PHYSICS 631: General Relativity
Homework # 1 Solution Key
Due April 14, 2015
1. [12 pts] Geometrized Units. Please express each of the following quantities in two ways: i) in mn , as
meters raised to some appropriate power, and ii) in kg n as kilograms raised to the appropriate power.
E.g. 2.94m−3 might be an answer.
(a) [4 pts] The momentum of an electron moving at 0.8c (Recall that p = mvγ)
Sol. It is probably worth reminding ourselves of the basic conversion factors involved:
1kg
=
7.41 × 10−28 m
1J
=
8.23 × 10−45 m
1s =
3 × 108 m
In Geometrized units:
v = 0.8
and γ is dimensionless:
γ = 1.66
so:
p = 9 × 10−58 m
p = 1.21 × 10−30 kg
(b) [4 pts] The age of the universe (13.8Gy)
Sol.
In distance, it’s fairly straightforward:
t = 4.35 × 1017 s = 1.31 × 1026 m
and
t = 1.76 × 1053 kg
(c) [4 pts] The orbital speed of the earth.
Sol.
This is a speed, so there are no units. The speed is 30km/s, so
v = 10−4
2. [18 pts] Schutz 1.3 (parts a-f)
Sol.
3. [24 pts] Schutz 2.1. Also, as part (h) please compute B α as a column vector.
Sol.
Noting:

5
 0 

Aµ = 
 −1 
−6

0
−2
4
1
 5
=
 4
−1
0
−2
5
−1
2
−2
2
−3
Bµ =

Cµν
0

3
0 

−2 
0
(a) [3 pts] Aα Bα
Aα Bα = 5 · 0 + 0 · (−2) + (−1) · 4 + (−6) · 0 = −4
(b) [3 pts] Aα Cαβ
Note that this will result in a row vector (1 lowered index). The first term will be:
Aα Cα0 = 5 · 1 + 0 · 5 + (−1) · 4 + (−6) · (−1) = 7
Working it all out, we get:
Aα Cαβ =
7
1
26
17
(c) [3 pts] Aγ Cγσ
I’m not going to dignify this with an answer, as it is clearly identical to the previous part.
(d) [3 pts] Aν Cµν
This has the same form as the last two parts, but we’re multiplying over a different index.
Aν Cµν =
−15
27
30
−2
(e) [3 pts] Aα Bβ
This produces a
1
1
tensor, with the values:

−10 20
0
0
2
−4
12 −24
0

0
Aα Bβ = 
 0
0

0
0 

0 
0
(f) [3 pts] Ai Bi
This is simply the trace of the lower submatrix of the previous part.
Ai Bi = −4
(g) [3 pts] Aj Bk
This is the 3x3 submatrix of part e).

0
0
0
−4 0 
Aα B β =  2
12 −24 0

(h) [3 pts] B α
This is somewhat more boring than I planned, since the first term is 0 and -0=0. However,

0
 −2 

Bα = 
 4 
0

4. [16 pts] Schutz 2.14
Sol.
The transformation is:

1.25
 0
µ
Λµ = 
 0
0.75

0 0 0.75
1 0
0 

0 1
0 
0 0 1.25
(a) [5 pts] The velocity is clearly in the z-direction. Since the terms are γ and vγ, we get the speed
from:
0.75
vz = −
= −0.6
1.25
I put the minus sign in there to indicate that positive velocities increase going into the barred
frame, so the barred frame must be going in the negative z direction.
(b) [5 pts] The inverse matrix is simple:

1.25

0
Λµµ = 

0
−0.75
which just comes from reversing the velocity.

0 0 −0.75

1 0
0


0 1
0
0 0 1.25
(c) [6 pts] In the barred frame:


1
 2 

Aµ = 
 0 
0
In the unbarred frame, we get:


1.25


2

Aµ = 


0
−0.75
5. [20 pts] Schutz 2.22
Sol.
(a) [10 pts] The four-momentum is:

4
 1 

pµ = 
 1 
0

The energy is just the zero component:
E=4
The rest mass can be computed via:
pµ pµ = −m2 = −16 + 2 = −14
so
m=
√
14 ' 3.74
and the velocity computed via:
vi =
pi
p0
so
~v = 0.25ˆi + 0.25ˆj
(b) [10 pts] The initial pair of particles has a 4 momentum:
 
5
 0 
α

pi = 
 1 
0
and since the total 4-momentum must be conserved, the 3rd particle produced is:


3
 −1/2 


pα
3 =
1 
0
The energy is:
E=3
The mass is:
m=
√
7.75 ' 2.78
The 3-velocity is:
1
1
~v = − ˆi + ˆj
6
3
The transformation matrix to the CM frame requires a boost of −0.2 in the y direction.
6. [10 pts] Schutz 2.30 (part b only)
Sol.
Using:
−pµ Uµ = E
This is simply:
E = −(−300 · 2 + 299 · 1) × 10−27 = 301 × 10−27
7. Read and discuss problem 1.17, but do not write it up for submission.