EN 615.748 - Introduction to RelativityJohns Hopkins University 18 February 2015Nam Nicholas Mai Homework Assignment 1

Problem 1: Exercise 1.18 - Velocity Composition for Hyperbolic Representation

Solution: (a) Introduce the velocity parameter u defined as

v tanhu, (1.1)where u P R and v P p1, 1q. Suppose, then, that we have a particle in O moving at some velocityw tanhU ; in another frame O, its velocity will be w1. Then, if O moves with velocity v with respect toO, then we can apply Equation 1.13 to have

w1 U ` u1` Uu

tanhU ` tanhu1` tanhU tanhu (1.2)

sinhUcoshU ` sinhucoshu1` sinhUcoshU sinhucoshu

(1.3)

sinhU coshu` sinhu coshUcoshU coshu` sinhU sinhu (1.4)

To simplify the last equality, we need to use the following two identities:

Identity 1.1 Hyperbolic Sinh of a Sum:

sinhU coshu` sinhu coshU sinh pU ` uq (1.5)Proof. Use the exponetial definition of sinh to expand the left side Equation 1.5 above:

sinhU coshu` sinhu coshU eU eU

2

eu ` eu2

` eu eu

2

eU ` eU2

(1.6)

eU`u ` eUu eU`u eUu

4` e

u`U ` euU eu`U euU4

(1.7)

2eU`u 2epU`uq

4(1.8)

eU`u epU`uq

2(1.9)

sinh pU ` uq (1.10)

Identity 1.2 Hyperbolic Cosh of a Sum:

coshU coshu` sinhU sinhu cosh pU ` uq (1.11)

EN 615.748 - Homework Assignment 1 2

Proof. Use the exponetial definition of sinh to expand the left side Equation 1.11 above:

coshU coshu` sinhU sinhu eU ` eU

2

eu ` eu2

` eU eU

2

eu eu2

(1.12)

eU`u ` eUu ` eU`u ` eUu

4` e

U`u eUu eU`u ` eUu4

(1.13)

2eU`u ` 2epU`uq

4(1.14)

eU`u ` epU`uq

2(1.15)

cosh pU ` uq (1.16)

Applying Identities 1.5 and 1.11 to Equation 1.4,

w1 sinhU coshu` sinhu coshUcoshU coshu` sinhU sinhu

sinh pU ` uqcosh pU ` uq tanh pU ` uq . (1.17)

Clearly, the parameters U and u add linearly, validating their use.

(b) A star S1 measures a second star S2 moving away from it at velocity v2,S1 0.9c v, implyingu2,S1 tanh1 v. The second star S2 measures a third star S3 moving alway from it at v3,S2 v,again implying u3,S2 tanh1 v. This continues for some N number of stars, where vN,SN1 v anduN,SN1 tanh1 v.

Let us examine the first three stars relative to each other. Using the relations above and the resultsfrom part (a), we have

v3,S1 tanh pu2,S1 ` u3,S2q tanh`tanh1 v ` tanh1 v tanh `2 tanh1 v (1.18)

u3,S1 2 tanh1 v (1.19)By the same argument, we can write v4,S1 as

v4,S1 tanh pu3,S1 ` u4,S3q tanh`2 tanh1 v ` tanh1 v tanh `3 tanh1 v (1.20)

u4,S1 3 tanh1 v (1.21)Now, we can use induction to prove the generalized statement,

vN,S1 tanhpN 1q tanh1 v (1.22)

uN,S1 pN 1q tanh1 v (1.23)Equations 1.23 and 1.22 are true for 1 N 4 (see above). Then, assume that for some k 4, bothequations are true. Then,

vk`1,S1 tanhpk ` 1 1q tanh1 v (1.24)

tanh pk 1q tanh1 v ` tanh1 v (1.25) tanh uk1,S1 ` uk,Sk1 (1.26)

EN 615.748 - Homework Assignment 1 3

The last equality is true by the velocity composition law proved in part (a). For large N , we have

vN,S1 tanhpN 1q tanh1 v 1 e2pN1q tanh1 v

1` e2pN1q tanh1 v (1.27)

1

e2 tanh

1 vpN1q

1`e2 tanh

1 vpN1q (1.28)

1 19pN1q

1` 19pN1q (1.29)

1 19pN1q

1` 19pN1q 1 19pN1q1 19pN1q (1.30)

1 2 p19qpN1q ` 192pN1q

1 192pN1q (1.31) 1 2 p19qpN1q (1.32)

The last approximation is taken since the second order terms are taken to be really small compared to thedominating constant and first order term.

Problem 2: Exercise 2.1 - Einstein Summation Notation

Solution: From the problem statement, we have the following for A and B

A

5016

, B 0 2 4 0

(2.1)

and C

C

1 0 2 35 2 2 0

4 5 2 21 1 3 0

(2.2)We evaluate each of the sections of the problem:

(a) AB A0B0 `A1B1 `A2B2 `A3B3 5 0` 0 2`1 4` 6 0 4(b) We have AC A0C0 ` A1C1 ` A2C2 ` A3C3 . This should yield a contravariant tensor D

with components:

D0 5 1` 0 5`1 4`6 1 7 (2.3)D1 5 0` 0 2`1 5`6 1 1 (2.4)D2 5 2` 0 2`1 2`6 3 26 (2.5)D3 5 3` 0 0`1 2`6 0 17 (2.6)

EN 615.748 - Homework Assignment 1 4

(c) Since is a dummy index and is free, this will be the same as (b).

(d) We have AC A0C0 `A1C1 `A2C2 `A3C3 . This should also yield a contravariant tensorE , but it will be dierent from D from part (b)

E0 5 1` 0 0`1 2`6 3 15 (2.7)E1 5 5` 0 2`1 2`6 0 27 (2.8)E2 5 4` 0 5`1 2`6 2 30 (2.9)E3 5 1` 0 1`1 3`6 0 2 (2.10)

(e) AB is simply the outer product of the two vectors, which produces a rank 2 tensor, which we willwrite as a matrix F

F

A0B0 A0B1 A0B2 A0B3A1B0 A1B1 A1B2 A1B3A2B0 A2B1 A2B2 A2B3

A3B0 A3B1 A3B2 A3B3

0 10 20 00 0 0 00 2 4 00 12 24 0

(2.11)

(f) AiBi A1B1 `A2B2 `A3B3 0 2`1 4` 6 0 4(g) This is simply the inner components from (e) with index values 1, 2, and 3.

F jk A1B1 A1B2 A1B3A2B1 A2B2 A2B3A3B1 A3B2 A3B3

0 0 02 4 012 24 0

(2.12)

Problem 3: Exercise 2.2 - Indices in Einstein Summation Notation

Solution:

(a) No free index. is the dummy index. 1 equation.

(b) is the free index. is the dummy index. 4 equations.

(c) , are free indices. , are dummy indices. 16 equations.

(d) , are free indices. No dummy indices. 16 equations. This turns out to be the definition of theEinstein tensor G . It can be thought of as the generalization of the gravitational field, whose sourceis given by the famous stress-energy tensor, T , in the field equations.

Problem 4: Exercise 2.14 - Lorentz Transformation Matrix

EN 615.748 - Homework Assignment 1 5

Solution: (a) From the problem statement, let the Lorentz transformation matrix from O into O denoted:

1.25 0 0 0.750 1 0 0

0 0 1 0

0.75 0 0 1.25

(4.1)The form here shows that O is moving in the z direction from O. The center of the matrix is the identitymatrix indicating that there is no relativistic change in x or y. Mathematically, one sees this as

txy

z

1.25 0 0 0.750 1 0 0

0 0 1 0

0.75 0 0 1.25

txy

z

0 0 v0 1 0 0

0 0 1 0

v 0 0

txy

z

, (4.2)so x x and y y. From Equation 4.2, we can clearly see that

v vz 0.75

0.751.25

0.6 rcs. (4.3)

(b) We can explicitly invert the Lorentz matrix above, but we can be even more clever with somesimple physical arguments. Since O and O are relative to each other, we know that cannot changebetween them. And since if v is O into 0, v has to be from the inverse. Therefore,

1 pvq 1.25 0 0 0.750 1 0 0

0 0 1 0

0.75 0 0 1.25

(4.4)

(c) To find the components of ~A O

p1, 2, 0, 0q in O, we simply multiply A by the inverse Lorentzmatrix from (b).

A pvqA

1.25 0 0 0.750 1 0 0

0 0 1 0

0.75 0 0 1.25

120

0

1.252

0

0.75

(4.5)

Problem 5: Exercise 2.32 - Derivation of Compton Scattering Equation from Four Momentum

Solution: We will apply conservation of the 4-momentum to derive the new frequency f of the scattered

photon. Let ~P denote the momentum of the particle (electron) and ~p denote the momentum of the photon.Then, we have know the following

~Pi m,~0

, ~pi hi p1, niq , ~pf hf p1, nf q (5.1)

EN 615.748 - Homework Assignment 1 6

From the definition of 4-momentum, we also have

~Pi ~Pi ~Pf ~Pf m2, ~pi ~pi ~pf ~pf 0 (5.2)Applying conservation of 4-momenta,

~Pi ` ~pi ~Pf ` ~pf ~Pf ~Pi ` ~pi ~pf (5.3)Taking the inner product of ~Pf with itself yields

~Pf ~Pf m2 ~Pi ` ~pi ~pf

~Pi ` ~pi ~pf

(5.4)

~Pi ~Pi ` 2~Pi p~pi ~pf q ` ~pi ~pi 2~pi ~pf ` ~pf ~pf (5.5) m2 ` 2~Pi p~pi ~pf q ` 0 2~pi ~pf ` 0 (5.6)

Therefore, we arrive at

2~Pi p~pi ~pf q 2~pi ~pf (5.7)2m,~0

rhi p1, niq hf p1, nf qs 2 rhi p1, niqs rhf p1, nf qs (5.8)

mh pi f q h2if p1 ni nf q (5.9)m pi f q hif p1 cos q (5.10)

i f ` hif1 cos

m

(5.11)

Dividing both sides of the equation by if we have the result

1

f 1

i` h

1 cos

m

(5.12)

Problem 6: Exercise A.1 - A 20-m pole is carried into a barn of length 10-m by a runner whose speedis such that the pole appears to be 10-m long in the barn frame. When the front end of the pole reachesthe closed end of the barn it appears that the pole fits just inside the barn for that instant and front barndoor can be closed. The rear barn door immediately then opens and runner goes through. From runnerspoint of view the pole is 20 m long and barn is only 5-m and the pole can never be enclosed in the barn.Explain the paradox by space time diagrams.

Solution: From the problem statement, the barn measures the pole to be 10 m when we know that thepole is actually 20 m in its rest frame. Therefore, we can deduce the speed at which the runner is running:

10 rms 20 rms c1 v2

c2 v

?3

2c (6.1)

This matches the corresponding calculation of the length of the barn measured by the runner in his restframe,

5 rms 10 rms c1 v2

c2 v

?3

2c. (6.2)

EN 615.748 - Homework Assignment 1 7

We can think of the barn as being observed to moving towards the runner although this is not necessary.

The gist of the paradox in this problem is simply the standard length contraction. That is, a movingobject will appear contracted to a fixed observer. In both cases above, the barn and runner are measuringlengths of objects moving with respect to themselves. The barn will see its own length and a contractedpole; likewise the runner will see the pole length and a contracted barn length. See the attached space-timediagrams to better understand the measurement intervals.

Note that the pole will still pass through the barn, despite the lengths being contracted. The frontend of the pole will never poke past the rear door.

Problem 7: Exercise A.2 - A satellite orbits the earth in the same direction in a circular orbit 200 kmfrom the surface above the equator. Calculate the number of seconds each day a clock on the satellite willbe slower compared to a clock at the equator. Note: we are only calculating the special relativistic eectand not the gravitational eect.

Solution: We will make some basic assumptions in the problem statement to keep things consistent. First,let us assume that the equatorial rotational velocity of the earth is given by

vC,eq RC,eq !C,eq 6.378137 106 rms 7.292115 105rad

s

465.101

ms

(7.1)

and the rotational velocity of the orbital satellite to be

vs cGMCRs

d

GMCRC `RsC

gffe6.67 1011 m3kgs2 5.9721986 1024 rkgs6.378137 106 ` 200 103 rms 7781.77

ms

(7.2)

Note that the satellite is not geosynchronous but rather in low-earth orbit. Now assume that there is athird clock at the center of the earth. This will be the reference clock. We know that the earth is rotatingso the clock on the equator will be dilated relative to the clock at the center of the earth; this is true alsofor the satellites clock. We want the dierence between the equatorial and orbital clocks. However, wewill have to calculate the dilations independently. Let the time to be Tc 86400 rss be the interval of theclock at the center of the earth. Then, the times measured by the two moving frames (equatorial orbit andsatelite) are expected to be slower relative to the clock at the center.

TC,eq Tcd1

465.101

3 1082

86399.99999989616 rss (7.3)

Ts Tcd1

7781.77

3 1082

86399.99997093312 rss (7.4)

Therefore, the dierence between the two clocks is given by

TC,eqs 86399.99999989616 86399.99997093312 28.963 rss (7.5)