Solutions for Problem Set for Ch.

23
(compiled by Nate Bode)
May 6, 2009
A-E
23.3 Geodesic Equation in an arbitrary coordinate system [by Keith
Matthews 05]

p
p = p

= p

) =
d
d
p

= 0
d
2
x

d
2
=

dx

d
dx

d
23.4 Constant of geodesic motion in a spacetime with symmetry [by
Alexander Putilin 99]
(a) Geodesic equation
p
p = 0,i.e.
p

p
;
= 0 (1)
(p
,

)p

=
dx

d
p

(2)
=
dp

= 0 (3)
which gives
dp

d
=
1
2
(g
,
+ g
,
g
,
)p

(4)
where in the brackets the rst and the third terms are antisymmetric over ()
so their contraction with the symmetric tensor p

is zero. Thus
dp

d
=
1
2
g
,
p

(5)
1
Take to be A and using g
,A
= 0, we nd
dp
A
d
= 0 (6)
namely p
A
is a constant of motion.
(b) Let x
j
(t) be the trajectory of a particle. Its proper time
d
2
= ds
2
= dt
2
_
1 + 2 (
jk
+ h
jk
)v
j
v
k

(7)
= dt
2
(1 + 2
jk
v
j
v
k
+ O(
v
4
c
4
)) (8)
thus
d = dt
_
1 + 2 v
2
= dt(1 +
1
2
v
2
) (9)
where we have omitted terms of order v
4
/c
4
(i.e. ||
2
). The 4-velocity is given
by
u

=
dx

d
=
dx

dt(1 +
1
2
v
2
)
(10)
=
dx

dt
(1 +
1
2
v
2
) (11)
thus in particular u
0
= 1 +
1
2
v
2
.
4-momentum: p

= mu

, and in particular p
0
= mu
0
= m(1 +
1
2
v
2
).
And the conserved quantity is then given by
p
t
= g
0
p

= g
00
p
0
= (1 + 2)m(1 +
1
2
v
2
) (12)
= m(m +
1
2
mv
2
) (13)
we see that p
t
is indeed the non-relativistic energy of a particle aside from an
additive constant m and an overall minus sign.
23.5 Action Principle for Geodesic Motion [by Xinkai Wu 00]
The action is given by:
S[x

()] =
_
1
0
(g

dx

d
dx

d
)
1/2
d (14)
2
S =
_
1
0
(g

dx

d
dx

d
)
1/2
d (15)
=
_
1
0
1
2
(g

dx

d
dx

d
)
1/2
(g

dx

d
dx

d
)d (16)
=
_
1
0
1
2
(g

dx

d
dx

d
)
1/2
_
g

dx

d
dx

d
+ g

dx

d
dx

d
+
g

dx

d
dx

d
_
d (17)
by renaming , and noticing g

= g

, we get:
S =
_
1
0
1
2
(g

dx

d
dx

d
)
1/2
{
g

dx

d
dx

d
+ 2g

dx

d
dx

d
}d (18)
Integrating the 2nd term in {...} by parts, we nd, after renaming some indices:
S =
_
1
0
(g

dx

d
dx

d
)
1/2
_
g

d
2
x

d
2
+
g

dx

d
dx

d

1
2
g

dx

d
dx

d

d ln(g

dx

d
dx

d
)
1/2
d
g

dx

d
_
x

d (19)
Noting that this is true for all variations x

we know S = 0 if and only if

d
2
x

d
2
+
g

dx

d
dx

d

1
2
g

dx

d
dx

d

d ln(g

dx

d
fracdx

d)
1/2
d
g

dx

d
(20)
Contracting both sides with g

, we get
0 =
d
2
x

d
2
+
1
2
g

_
2
g

dx

d
dx

d

g

dx

d
dx

d
_

d ln(g

dx

d
dx

d
)
1/2
d
dx

d
(21)
By renaming for the rst term in {...}, the above equation becomes
0 =
d
2
x

d
2
+
1
2
g

{
g

+
g

}
dx

d
dx

d

d ln(g

dx

d
dx

d
)
1/2
d
dx

d
(22)
3
which is just, using the expression for the Christoel symbols,
0 =
d
2
x

d
2
+

dx

d
dx

d

d ln(g

dx

d
dx

d
)
1/2
d
dx

d
(23)
Now lets reparametrize the world line, > s(), then the equation becomes,
0 = (
d
2
x

ds
2
+

dx

ds
dx

ds
)(
ds
d
)
2
+
dx

ds
[
d
2
s
d
2

d ln(g

dx

d
dx

d
)
1/2
d
ds
d
] (24)
Integrating [...] twice we readily nd that [...] vanishes for
s =
_
A(g

dx

d
dx

d
)
1/2
d + B , (25)
where A and B are arbitrary constants.
After this reparametrization, we get the familiar geodesic equation:
0 =
d
2
x

ds
2
+

dx

ds
dx

ds
(26)
23.7 Orders of magnitude of the radius of curvature [by Alexander
Putilin 99]
Eq. (23.43) tells us that, if a system has characteristic mass M and charac-
teristic length R, order of magnitude estimate gives,
1
R
2

GM
R
3
(27)
where R is the radius of curvature
R
_
R
3
M
in units G = c = 1 (28)
(a) near earths surfae: R R

6.4 10
6
m (earths radius), M M

4.4mm (earths mass), and R 2.4 10
11
m 1 astronomical unit.
(b) near suns surface: R R
sun
7 10
8
m, M M
sun
1.5km, and
R 5 10
11
m 1AU.
(c) near the surface of a white-dwarf star: R 5000km, M M
sun
1.5km,
and R 3 10
8
m
1
2
(sun radius).
(d) near the surface of a neutron star: R 10km, M M
sun
3km, and
R 20km.
(e) near the surface of a one-solar-mass black hole: M M
sun
1.5km,
R 2M 3km, and R 4km.
4
(f ) in intergalactic space: R 10(galaxy diameter) 10
6
light-year,
M (galaxy mass) 0.03 light-year (for Milky way), and R 610
9
light-years
Hubble Distance.
23.8 Components of Riemann in an arbitrary basis [by Xinkai Wu 02]
p

;
p

;
= R

(29)
we have
p

;
= (p

;
)
;
= (p

,
+ p

)
;
(30)
= (p

,
+ p

)
,
+

(p

,
+ p

(p

,
+ p

) (31)
interchaging and in the above expression and then taking the dierence, we
get
p

;
p

;
= (

,
+

)p

+ (32)
+(

+ (p

,
p

,
) + (

)p

,
(33)
= (

,
+

)p

+ (34)
+c

+ (p

,
p

,
) + c

,
(35)
where in the last step weve used c

(eq. (23.44)). We can see

,
p

,
=
e

e
p

(36)
=
[e

,e]
p

= c

e
p

(37)
= c

,
= c

,
(38)
where to get to the second line, weve used the fact that for any scalar f,

B
f

A
f = A

(B

f
;
)
;
B

(A

f
;
)
;
(39)
= A

f
;
+ A

;
f
;
B

f
;
B

;
f
;
(40)
= (A

;
B

;
)f
;
(41)
= [

A,

B]

f
;
(42)
=
[

A,

B]
f . (43)
(note f
;
= f
;
by the torsion free condition).
Thus we nally conclude that
R

,
+

(44)
5
23.9 Curvature of the surface of a sphere [by Alexander Putilin 99]
(a) We read o the metric components from the line element:
g

= a
2
, g

= a
2
sin
2
, g

= 0 (45)
g

=
1
a
2
, g

=
1
a
2
sin
2

, g

= 0 (46)
There are six independent connection coecients

= g

= g

1
2
g
,
= 0 (47)

= g

=
1
a
2
1
2
(g
,
+ g
,
g
,
) = 0 (48)

= g

1
2
(2g
,
g
,
) =
1
2a
2
(a
2
sin
2
)
,
= sin cos (49)

= g

1
2
(2g
,
g
,
) = 0 (50)

= g

1
2
(g
,
+ g
,
g
,
) =
1
2a
2
sin
2

(a
2
sin
2
)
,
= cot (51)

= g

1
2
g
,
= 0 (52)
(b) We can think of the Riemann tensor as a symmetric matrix R
[ij][kl]
with
indices [ij] and [kl]. Since R
ijkl
is antisymmetric in the rst and the second
pairs of indices, the only nontrivial component is [ij] = [], [kl] = []
R

= R

= R

= R

(53)
(c) Using eq. (23.57) and the fact that in a coordinate basis the c

s all vanish,
we get
R

,
+

(54)
=
1
2
(sin2)
,

(55)
= cos(2) (sin cos ) cot (56)
= sin
2
(57)
and thus
R

= g

= a
2
sin
2
(58)
(d) The new basis is related to the old by e

=
1
a
e

, e

=
1
a sin
e

. Thus by the
6
multilinearity of tensors in their slots, we have
g

=
1
a
2
g

= 1, g

=
1
a
2
sin
2

= 1, g

=
1
a
2
sin
g

= 0 . (59)
i.e. g

k
=

k
. We also have:
R

=
1
a
4
sin
2

=
1
a
2
(60)
R

k
= g
m n
R
m

j n

k
=
m n
R
m

j n

k
(61)
thus
R

= R

+ R

= R

=
1
a
2
(62)
R

= R

+ R

= R

=
1
a
2
(63)
R

= R

+ R

= 0 (64)
namely, R

k
=
1
a
2
g

k
.
R = R

k
g

k
=
1
a
2
g

j
=
2
a
2
(65)
23.10 Geodesic deviation on a sphere [by Alexander Putilin 99]
(a) The line element is given in problem 23.9: ds
2
= a
2
(d
2
+sin
2
d
2
). Then,
on the equator, =

2
, dl
2
= a
2
d
2
, l = a is the proper distance.
(b) Geodesic deviation eqn:
p

= R(..., p,

, p), with
p =
d
dl
=
1
a

, p

= 0, p

=
1
a
(66)
At =

2
, connection coecients vanish (see Ex. 23.9)

=
1
a
2
_

;
_
;
=
1
a
2
_

;
_
,
(67)

;
=

,
+

,
sin cos

(68)

;
=

,
+

,
+ cot

(69)
thus

=
1
a
2
_

,
sin cos

_
,
|
=

2
=
1
a
2

,
(70)

=
1
a
2
_

,
+ cot

_
,
|
=

2
=
1
a
2

,
(71)
7
!
"
"!"
"!#
n
Figure 1: geodesic deviation on a sphere
On the other hand

= R

=
1
a
2
R

=
1
a
2
R

(72)
=
sin
2

a
2

|
=

2
=
1
a
2

(73)
thus
1
a
2

,
=
1
a
2

d
2

d
2
=

(74)

=
1
a
2
R

= 0
d
2

d
2
= 0 (75)
(c) Initial conditions (note that the geodesics are parallel at = 0):

(0) = b,

(0) = 0;

(0) = 0,

(0) = 0 (76)
This gives

= A + B = 0. And

() = A

cos + B

sin = b cos (77)

2
)

2
, i.e. =

2
cos .
8
From Fig. (1) we see that the separation vectors points along -direction
(i.e.

= 0), and its magnitude is

= a(

2
) = a cos = b cos , which
is precisely what we got before.
23.12 Newtonian limit of general relativity [by Alexander Putilin 99]
(a) We are given that g

+ h

and |h

| 1. Proper time:
d
2
= g

dx

dx

dx

dx

dt
2
dx
2
dt
2
. (78)
where the last approximation is because in the non-relativistic limit, |dx|/|dt|
|v/c| 1). Thus d dt, and
u

=
dx

d

dx

dt
:u
0
=
dt
d
1, u
j
=
dx
j
d

dx
j
dt
= v
j
. (79)
(b) Geodesics eqn:
du

d
=

.
du
j
d

dv
j
dt

j
00
=
j00
=
1
2
(2g
j0,0
g
00,j
) (80)
= h
j0,0
+
1
2
h
00,j

1
2
h
00,j
(81)
where in the last step weve used |h
,t
| |h
,j
|.
dv
j
dt
= u

v
j
,

v
j
t
+ v
k
v
j
x
k
i.e.
d
dt
=

t
+v (82)
dv
j
dt
=
,j
h
00
= 2.
(c) We can write:

=
1
2
g

(g
,
+g
,
g
,
) =
1
2

(h
,
+h
,
h
,
)+O(h
2
) . (83)
And the Riemann tensor is:
R

,
+ O(
2
) (84)
=
1
2

(h
,
+ h
,
h
,
)
,

1
2

(h
,
+ h
,
h
,
)
,
+ O(h
2
)
=
1
2
(h

,
+ h

,
h

,
h

,
h

,
+ h

,
) + O(h
2
) (85)
Notice that in the last line the rst and fourth terms cancel. Thus we get
R

1
2
(h
,
+ h
,
h
,
h
,
) (86)
9
(d) R
j0k0
=
1
2
(h
j0,k0
+ h
k0,j0
h
jk,00
h
00,jk
). Recall that in non-relativistic
limit, time derivatives are small compared to spatial ones, thus the last term in
the brackets dominates. And we get
R
j0k0

1
2
h
00,jk
=
,jk
(87)
23.13 Gauge transformation in linearized theory [by Alexander Putilin
99]
(a) x

new
= x

old
+

,
g
new

(x
new
) =
x

old
x

new
x

old
x

new
g

(x
old
) (88)
Evaluate l.h.s. and r.h.s. up to linear order in

and h

:
l.h.s. =

+ h
new

(x
old
+ )

+ h
new

(x
old
) (89)
r.h.s. = (

,
)(

,
)g

(x
old
) (90)
= g

(x
old
) g

(x
old
)

,
g

(x
old
)

,
(91)

+ h
old

,
(92)

+ h
old

(x
old
)
,
(x
old
)
,
(x
old
) (93)
h
new

= h
old

,
(94)
(b)

h
new

= h
new

1
2
h
new

=

h
old

,

,
+

,
(95)
Lorentz gauge:

h
new,

= 0.

h
new,

=

h
old,

,


,
+

,
= 0 (96)
thus we need

,
=

h
old,

(97)
(c) In Lorentz gauge, all terms on the l.h.s. of eq. (23.102) vanish except the
rst one, thus it reduces to

h

,
= 16T

(98)
10
23.14 External Field of a Stationary, Linearized Source [by Keith
Matthews 05]
We start by examining the role of Gausss law. Because
x
j
x
k
= (x
j
)
, k
=
j
k
and T

,
= T
k
, k
= 0 we nd (T
j
x
k
)
, j
= T
k
. Then
_
V
T
k
d
3
x =
_
V
(T
j
x
k
)
, j
d
3
x =
_
S
T
j
x
k
d
j
= 0 (99)
where the third equality comes from choosing a surface of integration entirely
outside of the source where T = 0. Similarly (T
l
x
j
x
k
)
, l
= T
j
x
k
+ T
k
x
j
, so
_
V
(T
j
x
k
+ T
k
x
j
) d
3
x =
_
S
T
l
x
j
x
k
d
l
= 0. (100)
(a) There is a typo in (23.107),the x] is meant to be an x

. We make use of the

|
=
1
r
+
x x

r
3
+ . . . =
1
r
+
x
k
x

k
r
3
+ . . .
By inserting (23.109) into (23.107) for = 0 0
h
00
(x) = 4
_
T
00
(x

)
r
d
3
x

+ 4
_
T
00
(x

)
x
k
x

k
r
3
d
3
x

=
4
r
_
T
00
d
3
x

+
4x
k
r
3
_
T
00
x

k
d
3
x

=
4M
r
+
4D
k
x
k
r
3
We have the freedom to choose the origin of our coordinates to coincide with
the center of mass of our source. x
com
=
1
M
_
xd
3
x = 0 where = T
00
= T
00
.
Thus

D = Mx
com
= 0. This gives us the desired result.
(b) We have
h
0i
=
4
r
_
T
0i
d
3
x

+
4x
k
r
3
_
T
0i
x

k
d
3
x

The rst term is

k
(
1
r
)
=
4
r
2
D
k, 0
x
k

4
r
2
P
k
x
k
= D
k, 0
P
k
= 0
11
So P
k
= D
k, 0
= 0 makes the rst term go to zero.
We can insert (100) into the second term
4x
k
r
3
_
T
0i
x

1
2
(T
0i
x

k
+ T
0k
x

i
)d
3
x

=
2x
k
r
3
_
T
0i
x

k
T
0k
x

i
d
3
x

=
2x
m
r
3
_

j
i

n
m
(T
0j
x

n
T
0n
x

j
)d
3
x

=
2x
m
r
3
_
(
j
i

n
m

n
i

j
m
)T
0j
x

n
d
3
x

=
2x
m
r
3
_

imr

jnr
T
0j
x

n
d
3
x

=
2
r
3

ijk
S
j
x
k
(c) Using (99) we nd that the O(
1
r
) term is zero. We can construct a combi-
nation of divergences that will allow us to evaluate the second integral.
(T
il
x
j
x
k
)
, l
+ (T
lj
x
i
x
k
)
, l
(T
lk
x
i
x
j
)
, l
= 2T
ij
x
k
Then by applying Gausss law we nd that the second integral goes to zero also.
12