α 2 sec2 (a1 − β ) − α 2

= sec2 (a1 − β ) + α 2 tan2 (a1 − β ) e) To meet the four given conditions, we need a four1 + tan2 a2 1 + α 2 sin2 (a1 − β ) parameter family of geodesics. In addition to α and β = . = cos2 (a1 − β ) cos2 (a1 − β ) already introduced, rescaling the arc-length variable as σ = γ s + δ provides the two additional parameters. Our The last equality relies on the second initial condition, four-parameter family of geodesics is rewritten as    α tan δ α tan(a1 − β ) 2 σ tan  √ = α sin(a1 − β ). tan a = = 1  , q = β + arctan √ 2 + sec2 δ sec(a1 − β ) α 2 1+α   σ = γs + δ . q(s) : α tan σ  Now assume b1 6= 0; we consider the case b1 = 0 later on. 2  , q = arctan √ The last two initial conditions then imply α 2 + sec2 σ √ α α 2 + sec2 δ b2 The derivatives can be determined from formulas in the √ = b1 text on page 274: 1 + α 2 sec2 δ √ α 1 + α 2 sec(a1 − β ) √  1 2 2 =√   dq = γ 1 + α sec σ , 1 + α 2(1 + tan2 a2 )/ cos2 (a1 − β )  2 2 ds α + sec σ ′ q (s) : α cos(a1 − β )  dq2 α = .   √ =γ . 1 + tan2 a2 2 2 ds α + sec σ The pair of equations In terms of these expressions, the four initial conditions α sin(a1 − β ) = tan a2 , are 2 1 2 2 b   , α cos(a − β ) = (1 + tan a ) tan δ b1 a1 = β + arctan √ , 1 + α2 allow us to write    2 2 α tan δ 2 2 2 2 2 2 b a2 = arctan √ , α = tan a + (1 + tan a ) , α 2 + sec2 δ b1 √ b1 tan a2 1 + α 2 sec2 δ . tan(a1 − β ) = 2 , b1 = γ 2 2 b (1 + tan2 a2 )2 α + sec δ α We have thus determined α and β in terms of the given b2 = γ √ . 2 α + sec2 δ values a1 , a2 , b1 , b2 as s  2 2 It is not immediately obvious how to solve these equations b 2 a2 + (1 + tan2 a2 )2 α = ± tan , for α , β , γ , δ . The first initial condition gives us b1   b1 tan a2 1 tan δ 1 . β = a − arctan 2 tan(a − β ) = √ b (1 + tan2 a2 )2 1 + α2 Note that α 6= 0. The sign for α is chosen so that the or equation α sin(a1 − β ) = tan a2 p 1 tan δ = tan(a − β ) 1 + α 2. will be satisfied. Using these expressions, we can now

DVI file created at 16:06, 20 January 2011

64

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

determine γ and δ : √ √ b2 α 2 + sec2 δ b2 sec(a1 − β ) 1 + α 2 γ= = , α α p δ = arctan tan(a1 − β ) 1 + α 2 .

The geodesic equations are therefore

If b1

d 2 q1 + sinh q1 cosh q1 dt 2



dq2 dt

2

= 0,

2 1 d 2 q2 1 dq dq + 2 tanhq = 0. dt 2 dt dt

= 0, then the geodesic is a vertical great circle that passes through the poles of the sphere. It is not obvious 1 2 2 that the family q(s) contain these geodesics. However, On the first set of coordinate curves, q = t, q = k , so the discussion in § 5.2 of the text suggests that we take dq1 dq2 d 2 q1 d 2 q2 α → ∞. In this case, in the formula for q1 we have = 0, = 0. = 1, = 0, dt dt 2 dt dt 2 tan σ √ → 0, The geodesic equations reduce to 1 + α2 so q1 = β . In the formula for q2 we have α tan σ tan σ √ =p → tan σ , 2 2 α + sec σ 1 + sec2 σ /α 2

so q2 = arctan(tan σ ) = σ and dq2 /ds = γ . Therefore,

α=

1 (= ∞), b1

β = a1 ,

γ = b2 ,

δ = a2 .

6. To investigate geodesics, we first need to determine the Christoffel symbols. Because we begin with the metric intrinsically defined—rather than with an embedding (such as X in § 5.3)—we’ll start the computations with   1 ∂ g jl ∂ glk ∂ g jk Γ jk,l = − . + 2 ∂ qk ∂qj ∂ ql

0 + sinht cosht · 0 = 0,

0 + 2 tanht · 1 · 0 = 0,

so all these coordinate curves are geodesics. On the second set of coordinate curves, q1 = k1 , q2 = t, so dq1 =, dt

d 2 q1 = 0, dt 2

dq2 = 1, dt

d 2 q2 = 0. dt 2

The geodesic equations reduce to 0 + sinhk1 cosh k1 · 1 6= 0,

0 + 2 tanhk1 · 0 · 1 = 0.

The first expression will be zero only if sinh k1 = 0, implying that only the coordinate curve for which k1 = 0 is a geodesic.

The only nonzero derivative is

∂ g22 = −2 cosh q1 sinh q1 , ∂ q1

6.3 Curved Spacetime

so the only nonzero Christoffel symbols will have pre- 1. The geodesic equations for the hyperbolic plane are cisely two indices equal to 2: d 2 q1 2 dq1 dq2 1 1 1 1 1 0 = − , Γ12,2 = Γ21,2 = · −2 sinhq cosh q = − sinh q cosh q , dt 2 q2 dt dt 2 2 2   1 d 2 q2 1 dq1 1 dq2 Γ22,1 = · sinh q1 cosh q1 = sinh q1 cosh q1 , 0= + 2 − 2 , 2 dt 2 q dt q dt Γ jk,l = 0 otherwise. (text, page 283). If q1 = c, q2 = t, then all derivatives To obtain the Christoffel symbols of the second kind, we are zero except dq2 /dt = 1. The first geodesic equation is use therefore satisfied, but the second is not: the term g11 = 1, g12 = g21 = 0, g22 = − cosh−2 q1 , 2  1 1 dq2 and find = − ·1 − 2 q dt t Γ122 = g11 Γ22,1 + g12Γ22,2 = sinh q1 cosh q1 , fails to vanish, but all others do vanish. If, instead, q2 = et , Γ212 = Γ221 = g21 Γ12,1 + g22Γ12,2 then dq2 /dt = d 2 q2 /d 2t = et . The first geodesic equation − sinh q1 cosh q1 1 is still satisfied. The second is now satisfied, too, because = tanh q = − cosh2 q1 it becomes 1 1 Γijk = 0 otherwise. et + t · 0 − t · (et )2 = 0. e e

DVI file created at 16:06, 20 January 2011

6.3. CURVED SPACETIME

65

2. For the curve q(t) we have dq1 = r sin t, dt

dq2 = r cost. dt

The square of the length of q′ (t) is dqi dq j dt dt  2 2  1 2 dq dq 1 2 1 2 + g22(q , q ) = g11 (q , q ) dt dt
1 = 2 2 r2 sin2 t + r2 cos2 t r cos t 1 . = cos2 t

gi j (q1 , q2 )

Therefore the speed kq′ (t)k = 1/ cost is non-constant, so q(t) cannot be a geodesic. 3. a) For q(s) we have q1 = c + r tanhs,

q2 = r sech s,

dq2 dq1 = r sech2 s, = −r sech s tanh s, ds ds d 2 q1 = −2r sech2 s tanh s, ds2
d 2 q2 . = −r sech s sech2 s − tanh2 s 2 ds

The right-hand side of the first geodesic equation (from the solution to Exercise 1, above) is 2 · r sech2 s · (−r sech s tanh s) r sech s = −2r sech2 s tanh s + 2r sech2 s tanh s = 0.

−2r sech2 s tanh s −

The right-hand side of the second geodesic equation is
−r sech s sech2 s − tanh2 s
1 + r2 sech4 s − r2 sech2 s tanh2 s = 0. r sech s The geodesic equations are indeed satisfied.

b) Because g11 = g22 = 1/(q2 )2 = 1/r2 sech2 s, the square of the length of q′ (s) is gi j (q1 , q2 )

dqi

dq j

ds ds  2 2  1 2 dq dq + g22(q1 , q2 ) = g11 (q1 , q2 ) ds ds
1 = 2 r2 sech4 s + r2 sech2 s tanh2 s 2 r sech s = sech2 s + tanh2 s ≡ 1.

DVI file created at 16:06, 20 January 2011

In other words, q(s) is a unit-speed parametrization, so the length of the portion of q(s) from s = s1 to s = s2 is just |s2 − s1 |. Because the parameter domain for q(s) is the entire s-axis, the values of s1 and s2 , and hence the length |s2 − s1 |, can be arbitrarily large. Thus the whole geodesic has infinite length. 4. The first and second derivatives of the components of qβ (t) are the same as the corresponding components of q(t). Since the geodesic equations depend only on these first and second derivatives, if q(t) satisfies the geodesic equations then so must qβ (t), for every β . 5. The solution to Exercise 5e in § 6.2 (see Solution page 63) noted that a four-parameter family of geodesics is needed to meet the four initial conditions. In the present case we can take ( q1 = c + r tanh σ , σ = γs + δ ; q(s) : q2 = r sech σ , the parameters are c, r, γ , and δ . The derivative is  1 dq    = rγ sech2 σ , ′ q (s) : ds2    dq = −rγ sech s tanh s, ds

and the four initial conditions become the equations a1 = c + r tanh δ ,

b1 = rγ sech2 δ ,

a2 = r sech δ ,

b2 = −rγ sech δ tanh δ .

Now assume b1 6= 0 (we consider b1 = 0 below); then tanh δ b2 = − sinh δ . =− b1 sech δ

Therefore, δ = − arcsinh(b2 /b1 ); standard identities for the hyperbolic functions then give 1 sech δ = p , 1 + (b2/b1 )2

−b2 /b1 tanh δ = p . 1 + (b2/b1 )2

we can now solve the remaining initial conditions: p r = a2 / sech δ = a2 1 + (b2/b1 )2 ,

c = a1 − r tanh δ = a1 − sinh δ = a1 + b2 /b1 , √ −b2 −b1 1 + (b2/b1 )2 . γ= 2 = a tanh δ a2

If b1 = 0 then the geodesic is vertical so we write the geodesic in the form (cf. the solution to Exercise 1, above) ( q1 = c, q(s) : σ = γs + δ . q 2 = eσ ,

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

66

(We need only three parameters because one condition, As noted in the solution to Exercise 6, §6.1, we need namely b1 = 0, has been fixed and satisfied by making q1 determine only Rh j12, 1 ≤ h, j ≤ 2. Using constant.) The three remaining initial conditions are 1 g11 = g22 = 2 2 , g12 = g21 = 0, a1 = c, a2 = eδ , b2 = γ eδ ; (q ) the solutions are we find c = a1 , δ = ln a2 , γ = b2 /a2 . R1112 = g11 R1112 + g21R2112 = 0, R1212 = g11 R1212 + g21R2212 1 1 1 = 2 2 ·− 2 2 = − 2 4, (q ) (q ) (q )

6. a) We use the Christoffel symbols given in the text on pages 282–283 and make use of the fact noted in the solution to Exercise 5 of § 6.1 (Solutions page 57) that we need determine only the four components Rij12 , 1 ≤ i, j ≤ 2. The nonzero Christoffel symbols have the values Γ112 = Γ121 = Γ222 =

−1 , q2

Γ211 =

R2112 = g12 R1112 + g22R2112 1 1 1 = 2 2 · 2 2 = 2 4, (q ) (q ) (q )

1 , q2

R2212 = g12 R1212 + g22R2212 = 0.

and the nonzero derivatives have the values 1 ∂ Γ112 ∂ Γ121 ∂ Γ222 = = = 2 2, ∂ q2 ∂ q2 ∂ q2 (q )

−1 ∂ Γ211 = 2 2. ∂ q2 (q )

b) We find immediately that

Therefore, when we take i = j = 1, we have R1112 =

∂ Γ112 ∂ Γ111 − + Γ112Γ111 + Γ212Γ121 ∂ q1 ∂ q2 − Γ111 Γ112 − Γ211 Γ122 = 0.

When i = 1, j = 2, we have

∂ Γ122 ∂ Γ121 − + Γ122 Γ111 + Γ222 Γ121 ∂ q1 ∂ q2 − Γ121Γ112 − Γ221 Γ122   2  1 −1 2 −1 = 0− 2 2 +0+ − −0 (q ) q2 q2 1 =− 2 2. (q )

R1212 =

When i = 2, j = 1, we have

∂ Γ212 ∂ Γ211 − + Γ112 Γ211 + Γ212 Γ221 ∂ q1 ∂ q2 − Γ111 Γ212 − Γ211 Γ222 −1 −1 1 1 −1 = 0− 2 2 + 2 · 2 +0−0− 2 · 2 (q ) q q q q 1 = 2 2. (q )

R2112 =

Finally, when i = j = 2 we have R2212 =

∂ Γ222 ∂ q1

= 0.

−

∂ Γ221 ∂ q2

+ Γ122Γ211 + Γ222Γ221

− Γ121 Γ212 − Γ221 Γ222

DVI file created at 16:06, 20 January 2011

K=

R1212 −1/(q2)4 ≡ −1. = g 1/(q2 )4

7. a) By definition, X∗i · X∗j = (Xi∗ )t GX j∗ . But we can reformulate this expression two ways (using Gt = G): (Xi∗ )t GX j∗ = (Xi∗ )t (λ j X j∗ ) = λ j (Xi∗ )t X j∗ , (Xi∗ )t GX j∗ = (GXi∗ )t X j∗ = λi (Xi∗ )t X j∗ . It follows that (λ j − λi )(Xi∗ )t X j∗ = 0. When i 6= j, we know λ j − λi 6= 0, so (Xi∗ )t X j∗ = 0, and hence X∗i · X∗j = (Xi∗ )t GX j∗ = λ j (Xi∗ )t X j∗ = 0. For any scalar multiples of these vectors, the dot product must still be zero; thus Xi · X j = 0 for every i 6= j. b) When i = j, we have X∗i · X∗i = (Xi∗ )t GXi∗ = λi (Xi∗ )t Xi∗ . Because the factor (Xi∗ )t Xi∗ is the square of the ordinary Euclidean length of X∗i in R4 , it is positive. Therefore, the rescaling factors for the vectors Xi are all real, and we have

λ0 (X0∗ )t X0∗ 1 X∗0 · X∗0 = = 1, ∗ t λ0 (X0 ) λ0 (X0∗ )t X0∗ λα (Xα∗ )t Xα∗ 1 Xα · Xα = X∗α · X∗0 = = −1, ∗ t ∗ −λα (Xα ) Xα −λα (Xα∗ )t Xα∗ X0 · X0 =

for α = 1, 2, 3.

6.4. MAPPINGS

67

c) Any four linearly independent vectors in R4 constitute • spherical coordinates   a basis. To show {Xi } are linearly independent, suppose sin ϕ cos θ r cos ϕ cos θ −r sin ϕ sin θ we have a sum ai Xi = 0 (summation convention). Then, dMP =  sin ϕ sin θ r cos ϕ sin θ r sin ϕ cos θ  for each k = 0, 1, 2, 3, we have cos ϕ −r sin ϕ 0
0 = Xk · ai Xi = ai (Xk · Xi ) = ±ak . The determinant of dMP is r2 sin ϕ ; the matrix of cofactors In other words, a0 = a1 = a2 = a3 = 0, so the vectors are of dMP is  2 2  linearly independent and hence are a basis. r sin ϕ cos θ r2 sin2 ϕ sin θ r2 sin ϕ cos ϕ −r sin2 ϕ  ; d) Let us write Y = yi Xi (summation convention). Then r sin ϕ cos ϕ cos θ r sin ϕ cos ϕ sin θ −r sin θ r cos θ 0

Y · Y = y i Xi · y j X j = y i y j Xi · X j to get the inverse, divide by det dMP : = (y0 )2 X0 · X0 + (y1 )2 X1 · X1 + (y2 )2 X2 · X2   sin ϕ cos θ sin ϕ sin θ cos ϕ + (y3 )2 X3 · X3 (dMP )−1 =  cos ϕ cos θ /r cos ϕ sin θ /r − sin ϕ /r = (y0 )2 − (y1 )2 − (y2 )2 − (y3 )2 . − sin θ /r sin ϕ cos θ /r sin ϕ 0

6.4 Mappings

c) • polar coordinates We have

−y −y 1 ∂θ · = 2 , = ∂x 1 + (y/x)2 x2 x + y2 ∂θ 1 1 x · = 2 ; = 2 ∂y 1 + (y/x) x x + y2

1. The element in the kth row and ith column of dMP is ∂ f k /∂ ξ i . Given that f k (ξ j ) = aki ξ j , we have

∂ fk ∂ξ j j = akj i = akj δi = aki , i ∂ξ ∂ξ

therefore −1

! p p x/ x2 + y2 y/ x2 + y2

which is the element in the kth row and ith element of M. . d(M )Q = j −y/x2 + y2 x/x2 + y2 (The tensor δi is the Kronecker delta; it has the value 1 p when i = j and 0 otherwise; see the proof of ProposiNow set x2 + y2 = r, x = r cos θ , y = r sin θ ; then tion 6.3, text page 297.)   cos θ sin θ −1 d(M ) = = (dMP )−1 , Q 2. a) • polar coordinates The inverse is − sin θ /r cos θ /r ( p r = x2 + y2 , where Q = M(P). −1 M : θ = arctan(y/x). • spherical coordinates For compactness, we write •pspherical coordinates We have tan θ = y/x and tan ϕ = x2 + y2/z; hence  p 2 2 2  r = x + yp+ z , −1 2 M : ϕ = arctan( x + y2/z),   θ = arctan(y/x).

b) • polar coordinates   cos θ −r sin θ dMP = sin θ r cos θ

The determinant of dMP is r, and   sin θ cos θ . (dMP )−1 = − sin θ /r cos θ /r

DVI file created at 16:06, 20 January 2011

x x ∂r = , =p 2 2 2 ∂x r x +y +z

∂r y ∂r z = , = ; ∂y r ∂z r p for the derivatives of ϕ , we set ρ = x2 + y2 to obtain 1 ∂ϕ = 2 ∂x 1 + (x + y2 )/z2 ∂ϕ yz = 2, ∂y ρr 1 ∂ϕ = ∂z 1 + (x2 + y2 )/z2

·

x xz xz = 2, = 2 2 2 zρ ρ (z + x + y ) ρ r

·

−ρ −ρ = 2 . z2 r

With these identifications, we have   x/r y/r z/r
d M −1 Q = xz/ρ r2 yz/ρ r2 −ρ /r2  . −y/ρ 2 x/ρ 2 0

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

68

Using the map M to replace x, y, and z by their spherical Rotating by θ radians is accomplished by coordinate values, and noting that      r˙ cos θ − sin θ r˙ cos θ − θ˙ sin θ = . sin θ cos θ θ˙ r˙ sin θ + θ˙ cos θ x2 + y2 = r2 sin2 ϕ or ρ = r sin ϕ , we find d M −1





sin ϕ sin θ cos ϕ sin ϕ cos θ  cos ϕ cos θ /r cos ϕ sin θ /r − sin ϕ /r = Q − sin θ /r sin ϕ cos θ /r sin ϕ 0




= (dMP )−1 .

The composition of these actions is the product      cos θ − sin θ 1 0 cos θ −r sin θ = dMP , = 0 r sin θ cos θ sin θ r cos θ thus proving the claim about dMP .

5. a) direct calculation gives  αζ  3. a) With P = (r, θ ) = (2, π /3), the map dMP (X) from e cosh ατ eαζ sinh ατ dMP = αζ the solution to the previous exercise is e sinh ατ eαζ cosh ατ   √    αζ cosh ατ sinh ατ r˙ 1/2 − 3 = eαζ Hατ . = e √ dMP (X) = sinh ατ cosh ατ θ˙ 3/2 1 By factoring dMP (X) as √     r˙ 1 0 √1/2 − 3/2 , 0 2 θ˙ 3/2 1/2

b) We have det dMP = e2αζ because detHu = 1 for every u. Moreover, Hu−1 = H−u for every u, so  −1 −1 (dMP )−1 = eαζ Hαζ = e−αζ H−αζ . = e−αζ Hαζ

we see that (reading from right to left) dMP acts on the (˙r, θ˙ )-plane by first doubling θ˙ and then rotating the result c) The four partial derivatives in d(M −1 ) are Q by π /3 radians. ∂τ 1 1 1 z b) The four images are shown below. In the first row, · = = · , 2 z 2 − t 2) −1 −2 −3 ∂ t α 1 − (t/z) α (z the mesh sizes are 10 and 10 ; in the second, 10 1 −t −t 1 ∂τ and 10−4 . The last image certainly looks linear, with the · = = · , action of doubling in the second coordinate direction, fol∂z α 1 − (t/z)2 z2 α (z2 − t 2) lowed by rotation by 60◦ . 1 −t ∂ζ −2t α 2 = , · 2 2 2 = ∂t 2α α (z − t ) α (z2 − t 2 )

∂ζ 2zα 2 1 z · 2 2 2 = = ; ∂z 2α α (z − t ) α (z2 − t 2 )

hence  −t z  α (z2 − t 2) α (z2 − t 2 )  d(M −1 )Q =   z −t 2 2 2 2 α (z − t ) α (z − t ) 

Using the definitions of t and z in M, and noting that z2 − t 2 =

e2αζ , α2

we find 4. Stretching the (˙r , θ˙ )-plane by the factor r in the θ˙ is accomplished by      r˙ r˙ 1 0 = . ˙ 0 r rθ˙ θ

DVI file created at 16:06, 20 January 2011


eαζ /α · cosh αζ z = = e−αζ cosh αζ , α (z2 − t 2) α · e2αζ /α 2
eαζ /α · − sinh αζ −t = = −e−αζ sinh αζ . α (z2 − t 2) α · e2αζ /α 2

6.4. MAPPINGS Now − sinh(u) = sinh(−u) and cosh(u) = cosh(−u), so  
cosh(−αζ ) sinh(−αζ ) d M −1 Q = e−αζ sinh(−αζ ) cosh(−αζ ) = e−αζ H−αζ = (dMP )−1 .

6. a) With α = 0.1 and P = (τ , ζ ) = (−2, 1),   cosh(0.2), − sinh(0.2) dMP = e−0.2 H−0.1 = e0.1 − sinh(0.2) cosh(0.2)   1.13 −0.22 = . −0.22 1.13

69 8. a) With the standard Minkowski metric, the hyperbolic angle uJ between two vectors ξ and η satisfies

ξ t J1,1 η cosh uJ = p , p ±ξ t J1,1 ξ ±η t J1,1 η

where the signs are chosen to make the argument within each square root nonnegative. With the new metric ΓP = e2αζ J1,1 , the hyperbolic angle uΓ satisfies

ξ t ΓP η cosh uΓ = p √ ±ξ t ΓP ξ ±η t ΓP η

e2αζ ξ t J1,1 η p =p ±e2αζ ξ t J1,1 ξ ±e2αζ η t J1,1 η e2αζ ξ t J1,1 η p = = cosh uJ , p (eαζ )2 ±ξ t J1,1 ξ ±η t J1,1 η

where the signs are chosen the same way. Because the hyperbolic angle between two vectors is determined only up to sign (the angle from ξ to η is the negative of the angle from η to ξ ), we see uΓ = ±uJ . The angles are b) The four images are shown below. In the first row, equal because Γ = λ J , where λ = λ (P) = e2αζ > 0. P 1,1 the mesh sizes are 10−1 and 10−2; in the second, 10−3 and 10−4 . Even the first image resembles the image of the b) The quadratic forms associated with the two metrics linear map, above; the match is virtually complete in The satisfy last image. QΓ (ξ ) = ξ t ΓP ξ = e2αζ ξ t J1,1 ξ = e2αζ QJ (ξ ). A vector in lightlike in a given metric if its value on the quadratic form is zero; thus if ξ is lightlike in ΓP it is lightlike in J1,1 , and conversely. 9. The contravariant metric tensor is !   11 12   −2αζ 2 0 e−2αξ γ γ 0 e = . = 2 γ 21 γ 22 0 −e−2αζ 0 −e−2αξ To determine the Christoffel symbols of the first kind, note that the only nonzero partial derivatives of the γi j are 2 ∂ γ11 = 2α e2αξ , 2 ∂ξ

2 ∂ γ22 = −2α e2αξ ; 2 ∂ξ

therefore, as we noted earlier in the solution to Exercise 9a in §6.1 (Solutions page 59) either all three indices of Γi j,k have the value 2 or precisely one index does. That is, t = 7. Because dMP = eαζ Hατ , we have dMPt = eαζ Hατ αζ e Hατ as well. Then

Γ11,1 = Γ12,2 = Γ21,2 = Γ22,1 = 0,

and in the following only the nonzero terms are written:   ∂ γ11 2 1 t 2αζ 2αζ ΓP = (e )Hαζ J1,1 Hατ = e J1,1 , Γ11,2 = − 2 = −α e2αξ , 2 ∂ξ   because hyperbolic rotation preserves J1,1 (the Minkowski 2 1 ∂ γ11 Γ12,1 = Γ21,1 = = α e2αξ , metric). That is, 2 2 ∂ξ !      2αζ  2 2 αξ 2 1 ∂ γ22 e 0 e γ11 γ12 0 = −α e2αξ . Γ22,2 = Γ= = . = 2 2 αζ 2 2 αξ 2 ∂ξ γ21 γ22 0 −e 0 −e

DVI file created at 16:06, 20 January 2011

70

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

Because γ 12 = γ 21 = 0, the Christoffel symbols of the second kind follow the same pattern:

10. a) When we use the Γijk obtained in the previous exercise, the geodesic equations

Γ111 = Γ212 = Γ221 = Γ122 = 0,

j k d2ξ i i dξ dξ + Γ =0 jk du2 du du

while = γ Γ11,2 = −e

−2αξ 2

Γ222 = γ 22 Γ22,2 = −e

−2αξ 2

Γ211 Γ112

= Γ121

22

= γ Γ12,1 = e 11

−2αξ 2

· − αe

· αe

2αξ 2

2αξ 2

= α,

· − α e2αξ

2





reduce to = α, = α.

d2ξ 1 dξ 1 dξ 2 + 2α = 0, 2 du du du  1 2  2 2 d2ξ 2 dξ dξ + + = 0. α α 2 du du du

Therefore, the partial derivatives of all Γklj are zero. In terms of the original variables τ = ξ 1 , ζ = ξ 2 , these As mentioned in the solution to Exercise 9, § 6.1, the equations are only components of the mixed Riemann tensor we need d 2τ dτ dζ to find are Rij12 , 1 ≤ i, j ≤ 2. We now work these out in + 2α = 0, 2 detail. For i = j = 1, we have du du du    2 dτ 2 dζ d2ζ 1 1 ∂ Γ12 ∂ Γ11 +α = 0. +α 1 1 2 1 1 2 − + Γ12Γ11 + Γ12Γ21 R112 = du du du ∂ξ1 ∂ξ2 − Γ111 Γ112 − Γ211 Γ122 = 0.

b) The geodesic equations depend only on the first and second derivatives of ξ 1 (u) and ξ 2 (u). Translates leave derivatives unchanged, so translates of geodesics are still geodesics.

For i = 1, j = 2, we have

∂ Γ122 ∂ Γ121 − + Γ122 Γ111 + Γ222Γ121 ∂ξ1 ∂ξ2

R1212 =

− Γ121 Γ112 − Γ221 Γ122

= Γ222 Γ121 − Γ112Γ121

= 0.

For i = 2, j = 1, we have R2112 =

∂ Γ212 ∂ Γ211 − + Γ112 Γ211 + Γ212 Γ221 ∂ q1 ∂ q2 − Γ111Γ212 − Γ211Γ222 = Γ112 Γ211 − Γ211 Γ222 = 0

For i = j = 2, we have R2212 =

∂ Γ222 ∂ q1

−

∂ Γ221 ∂ q2

+ Γ122Γ211 + Γ222Γ221

− Γ121 Γ212 − Γ221 Γ222 = 0. For the covariant Riemann tensor, we have R1112 = γ11 R1112 + γ21R2112 = 0, R1212 = γ11 R1212 + γ21R2212 R2112 = γ12 R1112 + γ22R2112 R2212 = γ12 R1212 + γ22R2212

= 0, = 0, = 0.

Finally, because γ = det ΓP = −e4αξ , we find 2

K=

0 R1212 = 2 = 0. γ −e4αξ

DVI file created at 16:06, 20 January 2011

c) The metric γi j (P) in G is just the metric J1,1 on R carried back to the tangent plane T GP by the differential dMP . Thus all metric properties of G come from those in R. In particular, the image of geodesic in G must be a geodesic in R. 11. a) In terms of τ and ζ , the curve ξ ti is defined by

τ=

1 tanh−1 (α u), α

ζ=

1 ln(1 − α 2 u2 ); 2α

note that the formulas require |α u| < 1, or |u| < 1/|α |. The derivatives are 1 dτ , = du 1 − α 2u2 d2τ 2α 2 u = , 2 du (1 − α 2u2 )2

dζ −α u , = du 1 − α 2 u2 −α (1 + α 2u2 ) d 2ζ = . 2 du (1 − α 2u2 )2

In these terms, the first geodesic equation is −α u 1 2α 2 u + 2α · · = 0. (1 − α 2u2 )2 1 − α 2 u2 1 − α 2 u2 The second is −α (1 + α 2u2 ) 1 α 2 u2 +α · +α · = 0, 2 2 2 2 2 2 (1 − α u ) (1 − α u ) (1 − α 2u2 )2 so ξ ti is a geodesic.

6.4. MAPPINGS

71

Because e2αξ > 0, it is sufficient to check the sign of the simpler quadratic form (ξ˙ 1 )2 − (ξ˙ 2 )2 . For ξ ti , the tangent   1 1 1 vector is τ = tanh−1 ln(α 2 u2 − 1), , ζ=     α αu 2α dτ dζ 1 −α u ′ ξ ti (u) = = , , , du du 1 − α 2 u2 1 − α 2 u2 and now |α u| > 1, or |u| > 1/|α |. The domain therefore has two separate parts and ξ sp consists of two separate, so the simple quadratic form has the value disconnected curves; see the figure below. The derivatives 1 α 2 u2 1 are − = > 0. (1 − α 2u2 )2 (1 − α 2 u2 )2 1 − α 2 u2 dζ αu −1 dτ , , = = 2 2 The inequality holds because |α u| < 1 on the domain of du α 2 u2 − 1 du α u −1 ξ 2 2 2 2 2 ti (u); thus ξ ti is timelike. d τ 2α u −α (1 + α u ) d ζ = = , . For ξ sp , the tangent vector is du2 (α 2 u2 − 1)2 du2 (α 2 u2 − 1)2     −1 dτ dζ αu = , , The first geodesic equation is du du α 2 u2 − 1 α 2 u2 − 1 The curve ξ sp is defined by

2α 2 u

(α 2 u2 − 1)2

+ 2α ·

2

αu −1 · = 0. α 2 u2 − 1 α 2 u2 − 1

and the simple quadratic form has the value 1 (α 2 u2 − 1)2

The second is 1 −α (1 + α 2u2 ) α 2 u2 +α · 2 2 +α · 2 2 = 0, 2 2 2 2 (α u − 1) (α u − 1) (α u − 1)2 so ξ sp is a geodesic. In the figure on the left, below, ξ ti is the curve that passes through the origin. Above it is a vertical translate, and down to the left is a curve that has been translated both vertically and horizontally. For all three, α = 0.3. In the figure on the right, the curve with positive slope in the first and fourth quadrants is ξ sp when α = −0.3. For the other four, α = +0.3. The three with negative slope are ξ sp , a translation vertically downward, and a translation horizontally to the left. For these five, the parameter u is positive. The last curve, with positive slope, is the portion of ξ sp for which the u parameter is negative.

-4

4

4

2

2

2

-2

4

-4

2

-2

-2

-2

-4

-4

4

−

α 2 u2 (α 2 u2 − 1)2

=

−1 < 0. α 2 u2 − 1

This inequality holds because |α u| > 1 on the domain of ξ sp (u), so ξ sp is indeed spacelike. Proper time T on ξ ti is given by the integral T=

Z u 0

kξ ′ti kdu

where q kξ ′ti k = Q(ξ ′ti ) =

s

e2αξ . 1 − α 2 u2 2

On ξ ti , we know ξ 2 = ζ = ln(1 − α 2u2 )/2α , so e2αξ = 1 − α 2u2 2

and thus kξ ′ti k ≡ 1. It follows that u itself is a proper-time parameter on ξ ti . c) By what was explained in the solution to Exercise 10c, above, the timelike (resp. spacelike) geodesics in G should correspond to timelike (resp. spacelike) geodesics in R. To prove this, we map arbitrary translates of ξ ti and ξ sp using M. Along the way, we need to simplify expressions of the form

sinh tanh−1 x and cosh tanh−1 x .

−1 So x; that is, x = tanh y. Then sech y = p suppose y = tanh √ 2 b) A curve is timelike or spacelike at a point if its tangent 1 − tanh y = 1 − x2 and hence vector at that point is. By definition, a vector ξ = (ξ˙ 1 , ξ˙ 2 )
1 1 in T GP is timelike or spacelike depending on the sign of =√ cosh tanh−1 x = cosh y = , sech y 1 − x2 the quadratic form
x tanh y . =√ sinh tanh−1 x = sinh y =
i˙j 2αξ 2 1 2 2 2 ˙ ˙ ˙ sech y 1 − x2 QP (ξ ) = γi j ξ ξ = e (ξ ) − (ξ ) .

DVI file created at 16:06, 20 January 2011

72

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

Therefore, the image M((p, q) + ξ (u)) in the (t, z)-plane (where ξ is either a timelike or a spacelike geodesic) is given by

consequently,

eα (q+ζ ) t= sinh(α (p + τ )) α α q e eαζ (sinh(α p) cosh(ατ ) + cosh(α p) sinh(ατ )) , = α eα (q+ζ ) z= cosh(α (p + τ )) α α q e = eαζ (cosh(α p) cosh(ατ ) + sinh(α p) sinh(ατ )) . α When ξ = ξ ti , we know p eαζ = 1 − α 2u2 ,

ατ = tanh−1 (α u),

1 cosh(ατ ) = √ , 1 − α 2 u2 αu sinh(ατ ) = √ ; 1 − α 2 u2

consequently, eαζ cosh(ατ ) = 1,

eαζ sinh(ατ ) = α u,

eαζ sinh(ατ ) = ±1,

eαζ cosh(ατ ) = ±α u,

and the minus sign is chosen in both formulas precisely when α u is negative. Finally t = ±(α Au + B),

z = ±(α Bu + A).

The image is again a straight line with a linear parameter, and thus is a geodesic in R. Its slope is B/A, so it is spacelike. 12. a) We begin by parametrizing the coordinate line as ξ (u) = (ξ 1 , ξ 2 ) = (k, u). Then ξ ′ (u) = (0, 1) and 2 dξ i dξ j = γ22 · 12 = −e2αξ = −e2α u . du du p The curve is spacelike, so kξ ′ k = −Q(ξ ′ ) = eα u and its arc-length function is Z u eα u u eα u − 1 αu . = e du = s = s(u) = α 0 α 0

Q(ξ ′ ) = γi j

The inverse function is

1 ln(α s + 1), α t = A + α Bu, z = B + α Au, and thus η (s) = ξ (u(s)) = (k, ln(α s + 1)/α ). We verify where that kη ′ (s)k ≡ 1 by computing α q α q 2  2 2  e e dη 1 A= sinh(α p), B = cosh(α p). 2α u(s) ′ = −e Q(η ) = γ22 · α α ds αs + 1  2 The image is thus a straight line with a linear parameter; 1 it is a geodesic in R. Moreover, its slope is A/B and we = −(α s + 1)2 ≡ −1. αs + 1 have A = | tanh α p| < 1, b) In deciding whether η (s) is a geodesic, note that the B only nonzero derivatives involved are so the geodesic is timelike. −α 1 d2η 2 dη 2 For the images of ξ sp and its translates, we need simple = . = and 2 ds αs + 1 ds (α s + 1)2 expressions for

The first geodesic equation (from the solution to Exersinh tanh−1 (1/x) and cosh tanh−1 (1/x) . cise 10a) is immediately satisfied. The second equation is −1 2  If we set y = tanh p √ (1/x), then tanh y = 1/x, sech y = 1 −α 2 2 1 − (1/x) = x − 1/|x|, and so = 0; +α (α s + 1)2 αs + 1
1 |x| therefore η (s) is a geodesic. , cosh tanh−1 (1/x) = cosh y = =√ sech y x2 − 1 13. No parametrization of the horizontal line ζ = k will
tanh y |x|/x −1 turn it into a geodesic. The reason is that there is a unique sinh tanh (1/x) = sinh y = =√ . sech y x2 − 1 geodesic passing through a given point with a given velocity. Take the point to be (τ , ζ ) = (0, k) and take the velocWhen ξ = ξ sp , we know ατ = tanh−1 (1/α u) and ity to be (0, 1). The line ζ = k passes through this point; p because it is horizontal, it can be parametrized so its veln(α 2 u2 −1)/2 αζ = α 2 u2 − 1, e =e locity there is (0, 1). However, the curve (0, k) + ξ ti (u) is |α u|/α u |α u| the unique geodesic that satisfies these initial conditions, , cosh(ατ ) = √ ; sinh(ατ ) = √ 2 2 α u −1 α 2 u2 − 1 but this curve is not a horizontal line. u = u(s) =

and

DVI file created at 16:06, 20 January 2011

6.5. TENSORS

73

6.5 Tensors

∂ Γijl

∂ Γijk

+ Γ pjl Γipk − Γ pjk Γipl is a ∂ xk ∂ xl (1, 3) tensor by assembling the transformed version in the R G I,i a 1. We are given that T K,k L,l ↔ T J, j is a tensor of type (p, q). Greek frame (with variables like ξ ) of each term in the i Then, by definition, Roman frame (with variables like x ). In the process, nontensorial terms appear; we show that they cancel. To help R K,k G I,i ∂ xK ∂ ξ J ∂ xk ∂ ξ j clarify the process, we break it into a number of smaller T L,l = T J, j . ∂ ξ I ∂ xL ∂ ξ i ∂ xl steps. Step 1. The key to eliminating the nontensorial terms is To raise the covariant index l ↔ j, we multiply this tensor the identity by glh ↔ γ pm :  i ∂ ∂x ∂ 2 ξ g ∂ x p ∂ xi = − , l h R R R ∂ xk ∂ ξ a ∂ xk ∂ x p ∂ ξ a ∂ ξ g K,k lh K,k pm ∂ x ∂ x T T K,kh = g T = γ L L,l ∂ ξ p ∂ ξ m L,l which, in effect, “inverts” the more familiar expression l h G K J k j I,i ∂ x ∂ ξ ∂ x ∂ ξ  i pm ∂ x ∂ x =γ T ∂ 2 xi ∂ ξ f ∂ ∂x ∂ ξ p ∂ ξ m J, j ∂ ξ I ∂ xL ∂ ξ i |{z} ∂ xl . = k |{z} ∂ ξ f ∂ ξ a ∂ xk ∂x ∂ξa ∂ xh G ∂ xK ∂ ξ J ∂ xk To prove the new identity, we begin with the basic fact = γ pm δ pj m T I,i J, j ∂ξ ∂ ξ I ∂ xL ∂ ξ i δ pi = ∂ xi /∂ ξ a · ∂ ξ a /∂ x p and differentiate: G I,i ∂ xK ∂ ξ J ∂ xk ∂ xh  i   i a = γ jm T J, j ∂ ∂x ∂ξa ∂ ∂x ∂ξ ∂ xi ∂ 2 ξ a ∂ ξ I ∂ xL ∂ ξ i ∂ ξ m = + . 0= k ∂ x ∂ ξ a ∂ xp ∂ xk ∂ ξ a ∂ x p ∂ ξ a ∂ xk ∂ x p G I,im ∂ xK ∂ ξ J ∂ xk ∂ xh . = TJ Now replace the dummy index a 7→ g in the second term ∂ ξ I ∂ xL ∂ ξ i ∂ ξ m on the right, and then solve for the first term. The result is R K,kh G I,im  i This demonstrates that T L ↔ T J is a tensor of type ∂ ∂x ∂ xi ∂ 2 ξ g ∂ x p =− g k p · a. (p + 1, q − 1). The demonstration uses k a ∂x ∂ξ ∂ξ ∂x ∂x ∂ξ

∂ xl ∂ ξ j = δ pj , ∂ ξ p ∂ xl

2. We show Rijkl =

γ pm δ pj = γ jm .

Step 2. Transform

∂ Γijl ∂ xk

−

. The starting point is

b d i ∂ 2 ξ f ∂ xi a ∂ξ ∂ξ ∂x To lower the contravariant index k ↔ i, first note that the + Γijl = Γbd j ∂ x ∂ xl ∂ ξ a ∂ x j ∂ xl ∂ ξ f multi-indices I, J, K, L were not involved in the previous demonstration, and will not be involved in this one, either. (Corollary 6.2, text page 314). We use an “overline” to For clarity, then, we shall write the tensors without these denote a Christoffel symbol in the Greek frame. We have indices. Thus, we begin with a 2 b ∂ Γijl ∂ Γbd ∂ ξ c ∂ ξ b ∂ ξ d ∂ xi ∂ ξ d ∂ xi a ∂ ξ = + Γ R G ∂ xk ∂ ξ j bd ∂ xk ∂ ξ c ∂ xk ∂ x j ∂ xl ∂ ξ a ∂ xk ∂ x j ∂ xl ∂ ξ a T kl = T ij ,  i ∂ ξ i ∂ xl b 2 d b d i ∂ ξ ∂x ∂ ∂x a ∂ξ a ∂ξ ∂ξ + Γbd j k l · k + Γbd j l a ∂x ∂x ∂x ∂ξ ∂x ∂x ∂x ∂ξa and multiply by gkh ↔ γqm :  i 3 f i 2 f ∂ ξ ∂x ∂ ξ ∂ ∂x q m k j · + + G R R ∂ξ ∂ξ k j l j l k f i ∂x ∂ξ k ∂x ∂x ∂x ∂ξ ∂x ∂x ∂x ∂ξ f Tj , T lh = gkh T l = γqm k i ∂ xl h ∂ ξ ∂ x ∂ x a |{z} 2 b ∂ ξ d ∂ xi |{z} ∂ Γbd ∂ ξ c ∂ ξ b ∂ ξ d ∂ xi a ∂ ξ = + Γbd k j m j G j c a k l ∂ξ ∂ξ ∂ξ ∂x ∂x ∂x ∂ξ ∂ x ∂ x ∂ xl ∂ ξ a = γqm δiq h T ij l b ∂ 2 ξ d ∂ xi b d ∂ 2 ξ g ∂ x p ∂ xi ∂x ∂x a ∂ξ a ∂ξ ∂ξ j ∂ξm − Γbd j + Γbd j k l G ∂ξ j ∂ξm G ∂ ξ l a ∂x ∂x ∂x ∂ξ ∂ x ∂ x ∂ xk ∂ x p ∂ ξ a ∂ ξ g = T jm l . = γim T ij l h h ∂x ∂x ∂x ∂x ∂ 3ξ f ∂ xi ∂ 2 ξ f ∂ 2 ξ g ∂ x p ∂ xi + k j l − , R G ∂ x ∂ x ∂ x ∂ ξ f ∂ x j ∂ xl ∂ xk ∂ x p ∂ ξ f ∂ ξ g This demonstrates that T lh ↔ T jm is a tensor and is of using Step 1 twice. type (p − 1, q + 1).

DVI file created at 16:06, 20 January 2011

74

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

∂ Γijk . Step 3. Transform ∂ xl We can do this by copying Step 2 with the index substitutions k ↔ l, c ↔ d.

The first and third terms on the right simplify, owing to the factors

∂ Γijk

a

∂ xp ∂ ξ h = δeh ∂ ξ e ∂ xp

∂ ξ h ∂ xp = δ fh ; ∂ xp ∂ ξ f

and

2 b ∂ ξ c ∂ xi ∂ Γbc ∂ ξ d ∂ ξ b ∂ ξ c ∂ xi a ∂ ξ Γ + bc the result is ∂ xl ∂ ξ d ∂ xl ∂ x j ∂ xk ∂ ξ a ∂ xl ∂ x j ∂ xk ∂ ξ a b d c i b ∂ 2 ξ c ∂ xi b c ∂ 2 ξ g ∂ x p ∂ xi e a ∂ξ ∂ξ ∂ξ ∂x a ∂ξ a ∂ξ ∂ξ Γ pjl Γipk = Γbd Γec j − Γ + Γbc j l k bc ∂ x ∂ xl ∂ xk ∂ ξ a ∂x ∂x ∂x ∂ξa ∂ x j ∂ xk ∂ xl ∂ x p ∂ ξ a ∂ ξ g b d p ∂ 2 ξ g ∂ xi 3 f i 2 f 2 g p i e ∂ξ ∂ξ ∂x ∂ ξ ∂x ∂ ξ ∂ ξ ∂x ∂x + Γbd j + l j k − . ∂ x ∂ xl ∂ ξ e ∂ x p ∂ xk ∂ ξ g ∂ x ∂ x ∂ x ∂ ξ f ∂ x j ∂ xk ∂ xl ∂ x p ∂ ξ f ∂ ξ g i c 2 f a ∂ξ ∂x ∂ ξ ∂ Γijl ∂ Γijk + Γfc k − . Step 4. Compute the difference ∂ x ∂ ξ a ∂ x j ∂ xl ∂ xk ∂ xl 2 ∂ ξ f ∂ 2 ξ g ∂ x p ∂ xi In the third and fifth terms on the right in steps 2 + . ∂ x j ∂ xl ∂ x p ∂ xk ∂ ξ f ∂ ξ g and 3, the variables xk and xl appear symmetrically; therefore, those terms vanish when the difference is taken (and p Step 6. Transform Γ jk Γipl . dummy indices are adjusted): We can do this by copying Step 5 with the index substi! a a ∂ Γijl ∂ Γijk ∂ Γbd ∂ Γbc ∂ ξ b ∂ ξ c ∂ ξ d ∂ xi tutions k ↔ l, c ↔ d. The result is − = − ∂ xk ∂ xl ∂ξc ∂ ξ d ∂ x j ∂ xk ∂ xl ∂ ξ a i b c d e a ∂ξ ∂ξ ∂ξ ∂x p i Γ Γ = Γ Γ i i 2 b d 2 b c bc ed pl jk ∂ξ ∂x ∂ξ ∂x a ∂ ξ a ∂ ξ ∂ x j ∂ xk ∂ xl ∂ ξ a + Γbd k j − Γbc l j k ∂ x ∂ x ∂ xl ∂ ξ a ∂x ∂x ∂x ∂ξa b c p ∂ 2 ξ g ∂ xi e ∂ξ ∂ξ ∂x Γ + p i b d 2 g bc ∂ ξ ∂x ∂x a ∂ξ ∂ξ ∂ x j ∂ xk ∂ ξ e ∂ x p ∂ xl ∂ ξ g − Γbd j ∂ x ∂ xl ∂ xk ∂ x p ∂ ξ a ∂ ξ g d 2 e i a ∂ξ ∂x ∂ ξ + Γ p i b c 2 g ed ∂ ξ ∂x ∂x a ∂ξ ∂ξ ∂ xl ∂ ξ a ∂ x j ∂ xk + Γbc j ∂ x ∂ xk ∂ xl ∂ x p ∂ ξ a ∂ ξ g 2 ∂ ξ f ∂ 2 ξ g ∂ x p ∂ xi + . 2 f 2 g p i 2 f 2 g p i ∂ ξ ∂ ξ ∂x ∂x ∂ ξ ∂ ξ ∂x ∂x ∂ x j ∂ xk ∂ x p ∂ xl ∂ ξ f ∂ ξ g + . − j l k p ∂ x ∂ x ∂ x ∂ x ∂ ξ f ∂ ξ g ∂ x j ∂ xk ∂ xl ∂ x p ∂ ξ f ∂ ξ g Step 7. Compute the difference Γ pjl Γipk − Γ pjk Γipl . On the right only the first term is tensorial; the other six The result is one tensorial term, the first, and six nonwill cancel with terms in the remaining two terms of Rijkl , tensorial ones: which we now compute.

=

Step 5. Transform Γ pjl Γipk . The product of e

Γ pjl = Γbd

∂ ξ b ∂ ξ d ∂ xp ∂ 2ξ f ∂ xp + j l j l e ∂x ∂x ∂ξ ∂x ∂x ∂ξ f

and

∂ ξ h ∂ ξ c ∂ xi ∂ 2 ξ g ∂ xi + ∂ x p ∂ xk ∂ ξ a ∂ x p ∂ xk ∂ ξ g contributes four terms: a

Γipk = Γhc

∂ ξ b ∂ ξ d ∂ x p ∂ ξ h ∂ ξ c ∂ xi ∂ x j ∂ xl ∂ ξ e ∂ x p ∂ xk ∂ ξ a b d p ∂ 2 ξ g ∂ xi e ∂ξ ∂ξ ∂x + Γbd j ∂ x ∂ xl ∂ ξ e ∂ x p ∂ xk ∂ ξ g h c 2 f i ∂ xp a ∂ξ ∂ξ ∂x ∂ ξ + Γhc p k ∂ x ∂ x ∂ ξ a ∂ x j ∂ xl ∂ ξ f 2 ∂ ξ f ∂ 2 ξ g ∂ x p ∂ xi + j l p k . ∂x ∂x ∂x ∂x ∂ξ f ∂ξg e

a

Γ pjl Γipk = Γbd Γhc

DVI file created at 16:06, 20 January 2011

e

a

e

a

Γ pjl Γipk − Γ pjk Γipl = Γbd Γec − Γbc Γed


∂ ξ b ∂ ξ c ∂ ξ d ∂ xi ∂ x j ∂ xk ∂ xl ∂ ξ a

∂ ξ b ∂ ξ d ∂ x p ∂ 2 ξ g ∂ xi ∂ x j ∂ xl ∂ ξ e ∂ x p ∂ xk ∂ ξ g b c p ∂ 2 ξ g ∂ xi e ∂ξ ∂ξ ∂x − Γbc j ∂ x ∂ xk ∂ ξ e ∂ x p ∂ xl ∂ ξ g c 2 e d 2 e i i a ∂ξ ∂x ∂ ξ a ∂ξ ∂x ∂ ξ − + Γec k Γ ed ∂ x ∂ ξ a ∂ x j ∂ xl ∂ xl ∂ ξ a ∂ x j ∂ xk 2 f 2 g p i ∂ ξ ∂ ξ ∂x ∂x ∂ 2 ξ f ∂ 2 ξ g ∂ x p ∂ xi + j l p k − . ∂ x ∂ x ∂ x ∂ x ∂ ξ f ∂ ξ g ∂ x j ∂ xk ∂ x p ∂ xl ∂ ξ f ∂ ξ g e

+ Γbd

Step 8. In this final step, note that the six nontensorial terms obtained in Step 4 exactly cancel the six nontensorial terms in the previous step (when dummy indices are adjusted properly). Proposition 6.7 (text page 309) guarantees that the contraction R jl = Rhjhl is a tensor, covariant of rank 2.

6.5. TENSORS

75

3. Using overlines to denote expressions in the Greek With the substitutions obtained in part (a), we can rewrite frame, we are given that this as   K K ∂ ξ 1 ∂ ξ 1 22 ∂ξ1 ∂ξ1 11 SIJ11 ↔ SL11 and TIJ22 ↔ T L22 γ = 2 g + − 2 (−g12 ) ∂ x ∂ x2 ∂x ∂ x1 are tensors of type (p1 , q1 ) and (p2 , q2 ), respectively. This   ∂ξ1 ∂ξ1 ∂ ξ 1 ∂ ξ 1 11 means g + 1 − 2 (−g21 ) + 1 ∂x ∂x ∂ x ∂ x1 J1 ∂ ξ L1 J2 ∂ ξ L2 ∂ x ∂ x K K , TIJ22 = T L22 K , SIJ11 = SL11 K ∂ ξ 1 ∂ ξ 1 22 ∂ ξ 1 ∂ ξ 1 12 ∂ ξ 1 ∂ ξ 1 21 ∂ ξ 1 ∂ xI1 ∂ ξ 2 ∂ xI2 g + 1 g + 2 g = 2 ∂ x ∂ x2 ∂ x ∂ x2 ∂ x ∂ x1 and consequently multiplication gives 1 1 ∂ ξ 1 ∂ ξ 1 11 pq ∂ ξ ∂ ξ g = g . + J L J L 1 1 2 2 ∂ξ ∂x ∂ξ K K ∂x ∂ x1 ∂ x1 ∂ x p ∂ xq SIJ11 TIJ22 = SL11 T L22 K . I I ∂ ξ 1 ∂ x 1 ∂ ξ K2 ∂ x 2 For γ 12 (and thus for γ 21 = γ 12 ), we condense some of the This establishes that SIJ11 TIJ22

= UIJ11,I,J22

argument to K ,K ↔ U L11,L22

K K = SL11 T L22

is a tensor of type (p1 + p2 , q1 + q2 ). 4. We have

γ = det ΓP = det dMPt · GM(P) · dMP




= det dMPt · detGM(p) · det dMP = (det dMP )2 g.

5. a) We know (∂ ξ j /∂ xq ) = Therefore, since  1 ∂x ∂ξ1 dMP =   ∂ x2 ∂ξ1

d(M −1 )M(p) = (dMP )−1 .  ∂ x1 ∂ξ2  , ∂ x2  ∂ξ2

γ12 −1 ∂ xi ∂ x j gi j = 2 γ m ∂ξ1 ∂ξ2 g 1 −1 ∂ x ∂ x1 g11 −1 ∂ x1 ∂ x2 g12 = 2 + 2 m ∂ξ1 ∂ξ2 g m ∂ξ1 ∂ξ2 g 2 1 −1 ∂ x ∂ x g21 −1 ∂ x2 ∂ x2 g22 + 2 + 2 m ∂ξ1 ∂ξ2 g m ∂ξ1 ∂ξ2 g 2 1 2 ∂ ξ ∂ ξ 22 ∂ ξ ∂ ξ 1 12 ∂ ξ 2 ∂ ξ 1 21 = 2 g + 2 g + 1 g ∂ x ∂ x2 ∂ x ∂ x1 ∂ x ∂ x2 ∂ ξ 2 ∂ ξ 1 11 ∂ξ1 ∂ξ2 g = g pq p q . + 1 1 ∂x ∂x ∂x ∂x

γ 12 = −

For γ 22 , we have

γ11 1 ∂ xi ∂ x j g i j = 2 γ m ∂ξ1 ∂ξ1 g we can write 1 ∂ x1 ∂ x2 g12 1 ∂ x1 ∂ x1 g11    1 2 1 + = 1 ∂ x ∂ x ∂ξ ∂ξ m2 ∂ ξ 1 ∂ ξ 1 g m2 ∂ ξ 1 ∂ ξ 1 g  ∂ξ2 −∂ξ2   ∂ x1 ∂ x2  1 −1 2 1     1 ∂ x ∂ x g21 1 ∂ x2 ∂ x2 g22  ∂ ξ 2 ∂ ξ 2  = (dMP ) = m  ∂ x2 ∂ x1  , + 2 + m ∂ξ1 ∂ξ1 g m2 ∂ ξ 1 ∂ ξ 1 g − 1 ∂ξ ∂ξ1 ∂ x1 ∂ x2 2 2 2 ∂ ξ ∂ ξ 22 ∂ ξ ∂ ξ 2 12 ∂ ξ 2 ∂ ξ 2 21 = 2 g + 2 g + 1 g proving the claim. The components of the inverses of γi j ∂ x ∂ x2 ∂ x ∂ x1 ∂ x ∂ x2 and g pq are ∂ ξ 2 ∂ ξ 2 11 ∂ξ2 ∂ξ2 g = g pq p q . + 1 g g γ γ 1 22 22 12 12 ∂x ∂x ∂x ∂x γ 11 = , γ 12 = − , g11 = , g12 = − , γ γ g g Thus we have γ21 γ11 g21 22 g11 γ 21 = − , γ 22 = g21 = − , g = . ∂ξi ∂ξ j γ γ g g γ i j = g pq p q for all i, j, ∂x ∂x b) With m = det dMP , the transformation law for γ ↔ g i j pq so γ ↔ g is a contravariant tensor of rank 2. from Exercise 4 is γ = m2 g. Therefore, 6. a) To determine a covariant derivative on a surface, γ22 1 ∂ xi ∂ x j g i j 1 ∂ xi ∂ x j g = γ 11 = = 2 i j we need the Christoffel symbols of the second kind for γ m g ∂ξ2 ∂ξ2 m2 ∂ ξ 2 ∂ ξ 2 g that surface. In the present case, they are provided in the 1 ∂ x1 ∂ x1 g11 1 ∂ x1 ∂ x2 g12 text (page 272): = 2 + 2 m ∂ξ2 ∂ξ2 g m ∂ξ2 ∂ξ2 g Γ211 = sin(q2 ) cos(q2 ), Γ112 = Γ121 = − tan(q2 ), 1 ∂ x2 ∂ x2 g22 1 ∂ x2 ∂ x1 g21 + 2 . + 2 Γijk = 0 otherwise. m ∂ξ2 ∂ξ2 g m ∂ξ2 ∂ξ2 g

DVI file created at 16:06, 20 January 2011

γ 22 =

76

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

The covariant derivative of the contravariant vector ak is

by differentiating components of the original tensor. But suppose ϕ ↔ f is a tensor of type (0, 0); then

ak;i =

∂ ak + a j Γkij ∂ qi

ϕ (ξ 1 , . . . , ξ n ) = f (x1 (ξ 1 , . . . , ξ n ), . . . , xn (ξ 1 , . . . , ξ n )),

Given that a1 = 0, a2 = 1, all partial derivatives are zero so and the sum a j Γkij reduces to the single term Γki2 :

∂ϕ ∂ f ∂ xp . = ∂ξi ∂ xp ∂ ξ i |{z} |{z}

a1;1 = Γ112 = − tan(q2 ), a1;2 = Γ122 = 0,

a2;1 = Γ212 = 0,

a2;2 = Γ222 = 0.

b) Now suppose bi = f i (q1 , q2 ); then

∂ b1 ∂ q1 ∂ b1 b1;2 = 2 ∂q ∂ b2 b2;1 = 1 ∂q ∂ b2 b2;2 = 2 ∂q

b1;1 =

∂ b1 − b2 tan(q2 ), ∂ q1 ∂ b1 + bk Γ12k = 2 − b1 tan(q2 ), ∂q ∂ b2 + bk Γ21k = 1 + b1 sin(q2 ) cos(q2 ), ∂q ∂ b2 + bk Γ22k = 2 . ∂q + bk Γ11k =

There is no nontensorial term, so no counterbalancing term should be introduced. The equation shows that the gradient ∂ ϕ /∂ ξ i ↔ ∂ f /∂ x p is tensorial; therefore, the covariant derivative is simply the gradient. 9. By using twice the definition of Christoffel symbols of the first kind, we find

∂ gi j ∂ gi j − gl j Γlik − gil Γlk j = − Γik, j − Γk j,i ∂ xk ∂ xk   ∂ gi j 1 ∂ gi j ∂ gk j ∂ gik − + − = ∂ xk 2 ∂ xk ∂ xi ∂xj   1 ∂ gi j ∂ gk j ∂ gik = 0. + − − 2 ∂ xk ∂ xi ∂xj

gi j;k =

7. We are given that a p ↔ αi is a covariant vector field; Note that we can write gi j;k = 0 in the equivalent form hence ∂ξi a p = αi p . ∂ gi j ∂x = gl j Γlik + gil Γlk j . ∂ xk To show that the covariant derivative a p;q ↔ α; j is a covariant tensor of rank 2, we must establish that To analyze the contravariant metric, we begin by noting that δip = gi j g p j and hence ∂ξi ∂ξ j a p;q = αi; j p q . ∂x ∂x ∂ δ p ∂ gi j p j ∂ gp j 0 = ik = g + gi j k . We have (using Corollary 6.2) k ∂x ∂x ∂x   i ∂ ap ∂ ∂ ξ q a p;q = q − ar Γrpq = q αi p Therefore, using the fact that gqi gi j = δ j , we can write ∂x ∂x ∂x     i j k r 2 j r ∂ g pq ∂ξ ∂ ξ ∂x  l l qi p j ∂ gi j qi p j h ∂ξ ∂ξ ∂x g Γ + g Γ = −g g = −g g l j il + − αi r Γ jk p q ik k j ∂ xk ∂ xk ∂x ∂ x ∂ x ∂ ξ h ∂ x p ∂ xq ∂ ξ j |{z} |{z} |{z} qi p l = −g δl Γik − g p j δlq Γlk j = −gqiΓikp − g p j Γqk j . i 2 i ∂ αi ∂ ξ ∂ ξ = q p + αi p q This means ∂x ∂x ∂x ∂x j k 2 j ∂ ξ h ∂ξ ∂ξ ∂ g pq − αi Γ jk p q δhi − αi p q δ ji iq p pj q g;pq ∂x ∂x ∂x ∂x k = ∂ xk + g Γik + g Γk j = 0, i j ∂ αi ∂ ξ j ∂ ξ i h ∂ξ ∂ξ − αh Γi j p q = q p j as required. ∂ξ ∂x ∂x ∂x ∂x   i j i j ∂ αi ∂ξ ∂ξ ∂ξ ∂ξ h 10. We noted in the solution to Exercise 1, above (Solu= = αi; j p q . − αh Γi j ∂ξ j ∂ x p ∂ xq ∂x ∂x tions page 73) that the multi-indices I, J are unaffected in an analysis of something like gi j TJ,I j = TJI,i , so we omit At various points, dummy indices have been relabeled. them in our discussion. Thus our aim is to show that 8. The additional terms in a covariant derivative are intro

T;li = gi j T j ;l = gi j T j;l . duced to counterbalance the nontensorial terms generated

DVI file created at 16:06, 20 January 2011

6.5. TENSORS

77

To begin,
∂Ti ∂ + T j Γil j = l gi j T j + g j pTp Γil j l ∂x ∂x ∂ T ∂ gi j j T j + gi j l + g j pTp Γil j . = ∂ xl ∂x

Expanding the terms, and then permuting the indices j and k, we find

T;li =

A

ij g;l

At this point we use the equation = 0 (established in the solution to the previous exercise), in the form

∂ gi j j = −g p j Γipl − giq Γlq , ∂ xl to replace the first term in the expansion of T;li :

∂ Tj j + gi j l + g j pTp Γil j T;li = −g p j T j Γipl −giq T j Γlq ∂x | {z } | {z }  
∂ Tj − TpΓlpj = gi j T j;l . = gi j l ∂x

∂ Γi j ∂ am h ∂ 2 ah ∂ ai h i + Γ + Γ +a i j k j mk ∂ xk ∂ x j |∂ x{z ∂ xk |∂ x{z } } h

ah; j;k =

h + ai Γm i j Γmk −

B

∂ ah

Γ p − ai Γhip Γ pjk , p jk ∂ x | {z } | {z } C

ah;k; j =

∂ 2 ah ∂ x j ∂ xk

+

h i ∂ Γik h +a Γ ik ∂xj ∂xj

∂ ai

D

+

| {z } B

∂ am h Γ k mj |∂ x{z } A

∂ ah p h + ai Γm Γ − Γ − ai Γhip Γkpj . ik m j p kj ∂ x | {z } | {z } C

D

The second derivatives and the terms with corresponding labels are equal and thus cancel when the difference is With the exchange p ↔ j of dummy indices in the first taken; the result is term, we see that term cancels with the fourth. The ex! changes j ↔ p and q ↔ j then transform the second term ∂ Γhij ∂ Γhik h h i m h m h a; j;k − a;k; j = a + Γi j Γmk − Γik Γm j − into the form it has in the last line. ∂xj ∂ xk 11. To simplify the formulation of the covariant deriva= −Rhijk ai . tive of a tensor TJI (Definition 6.5, text page 317), we use the following abbreviation for expansion with respect to b) Let bi = giq bq ; because index-raising commutes with the contravariant indices: covariant differentiation, we have i ...i h ip b ıˆ hi ...i i1 = TJhI Γhm TJ 2 p Γhm + · · · + TJ 1 p−1 Γhm
giq bq;k;l − bq;l;k = bi;k;l − bi;l;k = −Rimkl bm . b the circumflex indicates an index is missing, and ıˆ is In I, that missing index. We use an analogous abbreviation for The last step follows from part (a). Therefore,

the covariant indices. Then bh;k;l − bh;l;k = ghi giq bq;k;l − bq;l;k = ghi bi;k;l − bi;l;k
∂ b b kˆ ıˆ = −ghi Rimkl bm = −Rhmkl bm . TJI SLK ;m = m (TJI SLK ) + TJhI SLK Γhm + TJI SLhK Γhm ∂x − ThIJbSLK Γhjˆm − TJI ShKbL Γhlm ˆ 13. By Definition 6.6 (text page 320), the covariant   I derivative of yl (t) = dxl /dt along xk (t) is ∂ TJ K hIb ıˆ I h S + T Γ − T Γ = L J hm hJb jˆm ∂ xm Dyl dyl dxk   K = + Γljk y j b kˆ hK K h I ∂ SL dt dt dt + SL Γhm − ShLb Γlm + TJ ˆ ∂ xm j dxk dx d 2 xl l + Γ = I K I K jk = TJ;m SL + TJ SL;m . dt 2 dt dt =0 12. a) We have because the curve xl (t) is itself a geodesic. Thus yl is h parallel along xk . a; j h h p ah; j;k = k + am Γ − a Γ ; j mk ;p jk ∂x   m   h 14. a) The tangent vector along the curve q(t) = ∂ ∂a ∂a h i h i m (q1 , q2 ) = (t, π /6) is y(t) = (y1 , y2 ) = (1, 0), and its co+ Γ + a Γ + a Γ = k i j i j mk ∂xj ∂xj ∂x variant derivative there is   h ∂a p i h + a Γip Γ jk − dq j dqk d 2 ql Dyl ∂ xp = 2 + Γljk . dt dt dt dt

DVI file created at 16:06, 20 January 2011

78

SOLUTIONS: CHAPTER 6. INTRINSIC GEOMETRY

Using

so ∆ = π . The figure below shows the vector z(t) for t = nπ /4, n = 0, 1, . . . , 8, rescaled for greater clarity. dq2 d 2 q1 d 2 q2 = = 0, = dt dt 2 dt 2 Γ111 = 0,

Γ211 = sin(π /6) cos(π /6) =

q2 π/2

√ 3 ; 4

π/6

we reduce the covariant derivative equations to Dy1 = 0, dt

π/2

√ 3 Dy2 = . dt 4

π

3π/2

2π

q1

−π/2 b) The parallel vector field z(t) = (z1 (t), z2 (t)) along q(t) must satisfy the parallel-transport equations and the c) The approach here is similar to the approach in the initial conditions (z1 (0), z2 (0)) = (1, 0). The parallel- previous part, except that now transport equations for zi (t) make use of the following Γ112 = Γ121 = − tan k, Γ211 = sin k cosk. facts: dq2 = 0, dt √ 3 −1 2 1 1 , Γ12 = Γ21 = − tan(π /6) = √ , Γ11 = 4 3 dq1 = 1, dt

The equations are then dz1 1 dz1 + Γ121 · z2 · 1 = − √ z2 = 0, dt dt 3 √ 2 2 dz dz 3 1 + Γ211 · z1 · 1 = + z = 0. dt dt 4 To solve these, we first form the combination √ d 2 z1 1 dz2 1 − 3 1 1 =√ =√ z = − z1 = 0, dt 2 4 3 dt 3 4

z1 (t) = A cos(t/2) + B sin(t/2), √ √ √ dz1 −A 3 B 3 2 = sin(t/2) + cos(t/2). z (t) = 3 dt 2 2 The initial conditions z1 (0) = 1, z2 (0) = 0 imply that A = 1, B = 0, and hence √ − 3 2 z (t) = sin(t/2). 2

In one circuit, z(0) = (1, 0),

dz1 dz1 + z2 Γ112 = − (tan k) z2 = 0, dt dt dz2 dz2 + z1 Γ211 = + (sin k cos k) z1 = 0. dt dt Combining these two equations the same way we did above, we get dz2 d 2 z1 = (tan k) = (tan k)(− sin k cos k)z1 = 0, dt 2 dt or (z1 )′′ = −(sin k)2 z1 . This implies z1 (t) = A cos(t sin k) + B sin (t sin k) , 1 dz1 tan k dt = −A cosk · sin (t sin k) + B cosk · cos (t sin k) .

z2 (t) =

or (z1 )′′ = −z1 /4, implying

z1 (t) = cos(t/2),

For simplicity we write z for zk . The parallel-transport equations are

The initial conditions z1 (0) = 1, z2 (0) = 0 then imply that A = 1 and B = 0, so z1 (t) = cos(t sin k),

z2 (t) = − cos k · sin(t sin k).

For the angle ∆, we have cos ∆ =

z(0) · z(2π ) , kz(0)kkz(2π )k

where v · w is computed in the metric as z(2π ) = (−1, 0),

DVI file created at 16:06, 20 January 2011

gi j (z(t))vi w j = gi j (t, k)vi w j = (cos2 k) · v1 w1 + (1) · v2w2 .

6.5. TENSORS

79

Consequently, since (z1 (0), z2 (0)) = (1, 0), (z1 (2π ), z2 (2π )) = (cos(2π sin k), − cos k · sin(2π sin k)), north

we find

z(π/2)

pole

z(0) · z(2π ) = (cos2 k) cos(2π sin k),

x2

kz(0)k = cos k,

while kz(2π )k = q (cos2 k) cos2 (2π sin k) + (− cosk)2 sin2 (2π sin k)

x1

circle at latitude k

= | cos k| = cosk.

Here cos k is positive because k is small. Hence cos ∆ =

(cos2 k) cos(2π sin k) = cos(2π sin k), cos k · cosk

north pole

or ∆ = 2π sin k. A similar analysis shows that, for any t, the angle between x1 (t, k) and z(t) is t sin k. d) In the illustrations at the right, we take k = π /2 − 1/4; this circle of latitude is about 14◦ away from the north pole. Because our figure is intended to represent a small portion of the surface itself, we must draw the basis vectors x1 and x2 to scale on this pcircle. Whereas x2 is always a unit vector, x1 has length g11 (t, k) = cos k ≈ 0.25. The upper figure shows x1 and x2 together with z(π /2), all drawn to the same scale (in which the lengths have been reduced for greater clarity). The lower figure shows just z(t), but with a different scaling to help reveal the change ∆ in z after one circuit. For our chosen k, ∆ ≈ 349◦, so the smaller angle between z(0) and z(2π ) is about 11◦ .

DVI file created at 16:06, 20 January 2011

z(2π) ∆

z(0)

Solutions: Chapter 7

General Relativity 7.1 The Equations of Motion

d) The Maxwell equations we use appear in the text on page 23. In light of the conclusion in part (c) above, we can write the Maxwell equation ρ = ∇· E as   ∂A ∂ 0 ρ = ∇· − − ∇A = − ∇· A − ∇2A0 . ∂t ∂t

1. a) Let us write H = (H 1 , H 2 , H 3 ); by definition, H1 =

∂ A2 ∂ A3 ∂ A3 ∂ A1 − , H2 = − , ∂z ∂y ∂x ∂z ∂ A1 ∂ A2 − . H3 = ∂y ∂x

For the second condition on A, we use the Maxwell equation ∇× H = J + ∂ E/∂ t in the form

∂E ∂ 2A ∂ = ∇× (∇× A) + 2 + (∇A0 ) ∂t ∂t ∂t We have used the definition of A and the solution to part (c). By the solution to Exercise 1b, § 1.4 (Solutions, page 4), we can replace ∇× (∇× A) and get J = ∇× H −

Therefore,

∂ H1 ∂ H2 ∂ H3 + + ∂x ∂y ∂z 2 2 2 3 ∂ A ∂ A ∂ 2 A3 ∂ 2 A1 ∂ 2 A1 ∂ 2 A2 − + − + − = ∂ x∂ z ∂ x∂ y ∂ y∂ x ∂ y∂ z ∂ z∂ y ∂ z∂ x =0

∇· H =

J = ∇(∇· A) − ∇2 A +

∂ 2A ∂ + (∇A0 ), ∂ t2 ∂t

as required.

e) This is Exercise 3, § 1.4; see Solutions, page 4. because “equality of mixed partials” causes the six terms 2. The Christoffel symbols Γi j,k are linear combinations cancel in pairs. of certain partial derivatives of the metric tensor !   b) If it is true that 3 e2αξ 0 γ00 γ03 =   3 γ30 γ33 0 −e2αξ ∂H ∂ ∂A , = − (∇× A) = − ∇× ∇× E = − ∂t ∂t ∂t 3 Most derivatives are zero. If we set ψ = e2αξ , then the only nonzero derivatives are then it follows that   ∂ γ00 ∂ γ33 ∂A ∂A = 2αψ , = −2αψ . . = ∇× E + 0 = ∇× E + ∇× ∂ξ3 ∂ξ3 ∂t ∂t A Christoffel symbol will therefore be zero unless its indices are {0, 0, 3}, in some order, or {3, 3, 3}. We have c) We apply the theorem from advanced calculus to the   1 ∂ γ03 ∂ γ30 ∂ γ00 vector field = −αψ , + − Γ00,3 = ∂A 2 ∂ξ0 ∂ξ0 ∂ξ3 F = E+   ∂t 1 ∂ γ00 ∂ γ03 ∂ γ03 Γ03,0 = Γ30,0 = = αψ , + − 0 2 ∂ξ3 ∂ξ0 ∂ξ0 to obtain a function ψ = −A for which   1 ∂ γ33 ∂ γ33 ∂ γ33 Γ33,3 = = −αψ . + − ∂A ∂A 0 0 2 ∂ξ3 ∂ξ3 ∂ξ3 = ∇ψ = −∇A , or E = − − ∇A . F = E+ ∂t ∂t 80

DVI file created at 16:06, 20 January 2011

7.1. THE EQUATIONS OF MOTION A Christoffel symbol of the second kind will also be zero unless its indices are {0, 0, 3}, in some order, or {3, 3, 3}. To compute them, we also need

γ 00 = 1/ψ ,

γ 33 = −1/ψ ,

γ i j = 0 otherwise.

81 We must now compute the equation of the tangent line to this graph at a point τ = τ0 , find its ζ -intercept, and then let τ0 → −∞. Because ζ ′ = − tanh ατ , the equation of the tangent line at τ = τ0 is

We then have Γ003 = Γ030 = γ 00 Γ03,0 + γ 03 Γ03,3 = α , Γ300 = γ 30 Γ00,0 + γ 33 Γ00,3 = α ,

ζ+

and the ζ -intercept is at

Γ333 = γ 30 Γ33,0 + γ 33 Γ33,3 = α . The geodesic equations for the geodesics ξ i (t) have the form dξ j dξ k d2ξ i = −Γijk , 2 dt dt dt and reduce to dξ 0 dξ 3 dξ 3 dξ 0 d2ξ 0 = −Γ003 − Γ030 2 dt dt dt dt dt dξ 0 dξ 3 , = −2α dt dt   3 2  2 d2ξ 3 dξ 0 dξ 3 3 = −Γ00 − Γ33 2 dt dt dt  0 2  3 2 dξ dξ = −α −α . dt dt 3. The figure in the text on page 340 shows the graphs of the family of functions   1 eα k ζk (τ ) = ln , α cosh α (τ − k)

and it indicates that the ζ -intercepts of these graphs approach the ζ -intercept β of their common asymptote as k → ∞; that is, β = lim ζk (0). k→∞

We have ζk (0) = (1/α ) ln(eα k / cosh α k), and eα k 2 eα k ; = = αk cosh α k (e + e−α k )/2 1 + e−2α k therefore ln 2 − ln(1 + e−2α k ) ln 2 = . k→∞ α α

β = lim

1 ln(cosh ατ0 ) = −(tanh ατ )(τ − τ0 ), α

ζ =−

1 ln(cosh ατ0 ) + τ0 tanh(ατ0 ). α

Keeping in mind that τ0 → −∞, we write cosh ατ0 =

e−ατ0 e−ατ0 + eατ0 = (1 + e2ατ0 ), 2 2

In these terms, the ζ -intercept is ln 2 − ln(1 + e2ατ0 ) − τ0 tanh ατ0 α ln 2 ln(1 + e2ατ0 ) = − + τ0 (1 + tanh ατ0 ) α α

ζ = τ0 +

Now 

eατ0 + e−ατ0 + eατ0 − e−ατ0 eατ0 + e−ατ0 2τ0 → 0 as τ0 → −∞, = 1 + e−2ατ0

τ0 (1 + tanh ατ0 ) = τ0



because the exponential function in the denominator dominates the linear function in the numerator. Finally, because ln(1 + e2ατ0 ) → 0 as τ0 → −∞, we see once again that the ζ -intercept of the tangent line approaches (ln 2)/α . 4. The solution to Exercise 14b, §2.2 (Solutions, page 13) expresses the inverse hyperbolic tangent as   1 1+w u = tanh−1 w = ln . 2 1−w Therefore,       t 1 + t/(a ± t) a±t +t 1 1 −1 tanh = ln = ln . a±t 2 1 − t/(a ± t) 2 a±t −t

With the plus sign, we have     Another way to find β is to use the fact that an asymp- 1 1 1 t ln a a + 2t −1 = = ln ln (a + 2t)− . tote is a “tangent at infinity.” Because the asymptote is α tanh a+t 2α a 2α 2α common to all the graphs ζ = ζk (τ ), we can work just With the minus sign, we have with the graph for which k = 0, that is, with       1 ln a −1 1 1 t 1 1 a = = = ζ = ζ0 (τ ) = ln ln(cosh ατ ). tanh−1 ln − ln (a − 2t). α cosh ατ α α a−t 2α a − 2t 2α 2α

DVI file created at 16:06, 20 January 2011

82

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

Because (ln a)/2α is a constant, we have confirmed that

confirming the geodesic equation for ξ 0 (cf. page 81 above). We also have  0 2  3 2 dξ dξ −α −α dt dt 1 1 1 1 −α · 2 = −α · 2 2 α (a ± 2t) α (a ± 2t)2

ξ 0 (t) = ±

1 ln(a ± 2t) + constant. 2α

For ξ 3 we can write  

ln α 2 ((a ± t)2 − t 2 ) = ln α 2 (a ± t + t)(a ± t − t) .

With the plus sign, this becomes
ln α 2 (a + 2t)a = ln(a + 2t) + ln(α 2 a),

and with the minus sign,
ln α 2 a(a − 2t) = ln(α 2 a) + ln(a − 2t).

We have thus confirmed

ξ 3 (t) =

1 ln(a ± 2t) + constant. 2α

To show that ξ (t) = we must verify that

(ξ 0 (t), ξ 3 (t))

=

confirming the geodesic equation for ξ 3 . 5. a) The given curve ξ ± (u) = (ξ 0 (u), ξ 3 (u)) is spacelike if dξ i dξ j 0 and either u > b or u < −b. Because u tanh(α (τ − k)) = b and α > 0, we see u > b ⇒ τ − k > 0,

u < −b ⇒ τ − k < 0.

These define the “±” choices for the curve ξ ± . Keeping in mind the parametric equation for ξ 3 = ζ , we write b2 (1 − tanh2 (α (τ − k)) b2 sech2 (α (τ − k)) u −b = = tanh2 (α (τ − k)) tanh2 (α (τ − k)) 2

2

=

b2 , sinh (α (τ − k)) 2

and consequently p u2 − b2 =

b , ± sinh(α (τ − k))

where the sign is chosen to agree with the sign of τ − k (or, equivalently, with the sign of ξ ± ). We finally obtain 
1  p 1 ζ= ln α 2 (u2 − b2) = ln α u2 − b2 2α  α 1 αb = ln = f± (τ ). α ± sinh(α (τ − k))

83 Using f± (τ ) = −(1/α ) ln(± sinh(α (τ − k)/α b), we find −α b f±′ (τ ) = . tanh(α (τ − k))

Because tanh(α (τ − k)) → ±1 as τ → ±∞ (recall α > 0), we see lim f ′ (τ ) τ →+∞ +

= −α b,

lim f ′ (τ ) τ →−∞ −

= +α b.

Thus ζ = f+ (τ ) has an asymptote of slope −α b = −α k2 at +∞, and ζ = f− (τ ) has an asymptote of slope +α b = +α k2 at −∞. b) Fix k1 and consider the one-parameter family of curves   b 1 ξb0 (u) = k1 + tanh−1 , α u u > b > 0, 1 3 2 2 2 ξb (u) = ln(α (u − b )), 2α The solution to Exercise 5 shows these curves are geodesics, and the solution to Exercise 6a shows they have the common asymptote τ = k1 . The vertical asymptote is therefore a singular solution to the geodesic equations (when the proper parametrization is used). To get the proper parametrization, suppose we start with ξ 0 (u) = k1 , ξ 3 (u) = ψ (u); then we must determine ψ so that the geodesic equation for ξ 3 is satisfied. The equation is

 0 2  3 2 c) The full eight-parameter family of spacelike geoded2ξ 3 dξ dξ + + = ψ ′′ + α (ψ ′ )2 = 0. α α sics is given by formulas analogous to those of the time2 du du du like family, using the formulas for ξ 0 and ξ 3 already given above, in part (a): To solve this, first let ψ ′ (u) = χ (u); then χ ′ + α χ 2 = 0.   This differential equation can be solved by the method of k2 1 ξ 0 (u) = k1 + tanh−1 , “separation of variables.” We write d χ /du = −α χ 2 as α u dχ = −α du. χ2

ξ 1 (u) = k3 u + k4,

ξ (u) = k5 u + k6, 1 ξ 3 (u) = ln(α 2 (u2 − k22 )), 2α u = k7 t + k8 . 2

6. a) To find the asymptotes of ξ ± , we can analyze the graphs ζ = f± (τ ) obtained in the previous exercise. First of all, f± (τ ) → ∞ as τ → k = k1 . Thus τ = ξ 0 = k1 is a vertical asymptote. For the asymptotes as τ → ±∞, note that we must use ζ = f+ (τ ) when τ → +∞ (because τ > k) but ζ = f− (τ ) when τ → −∞ (τ < k). In each case we show that the derivative of the relevant function has a finite limit as τ approaches +∞ or −∞, as appropriate.

DVI file created at 16:06, 20 January 2011

Integrating both sides, we get −

1 = − α u − c1 χ

or

χ (u) =

1 . α u + c1

Because ψ ′ = χ , we finally obtain

ξ 3 (u) = ψ (u) =

1 ln(α u + c1 ) + c2. α

7. Let Rθ : R2 → R2 be rotation by θ . Then (using τ = t) M has         y y η η = Rωτ , so = R −ω t z z ζ ζ

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

84 Therefore

9. The Christoffel symbols of the first kind involve the nonzero partial derivatives of the entries γi j . Taking symmetries into account, there are only four:

 τ = t,    ξ = x, M −1 :  η = y cos ω t + z sin ω t,    ζ = −y sin ω t + z cos ω t.

∂ γ00 = −2ω 2η , ∂ξ2 ∂ γ00 = −2ω 2ζ , ∂ξ3

In order to have compact expressions, let us set

For clarity, the results are expressed in terms of η and ζ . We conclude that the only nonzero Christoffel symbols involve (in some order) the indices {0, 0, 2}, {0, 0, 3}, or {0, 2, 3}. We find

YC = ωη cos ωτ , ZC = ωζ cos ωτ , Y S = ωη sin ωτ , ZS = ωζ sin ωτ ; then we can write 

1  0 dMP =  −Y S − ZC YC − ZS

Γ00,2 = ω 2 η ,

 0 0 0 1 0 0  . 0 cos ωτ − sin ωτ  0 sin ωτ cos ωτ

To determine (dMP )−1 (a function of the Greek variables), we begin with d(M −1 )M(P) = 

1  0  −ω y sin ω t + ω z cos ω t −ω y cos ω t − ω z sin ω t

 0 0 0 1 0 0  . 0 cos ω t sin ω t  0 − sin ω t cos ω t

As soon as we convert Roman to Greek variables, we have the inverse we want:   1 0 0 0  0 1 0 0   (dMP )−1 =   ωζ 0 cos ωτ sin ωτ  −ωη 0 − sin ωτ cos ωτ 8. The metric (γi j ) is given by 

1  0  J1,3 · dMP =  Y S + ZC −YC + ZS

and dMPt · J1,3 · dMP =

∂ γ03 = −ω , ∂ξ2 ∂ γ02 = ω, ∂ξ3

dMPt · J1,3 · dMP .

We have

 0 0 0  −1 0 0  0 − cos ωτ sin ωτ  0 − sin ωτ − cos ωτ

  1 − ω 2η 2 − ω 2ζ 2 0 ωζ −ωη  0 −1 0 0  . (γi j ) =   ωζ 0 −1 0  −ωη 0 0 −1

Γ02,0 = Γ20,0 = −ω 2 η , Γ00,3 = ω 2 ζ ,

Γ03,0 = Γ30,0 = −ω 2 ζ ,   1 ∂ γ03 ∂ γ02 Γ02,3 = Γ20,3 = = −ω , − 2 ∂ξ2 ∂ξ3   1 ∂ γ02 ∂ γ03 Γ03,2 = Γ30,2 = = ω, − 2 ∂ξ3 ∂ξ2   1 ∂ γ03 ∂ γ02 Γ23,0 = Γ32,0 = = 0. + 2 ∂ξ2 ∂ξ3 For the Christoffel symbols of the second kind, we have (not including Γkji when Γkij is given) Γ000 = γ 02 Γ00,2 + γ 03 Γ00,3 = ωζ · ω 2 η − ωη · ω 2 ζ = 0,

Γ100 = γ 12 Γ00,2 + γ 13 Γ00,3 = 0, Γ200 = γ 22 Γ00,2 + γ 23 Γ00,3

= (−1 + ω 2ζ 2 )ω 2 η − ω 2 ηζ · ω 2 ζ = −ω 2 η ,

Γ300 = γ 32 Γ00,2 + γ 33 Γ00,3

= −ω 2 ηζ · ω 2 η + (−1 + ω 2η 2 )ω 2 ζ = −ω 2 ζ ,

Γ002 = γ 00 Γ02,0 + γ 03 Γ02,3 = 1 · (−ω 2η ) − ωη · (−ω ) = 0, Γ102 = γ 10 Γ02,0 + γ 13 Γ02,3 = 0,

Γ202 = γ 20 Γ02,0 + γ 23 Γ02,3 = ωζ · (−ω 2 η ) − ω 2 ηζ · (−ω ) = 0,

Γ302 = γ 30 Γ02,0 + γ 33 Γ02,3

= −ωη · (−ω 2 η ) + (−1 + ω 2η 2 ) · (−ω ) = ω ,

Γ003 = γ 00 Γ03,0 + γ 02 Γ03,2 = 1 · (−ω 2ζ ) + ωζ · ω = 0, Γ103 = γ 10 Γ03,0 + γ 12 Γ03,2 = 0,

Γ203 = γ 20 Γ03,0 + γ 22 Γ03,2

= ωζ · (−ω 2 ζ ) + (−1 + ω 2ζ 2 ) · ω = −ω , Let γ jk be the matrix given in the text; direct evaluation 30 32 3 shows that γi j γ jk = δik , confirming that γ jk is the inverse Γ03 = γ Γ03,0 + γ Γ03,2 of the metric tensor. = −ωη · (−ω 2 ζ ) − ω 2ηζ · ω = 0.

DVI file created at 16:06, 20 January 2011

7.1. THE EQUATIONS OF MOTION

85

These calculations confirm that the only nonzero symbols Then  2 are dτ 2 = −ω η Γ200 = −ω 2 η , Γ203 = Γ230 = −ω du Γ300 = −ω 2 ζ ,

Γ302 = Γ320 = ω .

10. a) Because Γ0i j = Γ1i j = 0, the geodesic equations for ξ 0 = τ and ξ 1 = ξ have only the second-order term: d 2ξ d 2τ = 0, = 0. 2 du du2 For the other two variables, the geodesic equations each have two first-order terms, as follows:  2 dτ dτ dζ d2η 2 ω η − 2ω − = 0, 2 du du du du  2 dτ dτ dη d2ζ 2 + 2ω −ω ζ = 0. du2 du du du b) In R, the metric is constant, so all Christoffel symbols are zero. Consequently, the geodesic equations are simply

− ω 2 (a0 )2 (a2 u + b2)C − ω 2 (a0 )2 (a3 u + b3)S,

−2ω

dτ dζ = du du 2ω a0 a2 S + 2ω 2(a0 )2 (a2 u + b2)C − 2ω a3C + 2ω 2(a0 )2 (a3 u + b3)S,

and these terms cancel with the terms in d 2 η /du2 . Thus the geodesic equation for η is satisfied. For ζ , we need  2 dτ = −ω 2 ζ du

ω 2 (a0 )2 (a2 u + b2)S − ω 2(a0 )2 (a3 u + b3)C, dτ dη = +2ω du du 2ω a0 a2C − 2ω 2(a0 )2 (a2 u + b2)S + 2ω a3S

+ 2ω 2(a0 )2 (a3 u + b3)C; d 2 xi = 0, i = 0, 1, 2, 3. these terms cancel with those of d 2 ζ /du2 to confirm that du2 ζ satisfies its geodesic equation. The solutions are linear functions: xi = ai u + b1 , where i = 0, 1, 2, 3. The images of these lines in G are given by 11. a) Writing ξ j = (t, 0, η j , ζ j ), we find the following: ( the map M −1 . Thus, using η1′ = v cos ω t − ω vt sin ω t, ξ ′1 : C = cos(ω (a0 u + b0)), S = sin(ω (a0 u + b0)), ζ1′ = −v sin ω t − ω vt cos ω t, ( to make the formulas easier to view, we have η1′′ = −2ω v sin ω t − ω 2vt cos ω t, ξ ′′1 : τ = a0 u + b0 , ζ1′′ = −2ω v cos ω t + ω 2 vt sin ω t; ξ = a1 u + b1 , ( η2′ = −v cos ω t − ω (1 − vt) sin ω t, η = (a2 u + b2)C + (a3 u + b3)S, ξ ′2 : ζ2′ = v sin ω t + ω (vt − 1) cos ω t, ζ = −(a2 u + b2)S + (a3u + b3)C. ( η2′′ = 2ω v sin ω t − ω 2 (1 − vt) cos ω t, Because τ and η are linear functions of u, we see that their ξ ′′2 : ζ2′′ = 2ω v cos ω t − ω 2 (vt − 1) sin ω t. geodesic equations are immediately satisfied. To verify the geodesic equations for η and ζ , we note

dτ = a0 , du dη = a2C − ω a0(a2 u + b2)S + a3 S + ω a0(a3 u + b3)C, du d2η = −2ω a0a2 S + 2ω a0a3C − ω 2 (a0 )2 (a2 u + b2)C du2 − ω 2 (a0 )2 (a3 u + b3)S,

The geodesic equations for τ and ξ are immediately satisfied. For the others, we have

ω 2 η1 + 2ωζ1′ = ω 2 vt cos ω t − 2ω v sin ω t − 2ω 2 vt cos ω t = η1′′ , ω 2 ζ1 − 2ωη1′ = −ω 2 vt sin ω t − 2ω v cos ω t + 2ω 2vt sin ω t = ζ1′′ ,

ω 2 η2 + 2ωζ2′ = ω 2 (1 − vt) cos ω t + 2ω v sin ω t dζ 0 2 2 3 0 3 3 2 = −a S − ω a (a u + b )C + a C − ω a (a u + b )S, + 2ω 2(vt − 1) cos ω t = η2′′ , du d 2ζ ω 2 ζ2 − 2ωη2′ = ω 2 (vt − 1) sin ω t + 2ω v cos ω t = −2ω a0a2C − 2ω a0a3 S + ω 2(a0 )2 (a2 u + b2)S 2 du + 2ω 2(1 − vt) sin ω t = ζ2′′ . − ω 2 (a0 )2 (a3 u + b3)C. this confirms that ξ 1 and ξ 2 are geodesics.

DVI file created at 16:06, 20 January 2011

86

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

In the figure below, ξ 1 is the smaller curve and starts at the origin; ξ 2 starts at the point (η , ζ ) = (1, 0). Both curves turn to the right.

Within the brackets, the terms that are not positive squares either cancel or are positive by hypothesis. Therefore, η ′ ζ ′′ − η ′′ ζ ′ is negative for all t, as required. The condition a2 b3 − a3b2 > 0 is necessary. For example, the geodesic ξ (t) with a2 = b3 = 0, a3 = b2 = 1, and ω = 3/2 (shown below) does turn left for a short interval when t is near zero. 0.5

-3

-2

1

-1 -0.5

b) The tangent vector ξ ′j and the “curvature” vector ξ ′′j were both found in the previous part. They are plotted (at half-size) on the figure above.

-1.0

-1.5

c) By definition, ξ ′j ∧ ξ ′′j = η ′j ζ j′′ − η ′′j ζ j′ . A lengthy but The condition a2 b3 −a3 b2 > 0 guarantees that the paramestraightforward computation gives ter point on the geodesic (y, z) = (a2t +b2 , a3t +b3 ) moves η1′ ζ1′′ − η1′′ ζ1′ = −ω (2v2 + ω 2 v2t 2 ). clockwise with respect to the origin in the (y, z)-plane. 2 2 2 2 This expression is negative, because 2v + ω v t > 0 for all t and because we can take ω > 0 without loss of gen7.2 The Vacuum Field Equations erality. Another computation gives
For the first three exercises, we need the following quanη2′ ζ2′′ − η2′′ ζ2′ = −ω 2v2 + ω 2 (1 − vt)2 < 0, tities (taken, with modifications, from the solutions to Exfor a similar reason. ercises 4 and 5 in § 6.1 on Solutions pages 56, 57). 12. The curve ξ (t) is just a special instance of the general Γ211 = sin(q2 ) cos(q2 ), Γ121 = Γ112 = − tan(q2 ), geodesic analyzed in the solution of Exercise 10b, above, with R1212 = 1, R2112 = − cos2 (q2 ), R1112 = R2212 = 0. a0 = 1, b0 = a1 = b1 = 0. -2.0

The formulas for the general first and second derivatives 1. To verify, we calculate separately the second covariant of η and ζ found there, and using the abbreviations C and derivative of ∂ x/∂ q along x = (x1 , x2 ) and collection of S introduced there, along with terms involving the Riemann curvature tensor. We have 2 2 2 3 3 3   1 A = a t +b , A = a t +b , ∂x ∂ x ∂ x2 = (1, 0), = , reduce to ∂q ∂q ∂q

η ′ = (a2 + ω A3 )C + (a3 − ω A2)S, ζ ′ = (a3 − ω A2 )C − (a2 + ω A3)S,

η ′′ = ω (2a3 − ω A2)C − ω (2a2 + ω A3 )S

ζ ′′ = −ω (2a2 + ω A3 )C − ω (2a3 − ω A2)S.

A lengthy but straightforward calculation yields

η ′ ζ ′′ − η ′′ ζ ′ =   − ω (a2 + ω A3)(2a2 + ω A3)(2a3 − ω A2 )(a3 − ω A2)  = −ω (a2 + ω A3)2 + a2(a2 + ω A3) + a3(a3 − ω A2 )  + (a3 − ω A2)2  = −ω (a2 + ω A3)2 + (a3 − ω A2 )2 + (a2 )2 + (a3 )2  + ω a2 a3t − ω a3a2t + ω (a2 b3 − a3b2 ) .

DVI file created at 16:06, 20 January 2011

and also dx = dq



dx1 dx2 , dq dq



= (0, 1).

Therefore,

∂ x1 dx2 D ∂ x1 d ∂ x1 = + Γ112 dt ∂ q dt ∂ q ∂ q dt = 0 − tan(t) · 1 · 1 = − tan(t),

d ∂ x2 D ∂ x2 ∂ x1 dx1 = + Γ211 = 0 − Γ211 · 1 · 0 = 0. dt ∂ q dt ∂ q ∂ q dt The new contravariant vector field F 1 (t) =

D ∂ x1 = − tan(t), dt ∂ q

F 2 (t) =

D ∂ x2 = 0, dt ∂ q

7.2. THE VACUUM FIELD EQUATIONS

87

has, for its covariant derivative along x, the second covari- To compute the covariant derivatives along zi (s) we need ant derivative of ∂ x/∂ q. We have the values of the Christoffel symbols along zi (s); but they are 2 DF 1 dF 1 1 1 dx = + Γ12 · F · Γ211 (z1 , z2 ) = sin(z2 ) cos(z2 ) = 0, dt dt dt = − sec2 (t) + (− tan(t))(tan(t)) · 1 = −1, Γ1 (z1 , z2 ) = Γ1 (z1 , z2 ) = − tan(z2 ) = 0. 12

DF 2 dF 2 dx1 = + Γ211 · F 1 · dt dt dt = 0 + (sin(t) cos(t)) · (− tan(t)) · 0 = 0.

21

This means that covariant derivatives along zi (s) are just ordinary derivatives, so the geodesic separation equations reduce to

That is,

d 2 w2 w2 d 2 w1 D2 ∂ x 2 D2 ∂ x 1 = 0 and = − . = −1, = 0. ds2 ds2 r2 dt 2 ∂ q dt 2 ∂ q For each index i, all terms but one in the Riemann curva- Linear functions solve the first differential equation, and ture tensor have at least one factor equal to zero; we have linear combinations of sin(s/r) and cos(s/r) solve the second. l 2 j 2 k 1 1 dx ∂ x dx 1 dx ∂ x dx R jkl = R212 = 1 · 1 · 1 · 1 = 1, dt ∂ q dt dt ∂ q dt 4. a) From the solution to Exercise 6, § 6.2 (Solutions j l 2 2 k 1 page 64), the only nonzero Christoffel symbols are dx dx dx dx ∂ x ∂ x R2jkl = R2212 = 0 · 1 · 1 · 1 = 0. dt ∂ q dt dt ∂ q dt Γ122 = sinh(q1 ) cosh(q1 ), Γ221 = Γ212 = tanh(q1 ). Thus we see The only nonzero derivatives of these are l j k D2 ∂ x i i dx ∂ x dx + R = 0, i = 1, 2. jkl dt 2 ∂ q dt ∂ q dt ∂ Γ122 ∂ Γ212 2 1 2 1 = cosh (q ) + sinh (q ), = sech2 (q1 ). ∂ q1 ∂ q1 2. Along the equator q2 = 0, all Christoffel symbols of the second kind equal zero. By definition, (q1 , q2 ) are Then, using the remarks about the solution to Exercise 5, Fermi coordinates there. Along the prime meridian (the § 6.1 (Solutions page 57), we obtain the essential elements q2 -axis), the coordinate lines q2 = constant are not geode- of the mixed Riemann curvature tensor as follow. (Each sics (they are parallels of latitude). Therefore, by Corol- nonzero contribution is marked with a bracket.) lary 7.2 (text page 355), (q1 , q2 ) cannot be Fermi coordinates along the prime meridian.

R1112 =

3. a) We have dz1 /ds = 1/r, dz2 /ds = 0; therefore, K hj

= Rh1 j1



dz1 ds

2

=

Rh1 j1 r2

R1212 =

.

We know immediately (because of the skew-symmetry Rhjlk = −Rhjkl ) that Rh111 = 0, so K11 = K12 = 0. For j = 2, we have R1121 = −R1112 = 0, (because

z2

R2121 = −R1112 = cos2 (z2 ) = 1,

= 0); therefore

K21 =

R1121 = 0, r2

K22 =

R2121 1 = 2. r2 r

The geodesic separation equations along the are therefore D2 w1 = 0 and ds2

equator zi (s)

D2 w2 w2 = − . ds2 r2

DVI file created at 16:06, 20 January 2011

∂ Γ112 ∂ Γ111 − + Γ112Γ111 + Γ212 Γ121 ∂ q1 ∂ q2 − Γ111Γ112 − Γ211Γ122 = 0, ∂ Γ122 ∂ Γ121 − 2 + Γ122Γ111 + Γ222Γ121 ∂ q1 ∂q | {z }

− Γ121Γ112 − Γ221Γ122 | {z }

= cosh2 (q1 ) + sinh2 (q1 ) − tanh(q1 ) · sinh(q1 ) cosh(q1 ) = cosh2 (q1 ), R2112 =

∂ Γ212 ∂ Γ211 − 2 + Γ112Γ211 + Γ212Γ221 | {z } ∂ q1 ∂q | {z } − Γ111Γ212 − Γ211Γ222

= sech2 (q1 ) + tanh2 (q1 ) = 1, R2212 =

∂ Γ222 ∂ Γ221 − + Γ122Γ211 + Γ222 Γ221 ∂ q1 ∂ q2 − Γ121Γ212 − Γ221Γ222 = 0.

88

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

On the curve xi (τ , q) we have dx1 /d τ = 1, dx2 /d τ = 0. The geodesic separation tensor on xi is therefore

then

K hj

= Rh1 j1



dx1 dτ

2

= Rh1 j1 .

As in the solution to Exercise 3a, above, K11 = K12 = 0 follows immediately, while K21 = R1121 = −R1112 = 0,

K22 = R2121 = −R2112 = −1.

b) To compute the covariant derivatives of ∂ x j /∂ q along xi , we need to evaluate the Christoffel symbols there: Γ122 = sinh(x1 ) cosh(x1 ) = sinh(τ ) cosh(τ ), Γ221 = Γ212 = tanh(x1 ) = tanh(τ ).

D dτ D F2 = dτ

+ Γ112 · w1 ·

dq2 dq1 + Γ121 · w2 · ds ds

+ w1 tanh(s) − w2 sech(s), + Γ211 · w1 ·

dq1 dq2 + Γ222 · w2 · ds ds

+ w1 sech(s) + w2 tanh(s).

Straightforward calculations then confirm that DF 1 DF 2 D2 w1 D2 w2 and = = 2 ds ds ds ds2 have the forms stated in the exercise. b) We use the elements of the mixed Riemann curvature tensor obtained in the solution to Exercise 6a of § 6.3 (Solutions page 66), evaluated on the curve q:

Now ∂ x1 /∂ q = 0, ∂ x2 /∂ q = 1; therefore F1 =

Dw1 dw1 = ds ds dw1 = ds 2 2 dw Dw = F2 = ds ds dw2 = ds

F1 =

d ∂ x1 ∂ x1 ∂ x2 dx2 = + Γ122 = 0, ∂q dτ ∂ q ∂ q dτ d ∂ x2 ∂ x2 ∂ x2 dx1 = + Γ221 = tanh(τ ). ∂q dτ ∂ q ∂ q dτ

Next, dF 1 dx2 DF 1 = + Γ122F 2 = 0, dτ dτ dτ dF 2 dx1 DF 2 = + Γ221F 2 = sech2 (τ ) + tanh2 (τ ) = 1. dτ dτ dτ We can now see that the geodesic separation equations are satisfied: D2 ∂ x 1 ∂ x1 ∂ x2 + K11 + K21 = 0 + 0 + 0 = 0, 2 dτ ∂ q ∂q ∂q D2 ∂ x 2 ∂ x1 ∂ x2 + K12 + K22 = 1 + 0 − 1 = 0. 2 dτ ∂ q ∂q ∂q

cosh2 (s) = −R1212 = R1221 , r2 R1112 = R2212 = 0.

−R2121 = R2112 = Therefore K11

 2 2 dq2 dq 1 + R212 ds ds ds

dq = R1112

1

− cosh2 (s) 2 · r sech2 (s) tanh2 (s) = − tanh2 (s), r2  1 2 dq dq2 dq1 1 1 K2 = R121 + R1221 ds ds ds =

cosh2 (s) · −r sech(s) tanh(s) · r sech2 (s) r2 = − sech(s) tanh(s),  2 2 dq1 dq2 dq K12 = R2112 + R2212 ds ds ds =

cosh2 (s) · r sech2 (s) · −r sech(s) tanh(s) r2 = − sech(s) tanh(s),  1 2 dq dq2 dq1 K22 = R2121 + R2221 ds ds ds =

5. a) From the text, page 282, we have Γ112 =

−1 , q2

Γ211 =

1 , q2

Γ222 =

−1 ; q2

therefore, on q, these have the values Γ112 = Γ222 =

− cosh(s) , r

Γ211 =

= cosh(s) . r

To compute covariant derivatives along q, we also need dq1 = r sech2 (s), ds

dq2 = −r sech(s) tanh(s); ds

DVI file created at 16:06, 20 January 2011

− cosh2 (s) 2 · r sech4 (s) = − sech2 (s). r2

c) For the family xk (s, c) we have w1 = ∂ x1 /∂ c = 1, w2 = ∂ x2 /∂ c = 0. Therefore, the equations in part (a) reduce to D2 w1 = tanh2 (s), ds2

D2 w2 = tanh(s) sech(s). ds2

7.2. THE VACUUM FIELD EQUATIONS From part (b), K 1j w j = K11 = − tanh2 (s),

K 2j w j = K12 = − sech(s) tanh(s). D2 wi /ds2 + K ij w j

This confirms

d) For the family w1 =

xk (s, r),

= 0 for i = 1, 2.

w2 =

For the hyperbolic plane, we use the values of Rhijk given in the solution to Exercise 6a in § 6.3 (Solutions page 66): R1212 =

−1 = −R1221 , (q2 )2

R2112 =

Rhijk = 0

we have

∂ x1 = tanh(s), ∂r

89

∂ x2 = sech(s). ∂r

The equations in part (a) now reduce to D2 w1 = −2 sech2 (s) tanh(s) + 2 sech2 (s) tanh(s) ds2 + 2 sech2 (s) tanh(s) + tanh3 (s) − tanh(s) sech2 (s) = tanh(s),

D2 w2 = sech(s) tanh2 (s) − sech3 (s) + 2 sech3 (s) ds2 − 2 sech(s) tanh2 (s) + tanh2 (s) sech(s) sech(s) tanh2 (s) = sech(s).

K 1j w j = K11 w1 + K21 w2 = − tanh2 (s) tanh(s)

− sech(s) tanh(s) sech(s) = − tanh(s),

K 2j w j = K12 w1 + K22 w2 = − sech(s) tanh2 (s) − sech3 (s) = − sech(s).

This confirms D2 wi /ds2 + K ij w j = 0 for i = i, 2.

otherwise.

Therefore,   2     R121 R1112 −1/(q2 )2 0 R11 R12 = = R21 R22 R2221 R1212 0 −1/(q2 )2 The discussion of the “cosmological constant” in § 7.3 indicates how global curvature can effect the Ricci tensor even in the absence of matter. The sphere and the hyperbolic plane are both globally curved. b) We have Rhk = gh j R jk = gh1 R1k + gh2R2k . Therefore, for the sphere, R11 = g11 R11 = R22 = g22 R22 =

From part (b),

+1 = −R2121 , (q2 )2

1 r2 cos2 (q2 )

· cos2 (q2 ) =

1 , r2

1 1 ·1 = 2. r2 r

Consequently, the scalar curvature is R = Rii = 2/r2 . For the hyperbolic plane, −1 = −1, (q2 )2 −1 R22 = g22 R22 = (q2 )2 · 2 2 = −1, (q ) R11 = g11 R11 = (q2 )2 ·

6. Let p and q be arbitrary integers between 1 and n. Let vi = 1 if i = p and vi = 0 otherwise; similarly, let w j = 1 so R = Rii = −2. if j = q and w j = 0 otherwise. Then 8. From Exercise 4a, above, we have A pq = Ai j vi w j = Bi j vi w j = B pq ; R1212 = cosh2 (q1 ) = −R1221 , since 1 ≤ p, q ≤ n were arbitrary, we have established the R2112 = 1 = −R2121 , Rhijk = 0 otherwise. claim. 7. a) For a 2-dimensional surface, the Ricci tensor is Rik = Rhihk = R1i1k + R2i2k For the sphere, we use the values of Rhijk given before the solution to Exercise 1 in this section (Solutions page 86): R1212 = 1 = −R1221 ,

R2112 = − cos2 (q2 ) = −R2121 ,

Rhijk = 0 otherwise.

Therefore,   2    2 2  R121 R1112 R11 R12 cos (q ) 0 = = R21 R22 0 1 R2221 R1212

DVI file created at 16:06, 20 January 2011

Therefore,   2     −1 0 R121 R1112 R11 R12 = = R21 R22 R2221 R1212 0 cosh2 (q1 ) Note that Ri j = −gi j ; using g11 = 1, g22 = −1/ cosh2 (q1 ), we then have R11 = g11 R11 = −1, −1 R22 = g22 R22 = · cosh2 (q1 ) = −1, cosh2 (q1 ) and finally R = Rii = −2.

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

90

9. a) We make some preliminary observations. First, be- b) Becauseζ (0) = 0, we have e2αζ = 1 and the metric cause Γ0i j = Γ1i j = 0, we have R0i jk = R1i jk = 0. Second, in the tangent plane T G(0,0) is the Minkowski metric J1,1 . because Let Q be the quadratic form for this metric; then Γ200 = −ω 2 η ,

Γ203 = Γ230 = −ω ,

Γ300 = −ω 2 ζ ,

Γ302 = Γ320 = ω ,

are the only nonzero Christoffel symbols, and

∂ Γ200 ∂ Γ200 = = −ω 2 , ∂ξ2 ∂η

∂ Γ300 ∂ Γ200 = = −ω 2 , ∂ξ3 ∂ζ

Q(e0 ) = 1,

Q(e3 ) = −1.

Therefore e0 is a timelike unit vector and e3 is a spacelike unit vector. Let V l (T ) denote the parallel translate of either e0 or e3 along ζ (T ) = (τ (T ), ζ (T )) = (ξ 0 (T ), ξ 3 (T )). Then V l satisfies the differential equation

dV l dξ k DV l are the only nonzero derivatives, the definition of R2jkl or = + Γljk V j . 0= dT dT dT R3jkl requires that two of the covariant indices j, k, l must be 0. The only possibilities are We have d ξ 0 /dT = τ ′ and d ξ 3 /dT = ζ ′ , from above; furthermore, the nonzero Christoffel symbols are ∂ Γ200 2 3 2 2 + Γ02 Γ30 = ω + ω · (−ω ) = 0, R002 = − ∂ξ2 Γ003 = Γ030 = Γ300 = Γ333 = α . 3 ∂Γ + Γ203 Γ320 = ω 2 + (−ω ) · ω = 0. R3003 = − 00 Therefore, the differential equations take the form ∂ξ3 dV 0 dξ 3 dξ 0 Therefore, = −Γ003 V 0 − Γ030 V 3 dT dT dT α 2T α R00 = R2020 + R3030 = −R2002 − R3003 = 0. = V0 − V 3, 1 − α 2T 2 1 − α 2T 2 For all other elements, we have dV 3 dξ 0 dξ 3 = −Γ300 V 0 − Γ333 V 3 dT dT dT Rik = Rhihk = R2i2k + R3i3k = 0; 2 α α T =− V0 + V 3. the Ricci tensor is identically zero. 1 − α 2T 2 1 − α 2T 2

b) The τ -axis is a particular instance of a geodesic that Let V ⊕ = V 0 + V 3 , V ⊖ = V 0 − V 3 ; then was established in the solution to Exercise 10b, § 7.1 (So−α (1 − α T ) α 2T − α ⊕ dV ⊕ lutions page 85; a1 = a2 = a3 = b1 = b2 = b3 = 0). HowV = = V ⊕, ever, G’s coordinates are not Fermi coordinates along the dT 1 − α 2T 2 (1 − α T )(1 + α T ) τ -axis because some Christoffel symbols (namely Γ203 and dV ⊖ α 2T + α ⊖ −α (1 + α T ) = V ⊖. V = Γ302 ) are nonzero there. dT 1 − α 2T 2 (1 − α T )(1 + α T )

10. a) The curve ζ (T ) is one of the timelike geodesics These differential equations are easier to solve; in particgiven on page 338 of the text, with k1 = 0, k2 = 1/α , ular, by separating variables we get t = T . To show that T is proper time, we must show α −α dV ⊖ dV ⊕ kζ ′ (T )k2 = 1. Because e2αζ = 1 − α 2 T 2 , we have = = dT, dT. ⊕ V 1 + αT V⊖ 1 − αT
′ 2 ′ 2 ′ 2 2 2 kζ (T )k = (τ ) − (ζ ) (1 − α T )   The solutions are 1 −α T 2 2 = (1 − α T ) ≡ 1, − 1 − α 2T 2 1 − α 2 T 2 ln(V ⊕ ) = − ln(1 + α T ) + ln(A), ln(V ⊖ ) = − ln(1 − α T ) + ln(B),

as required. The curve with α = 1 is sketched below. -2

1

-1

2

-0.5

-1.0

-1.5

DVI file created at 16:06, 20 January 2011

or

B A , V⊖ = , 1 + αT 1 − αT where A and B are arbitrary constants, to be determined by the initial conditions on V l . V⊕ =

7.2. THE VACUUM FIELD EQUATIONS For e0 , V 0 (0) = 1, V 3 (0) = 0, so V ⊕ (0) = V ⊖ (0) = 1 and 1 , 1 − α 2T 2 −α T V 3 (T ) = V ⊕ − V ⊖ = . 1 − α 2T 2

V 0 (T ) = V ⊕ + V ⊖ =

For e3 , V 0 (0) = 0, V 3 (0) = 1, so V ⊕ (0) = 1, V ⊖ (0) = −1 and V 0 (T ) =

−α T , 1 − α 2T 2

V 3 (T ) =

1 . 1 − α 2T 2

c) From the solution just obtained,   1 −α T . , v= 1 − α 2T 2 1 − α 2T 2

91 implying k7 = 1, and the geodesic we seek is   1 T 0 −1 , ξ (S) = tanh α S + 1/α 1 ξ 3 (S) = ln(α 2 ((S + 1/α )2 − T 2 )). 2α To determine the solution of the geodesic equations ab initio, let xi = d ξ i /dS, i = 0, 3. In terms of xi , the geodesic equations are dx0 = −2α x0 x3 , dS

dx3 = −α (x0 )2 − α (x3 )2 . dS

Now let x⊕ = x0 + x3 , x⊖ = x0 − x3 ; then dx⊕ = −α (x⊕ )2 , dS

dx⊖ = α (x⊖ )2 . dS

Because v is spacelike, the geodesic we seek is spacelike. The method of separation of variables gives The entire eight-parameter family of spacelike geodesics 1 1 was found in the solution to Exercise 5c, §7.1 (Solutions , x⊖ = . x⊕ = page 83, with t = S). The part we need is c1 + α S c2 − α S   The initial conditions 1 k2 ξ 0 (u) = tanh−1 , α u −α T 1 x0 (0) = , x3 (0) = , 1 2T 2 2T 2 3 2 2 2 1 − α 1 − α ξ (u) = ln(α (u − k2 )), 2α become u = k7 S + k8 . 1 1 1 1 = x⊕ (0) = , = x⊖ (0) = ; The initial conditions η (0) = ζ (T ) translate into u = k8 αT + 1 c1 αT − 1 c2 and   hence 1 1 k2 = tanh−1 (α T ), tanh−1 1 1 α k8 α , x⊖ = . x⊕ = α T + 1 + α S α T − 1 − αS 1 1 ln(α 2 (k82 − k22 )) = ln(1 − α 2T 2 ). 2α 2α In terms of d ξ 0 /dS = x0 and d ξ 3 /dS = x3 , we have Hence dξ 0 dξ 3 −(1 + α S) αT k2 , = = 2 2 2 2 2 = α T, α (k8 − k2 ) = 1 − α T , dS (α T )2 − (1 + α S)2 dS (α T )2 − (1 + α S)2 k8 We rewrite the first equation as or k2 = α T k8 ,

α 2 k82 (1 − α 2T 2 ) = 1 − α 2 T 2 .

αT −T dξ 0 1 , = 2 2 = dS α (T − (S + 1/α )2) α (S + 1/α )2 − T 2

Thus we can take k8 = 1/α and k2 = T . To determine k7 , we use the initial condition η ′ (0) = v. In fact, we need and find (setting the constant of integration equal to zero) only   1 T 0 −1 d ξ 0 du −k2 1 dξ 0 . ξ = tanh = = · · k7 α S + 1/α dS du dS α u2 − k22 When S = 0,

1 −α T k7 · − α T d ξ 0 −k2 · 2 = , = · k7 = 2 2 2 dS α k8 − k2 1−α T 1 − α 2T 2

DVI file created at 16:06, 20 January 2011

We rewrite the second equation as

dξ 3 1 α (S + 1/α ) 2(S + 1/α ) , = 2 = 2 2 dS α ((S + 1/α ) − T ) 2α (S + 1/α )2 − T 2

92

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

and find (incorporating an additive integration constant in the form ln c3 /2α into the main expression)

with

ξ3 =


1 ln c3 ((S + 1/α )2 − T 2 ) . 2α

tanh(ατ ) = Then

αT sinh(ατ ) = p (α S + 1)2 − α 2 T 2

The initial condition is


1 1 ln(1 − α 2T 2 ) = ξ 3 (0) = ln c3 (1/α )2 − T 2 , 2α 2α

implying c3 = α 2 :

ξ3 =


1 ln α 2 ((S + 1/α )2 − T 2 ) . 2α

αT . αS + 1

Now

q αζ = ln (α S + 1)2 − α 2 T 2 ,

so

eαζ = and hence

q (α S + 1)2 − α 2 T 2 ,

αT . eαζ d) We constructed (T, S) as Fermi coordinates. On the Therefore, the curve T = constant is the graph of geodesic C, given as ζ (T ) = (τ (T ), ζ (T )), we have   αT 1 tanh(ατ ) = α T. τ = sinh−1 αζ . α e An identity for hyperbolic functions then provides In the figure below, α is chosen to equal 1; the curves p p 1 S = constant are the ones that are horizontal when τ = 0. 2 = 1 − tanh (ατ ) = 1 − α 2 T 2 ; cosh(ατ ) 1.0 sinh(ατ ) =

hence − ln(cosh(ατ )) 1 = ln(1 − α 2T 2 ) = ζ . α 2α That is, C is the graph of ζ = − ln(cosh(ατ ))/α , and it is also the curve for which S = 0. For S = constant 6= 0, the analysis is similar. tanh(ατ ) =

αT T = . S + 1/α αS + 1

The hyperbolic function identity yields p q (α S + 1)2 − α 2 T 2 1 2 = 1 − tanh (ατ ) = . cosh(ατ ) αS + 1 Therefore,

0.5

0.0

-0.5

-1.0 -1.0

-0.5

0.0

0.5

1.0

7.3 The Matter Field Equations

1. By the intrinsic definition of the Christoffel symbols 1 ln( − ln(cosh(ατ )) α S + 1) of the first kind, = ln((α S + 1)2 − α 2 T 2 ) − , α α |2α {z } ∂ g jh ∂ gi j ∂ gih 2Γih, j = + h − , ζ ∂ xi ∂x ∂xj or ∂ gih ∂ g ji ∂ g jh + h − , 2Γ jh,i = − ln(cosh(ατ )) ln(α S + 1) ∂xj ∂x ∂ xi ζ= + . α α ∂ gi j ∂ g ji ∂ gi j 2Γih, j + 2Γ jh,i = + h =2 h, h This is a vertical translation of C by ∆ζ = ln(α S + 1)/α , ∂x ∂x ∂x For the curves T = constant, we make use of the idenor Γih, j + Γ jh,i = ∂ gi j /∂ xh . Therefore, tity tanh(ατ ) sinh(ατ ) = q , ∂ gi j j gi j h = gi j Γih, j + gi j Γ jh,i = Γiih + Γ jh = 2Γkkh . 1 − tanh2 (ατ ) ∂x

DVI file created at 16:06, 20 January 2011

7.3. THE MATTER FIELD EQUATIONS 2. a) We use the fact that g21 = g12 ; then g22 1 ∂ g −g21 1 ∂ g = , g12 = = , g g ∂ g11 g g ∂ g12 −g12 1 ∂ g g11 1 ∂g g21 = = , g22 = = , g g ∂ g21 g g ∂ g22 g11 =

b) The solutions to Exercise 1 and 2a together with the chain rule (∂ g/∂ xh = ∂ g/∂ gi j · ∂ gi j /∂ xh ) establish 1 1 ∂ g ∂ gi j 1 1 ∂g 1 ∂ gi j = Γkkh = gi j h = 2 ∂x 2 g ∂ g i j ∂ xh 2 g ∂ xh

93 c) This is Theorem 5 in Ch. X, § 2 of A Survey of Modern Algebra. d) The symmetry of G allows us to write the result of part (c) as 1 ∂g 1 . g i j = ∆i j = g g ∂ gi j The rest of the argument in the solution to Exercise 2 then goes through unchanged, leading once again to √
1 ∂ α h −g . ah;h = √ −g ∂ xh

Then, quite generally, 1 ∂g ∂ ln |g| = , h g ∂ xh ∂x

so

Γkkh =

1 ln |g| . 2 ∂ xh

We are given that g < 0, so |g| = −g and √ 1 ln |g| = ln −g, 2 so Γkkh

√ √ ∂ ln −g 1 ∂ −g =√ . = ∂ xh −g ∂ xh

c) By the definition of the covariant derivative and by the solution to part (b), √ ∂ αh ∂ αh α k ∂ −g k h √ + + . Γ = α;hh = α hk ∂ xh ∂ xh −g ∂ xk It is also true (by the product rule) that √
√ √ 1 ∂ α h −g ∂ α h −g α h ∂ −g √ √ √ = . + −g −g ∂ xh ∂ xh ∂ xh −g Hence, with a change k → h in the dummy index, we get √
1 ∂ α h −g h a;h = √ . −g ∂ xh

4. a) By definition, Rik = Rhihk =

∂ Γhik ∂ Γhih p h − + Γikp Γhph − Γih Γ pk ∂ xh ∂ xk

For the second term on the right, we have (Exercise 2b) √ √ ∂ Γhih ∂ ln −g ∂ 2 ln −g Γhih = , so = . ∂ xi ∂ xk ∂ xi ∂ xk Replacing Γhph in the third term in a similar way, we obtain Rik =

√ √ ∂ Γhik ∂ 2 ln −g p ∂ ln −g p h − + Γ − Γih Γ pk . ik ∂ xh ∂ xi ∂ xk ∂ xp

b) We must show that the terms are unchanged when we exchange i ↔ k. For the second term, this is obvious. For the first and third, we have Γki∗ = Γik∗ . For the last term, we use the following sequence of equalities that finish by exchanging the dummy indices p ↔ h: p h p h p p h Γkh Γ pi = Γhk Γip = Γhip Γhk = Γih Γ pk .

All term remain unchanged, so Rki = Rik . c) If g ≡ −1, then the Ricci tensor reduces to Rik =

∂ Γhik p h − Γih Γ pk . ∂ xh

3. a) This is a standard result in linear algebra; see, for example, the classic text A Survey of Modern Algebra by 5. We are given that G is the rest frame for the swarm and Birkhoff and MacLane, Ch. X, § 1. that νe is the numerical density of the swarm at rest. [To b) Each matrix Gip , p = 1, . . . , 4, excludes the entire ith distinguish density ν (Greek nu) from velocity v (Roman row of G, and therefore excludes the element gi j in that vee), we continue to put a tilde over the nu; see the comrow. Therefore, ∆ip = ± detGip is independent of that el- ment on page 18 above.] Suppose C’s 3-velocity in G’s ement, so ∂ ∆ip /∂ gi j = 0. If frame is V; the text shows that √ the numerical density of the swarm in C’s frame is νe / 1 − V 2 (where V denotes ip g = gip ∆ , sum on p but not i, the Euclidean length of V in the usual way). In G, we can write C’s proper 4-velocity in G as then we must have   ∂ gip ip ∂g 1 1 = ∆ + gip · 0 = δ j p ∆ip = ∆i j . UG = √ ,√ V . ∂ gi j ∂ gi j 1 −V2 1 −V2

DVI file created at 16:06, 20 January 2011

94

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

In G, the particle density 4-vector is

of 4-velocity U all have the dimensions of velocity. Because the dimensions of µ and νe are mass and length−3, respectively, the dimensions of each T i j are

NG = (νe , 0),

mass × velocity2 energy because the swarm is at rest there. Their Minkowski inner = . 3 product is length length3 νe NG · UG = √ , 1 −V2 b) Energy has dimensions of mass × velocity2 while momentum has dimensions of mass × velocity. Because T α 0 the numerical density of the swarm in C’s frame. 3 Now suppose R moves with 3-velocity v with respect has dimensions of energy/length , it represents spatial α density of energy. By contrast, T 0 /c has dimensions of to G; then the boost Bv maps G to R, and momentum/length3 and thus represents spatial density of of momentum in the α -direction. UR = Bv (UG ), NR = Bv (NG ). By expressing the dimensions of T i j in the form Because Minkowski inner product is invariant under 1 kg  m 2 kg × m/sec boosts, numerical density of the swarm in C’s frame is × = × 2 3 m sec sec m νe we see that it represents the flow of momentum across a √ = NG · UG = Bv (NG · UG ) = NR · UR , 1 −V2 plane (namely, a plane perpendicular to the j-direction). Therefore, c · T 0β represents the flow of energy across a as we had to show. plane perpendicular to the β -direction. 6. a) By definition, the energy–momentum tensor of a swarm of noninteracting particles is defined as T i j = pi n j where P = (p0 , p1 , p2 , p3 ),

N = (n0 , n1 , n2 , n3 )

are the 4-vectors of momentum and particle density of the swarm, respectively, in a particular inertial frame. If the proper 4-velocity of the swarm is U = (u0 , u1 , u2 , u3 ),

µ is individual rest mass, and νe is rest density, then pi = µ ui

It follows that

and ni = νeui .

T i j = µ νeui u j ,

8. a) First of all, the components qi of an event all have the dimensions of length; the components gi j of the metric tensor are dimensionless. This can be seen from the computation s Z b a

gi j

dqi dq j dt dt dt

of the length of the curve qi (t). Therefore g and the gi j are dimensionless; neither raising nor lowering an index alters dimensions. The Christoffel symbols, as combinations of ∂ gi j /∂ qk , have the dimensions of length−1 . The Riemann curvature tensor, involving products of Christoffel symbols and derivatives of them by the ql , has the dimensions of length−2 . The same is true of the Ricci tensor. The text, page 171, reports that the potential function Φ has the dimensions of velocity2 . Therefore, after two spatial derivatives are taken, ∇2 Φ has the dimensions of time−2 .

i j is diand this expression holds whether traditional or dimen- 9. a) The solution to Exercise 8 established that g−2 mensionless and Ri j has the dimensions of length . The sionally homogeneous coordinates are used. same must be true of the contraction R. The solution to b) The previous result makes it immediate that T i j is Exercise 8 also notes that lowering an index does not alter symmetric: dimensions, so Ti j has the dimensions

T ji = µ νeu j ui = µ νeui u j = T i j .

7. a) When the components x0 , . . . , x3 of 4-position X all have the dimensions of length, the components u0 , . . . , u3

DVI file created at 16:06, 20 January 2011

mass × length2 /time2 mass energy = = . 3 length length3 length × time2

Because Λ is introduced in the equation ∇2 Φ − ΛΦ = 4π Gρ , it must have the dimensions length−2 .

7.3. THE MATTER FIELD EQUATIONS

95

b) The text, on page 375, establishes that the portion b) Let θ , ϕ , and ω be q1 , q2 , and q3 , respectively, with Ri j − 21 gi j R has zero divergence. For the other portion we xi = ∂ x/∂ qi . Then have  x1 = −r sin θ cos ϕ cos ω ,  (gi j Λ);k = gi j;k Λ = 0   y = r cos θ cos ϕ cos ω , 1 x1 : by the solution to Exercise 9 of § 6.5 (Solutions page 76).  z 1 = 0,  This implies zero divergence for the second portion and   u1 = 0; hence for the entire left-hand side.  x2 = −r cos θ sin ϕ cos ω ,  c) With Ti j ≡ 0, raising an index in the equation gives   y = −r sin θ sin ϕ cos ω , 2 x2 : 0 = Rhi − 12 δih R − δih Λ.  z   2 = r cos ϕ cos ω ,  u2 = 0; Contraction then gives (with Rii = R, δii = 4)   x3 = −r cos θ cos ϕ sin ω ,  y = −r sin θ cos ϕ sin ω , 0 = R − 2R − 4Λ = −R − 4Λ, 2 x3 : z2 = −r sin ϕ sin ω ,   or R = −4Λ. Therefore, still with Ti j ≡ 0, we have  u2 = r cos ω .
1 Ri j = gi j 2 R + Λ = gi j (−2Λ + Λ) = −gi j Λ. Therefore the metric (ai j ) is given by the 3 × 3 diagonal matrix d) The equation Ri j = gi j Λ just obtained declares that the  2 2  r cos ϕ cos2 ω empty universe (Ti j ≡ 0) is not flat (i.e., Ri j 6= 0) if Λ 6= 0.  . r2 cos2 ω 4 e) For simplicity, let κ = 8π G/c , so the field equations r2 have the form c) A 4 × 3 matrix M defines a linear map M : R3 → R4 . The image of the unit cube is a parallelepiped in R4 whose edges are the three columns of M. Theorem 7 of Ch. X, Raising an index and contracting gives § 3 of Birkhoff and MacLane’s A Survey of Modern Algebra declares that the square of the volume of that paralR − 2R − 4Λ = κ T. lelepiped is det Mt M. (Note: in the book, our linear map Thus −R− 4Λ = κ T , so R+ 2Λ = −κ T − 2Λ. This allows M acts on a row vector as multiplication by Mt on the us to rewrite the field equations as right; to compensate, we have interchanged M and Mt in √ t the statement of the theorem.) Thus, det M M is the volRi j − 12 gi j (−κ T − 2Λ) = κ Ti j , ume magnification factor of the map M. Let X be the 4 × 3 matrix whose columns √ are the vecor tors x , x , x , evaluated at a point q. Then detX t X is 1 2 3
1 Ri j + gi j Λ = κ Ti j − 2 gi j T . the local volume magnification factor for the map x in the 3-dimensional tangent space to S3 (r) at x(q). The entry in the ith row and jth column of X t X is xi ·xp j ; in other words, √ 10. a) We have the local magnification factor for x is det(ai j ) = a. Therefore, if D is any region in Ω, x2 + y2 = r2 cos2 ϕ cos2 ω ; ZZZ √ a dθ dϕ dω . vol D = therefore D x2 + y2 + z2 = r2 cos2 ω Now a = r6 cos2 ϕ cos4 ω , so when D = Ω we have and finally Z 2π Z π /2 Z π /2 x2 + y2 + z2 + u2 = r2 . V = vol Ω = r3 d θ cos ϕ d ϕ cos2 ω d ω −π /2 −π /2 0 3 This implies that the image of x lies in S (r). The ranges = 2π r3 · 2 · 12 π = 2π 2 r3 . of θ , ϕ , and ω guarantee that x is onto. Ri j − 12 gi j (R + 2Λ) = κ Ti j .

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 7. GENERAL RELATIVITY

96

11. a) The Minkowski metric on E = R × S3(r) will just because of the symmetry Rki = Rik , it is sufficient to conbe the unit metric on “time” factor R combined with neg- sider i ≤ k. There are no nonzero terms in which at least ative of the metric (ai j ) on the “space” factor S3 (r). This one index is 0; thus all R0k = 0. Now consider i = 1. When k = 1 the second derivative yields the given (gi j ). of L is zero and we have b) The only nonzero derivatives are ∂ Γ211 ∂ Γ311 ∂L ∂L R = + + Γ211 2 + Γ311 3 ∂ g11 11 2 2 2 3 ∂ x ∂ x ∂ x ∂ x = 2r cos ϕ sin ϕ cos ω , ∂ x2 1 2 1 3 2 1 − Γ12 Γ11 − Γ13 Γ11 − Γ11 Γ21 − Γ311 Γ131 ∂ g11 2 2 = 2r cos ϕ cos ω sin ω , = cos2 ϕ − sin2 ϕ + cos2 ϕ cos2 ω − cos2 ϕ sin2 ω ∂ x3 ∂ g22 − sin2 ϕ − 2 cos2 ϕ sin2 ω + 2 sin2 ϕ + 2 cos2 ϕ sin2 ω = 2r2 cos ω sin ω .
3 ∂x = cos2 ϕ 1 + cos2 ω − sin2 ω = 2 cos2 ϕ cos2 ω Therefore, 2 = − 2 g11 . 2 2 r Γ11,2 = −Γ12,1 = −r cos ϕ sin ϕ cos ω , When k = 2 or 3, then Γh1k = 0 when h = 2 or 3, so the first and third terms in the formula for R1k are zero. The Γ22,3 = −Γ23,2 = −r2 cos ω sin ω . second term continues to be zero. The fourth term has the To determine the Christoffel symbols of the second kind form p Γ1h Γhpk . we need Γ11,3 = −Γ13,1 = −r2 cos2 ϕ cos ω sin ω ,

g11 =

−1 , 2 2 r cos ϕ cos2 ω

g22 =

−1 , 2 r cos2 ω

g33 =

−1 ; r2

then Γ211 = g22 Γ11,2 = sin ϕ cos ϕ , Γ311 = g33 Γ11,3 = cos2 ϕ sin ω cos ω , = g Γ22,3 = sin ω cos ω , − sin ϕ = − tan ϕ , Γ121 = Γ112 = g11 Γ12,1 = cos ϕ − sin ω Γ131 = Γ113 = g11 Γ11,3 = = − tan ω , cos ω − sin ω Γ232 = Γ223 = g22 Γ23,2 = = − tan ω . cos ω Γ322

33

c) We have g = −a = −r6 cos2 ϕ cos4 ω and √ L = ln −g = ln(r3 cos ϕ cos2 ω ) = 3 ln r + ln(cos ϕ ) + 2 ln(cos ω ). In preparation for computing Rik we note

∂L = − tan ϕ , ∂ x2 ∂ 2L = − sec2 ϕ , ∂ (x2 )2

∂L = −2 tan ω , ∂ x3 ∂ 2L = −2 sec2 ω , ∂ (x3 )2

while all other first and second derivatives of L are zero. We now compute Rik using the formula Rik =

∂ Γhik ∂ 2L ∂L p h − + Γikp p − Γih Γ pk ; ∂ xh ∂ xi ∂ xk ∂x

DVI file created at 16:06, 20 January 2011

If p = 1, then h must be 2 or 3 for the first factor to be nonzero, but then the second factor is zero. If p = 2 or 3, then h must be 1 to make the first factor nonzero, but then the second factor is zero. Thus R1k = 0; because g1k = 0 also, R1k = −2g1k /r2 . Now consider i = 2. When k = 2 we have R22 =

∂ Γ322 ∂ 2L ∂L − + Γ322 3 ∂ x3 ∂ (x2 )2 ∂x − Γ121 Γ112 − Γ223Γ322 − Γ322Γ232

= cos2 ω − sin2 ω + sec2 ϕ − 2 sin2 ω

2 g22 . r2 When k = 3, the first and second terms are zero, and we have ∂L R23 = Γ223 2 − Γ121 Γ113 = tan ω tan ϕ − tan ϕ tan ω = 0. ∂x − tan2 ϕ + 2 sin2 ω = 2 cos2 ω = −

Because g23 = 0, we have R23 = −2g23/r2 . Finally, consider i = 3 and k = 3. We have R33 = −

∂ 2L − Γ131 Γ113 − Γ232Γ323 ∂ (x3 )2

= 2 sec2 ω − tan2 ω − tan2 ω = 2. Because g33 = −r2 , we have R33 = −2g33/R2 . Thus R00 = 0 and Ri j = −2gi j /r2 for all (i, j) 6= (0, 0). 12. a) For the mixed tensor we have ( ρ c2 , 0 0h 00 T j = T gh j = T g0 j = 0,

j = 0, j 6= 0

7.3. THE MATTER FIELD EQUATIONS If i 6= 0, then T ji = T ih gh j = 0. For the covariant tensor, T0 j = g0h T jh

= g00 T j0

=

(

ρ c2 , 0,

j = 0, j 6= 0.

If i 6= 0, then Ti j = gih T jh = gii T ji = 0. In summary, T00 = T00 = ρ c2 ,

T ji = Ti j = 0 otherwise.

The Laue scalar is T = Tii = ρ c2 . b) First take i = j = 0; then the matter field equation reduces to 0+Λ·1= or


8 π G ρ c2 8π G , ρ c2 − 12 · 1 · ρ c2 = 4 · 4 c c 2 Λ=

4 π Gρ . c2

For (i, j) 6= (0, 0), we use Ri j = −2gi j /r2 and the matter field equation becomes   ρ c2 8π G 4 π Gρ 2 gi j = − 2 gi j . − 2 + Λ gi j = 4 · − r c 2 c When i = j, we can divide by gi j and get −

2 4 π Gρ +Λ = − 2 . r2 c

DVI file created at 16:06, 20 January 2011

97

Solutions: Chapter 8

Consequences 8.1 The Newtonian Approximation

2. a) Let B = O(a) as a → 0; then O(B) = O(a) and

1 = 1 − B + B2 + · · · = 1 + O(B), 1. The determinant of (gi j ) consists of products of four 1+B factors, with one factor taken from each row and each col√ 1 + B = 1 + 12 B + · · · = 1 + O(B). umn. One possibility is to choose the factors on the main diagonal. The resulting product is These translate to  ! (2) 3 p g 1 1 1 + O(a) = 1 + O(a). = 1 + O(a), ∏ Jii + cii2 + O c4 1 + O(a) i=0   (2) (2) (2) (2) g g g g 1 b) By part (a), = −1 − 002 + 112 + 112 + 112 + O 4 . c c c c c s     Every other such product must exclude at least two of the 1 dt 1 = 1+O 2 = 1+O 2 , main diagonal entries, and thus has order at least 1/c4 . dτ c c Therefore,   and (2) (2) (2) (2) g − g11 − g22 − g33 1   . + O −g = 1 + 00 1 1 dτ 1 c2 c4 . = = 1 + O = 2) 2 dt dt/d τ 1 + O(1/c c For the logarithm function we have ln(1 + A) = A + O(A2) as A → 0;

3. Page 392 of the text notes the following:     1 ∂ gkl 1 Γk0,k = O 3 , , = O c ∂ x0 c3

therefore, if we let A=

  (2) (2) (2) (2) g00 − g11 − g22 − g33 1 , + O 2 c c4

then A2 = O(1/c4 ) and √ ln −g = 21 ln(1 + A) =

(2) (2) (2) (2) g00 − g11 − g22 − g33 2c2

+O



1 c4

while, in general





1 Γkl,i = O 2 c .

gi j g jk = J i j J jk + O



1 c2



= δki + O

,

  ∂ gkl 1 =O 2 , α ∂x c

(We keep the convention that Roman indices i, j, . . . run from 0 to 3 while Greek indices α , β , . . . run from 1 to 3.) We have

Finally, the inverse of g jk has the form   1 ij ij g = J +O 2 c because



Γhh0 = ghk Γho,k = ghh Γh0,h + ghl Γh0,l , 

1 c2



l 6= h.

The nondiagonal element ghl has order 1/c2 so every ghl Γh0,l has order at least 1/c4. The diagonal element ghh has order 1, so the term ghh Γh0,h has order 1/c3; it dominates Γhh0 , which thus has order 1/c3 .

.

98

DVI file created at 16:06, 20 January 2011

8.1. THE NEWTONIAN APPROXIMATION

99

When (k, l) 6= (0, 0) we can write

b) We must simply modify the argument in STEP 2, text page 391: Γhkl = ghh Γkl,h + gh j Γkl, j , j 6= h.    dx0 dx0 dt 1 hh h j 2 = c + O(1), = = c 1 + O Because g has order 1 and g has order 1/c , while d τ dt d τ c 2    Γkl,h and Γkl, j have order 1/c , the first term dominates  dt d 2 x0 d 1 and gives Γhkl order 1/c2 . c + O(1) = O(1). = = O(1) 1 + O dτ 2 dt dτ c 4. We modify STEP 1 of the discussion in the text, pages In the Newtonian approximation, the first relativistic 390–391, by incorporating the new assumption x˙α = O(c) 0 k l 2 equation of motion is ignored as negligible because all along with x˙ = c. We now have x˙ x˙ = O(c ) for all k, l, terms have order at most 1/c. It then becomes possible so k l to identify the remaining three with the three Newtonian gkl x˙ x˙ = g00 + O(1) = 1 + O(1). 2 equations. c In the present circumstances, the first relativistic equaTherefore, tion is not negligible because the left-hand side, at least, s has order 1. There are now four four significant relativisk l p gkl x˙ x˙ 1 + O(1) dt = (1 + O(1)) dt. dτ = dt = tic equations, so it is impossible to match them with the c2 three Newtonian equations. In this way the Newtonian As a consequence, the difference between the proper-time approximation breaks down. and the world-time derivatives of the spatial variables is c) To find the dominant terms in Γα it will be enough to 00 not negligible. We have modify STEP 3 of the text (page 392). The new facts we dxα dt dxα = = x˙α (1 + O(1)) = x˙α + O(1), dτ dt d τ  dt d 2 xα d α = x¨α + O(1). x˙ + O(1) = 2 dτ dt dτ 5. a) To determine the new relation between d τ and dt, we still use the metric expression x˙α x˙α X˙ β gkl x˙k x˙l = g00 + 2g0α + gαβ 2 . 2 c c c If we now assume   1 (1) 1 gkl = Jkl + g jk + O 2 , c c

then the metric expression becomes (using x˙α = O(1))     1 1 1 +O 2 +O 2 +2 ·O 1+ c c c c c   (1) g 1 = 1 + 00 + O 2 . c c √ Therefore, because 1 + x = 1 + x/2 + O(x2), we have s s   (1) g00 gkl x˙k x˙l 1 1 + = + O 2 c c c2   (1) g 1 = 1 + 00 + O 2 . 2c c (1) g00





(1) g 0α

(1)

Thus d τ = (1 + A/c + O(1/c2)) dt, where A = g00 /2.

DVI file created at 16:06, 20 January 2011

take into account are     1 ∂ gkl ∂ gkl 1 , =O 2 , =O 0 α ∂x c ∂x c

  1 g = J +O c ij

ij

Then Γα00 = gαα Γ00,α + gα 0Γ00,0 + gαβ Γ00,β       ∂ g0α ∂ g00 1 1 = −1 + O 2 0 − α c 2 ∂x ∂x       ∂ g0β ∂ g00 1 1 1 1 ∂ g00 2 + O − +O c 2 ∂ x0 c 2 ∂ x0 ∂ xβ   (1) 1 1 ∂ g00 +O 2 , = α 2c ∂ x c and Γα0β = gαα Γ0β ,α + gα 0Γ0β ,0 + gαγ Γ0β ,γ       1 1 ∂ gαβ ∂ g0α ∂ g0β = −1 + O − + c 2 ∂ x0 ∂ xα ∂ xβ   1 1 ∂ g00 +O c 2 ∂ xβ     1 1 ∂ gγβ ∂ g0β ∂ g0β +O − + c 2 ∂ x0 ∂ xγ ∂ xγ   (1)   (1) ∂ g 0α  1 1  ∂ g 0β + O = − . α β 2c ∂ x c2 ∂x

With these facts, the relativistic equation  α 2 0 β β γ dx d 2 xα α α dx dx α dx dx = −Γ − 2Γ − Γ 00 0β βγ dτ 2 dτ dτ dτ dτ dτ

SOLUTIONS: CHAPTER 8. CONSEQUENCES

100 assumes the form x¨α = −

(1)

c ∂ g00 1 − α 2 ∂x 2



(1) ∂ g 0β  ∂ xα

−

and



  (1) ∂ g 0α  β 1 + O x ˙ β c ∂x

γ Rαγβ

This differs from the Newtonian expression two ways: in the presence of the factor c in the first term on the right, and in the presence of a second term involving x˙β . 6. a) P = (mc, mv1 , mv2 , mv3 ) and N = (nc, nv1, nv2 , nv3 ), with mn = ρ , vα = O(1); therefore T

00

= ρc , 2

T

α0

=T

0α

γ

∂ Γαγ γ γ p − + Γ p Γ γ − Γαγ Γαβ γ ∂x  ∂ xβ αβ p    1 1 1 1 = O 2 +O 2 +O 4 −O 4 c c c c   1 =O 2 , c =

Thus Rik = O(1/c2 ) for all i, k.

α

= ρ cv = O(c),

T αβ = ρ vα vβ = O(1).

8.2 Spherically Symmetric Fields

To determine T = T00 + Tαα , we need T00 = g00 T 00 + g0β T 0β
= 1 + O(1/c2) ρ c2 + O(1/c2) · O(c)

1. The sum of dx2 , dy2 and dz2 produces the following terms, with coefficients as indicated:

= ρ c2 + O(1),

Tαα = gαα T αα + gα 0T 0α + gαγ T γβ 2

γ ∂ Γαβ

2

= O(1) + O(1/c ) · O(c) + O(1/c ) = O(1).

ρ c2 + O(1).

Therefore, T = The computations assume a weak gravitional field, so gi j = Ji j + O(1/c2).

dr2 : sin2 ϕ cos2 θ + sin2 ϕ sin2 θ + cos2 ϕ = 1, dr d ϕ : 2r sin ϕ cos ϕ cos2 θ + 2r sin ϕ cos ϕ − 2r cos ϕ sin ϕ = 0,

dr d θ : −2r sin2 ϕ sin θ cos θ + 2r sin2 ϕ sin θ cos θ = 0,

d ϕ 2 : r2 cos2 ϕ cos2 θ + r2 cos2 ϕ sin2 θ + r2 sin2 ϕ = r2 ,

d ϕ d θ : −2r2 cos ϕ sin ϕ cos θ sin θ b) The text (page 394) considers R00 but raises the ques+ 2r2 cos ϕ sin ϕ sin θ cos θ = 0, tion of the value of R0000 . We note Rijkk = 0 for all i, j, k, d θ 2 : r2 sin2 ϕ sin2 θ + r2 sin2 ϕ cos2 θ = r2 sin2 ϕ . so R0000 = 0 as claimed. Thus we find   (2) 1 1 3 ∂ 2 g00 Thus . + O R00 = 2 ∑ 2c α =1 ∂ (xα )2 c3 x2 + dy2 + dz2 = dr2 + r2 d ϕ 2 + r2 sin2 ϕ d θ 2 . β i For Rα 0 = Rα i0 = Rαβ 0 we have β Rαβ 0

β

∂ Γβα 0 ∂ Γαβ β β = + Γp Γ − Γp Γ − ∂ xβ  ∂ x0 α 0 pβ αβ α 0  1 1 1 1 = O 2 +O 3 +O 4 −O 4 c c c c   1 =O 2 . c γ

For Rαβ = Riα iβ = Rα0 0β + Rαγβ , γ 6= β , we have

∂ Γ0αβ

∂ Γ0 p 0 − αβ0 + Γαβ Γ0p0 − Γαp 0 Γαβ 0 ∂x  ∂ x       1 1 1 1 = O 3 +O 2 +O 4 −O 4 c c c c   1 =O 2 , c

Rα0 0β =

DVI file created at 16:06, 20 January 2011

2. To compute the Christoffel symbols of the second kind, we need the following expressions for γ hh = 1/γhh (which follow from 1/(1 + A) = 1 − A + O(A2)).

γ

00

=

γ 11 = γ 22 = γ 33 =

  1 2Ψ(r) 1 = 1− 2 +O 3 , 1 + 2Ψ(r)/c2 + O(1/c3) c c   −1 P(r) 1 = −1 − 2 + +O 3 , 1 − P(r)/c2 + O(1/c3) c c −1 , r2 −1 . r2 sin2 ϕ

As noted in the text, Γhjk will involve just the single term γ hh Γ jk,h because γ hi = 0 if h 6= i. For the Christoffel sym-

8.2. SPHERICALLY SYMMETRIC FIELDS bols of he first kind, we have: 1 ∂ γ00 2 ∂ξ1 1 ∂ γ11 Γ11,1 = 2 ∂ξ1 1 ∂ γ22 Γ12,2 = 2 ∂ξ1 1 ∂ γ33 Γ13,3 = 2 ∂ξ1 Γ10,0 =

1 ∂ γ00 2 ∂r 1 ∂ γ11 = 2 ∂r 1 ∂ γ22 = 2 ∂r 1 ∂ γ33 = 2 ∂r =

  Ψ′ (r) 1 = 2 +O 3 , c c   P′ (r) 1 , + O = 2c2 c3 = −r, = −r sin2 ϕ .

Therefore, Γ010 = γ 00 Γ10,0     ′   2Ψ(r) Ψ (r) 1 1 = 1− 2 +O 3 + O c c c2 c3   1 Ψ′ (r) = 2 +O 3 , c c

= γ Γ11,1     ′   P(r) P (r) 1 1 = −1 − 2 + +O 3 +O 3 c c 2c2 c   1 −P′ (r) +O 3 , = 2c2 c −1 1 Γ212 = γ 22 Γ12,2 = 2 · (−r) = , r r −1 1 3 33 Γ13 = γ Γ13,3 = 2 2 · (−r sin2 ϕ ) = r r sin ϕ Γ111

11

We list all other Christoffel symbols in R11 in order to show they are zero. First, consider Γh11 = γ hh Γ11,h with h 6= 1. Because γhi = 0 for all h 6= i and γ11 depends on ξ 1 alone, we have   γ 00 ∂ γ10 ∂ γ11 00 0 2 1− = 0, Γ11 = γ Γ11,0 = 2 ∂ξ ∂ξ0   γ 22 ∂ γ12 ∂ γ11 2 1− Γ211 = γ 22 Γ11,2 = = 0, 2 ∂ξ ∂ξ2   γ 33 ∂ γ13 ∂ γ11 2 1− = 0. Γ311 = γ 33 Γ11,3 = 2 ∂ξ ∂ξ3

101 Every term off the main diagonal involves either Γh11 with h 6= 1 or else a symbol with three different indices and is thus zero. The remaining four terms are
2
2
2
2 Γ010 + Γ111 + Γ212 + Γ313 . The first two have order 1/c4 ; each of the remaining two is 1/r2 . 3. a) Using ln(1 + A) = A + O(A2), we have      2Ψ 1 1 2Ψ ln(γ00 ) = ln 1 + 2 + O 3 = 2 +O 3 , c c c c      P 1 1 P ln(−γ11 ) = ln 1 − 2 + O 3 = 2 +O 3 . c c c c Therefore, because −γ = r4 sin2 ϕ · γ00 · −γ11 , we have √ ln −γ = 12 ln(−γ ) = 2 ln r + ln sin ϕ + 12 ln(γ00 ) + 21 ln(−γ11 )   Ψ P 1 = 2 ln r + ln sin ϕ + 2 − 2 + O 3 c 2c c

b) The partial derivatives we need are   √ P′ 1 2 Ψ′ ∂ ln −γ = + 2 − 2 +O 3 , ∂ξ1 r c 2c c   √ 1 −2 Ψ′′ P′′ ∂ 2 ln −γ = 2 + 2 − 2 +O 3 . 1 2 ∂ (ξ ) r c 2c c Then R11 becomes √ √
2
2 ∂ Γ111 ∂ 2 ln −γ 1 ∂ ln −γ − + Γ11 − Γ212 − Γ313 ∂ξ1 ∂ (ξ 1 )2 ∂ξ1   −P′′ 2 Ψ′′ P′′ P′ 2 1 = + − + − − + O 2c2 r2 c2 2c2 rc2 r2 c3   Ψ′′ P′ 1 = − 2 − 2 +O 3 . c rc c 4. At the outset we can express the possibilities as the entries in the following four 4 × 4 symmetric matrices: Γ0i j ,

Γ1i j ,

Γ2i j ,

Γ3i j ,

0 ≤ i ≤ j ≤ 3. This result also means that the only nonzero terms of the p h has 10 distinct entries. We have already found that sum Γ11 Γ ph occur when p = 1; the sum itself is computed Each Γ001 , Γ111 , Γ212 , and Γ313 are nonzero. p h in the text. The sum Γ1h Γ p1 expands into the following We now determine systematically which Γijk are zero. sixteen terms: In each matrix, 3 of the 10 entries are immediately zero because their three indices are different. To proceed, we Γ010 Γ001 Γ011 Γ101 Γ012 Γ201 Γ013 Γ301 consider various i, j index pairs, starting with the diagonal Γ110 Γ011 Γ111 Γ111 Γ112 Γ211 Γ113 Γ311 elements i = j. We have   Γ210 Γ021 Γ211 Γ121 Γ212 Γ221 Γ213 Γ321 1 ∂ γii hh hh ∂ γhi h . − γ Γ = γ Γ = ii,h ii Γ310 Γ031 Γ311 Γ131 Γ312 Γ231 Γ313 Γ331 ∂ξi 2 ∂ξh

DVI file created at 16:06, 20 January 2011

SOLUTIONS: CHAPTER 8. CONSEQUENCES

102 If h 6= i, the first term is zero. In detail, Γhii will be zero in the following cases:

For Γ323 , we have Γ323 = γ 33 Γ23,3 =

(i, h) = (0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 2), (2, 3), (3, 0),

(3, 3).

Now take i = 0 < j = h; then   γ hh ∂ γ0h ∂ γ0h Γh0h = γ hh Γ0h,h = = 0. − 2 ∂ξh ∂ξh That is, Γ101 = Γ202 = Γ303 = 0. Next, when i = h = 0 < j, Γ00 j That is,

γ 00 ∂ γ00 = 0 if j = 2, 3. = γ Γ0 j,0 = 2 ∂ξ j

Γ002

00

= Γ003

= 0. When i = h = 1 < j,

Γ11 j = γ 11 Γ1 j,1 =

γ 11 ∂ γ11 = 0. 2 ∂ξ j

That is Γ112 = Γ113 = 0. When i = h = 2, j = 3, Γ223 = γ 22 Γ23,2 =

γ 22 ∂ γ11 = 0. 2 ∂ξ3

γ 33 ∂ γ33 cos ϕ = . 2 ∂ξ2 sin ϕ

Modulo terms of order 1/c3 , the distinct nonzero Christoffel symbols are Ψ′ , c2 Ψ′ Ψ′ Γ100 = 2 , Γ111 = − 2 , c c Γ001 =

2rΨ , c2 2rΨ sin2 ϕ Γ133 = −r sin2 ϕ − , c2

1 Γ212 = , r 1 3 Γ13 = , r

Γ122 = −r −

Γ233 = − sin ϕ cos ϕ , Γ323 = cot ϕ .

5. To obtain the geodesic equations, we use the solution to the previous exercise. Because ξ 0 = ct(τ ), we have d ξ 0 /d τ = c · dt/d τ and d 2 ξ 0 /d τ 2 = c2 · d 2t/d τ 2 . The first geodesic equation is c2

Ψ′ (r) dt dr d 2t ·c = −2 dτ 2 c2 dτ dτ

or

d 2t =0 dτ 2

2 1 This shows that, altogether, 31 of the 40 distinct Christof- up to terms of order 1/c . The equation for ξ = r is 0 1 2 3      2   fel symbols are zero. Thus, besides Γ01 , Γ11 , Γ12 , and Γ13 , dt 2 Ψ′ dr 2 dϕ 2Ψ Ψ′ d2r there are at most five more that are nonzero. We have, with + 2 +r 1+ 2 =− 2 c 2 d τ c d τ c d τ c dτ h = 1, i = j 6= 1,   2    dϕ 2Ψ 1 γ 11 ∂ γii +O 3 + r sin2 ϕ 1 + 2 1 . Γii = − c dτ c 2 ∂ξ1  2  2  2 dt dϕ dθ ′ 2 = −Ψ (r) +r + r sin ϕ According to Theorem 8.3, P(r) = 2Ψ(r)+O(1/c); theredτ dτ dτ fore, we can now write   up to terms of order 1/c2 . The equations for ξ 2 = ϕ and 2Ψ(r) 1 11 γ = −1 − 2 + O 3 , ξ 3 = θ are c c  2 d2ϕ 2 dr d ϕ dθ and hence , =− + sin ϕ cos ϕ 2   d τ r d τ d τ dτ Ψ′ (r) 1 1 Γ00 = 2 + O 3 , 2 dr d θ d 2θ dϕ dθ c c =− − 2 cot ϕ . 2   dτ r dτ dτ dτ dτ 1 2rΨ(r) 1 +O 3 , Γ22 = −r − c2 c 6. a) Because ϕ and θ are constant,   2rΨ(r) sin2 ϕ 1 1 2 Γ33 = −r sin ϕ − +O 3 . d 2ϕ d 2θ dϕ dθ c2 c = 2 = = 2 = 0. dτ dτ dτ dτ For Γ233 , we have Therefore, the geodesic equations for ϕ and θ are identically zero, and the other two are (up to terms of order γ 22 ∂ γ33 Γ233 = γ 22 Γ33,2 = − 1/c2) 2 ∂ξ2  2 1 d 2t d2r dt ′ = 2 · (−2r2 sin ϕ cos ϕ ) = − sin ϕ cos ϕ . = 0, = −Ψ (r) 2 2 2r dτ dτ dτ

DVI file created at 16:06, 20 January 2011

8.2. SPHERICALLY SYMMETRIC FIELDS Because these equations involve only t and r, they have solutions. In fact, t = kτ + b and the only equation left to solve is d2r k2 GM 2 ′ . = −k Ψ (r) = dτ 2 r2

103 8. In the solution to Exercise 4, we showed that only 9 Christoffel symbols out of the 40 distinct possibilities were nonzero. The result followed from the fact that the weak spherical metric had the general form

γ00 = f (r),

γ11 = g(r),

γ22 = −r2 ,

b) The goal is to show that ϕ (t) ≡ π /2 satisfies the γ33 = −r2 sin2 ϕ , γi j = 0 otherwise, geodesic equations, assuming that (t(τ ), r(τ ), θ (τ )) also satisfy them. The equation for ϕ is identically zero and and f (0) = 1, g(0) = −1. Because the Schwarzschild metric has the same general form, the indices of its the other three equations reduce to nonzero Christoffel symbols are the same as the indicies d 2t of the nonzero symbols of the weak spherical metric. For = 0, dτ 2 the Schwarzschild metric, we note first that  2  2  11 d d 2r dt θ ′ γ ∂ γ11  + r , = −Ψ (r)  , i = 1,  dτ 2 dτ dτ 2 ∂ξ1 Γ1ii = γ 11 Γii,1 = 2 11 d θ  2 dr d θ γ ∂ γii  − . =− , i 6= 1. 2 dτ r dτ dτ 2 ∂ξ1

Because ϕ is absent from these three equations, they have In particular, then, solutions for (t(τ ), r(τ ), θ (τ )). γ 11 ∂ γ00 e−Q ′ T
T ′ T −Q Γ100 = − Te = e = , Suppose instead that ϕ (0) = ϕ0 6= π /2; we must show 2 ∂r 2 2 ϕ (τ ) ≡ ϕ0 is not part of a solution to the geodesic equa
Q′ γ 11 ∂ γ11 −e−Q tions. On the one hand, d 2 ϕ /d τ 2 must still be zero. On Γ111 = −Q′ eQ = , = 2 ∂r 2 2 the other hand, the equation for ϕ is 11 ∂ γ −Q γ e 22  2 = (−2r) = −re−Q , Γ122 = − d 2ϕ dθ 2 ∂r 2 ϕ ϕ = 6 0. = sin cos 0 0 dτ 2 dτ γ 11 ∂ γ33 e−Q = (−2r sin2 ϕ ) = −re−Q sin2 ϕ . Γ133 = − 2 ∂ r 2 Note: while ϕ0 = 0 or ϕ = π would make the right-hand side equal to zero, each choice gives a radial line found For the other nonzero Christoffel symbols, we have already in part (a) and thus excluded here. γ 00 ∂ γ00 e−T ′ T
T ′ Te = , Γ001 = γ 00 Γ01,0 = = 2 ∂ξ1 2 2 7. a) First of all, the geodesic equation for t implies that 22 −2 γ ∂ γ22 −r 1 dt/d τ = k, a constant. Now let r = r and ϕ = π /2; then Γ212 = γ 22 Γ12,2 = (−2r) = , = the equation for r reduces to 2 ∂ξ1 2 r −2 −2 33   2 
γ ∂ γ33 −r sin ϕ k2 GM dθ dθ 2 −2r sin2 ϕ = Γ313 = γ 33 Γ13,3 = 1 0 = −k2 Ψ′ (r) − r = , r − 2 ∂ ξ 2 dτ dτ r2 1 = , or  2 r 2 dθ k GM . = γ 22 ∂ γ33 r−2 dτ (−2r2 sin ϕ cos ϕ ) Γ233 = − = r3 2 ∂ϕ 2 The right-hand side is a positive constant—call it ω 2 . = − sin ϕ cos ϕ , Thus d θ /d τ = ω (where ω may be positive or negative). γ 33 ∂ γ33 3 33 b) We have θ = ωτ + ω0 , t = kτ + t0 . One period, T , is Γ23 = γ Γ23,3 = 2 ∂ϕ defined as the change in t when θ changes by 2π . That is, −2 sin−2 ϕ
cos ϕ −r T = ∆t, 2π = ∆θ = ω ∆τ , and therefore = . −2r2 sin ϕ cos ϕ = 2 sin ϕ 2π k T = ∆t = k∆τ = . ω 9. a) We have γ = γ00 · γ11 · γ22 · γ33 , so q
√ c) From part (a) and d θ /d τ = ω we have −γ = −eT (−eQ ) (−r2 ) −r2 sin2 ϕ r3 =

k2 GM GM 2 = 2T . ω2 4π

DVI file created at 16:06, 20 January 2011

= r2 e

T +Q 2

sin ϕ .

SOLUTIONS: CHAPTER 8. CONSEQUENCES

104 Therefore,

Therefore,

√ T +Q L = ln −γ = 2 ln r + + ln sin ϕ . 2

R22 = −e−Q + rQ′ e−Q +

b) The formula for Rik involves various first and sec√ ond partial derivatives of L = ln −γ . The only nonzero derivatives are

∂L 2 T ′ + Q′ , = + 1 ∂ξ r 2 −2 T ′′ + Q′′ ∂ 2L = + , ∂ (ξ 1 )2 r2 2

∂L cos ϕ = ∂ξ2 sin ϕ 2 −1 ∂ L = . ∂ (ξ 2 )2 sin2 ϕ

We then have

1 − re−Q sin2 ϕ

cos2 ϕ sin2 ϕ

+ 2e−Q −

= 1 − e−Q + re−Q ·



2 T ′ + Q′ + r 2



Q′ − T ′ . 2

For R33 we have R33 =

∂ Γ133 ∂ Γ233 ∂L ∂L p h + + Γ133 1 + Γ233 2 − Γ3h Γ p3 1 2 ∂ξ ∂ξ ∂ξ ∂ξ

The last term expands as

∂ Γ100 ∂L p h + Γ100 1 − Γ0h Γ p0 R00 = 1 ∂ξ ∂ξ

p h Γ3h Γ p3 = Γ133 Γ313 + Γ331 Γ133 + Γ233 Γ323 + Γ332 Γ233

= −2e−Q sin2 ϕ − 2 cos2 ϕ . In the last term, the only nonzero possibilities occur when p h p and h equal 0 or 1: Γ0h Γ p0 = Γ001 Γ100 + Γ100Γ010 . Therefore  ′′   ′ ′ ′ ′ T T T 2 T +Q R33 = −e−Q sin2 ϕ + rQ′ e−Q sin2 ϕ − cos2 ϕ + sin2 ϕ R00 = eT −Q + (T ′ − Q′ ) + +   2 2 2 r 2 cos ϕ 2 T ′ + Q′ −Q 2    − re sin ϕ − sin ϕ cos ϕ · + ′′ (T ′ )2 T ′ Q′ T ′ T′ T′ r 2 sin ϕ T −Q T =e . + − + −2 · 2 2 2 4 4 r −Q 2 2 + 2e sin ϕ + 2 cos ϕ   For R11 we have Q′ − T ′ = sin2 ϕ 1 − e−Q + re−Q · = sin2 ϕ · R22 . 2 1 2 ∂ Γ11 ∂ L ∂L p h R11 = − + Γ111 1 − Γ1h Γ p1 1 1 2 ∂ξ ∂ (ξ ) ∂ξ To see that all the other terms of the Ricci tensor are zero (recall R ji = Ri j ), we begin with The last term expands into four terms in which p = h: p h Γ1h Γ p1 = Γ010 Γ001 + Γ111Γ111 + Γ212Γ221 + Γ313Γ331

=

2 (T ′ )2 (Q′ )2 + + 2. 4 4 r

Thus R11 =

2 T ′′ + Q′′ Q′ Q′′ + 2− + 2 r 2 2

(T ′ )2 (Q′ )2 2 − − 2 4 4 r T ′ Q′ Q′ T ′′ (T ′ )2 = + − − . 4 r 2 4



2 T ′ + Q′ + r 2



R0α =

All derivatives with respect to ξ 0 are zero. For the last term we must have (p, h) = (0, 1) or (1, 0). However, Γ10α = Γ01α = 0, so the last term is zero. Hence R0α = 0. Next we consider R1α , with α = 2, 3. We have

∂ Γ212 = 0, ∂ξ2

−

For R22 we have

∂ Γ122 ∂ 2L ∂L p h − + Γ122 1 − Γ2h Γ p2 R22 = ∂ξ1 ∂ (ξ 2 )2 ∂ξ The last term is p h Γ2h Γ p2 = Γ221 Γ122 + Γ122 Γ212 + Γ323 Γ332

= −2e−Q +

cos2 ϕ sin2 ϕ

DVI file created at 16:06, 20 January 2011

∂ Γ00α ∂L p h + Γ00α 0 − Γ0h Γ pα . 0 ∂ξ ∂ξ

so R1α = Γ21α

∂ Γ313 = 0, ∂ξ3

∂L p h − Γ1h Γ pα . ∂ξ2

The last term collapses to Γ313 Γ33α . When α = 2 we have R12 =

1 cos ϕ 1 cos ϕ − = 0. r sin ϕ r sin ϕ

When α = 3, both terms are zero so R13 = 0. The last component of the Ricci tensor to check is R23 =

∂ Γ323 ∂L p h + Γ323 3 − Γ2h Γ p3 = 0. ∂ξ3 ∂ξ

8.2. SPHERICALLY SYMMETRIC FIELDS

105

10. The proof of Theorem 8.4 establishes that Q = −T , implying R22 = 1 − eT − rT ′ eT .

12. a) Because the nonzero Christoffel sumbols of the Schwarzschild and weak spherical metrics have the same indices, the nonzero terms in their geodesic equations correspond in the same way. The Christoffel symbols here T The proof also shows that e = 1 + a/r, with a arbitrary. involve T , Q = −T , eT , and their derivatives. The text Therefore,
′ a shows that eT = 1 − rM /r; hence T ′ eT = eT = − 2 ,   r T′ 1 1 rM 1 Γ010 = , − = 2 = and so 2 2r (1 − rM /r) 2 r − rM r     a a   R22 = 1 − 1 + − r · − 2 = 0. T ′ T −Q T ′ 2T 1 rM r − rM 2 1 r r Γ00 = e = e = 2 2 2 r(r − rM ) r r (r − r ) M M 11. a) We have = , 2r3  ′′    ′ 2 ′ ′ ′ ′ ′ T (T ) TQ T Q T 1 1 1 + − + R00 = γ 00 R00 = e−T · eT −Q , Γ111 = =− = − 2 4 4 r 2 2 2 r r − rM   ′′  rM  (T ′ )2 T ′ Q′ T ′ T = rM − r, Γ122 = −re−Q = −reT = −r 1 − , + − + = e−Q r 2 4 4 r   Γ133 = Γ122 sin2 ϕ = (rM − r) sin2 ϕ , T ′′ (T ′ )2 T ′ Q′ Q′ 1 11 −Q R1 = γ R11 = −e − + + − 1 2 4 4 r Γ212 = , Γ233 = − sin ϕ cos ϕ ,   ′′ r (T ′ )2 T ′ Q′ Q′ T 1 cos ϕ , + − − = e−Q Γ313 = , . Γ323 = 2 4 4 r r sin ϕ  ′   ′ 1 Q T R22 = γ 22 R22 = − 2 1 − e−Q + re−Q − First we consider the geodesic equations that involve r 2 2 ξ 0 /c. One of them is the equation for t: t =   1 e−Q e−Q Q′ T ′ =− 2 + 2 − − dt dr d 2t r r r 2 2 c2 2 = −2Γ001 · c d τ d τ dτ −1 −1   R33 = γ 33 R33 = 2 2 R22 sin2 ϕ = 2 R22 = R22 . 2t d 1 1 dt dr 1 r r sin ϕ or . − = 2 dτ c r r − rM d τ d τ i We calculate R = Ri in stages. We have The other is the equation for r:  ′ 2 ′ ′ ′ ′ (T ) T Q T Q  2   R00 + R11 = e−Q T ′′ + , − + − d2r dt 2 dr 1 1 2 2 r r − Γ11 = −Γ00 c 2  ′  dτ dτ dτ T Q′ 2 2  2  2 − + 2 , R22 + R33 = 2R22 = − 2 + e−Q dθ dϕ r r r r − Γ133 − Γ122 dτ dτ and then    2  2 1 dr dt 1 1 2 rM (rM − r)   − =c + ′ )2 ′ Q′ ′ ′ 2 (T T 2T 2Q 2 3 2r dτ 2 r − rM r dτ R = − 2 +e−Q T ′′ + − + − + 2 .  2  2 r 2 2 r r r dθ dϕ + (r − rM ) sin2 ϕ . + (r − rM ) d τ dτ 0 1 b) The vacuum field equations R0 = R/2 and R1 = R/2 imply R00 = R11 . From this we get T ′ /r = −Q′ /r and thus The remaining two equations are T + Q = k. Arguing as on page 404 of the text, we can  2 d2ϕ 2 dr d ϕ dθ rescale t to conclude that k = 0 and Q = −T . Substituting + sin ϕ cos ϕ , = − 2 2 d τ r d τ d τ dτ Q = −T into the field equation R2 = R/2 yields cos ϕ d ϕ d θ 2 dr d θ d 2θ −2 . =− 2T ′ ′′ ′ 2 dτ 2 r dτ dτ sin ϕ d τ d τ = 0. T + (T ) + r This is essentially the same differential equation that the text obtained and used to deduce that eT = 1 + a/r.

DVI file created at 16:06, 20 January 2011

b) As in Exercise 6, we have d ϕ /d τ = d θ /d τ = 0 and d 2 ϕ /d τ 2 = d 2 θ /d τ 2 = 0. The geodesic equation for t is

SOLUTIONS: CHAPTER 8. CONSEQUENCES

106

unchanged; it involves only t and r. The equation for r 14. a) We have reduces to     rM −rM rM 2 ′       2 2 f ( + 2 1 + · ρ) = 1 + ρ 1 d 2r 1 dr dt 1 4 ρ 4 ρ 4ρ 2 2 rM (rM − r) + . =c −     2 3 dτ 2r dτ 2 r − rM r dτ rM rM 2rM = 1+ 1+ − 4 ρ 4 ρ 4ρ It likewise involves only t and r. The equations for ϕ and    θ collapse to zero. Thus any solution to the equations for rM rM 1− . = 1+ t and r, together with ϕ = ϕ and θ = θ define a (radial) 4ρ 4ρ solution to the four geodesic equations. Therefore f ′ (ρ ) = 0 if either factor is zero; in particular, c) As in the solution to Exercise 6 (b), the aim is to show rM rM that the function ϕ (τ ) ≡ π /2 can be part of a solution to = 0 when ρ = . 1− 4ρ 4 the geodesic equations. The equation for t is unchanged, while the equation for r becomes When ρ = rM /4, we have rM /4ρ = 1 and thus       rM d 2r 1 1 dr 2 dt 2 1 (1 + 1)2 = rM . f (rM /4) = 2 rM (rM − r) + − = c 4 dτ 2 2r3 dτ 2 r − rM r dτ  2 b) In the expression above for f ′ (ρ ), the first factor is dθ + (r − rM ) . positive for all ρ > 0 and the second factor is positive if dτ rM rM 1− > 0, or ρ > . Because cos ϕ ≡ 0, the equation for ϕ collapses to zero 4ρ 4 and the equation for θ reduces to c) As ρ → ∞, the factor (1 + rM /(4ρ ))2 tends monotonically to 1, while the factor ρ tends to ∞; hence the product f (ρ ) tends to ∞. Thus f is 1–1 on [rM /4, ∞) and the image Because the geodesic equations for t, r, and θ are inde- of this set is [r , ∞). M pendent of ϕ , any solution t(τ ), r(τ ), θ (τ ) of those three equations can be combined with ϕ (τ ) ≡ π /2 to define 15. There are two basic results we need for A ≈ 0: 1 a geodesic. In other words, a geodesic that starts in the = 1 − A + O(A2), (1 + A)2 = 1 + 2A + O(A2). equatorial plane ϕ = π /2 stays there for all time. 1+A 2 dr d θ d 2θ =− . 2 dτ r dτ dτ

13. As in the solution to Exercise 7, we look for a solution to the geodesic equations of the form r(τ ) = r, ϕ (τ ) = π /2. With these assumptions, the equation for t takes the simple form d 2t/d τ 2 = 0, implying dt/d τ = k, a constant. The equation for r then takes the form  2 dθ d 2r 2 rM (rM − r) 2 k + (r − rM ) 0= 2 =c , 3 dτ dτ 2r or 

dθ dτ

2

rM c2 2 = k = ω 2 , a positive constant. 2r3

As in the solution to Exercise 7, we have d θ /d τ = ω , the period T of the motion is T = 2π k/ω , and  2 rM c2 T rM c2 2 GM 2 r3 = T = T . = 2 2π 8π 2 4π 2 Thus there are circular orbits in the Schwarzschild metric, and they obey a Kepler’s law. The last equality follow from rM = 2GM/c2 , and puts Kepler’s law in exactly the form it has in the weak spherical metric.

DVI file created at 16:06, 20 January 2011

Thus, rM     rM rM 1 4ρ 1 − = 1 − + O rM 2 4 ρ 4 ρ ρ 1+ 4ρ   rM 1 = 1− +O . 2ρ ρ2

1−

Therefore 

rM 2   2 rM 1 4ρ   = 1 − + O   rM 2ρ ρ2 1+ 4ρ   rM 1 +O = 1− . ρ ρ2 1−

Also,  2    1 rM 4 rM +O 1+ = 1+ 4ρ 2ρ ρ2   1 rM . +O = 1+ ρ ρ2

8.4. PERIHELION DRIFT

107

Theorem 8.5 expresses the metric ds2 using the factors     rM 4 1 − rM /4ρ 2 and 1+ . 1 + rM /4ρ 4ρ

8.4 Perihelion Drift

1. We know kp × qk is the area of the parallelogram determined by p and q. If we take the base of the parallelCorollary 8.1 then assumes that ρ is large. This makes ogram along the vector p, then the base has length kpk 1/ρ 2 negligible and allows us to replace the two factors and the altitude has length kqk sin θ , where θ is the angle from p to q. Therefore used in the theorem by
rM rM kp × qk2 = kpk2 kqk2 sin2 θ = kpk2 kqk2 1 − cos2 θ and 1 + , 1− ρ ρ = p2 q2 − (p · q)2 . yielding the statement of the Corollary. 2. a) The description provided in Proposition 8.2 (text page 423) continues to apply here, with just a couple of 8.3 The Bending of Light modifications. The new curve is the ellipse that is the reflection across the y-axis of the ellipse with eccentric1. We have s = ct(1 + A) where A = O(1/c). Therefore ity |e|, because the two curves have the same semimajor ct = s/(1 + A) = s(1 + O(A)) = s(1 + O(1/c)) and hence and semiminor axis lengths and they share a focus at the     origin. The center is at the point where x = −ke/(1 − e2); dt dt 1 1 1 or c = 1+O = +O 2 . this is negative for the original curve but positive for the ds c ds c c new one. The argument in the text (page 415) then gives      b) When e = ±1, the Cartesian equation becomes dθ 1 dθ d θ dt dθ 1 1 1 = = = +O 2 +O 2 y2 = k2 ∓ 2kx. ds dt ds dt c c c dt c   This is the equation of a parabola whose symmetry axis is 1 1 ∂γ +O 2 ; =− the x-axis. We can rewrite it in the form c ∂n c k y2 we have used the fact d θ /dt = −∂ γ /∂ n that was estabx=± ∓ . 2 2k lished on the same page. Because we always assume k > 0, this is a parabola that 2. The conditions z(0) = 0, z′ (0) = 0 in Cartesian coor- opens to the left when e = +1, to the right when e = −1. dinates become ϕ (0) = π /2, ϕ ′ (0) = 0 in spherical coor- It is a familiar fact that the focus of the parabola x = ay2 dinates. We saw in the solution to Exercise 6 (b), § 8.2 is on the x-axis at x = 1/4a, so the focus of x = ∓y2 /2k is (Solutions page 103) that a geodesic in the weak spherical at x = ∓k/2. Because our curve is the horizontal translate metric with these initial conditions had ϕ (t) = π /2, that of x = ∓y2 /2k by ±k/2, the focus of our curve is at the is z(t) = 0, for all t. origin. 3. The substitution y = X tan u (so that u = ±π /2 when c) Because e2 − 1 > 0, we can rewrite the Cartesian form y = ±∞) yields of the equation, as given in the proof of Proposition 8.2, Z ∞

dy = −∞ (X 2 + y2 )3/2 =

Z π /2 X sec2 u −π /2

X 3 sec3 u

Z 1 π /2

X2

−π /2

in the form

du

cos u du =

2 . X2

The same substitution also yields Z ∞

3y2 dy = 2 −∞ (X + y2 )5/2 = =

Z π /2 3X 2 tan2 u · X sec2 u −π /2

1 X2

Z π /2

−π /2

X 5 sec5 u

2  ke 1 x − − 2 2 y2 = 1. 2 2 2 2 k /(e − 1) e −1 k /(e − 1) 1

du

3 sin2 u · cosu du

π /2 2 sin3 u . = X 2 −π /2 X 2

DVI file created at 16:06, 20 January 2011

This describes a hyperbola whose center is at the point (ke/(e2 − 1), 0). Its asymptotes satisfy the equation 2  1 ke 1 − 2 2 x − y2 = 0. k2 /(e2 − 1)2 e2 − 1 k /(e − 1) We can rewrite this as the pair of linear equations   p ke y = ± e2 − 1 x − 2 , e −1 √ implying that the asymptotes have slopes ± e2 − 1.

SOLUTIONS: CHAPTER 8. CONSEQUENCES

108 It is known that the foci of the hyperbola (x − p)2 (y − q)2 − =1 a2 b2

p √ Because b2 − 4ac = 4 − 4(e2 − 1)(−1) = 2e, the evaluation takes the form p   1+e w + e2 − 1 1 e2 − (w − 1)2 arcsin + (1 − e2)w ew (1 − e2)3/2

lie at distance aε from the center (p, q), where ε = 1−e p 1 + b2/a2 is the eccentricity of the hyperbola. In our The first term is zero at both endpoints. For the second case, b2 /a2 = e2 − 1, so ε = |e| and aε = k|e|/(e2 − 1). term, we have Because the center of our hyperbola is on the x-axis at 1 + e + (e + 1)(e − 1) w + e2 − 1 ke/(e2 − 1), one focus is at the origin. The original curve = r = k/(1 + e cos θ ) is the half of the hyperbola that lies ew e(1 + e) 1+e closer to the origin. 1 + (e − 1) = +1, = 3. This integral has no obvious antiderivative, though e Mathematica can supply a value. The following approach and may be somewhat more transparent: it transforms the in w + e2 − 1 1 − e + (e + 1)(e − 1) tegral into one of the standard forms in a table of integrals = ew e(1 − e) (Numbers 205 and 203, CRC Tables, 12th edition, 1959). 1−e First note that the integrand is symmetric about θ = π , 1 − (e + 1) = −1. = because e Therefore, cos(π − θ ) = cos(π + θ ) = − cos(θ ).   1+e w + e2 − 1 It follows that arcsin = π, ew Z π Z 2π 1−e dθ dθ =2 , 2 and thus (1 + e cos θ )2 0 0 (1 + e cos θ ) Z 2π dθ 2π so we will work with the integral over the domain [0, π ]. . = 2 (1 + e cos θ ) (1 − e2)3/2 0 Next we make the transformation v = e cos θ , so that dv = −e sin θ d θ , or

dθ =

−dv −dv =√ . e sin θ e2 − v2

Furthermore, because v = e when θ = 0 and v = −e when θ = π , we have Z π 0

dθ = (1 + e cos θ )2

Z −e

−dv √ e (1 + v)2 e2 − v2 Z e dv √ = 2 −e (1 + v) e2 − v2

The further substitution w = 1 + v transforms this into Z 1+e 1−e

dw p = 2 2 w e − (w − 12)

Z 1+e

1−e w2

dw √ . 2 e − 1 + 2w − w2

This is of the form Z

dx √ , x2 X

X = a + bx + cx2,

with a = e2 − 1 < 0, b = 2, and c = −1. The table of integrals evaluates this integral as √   X b bx + 2a − + arcsin √ . ax 2(−a)3/2 x b2 − 4ac

DVI file created at 16:06, 20 January 2011

Because the semimajor axis length is a = J 2 /µ (1 − e2 ) (this a is not related to the a in X, above), we can finally write T=

J3 µ2

Z 2π 0

2π a3/2 dθ 2π 3/2 = 1/2 = √ a , 2 (1 + e cos θ ) µ GM

which agrees completely with the earlier calculation of the period of a circular orbit in the Exercises 7 (c) and 13 in § 8.2 (Solutions pages 103, 106). 4. If the eccentricity e equals 0, then r=

J 2 /µ = J 2 / µ = constant, 1 + 0 · cos θ

so the orbit is a circle. The angular momentum constant r2 θ˙ = J then implies

µ2 J θ˙ = 2 = 3 . r J Thus angular velocity is constant. (Incidentally, this implies that the period of the orbit is T=

2π J 3 2π = 1/2 r3/2 , µ2 µ

giving Kepler’s third law once again.)

8.4. PERIHELION DRIFT

109

5. If du/d θ ≡ 0, then u = 1/r is a constant. The orbits 8. a) The eccentricity e of an orbit and the length a of its are circles. semimajor axis are independent quantities; therefore, perihelion drift as a function of a has the form (text page 430) 6. a) The result follows from the linearity of the differ6π GM constant 2π D = 2 = . entiation operator. We have c a(1 − e2) a L(v + w) = (v + w)′′ + (v + w) = v′′ + v + w′′ + w = L(v) + L(w). Also,

In Exercise 3, above, we showed T = constant × a3/2 ; therefore, 2π D constant . = T a5/2

b) According to part (a), all orbits with the same fixed eccentricity have a perihelion drift over one earth century, namely 2π D/T , that depends only on the length a of the Thus L is linear. Hence, if L(vA ) = A and L(vB ) = B, then semimajor axis: constant . perihelion drift in one earth century = L(vA + vB) = L(vA ) + L(vB ) = A + B. a5/2 Therefore, because semimajor axes of the earth and Merb) The function vA is a constant, namely vA = A. There- cury are related by fore v′′A = 0 and thus L(vA ) = vA = A. For vB , let k = aearth ≈ 2.5 × aMercury, 3eµ 3 /J 4 ; then L(av) = (av)′′ + (av) = av′′ + av + a(v′′ + v) = aL(v).

v′B = k sin θ + kθ cos θ ,

v′′B = 2k cos θ − kθ sin θ .

constant

Therefore, L(vB ) = 2k cos θ − kθ sin θ + kθ sin θ = 2k cos θ , as required. For vC = k cos2θ , where k = −e2 µ 3 /2J 4, we have vC′′ = −4vC , so L(vC ) = −4vC + vC = −3vC =

we see that the perihelion drift of the earth over one century is

3e2 µ 3 cos2θ , 2J 4

as required.

5/2 aearth

=

constant 1 × = 0.1 × 43′′ = 4.3′′ . 2.55/2 a5/2 Mercury

c) In the constant in the expression for perihelion drift, eccentricity appears only in the factor 1/(1 − e2). Because the earth’s perihelion drift (assuming its orbit had the eccentricity of Mercury’s) is only 1/10th of Mercury’s, the factor 1/(1 − e2 ) must be made 10 times larger. Equivalently, 1 − e2 has to be reduced by the factor 1/10. For Mercury, 1 − e2 = 0.9577, so the new value should be 0.09577. Therefore, th earth’s eccentricity would have to be √ e = 1 − 0.09577 = 0.95.

7. a) This question was previously raised in Exercise 6 (b) of § 8.2 (Solutions page 103). The geodesic equation Shown below are ellipses with e = 0.2 and 0.95; their for ϕ (τ ) in the weak spherical metric is semimajor axis lengths are roughly equal.
2 2 10 ϕ ′′ = − r′ ϕ ′ + sin ϕ cos ϕ θ ′ r

If ϕ (τ ) ≡ π /2, then ϕ ′ = ϕ ′′ = 0 and the second term on the right side of the geodesic equation has the factor cos π /2 = 0. Thus ϕ (τ ) ≡ π /2 is a solution to the geodesic equation.

b) The solution ϕ (τ ) ≡ π /2 satisfies the initial conditions ϕ (0) = π /2 and ϕ ′ (0) = 0. By the uniqueness of solutions to second order differential equations, ϕ (τ ) ≡ π /2 is the only solution that satisfies these initial conditions. That is, any solution ϕ (τ ) that satisfies the given initial conditions must have ϕ (τ ) = π /2 for all τ .

DVI file created at 16:06, 20 January 2011

5

-20

-15

-10

5

-5

-5

-10

http://www.springer.com/978-0-387-98641-8