Grav. o. Kosm. Exercises No. 5
Notes on the Riemann tensor
Parallel transporting a vector around a loop in some general space with metric gµν ,
will change the vector. This change corresponds to what we will call curvature, and is
represented by the Riemann tensor,
[∇ρ , ∇λ ]Aµ = Rµσρλ Aσ
(1)
and it can be calculated by the Christoffel symbols and the metric
Rµσρλ = ∂ρ Γµσλ − ∂λ Γµσρ + Γµκρ Γκσλ − Γµκλ Γκσρ .
(2)
It is however not trivial to realize the symmetries of this tensor, so we will instead be
using the one with lowered first index
Rνσρλ = gµν Rµσρλ .
(3)
This is anti-symmetric in the first two and the last two indices, and symmetric in the
interchange of the two first with the two last
Rνσρλ = −Rσνρλ ,
Rνσρλ = −Rνσλρ ,
Rνσρλ = Rρλνσ .
(4)
As Joe’s notes tells us, in D = 3 there are 6 of them that are independent, and in D = 4
there are 20 of them that are independent. This saves some time, in D = 4 it is still a lot
of them, and we will have to use tricks every time to make it manageable.
But is is good to know how to identify these. Take D = 3, where we have to repeat
at least one index (since there are 3 different ones, and 4 positions) and max repeat it 2
times (otherwise the anti-symmetry will make it zero), that is
1221 1231
1331 2132
2332 3123
(5)
are the possible ones, where the numbers represent a specific coordinate, for example
1
(x, y, z) = (1, 2, 3). In D = 4 the 20 independent ones are
1221
1331
1441
2332
2442
3443
1231 1234
1241 1324
1341
2132
2142
2342
3123
3143
3243
4124
4134
4234
(6)
with numbers representing a specific coordinate, for example (t, r, θ, φ) = (1, 2, 3, 4).
From this tensor we can construct the Ricci tensor,
Rµν = Rαµαν = g αβ Rαµβν
(7)
R = g µν Rµν .
(8)
and the Ricci scalar
These together are combined to what is known as the Einstein tensor
1
Gµν = Rµν − gµν R
2
2
(9)
5.1
Consider the metric
ds2 = R2 cos2 θdψ 2 + dθ2 + sin2 θdφ2
(10)
a) Find Christoffel symbols.
b) Find Riemann tensor.
c) Find Ricci tensor.
d) Find Ricci scalar.
Solution
a) The Christoffel symbols are
Γθ ψψ = cos θ sin θ
Γψψθ = − tan θ
Γφθφ = cot θ
(11)
Γθ φφ = − cos θ sin θ
b) The independent Riemann tensors we want to check are the ones with components
ψθθψ ψθφψ
ψφφψ θψφθ
θφφθ φψθψ
(12)
after some calculation these components are found to be
Rψθθψ
Rψφφψ
Rθφφθ
Rψθφψ
Rθψφθ
Rφψθφ
= −R2 cos2 θ
= −R2 sin2 θ cos2 θ
= −R2 sin2 θ
=0
=0
= 0.
(13)
c) The Ricci tensor is then
Rµν = g αβ Rαµβν = g ψψ Rψµψν + g θθ Rθµθν + g φφ Rφµφν
(14)
where we used that the metric is diagonal so that the sum gets simplified. Since we only
had Riemann tensor components on the form abba, the Ricci tensor will only have aa
components, hence we need to calculate three of them. These are
Rψψ = 2 cos2 θ
Rθθ = 2
Rφφ = 2 sin2 θ.
3
(15)
d) The Ricci scalar then is
R = g µν Rµν =
4
6
R2
(16)
5.2
Consider the metric
ds2 = −dt2 + dx2 + dy 2 e2βz + dz 2
(17)
a) Find Christoffel symbols.
b) Find Riemann tensor.
c) Find Ricci tensor, Ricci scalar, Einstein tensor.
Solution
Before one starts calculating one should notice some things about this metric. First the
metric is four dimensional, and when we get to the Riemann tensor we need to calculate
20 independent components, which is a bit too much. However the second thing we notice
is that this metric is quite simple, up to a sign the tt, xx and yy are the same
− gtt = gxx = gyy = e2βz
(18)
so lets invent a new index i, j, . . . = (t, x, y) and we only need to keep track of the sign for
the tt component.
a) In the method we used to easy calculate the Christoffel components we tried all the
different combinations of the two lower indices. With this new index we only need to check
ij, iz and zz, and we find
Γz ii = −βgii ,
Γi zi = β.
(19)
b) With the clever index the 20 components reduce down to
ijji ijki ijkz
izzi ijzi
zijz
(20)
Rijji = gii β 2 gjj
Rizzi = gii β 2
(21)
The calculation gives that
are the only non-vanishing. If we spell them out they are
Rtxxt = −β 2 e4βz ,
Rtyyt = −β 2 e4βz ,
Rxyyx = β 2 e4βz ,
Rtzzt = −β 2 e2βz ,
Rtxxt = β 2 e2βz ,
Rtyyt = β 2 e2βz .
5
(22)
c) The metric was diagonal and the Riemann tensor only have components on the form
abba so the Ricci tensor is diagonal. The components are
Rtt = 3β 2 e2βz ,
Rxx = −3β 2 e2βz ,
Ryy = −3β 2 e2βz ,
Rzz = −3β 2 .
(23)
The Ricci tensor is
R = −12β 2 .
(24)
Then the Einstein tensor is diagonal, since the Ricci and metric is, and hence we get
Gtt = −3β 2 e2βz ,
Gxx = 3β 2 e2βz ,
Gyy = 3β 2 e2βz ,
Gzz = 3β 2 .
6
(25)
5.3
Consider a Riemann tensor on the form
Rµνσκ = λ (gµσ gνκ − gµκ gνσ )
(26)
a) Show that this expression have the proper symmetries.
b) Find the Ricci tensor, Ricci scalar and Einstein tensor.
c) Show that the Einstein tensor satisfies the Bianchi identity.
Solution
a) The symmetries to show are
Rµνσκ = −Rνµσκ ,
Rµνσκ = −Rµνκσ ,
Rµνσκ = Rσκµν .
(27)
To see that these symmetries are there, just shuffle around the metrics and extract some
minus signs
Rµνσκ = λ (gµσ gνκ − gµκ gνσ )
= −λ (gµκ gνσ − gµσ gνκ )
= −Rµνκσ
Rµνσκ = λ (gµσ gνκ − gµκ gνσ )
= −λ (gµκ gνσ − gµσ gνκ )
= −λ (gνσ gµκ − gνκ gµσ )
= −Rνµσκ
Rµνσκ = λ (gµσ gνκ − gµκ gνσ )
= λ (gσµ gκν − gκµ gσν )
= Rσκµν
(28)
b) For the Ricci tensor we simply trace the first and third index of the Riemann tensor
Rµν = g αβ Rαµβν = 3λgµν .
(29)
Note, every space with Rµν ∝ gµν is called an Einstein space. Trace this to get the Ricci
scalar
R = g µν Rµν = 12λ.
(30)
The Einstein tensor then is
1
Gµν = Rµν − gµν R = −3λgµν .
2
7
(31)
c) Does this Einstein tensor satisfy the Bianchi identity
∇µ Gµν = 0?
(32)
Yes. As we found in b), the Einstein tensor is proportional to the metric, and the metric
is covariantly constant
∇µ gνρ = 0
(33)
the trace of this must also be zero.
8
5.4
Consider the Reissner-Nordström metric
ds2 = −f (r)dt2 + f −1 (r)dr2 + r2 dθ2 + sin2 θdφ2
(34)
with
2M G GQ2
+ 2 .
(35)
r
r
The following can be done in terms of the function f (r).
a) Find Christoffel symbols.
b) Find Riemann tensor.
c) Find Ricci tensor and scalar, and the Einstein tensor.
√
d) Analyze the invard and outward trajectories for a light ray when Q = M G.
e) Study the Einstein tensor components as Q → 0.
f (r) = 1 −
Solution
a) The Christoffel symbols are
G (Q2 − M r) (r2 + G (Q2 − 2M r))
1
,
Γr tt = f f 0 = −
2
r5
1 f0
G (−Q2 + M r)
Γt tr =
=
,
2f
r (r2 + G (Q2 − 2M r))
1 f0
Γr rr = −
= (G(Q2 − M r))/(r(r2 + G(Q2 − 2M r))),
2f
1
Γθ rθ = ,
r
1
φ
Γ rφ = ,
r
GQ2
Γr θθ = −rf = 2GM −
− r,
r
Γφθφ = cot θ,
Γ
(r2 + G (Q2 − 2M r)) sin2 θ
= −rf sin θ = −
,
r
= − cos θ sin θ.
2
r
φφ
Γθ φφ
9
(36)
b) The independent Riemann tensor components that are non-zero are
1
G (3Q2 − 2M r)
Rtrrt = − f 00 = −
,
2
r4
1
G (Q2 − M r) (r2 + G (Q2 − 2M r))
Rtθθt = − rf f 0 =
,
2
r4
G (Q2 − M r) (r2 + G (Q2 − 2M r)) sin2 θ
1
,
Rtφφt = − rf f 0 sin2 θ =
2
r4
G (Q2 − M r)
1 0
rf = − 2
,
Rrθθr =
2f
r + G (Q2 − 2M r)
1 0 2
G (Q2 − M r) sin2 θ
Rrφφr =
rf sin θ = − 2
,
2f
r + G (Q2 − 2M r)
Rθφφθ = r2 (f − 1) sin2 θ = G Q2 − 2M r sin2 θ.
(37)
c) The non-zero components of the Ricci tensor are
1
GQ2 (r2 + G (Q2 − 2M r))
1
,
Rtt = f f 00 + f f 0 =
2
r
r6
1 f 00 1 f 0
GQ2
Rrr = −
−
=− 2 2
,
2 f
rf
r (r + G (Q2 − 2M r))
GQ2
Rθθ = (1 − f ) − rf 0 = 2 ,
r
GQ2
Rφφ = ((1 − f ) − rf 0 ) sin2 θ = 2 sin2 θ.
r
(38)
The Ricci scalar is
R=−
1 2 00
r f + 4rf 0 2f − 2
2
r
(39)
Put in the expression for f (r) to discover that
R = 0.
(40)
This also implies that the Einstein tensor is
1
Gµν = Rµν − gµν R = Rµν .
2
(41)
d) A light ray travels along null geodesics and hence their movement is governed by
1 2
dr + r2 dΩ2 .
f
√
We will assume no angular motion, dΩ = 0 and Q = M G gives
2
2M G GQ2
2M G M 2 G2
MG
(r − M G)2
f (r) = 1 −
+ 2 =1−
+
=
1
−
=
.
r
r
r
r2
r
r2
0 = −f dt2 +
10
(42)
(43)
This is known as an extremal black hole, it is ”as massive as it is charged”. This have some
nice features, for example one can put two of these next to each other and they will not
feel any net force, because the electromagnetic repulsion is as great as the gravitational
attraction.
Multiplying (42) with dtf2 to get
dr
dt
2
=
(r − GM )4
.
r4
(44)
It is not so easy to solve this and study the behavior, instead we can look at the velocity
2
and acceleration a = ddt2r
v = dr
dt
v± = ±
(r − GM )2
,
r2
the acceleration is given by the derivative w.r.t. t
d
2(r − GM )
2GM
(r − GM ) dr
a± = v± = ±
=
(r − GM )3 .
1−
2
dt
r
r
dt
r5
(45)
(46)
Lets consider the v+ solutions. These have positive velocity, hence are traveling outwards (towards larger r), even for r < GM . However a+ is negative, hence slowing down,
and as r → GM , v+ → 0+ and a+ → 0− , so it stops at the horizon, at r = GM .
Lets look at the v− solutions. Negative velocity means going into, so lets look at the
light ray coming from the outside of the horizon, r > GM . a− is positive, that is v− is
increasing, and as r → GM , v− → 0− and a− → 0+ and it again stops at the horizon.
e) Above we got that Gµν = Rµν . And looking at the expressions we see that every
component is zero if we take Q → 0.
extra) The conclusion in e) is not completely correct. Even if Q = 0, there is still a
source at r = 0: a massive point. How does this work? Let’s look at something you might
know. For a charged particle in Maxwell’s electrodynamics, we have to solve
∂r (r−2 ∂r At (r)) = 0
(47)
where At is the temporal component of the vector potential Aµ . The solution to this
equation is
A
(48)
At (r) = − + B
r
where B is a gauge choice, usually set to zero (zero potential at infinity). A however, is
related to the charge. How do we see this? Consider
Z
∂r (r2 ∂r (−A/r))dr = A
(49)
11
according to the fundamental theorem of calculus. However, this cannot be true if the
right hand side is identically zero, but it is not, there is a point source there. Hence, the
equation we really solve is
∂r (r−2 ∂r At (r)) = Qδ(r).
(50)
The same happens here, it appears that it is zero, but if one where to integrate, one
would get a constant. Look for example at Grr
Grr =
−1 + f (r) + rf 0 (r)
1
1
−2M G
0
0
=
(rf
(r)
−
r)
=
(−2M
G)
=
δ(r). (51)
r2 f (r)
r2 f (r)
r2 f (r)
r2 f (r)
12