B
Appendix B Useful results
B.1 Linear algebra
The inverse of a 2 × 2 matrix can be found from (B.1). Note the final factor in parentheses
is the determinant.
a b
d −b
1 0
=
(ad − bc).
(B.1)
c d
−c a
0 1
B.2 Series approximations
The follow series are obtained via a Taylor series expansion about x = 0. The first one
appears very often so we christen it “binomial series”:
1
1
(1 + x)a = 1 + ax + a(a − 1)x 2 + a(a − 1)(a − 2)x 3 · · ·
2
6
1
= 1 − x + x2 − x3 + · · ·
1+x
binomial series (B.2)
(B.3)
Letting x → −x in (B.3) immediately gives
1
= 1 + x + x2 + x3 + · · ·
1−x
√
1
1
1
1 + x = 1 + x − x2 + x3 + · · ·
2
8
16
(B.4)
(B.5)
Letting x → −x in (B.5) immediately gives
√
1
1
1
1 − x = 1 − x − x2 − x3 + · · ·
2
8
16
5
35 4
63 5
3
1
1
x −
x ···
= 1 − x + x2 − x3 +
√
2
8
16
128
256
1+x
(B.6)
(B.7)
Letting x → −x in (B.7) immediately gives the series we will use repeatedly to
approximate the Lorentz factor; just let (x = v 2 ):
√
295
1
5
35 4
63 5
3
1
x +
x ···
= 1 + x + x2 + x3 +
2
8
16
128
256
1−x
(B.8)
296
Useful results
x2
x3
x4
+
−
+ ···
2
3
4
(B.9)
a2 2 a3 3 a4 4
x + x + x + ···
2
6
24
(B.10)
ln(1 + x) = x −
eax = 1 + ax +
B.3 Transformations between spherical polar and Cartesian
coordinates
B.3.1 Upper indices: Cartesian to polar (lower indices polar to Cartesian)
r=
x 2 + y 2 + z2
θ = arccos(z/r)
φ = arctan(y/x)
(B.11)
∂θ
1
= cos θ cos φ
∂x
r
∂θ
1
= cos θ sin φ
∂y
r
∂θ
1
= − sin θ
∂z
r
∂r
= sin θ cos φ
∂x
∂r
= sin θ sin φ
∂y
∂r
= cos θ
∂z
∂φ
1 sin φ
=−
∂x
r sin θ
∂φ
1 cos φ
=
∂y
r sin θ
∂φ
=0
∂z
(B.12)
B.3.2 Upper indices: polar to Cartesian (lower indices Cartesian to polar)
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
∂x
= sin θ cos φ
∂r
∂y
= sin θ sin φ
∂r
∂z
= cos θ
∂r
∂x
= r cos θ cos φ
∂θ
∂y
= r cos θ sin φ
∂θ
∂z
= −r sin θ
∂θ
(B.13)
∂x
= −r sin θ sin φ
∂φ
∂y
= r sin θ cos φ
∂φ
∂z
=0
∂φ
(B.14)
B.4 Selection of spacetimes
Below we summarize the important spacetimes studied herein by listing their line element,
Christoffel symbols, and important tensors. Components that are not listed and are not
related by symmetry to one of those listed are zero.
297
Selection of spacetimes
B.4.1 Rindler spacetime
The line element is
ds 2 = −a 2 dλ2 + da 2 .
(B.15)
This metric applies in flat (Minkowski) spacetime and was derived in Exercise 5.21 for
a set of uniformly accelerating observers. Because it’s a flat spacetime it follows that
the Riemann tensor vanishes, and thus the Ricci tensor and scalar and Einstein tensor all
vanish.
Christoffel symbols
λ
λa
=
1
a
a
λλ
= a.
(B.16)
B.4.2 Static spherically symmetric spacetimes
See Exercise 6.35. The line element is
ds 2 = −e2(r) dt 2 + e2Λ(r) dr 2 + r 2 dθ 2 + r 2 sin2 θ dφ 2 .
(B.17)
Christoffel symbols
= e−2Λ e2 r
φφ
= −e−2Λ r sin2 θ
θ
φ
rφ
=
tr
=
θθ
= −e−2Λ r
r
φφ
= − sin(θ ) cos(θ )
t
r
θ
r
tt
1
r
rr
rθ
φ
θφ
=Λ
1
=
r
cos(θ )
=
,
sin(θ )
(B.18)
where ≡ d/dr and Λ ≡ dΛdr.
Ricci tensor and Ricci scalar
See SP6.8:
2
Rtt = −e(2φ−2Λ) Λ − 2 − −
r
−2Λ
[1 − r(Λ − )]
Rθθ = − −1 + e
2Λ
Rrr = − −Λ + 2 + −
r
2
−2Λ
Rφφ = − sin θ e
[1 + r( − Λ )] − 1
(B.19)
and
2( − Λ ) 1 − e2Λ
+
.
R = −2e−2Λ −Λ + 2 + +
r
r2
(B.20)
298
Useful results
Einstein tensor
1 2φ d −2Λ
e
)
,
r(1
−
e
r2
dr
Λ
2 −2Λ
2
=r e
−Λ −
,
+ ( ) +
r
r
1
2
(1 − e2Λ ) + ,
2
r
r
Gtt =
Grr =
Gθθ
Gφφ = sin2 θ Gθθ .
(B.21)
as in Schutz Eq. (10.14)–(10.17).
B.4.3 Schwarzschild spacetime
The line element is
2M
2M −1 2
dr + r 2 d2 ,
dt 2 + 1 −
ds 2 = − 1 −
r
r
Schutz Eq. (10.36)
(B.22)
with d2 = dθ 2 + sin2 θ dφ 2 (same as metric (ii) of Exercise 7.7). This metric applies in
the vacuum around a static spherically symmetric source. Because it’s a vacuum spacetime
it follows from the Einstein equations (see SP9.2) that Rαβ = Gαβ = 0 and R = 0.
Christoffel symbols
M
r2
2M
r
−1
2M
r
tr
θθ
= −r + 2M
r
φφ
= (−r + 2M) sin2 θ
φφ
= − sin(θ ) cos(θ )
φ
rφ
=
r
θ
1−
r
tt
=
M
r2
=
t
1−
rr
=−
rθ
=
r
θ
1
r
φ
θφ
M
r2
1−
2M
r
1
r
cos(θ )
=
.
sin(θ )
−1
(B.23)
B.4.4 Weak gravitational field
See Exercise 7.2. The line element is:
ds 2 = −(1 + 2φ)dt 2 + (1 − 2φ) dx 2 + dy 2 + dz2 .
Schutz Eq. (7.8)
(B.24)
Christoffel symbols
i
tt
i
jk
= φ,i + O(φ 2 ),
= δj k δ il φ,l − δji φ,k − δki φ,j + O(φ 2 ),
where i, j , k ∈ {x, y, z}.
i
tj
= −φ,t δ i j + O(φ 2 ),
(B.25)
299
Selection of spacetimes
Ricci tensor and Ricci scalar
Rtt = 3φ,tt + φ,xx + φ,yy + φ,zz + O(φ 2 )
Rti = 2φ,ti + O(φ 2 )
Rii = −φ,tt + φ,xx + φ,yy + φ,zz + O(φ 2 )
Rij = 0 + O(φ 2 )
when i j . (B.26)
R = −6φ,tt + 2(φ,xx + φ,yy + φ,zz ) + O(φ 2 ).
(B.27)
Einstein tensor
Gtt = 2(φ,xx + φ,yy + φ,zz ) + O(φ 2 )
Gti = 2φ,it + O(φ 2 )
Gii = 2φ,tt + O(φ 2 )
Gij = 0 + O(φ 2 )
when i j .
(B.28)
B.4.5 Post-Newtonian spherical rotating star
The line element was derived in Exercise 8.19, see eqn. (8.59):
2M
sin2 θ
2M
2
2
dt − 4J
dt dφ + 1 +
(dr 2 + r 2 dθ 2 + r 2 sin2 θ dφ 2 ).
ds = − 1 −
r
r
r
(B.29)
This metric applies in the vacuum around a spherical source that rotates. Because it’s a
vacuum spacetime it follows from the Einstein equations (see SP9.2) that Rαβ = Gαβ = 0
and R = 0.
Christoffel symbols
t
tr
t
rφ
r
r
r
tt
rr
φφ
θ
rθ
φ
tr
φ
rφ
2J 2 cos2 θ + 2M 2 r 2 + r 3 M − 2J 2
r(4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2 )
rJ sin2 θ (3r + 4M)
=
4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2
M
=
r(r + 2M)
M
=−
r(r + 2M)
r(M + r) sin2 θ
=−
r + 2M
M +r
=
r(r + 2M)
J
=
2
2
4(J sin θ − M 2 r 2 ) + r 4
2J 2 sin2 θ + 2M 2 r 2 + r 3 M − r 4
=−
r[4(J 2 sin2 θ − M 2 r 2 ) + r 4 ]
=−
tθ
=−
tφ
=−
t
r
r
θθ
θ
tφ
4J 2 sin θ cos θ
4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2
J sin2 θ
r(r + 2M)
r(M + r)
=−
r + 2M
2J sin θ cos θ
= 2
r (r + 2M)
φφ
= − sin θ cos θ
φ
tθ
=
φ
θφ
= cot θ
θ
2J cos θ (2M − r)
sin θ[4(J 2 sin2 θ − M 2 r 2 ) + r 4 ]
(B.30)
300
Useful results
B.4.6 Kerr spacetime
The line element is:
− a 2 sin2 θ 2
2Mr sin2 θ
dt
−
2a
dt dφ
ρ2
ρ2
(r 2 + a 2 )2 − a 2 sin2 θ
ρ2 2
2
2
+
sin
θ
dφ
+
dr + ρ 2 dθ 2 ,
ρ2
ds 2 = −
(B.31)
where M and a are constants and ≡ r 2 − 2Mr + a 2 , ρ 2 ≡ r 2 + a 2 cos2 θ. (This
is metric (iii) of Exercise 7.7). This metric applies in the vacuum around a source that
rotates. Because it’s a vacuum spacetime it follows from the Einstein equations (see SP9.2)
that Rαβ = Gαβ = 0 and R = 0.
Christoffel Symbols
t
tr
t
tθ
t
rφ
t
θφ
r
r
tφ
r
r
r
r
tt
rr
rθ
θθ
φφ
θ
tt
θ
tφ
M(r 2 − a 2 cos2 θ )(r 2 + a 2 )
ρ4 2
2a sin θ cos θ Mr
=−
ρ4
2
aM sin θ [a 4 cos2 θ − a 2 r 2 (1 + cos2 θ ) − 3r 4 ]
=
ρ4
2 sin3 θ cos θ Ma 3 r
=
ρ4
M(r 2 − a 2 cos2 θ )
=
ρ6
=
= −a sin2 θ
r
tt
M(a 2 cos2 θ − r 2 ) + a 2 r sin2 θ
ρ2
a 2 sin θ cos θ
=−
ρ2
r
=− 2
ρ
sin2 θ [rρ 4 + M(a 2 cos2 θ − r 2 )a 2 sin2 θ]
=−
ρ6
2Ma 2 r cos θ sin θ
=−
ρ6
2
2
a +r θ
=−
tt
a
=
−pb2
qa 2 sin(2θ )
=
a sin2 θ
=
(pb2 + 2qr)
qa 3 sin2 θ sin(2θ )
=−
p
=−
ap sin2 θ
=
r
M −r
=
+
2
a sin(2θ )
=−
2
r
=−
sin2 θ
=−
(r + pa 2 sin2 θ )
qa 2 sin(2θ )
=
2
qab2 sin(2θ )
=−
2
=
301
Selection of spacetimes
θ
rr
θ
rθ
θ
θθ
θ
φφ
a 2 sin θ cos θ
ρ2
r
= 2
ρ
a 2 sin(2θ )
2
r
=
=
=
r
=
=
rθ
sin θ cos θ 4 2
=−
[ρ (r + a 2 )
ρ4
+ 2Mra 2 sin2 θ (2ρ 2 + a 2 sin2 θ )]
aM(r 2 − a 2 cos2 θ )
ρ4
φ
tr
=
φ
tθ
=−
φ
rφ
=
rρ 4 − 2Mr 2 ρ 2 + a 2 M sin2 θ (a 2 cos2 θ − r 2 )
ρ4
φ
θφ
=
cot θ 4
(ρ + 2Mra 2 sin2 θ )
ρ4
r
sin(2θ ) 2
=−
b − 2a 2 sin2 θ q
2
a 2 sin2 θ
× 2+
=−
2aMr
cot θ
ρ4
rθ
=
ap
2qa cot θ
sin2 θ
r
(1 + 2q) + a 2 p
cot θ
=
(1+2q)(b2 −2qa 2 sin2 θ )
2qa 2 b2 sin2 θ
−
=
(B.32)
The second equality for each αμν in eqn.(B.32) uses the notation of Frolov and Novikov
(1998, Appendix D), with = ρ 2 , q = −Mr/, p = M(a 2 cos2 θ −r 2 )/ 2 , b2 = r 2 +a 2
when the total charge Q = 0.
B.4.7 Robertson–Walker spacetime
The line element is:
1
ds = −dt + R (t)
dr 2 + r 2 (dθ 2 + sin2 θ dφ 2 )
1 − kr 2
2
2
2
Schutz Eq. (12.13)
(B.33)
with three possible values of k:
k=1
“closed” or “spherical” universe
k=0
spatially “flat” universe
k = −1
“open” or “hyperbolic” universe
(This is metric (iv) of Exercise 7.7.)
302
Useful results
Christoffel symbols
t
rr
r
r
tr
φφ
θ
tθ
φ
tφ
R Ṙ
1 − kr 2
Ṙ
=
R
= −r(1 − kr 2 ) sin2 θ
=
t
θθ
rr
=
rθ
=
r
Ṙ
R
Ṙ
=
R
=
= R Ṙr 2
θ
φ
rφ
t
φφ
kr
1 − kr 2
θθ
= −r(1 − kr 2 )
φφ
= − sin θ cos θ
r
1
r
1
=
r
= R Ṙr 2 sin2 θ
θ
φ
θφ
= cot θ
(B.34)
Ricci tensor and Ricci scalar
Ricci tensor:
3R̈
R
2
= r R R̈ + 2Ṙ 2 + 2k
R R̈ + 2Ṙ 2 + 2k
1 − kr 2
Rtt = −
Rrr =
Rθθ
Rφφ = Rθθ sin2 θ.
Ricci scalar:
R αα
Einstein tensor:
R̈
k
Ṙ 2
=6
+ 2+ 2
R
R
R
.
Ṙ 2
k
Gtt = 3
+ 2
R2
R
2
Gθθ = −r 2R R̈ + Ṙ 2 + k
(B.35)
Grr =
(B.36)
−2R R̈ − Ṙ 2 − k
1 − kr 2
Gφφ = Gθθ sin2 θ.
(B.37)