GR Lecture Notes October 27th, 2014
Bobby Bond
October 27, 2014
Schwarschild Solution
Gµν (g) = 8πGN Tµν
(1)
Looking for vacuum solutions, set Tµν = 0
1
Gµν = Rµν − Rgµν = 0 → Rµν = 0
2
(2)
Looking for Static, Spherically Sym.
ds2 = h(R)dt2 + k(R)dR2 + f (R)dΩ2
(3)
Let r2 = f (R), which leads us to the following equation
ds2 = −gtt (r)dt2 + grr (r)dr2 + r2 dΩ2
(4)
We can now solve for elements of G
Gtt = −r
0 + g2 − g
grr
1
rr
rr
2
= grr
(r(
− 1))0
2
r grr
grr
(5)
In EM ∇ · E = 4πε0 ρ and ∇ · B = 0. These are constraints.
In GR G00 = 4πGN T00 and G0i = 8πGN T 0i . These are constraints.
If we look at Gij → Π̇ij : we get a time derivative of momentum.
Combining equation (2) and (5) together we can see that Gtt = 0 and thus
r(
1
2GN M
− 1) = c = rs =
grr
c2
grr =
1
≈ 1 − 2φ
1 − rrs
∀t
(6)
(7)
To solve for gtt we will look at the following
Gtt − Grr = (
1 0
) =0
grr gtt
1
(8)
1
1
rs
D
=
= D(1 − )
→ gtt =
grr gtt
D
grr
r
2
Looking at the dimensions we can see that D ⇒ [c ]
Calculating Gθθ we get the following form
Gθθ = [. . .]0 + [. . .]0 + [. . .]0
(9)
(10)
The first term in equation (10) implies the following
[2r(
r2 0
4
1
− 1) + gtt
+ r3 Λ] = F
grr
D
3
(11)
You can easily check that equation (11) is a constant by plugging in grr and gtt .
Equation (11) also follows from conservation, ∇µ Gµν = 0.
Now that we solved for grr and gtt we can plug them into equation (4) to get the following
ds2 = −(1 −
rs 2
dr2
)dt +
+ r2 dΩ2
r
1 − rrs
(12)
There are two issues with equation (12)
1) at r = rs gtt = 0 grr = ∞, Metric is ”Bad” ← Coord. Singularity
2) at r = 0 gtt = ∞ grr = 0, Metric is ”Bad” ← Real Physical Singularity (Curvature
invariants becomes singular.
Look at r < rs → −gtt > 0 grr < 0. Light cone looks totally unfamiliar, r becomes timelike
and t becomes spacelike. Chart from Introduction To General Relativity by Sean Carroll
on page 221.
If you drop Static, Grt = 0 → grr
˙ = 0 but other equations will hold. Thus in a vacuum
Spherical sym. implies static. This is known as Birkhoff’s Theorem.
2