GR October 13th Lecture
Bobby Bond
October 13, 2014
From the previous lectures we discussed the following equations
~a = −∇φ
(1)
∇2 φ = 4πGρ
(2)
−∇~a = 4πGρ
(3)
We want to show the connection between these equations to the Einstein equation. In
order to show the connection we need to replace the ~a equation with the geodesic equation,
U µ ∇µ U ν = 0. We will also need the stress tensor for the perfect fluid.
T µν = (ρ + p)U µ U ν + pg µν
(4)
The limits we will look at for this example are the following:
dt
1) Slow Motion ( dx
dτ << c dτ )
2) Sources are stationary (∂0 gµν = 0)
Now lets look at the geodesic equations:
ν
σ
d2 xµ
µ dx dx
+
Γ
=0
νσ
dτ 2
dτ dτ
i
j
d2 x0
d2 x0
0 dx dx
+
Γ
=
=0
ij
dτ 2
dτ dτ
dτ 2
Equation (5) used the fact that since
dt
is a constant.
shows use that dτ
dxi
dτ
<<
dt
dτ
(5)
we assume it goes to zero. This in turn
j
j
k
d2 xi
i dt dt
i dx dt
i dx dx
+
Γ
+
Γ
+
Γ
=0
00
j0
jk
dτ 2
dτ dτ
dτ dτ
dτ dτ
d 2 xi
d2 xi
i dt dt
+
Γ
=
0
⇒
+ Γi00 = 0
00
dτ 2
dτ dτ
dt2
1
(6)
Now lets use the equation for Γµνσ and remember the second limit we are using (stationary
sources ∂0 gµν = 0).
∂gµ0 ∂g0µ ∂g00
1
Γi00 = g iµ ( 0 +
−
)
2
∂x
∂x0
∂xµ
1
∂g00
Γi00 = g ij (− j )
(7)
2
∂x
We can now combine equations (6) and (7)
We are only looking at 1st
1 ∂g00
d2 xi
=
(8)
2
dτ
2 ∂xi
order. Using g00 = η00 + h00 , where h00 = −2φ we then get
d2 xi
1 ∂h00
=
= −∇i φ
(9)
2
dτ
2 ∂xi
µ
µ
but ∇ν ∇µ Rνρσ
has too many
We want (∂ρ ∂σ g)µν ∝ Tµν . The inside should be Rνρσ
derivatives. This leads use to the following possible equation
(Rνσ + αRgµν ) = κTµν
(10)
We know that ∂µ T µν = 0 and ∇µ T µν = 0 which implies
∇µ (Rµν + αRg µν ) = κ∇µ T µν = 0
This tells us that α =
− 21 .
(11)
We will now define the following
1
(12)
Gµν = Rµν − Rg µν
2
All we have left to do is to find κ. Using equation (10) we get the following 3 equations
1
R00 − Rg 00 = κ(T 00 = ρ)
2
1
R0i − Rg 0i = κT 0i = 0
2
1
Rij − Rg ij = κT ij = 0
2
Looking at the trace we can see that R − 2R = κT ⇒ R = −κT = κρ
1
1
R00 = T 00 κ − T g 00 = κρ
2
2
Now lets look at the definition of R00 to find κ
1
1
1
1
i
R00 = R0i0
= ∂i ( g iλ (∂0 gλ0 +∂0 g0λ −∂λ g00 )) = − δji ∂i ∂j g00 = − ∇2 g00 = − ∇2 h00
2
2
2
2
(13)
(14)
(15)
(16)
= ∇2 φ
(17)
Using equation (17) we can see that κ = 8πG. We are finally left with the Einstein Equation
Gµν = 8πGT µν
2
(18)