GR December 8 Lecture Notes
Bobby Bond
December 8, 2014
We will look at Degrees of Freedom by setting the perturbed metric components to
certain values
gµν = ηµν + hµν
(1)
We will now set the values of h to the following
h00 = −2Φ
(2)
h0i = +wi
(3)
hij = 2Sij − 2Ψδij
(4)
Both Φ and Ψ are scalars (up to Spin 0). wi is a vector (up to Spin 1). Sij is a traceless 2
tensor (up to Spin 2). Plugging this back into equation (1) and calculating ds2 we get the
following
ds2 = −(1 − 2Φ)dt2 + 2wi dxi dt + ((1 − 2Ψ)δij + 2Sij )dxi dxj
(5)
We will now define a new quantity
p0 =
dt
= “E”
dλ
(6)
µ
Note equation (6) is not p0 = mg0µ dx
dλ = −E
pi =
dxi
dt dxi
=
= Ev i
dλ
dλ dt
(7)
Now let’s look at the time derivatives of E and pi
1
dE
= −E[∂0 Φ + 2(∂k Φ)v k − (∂(i wj) − ∂0 hij )v j v k ] 6= 0
dt
2
(8)
Equation 8 is not 0 because it does not capture all of the gravitational energy.
1
dpi
= −E[∂i Φ + ∂0 wi + 2(∂[i wj] + ∂0 hij )v j + (∂(j hk)i − ∂i hjk )v j v k ]
dt
2
1
(9)
Now we can define some new quantities
Gi = −∂i Φ − wi
(10)
−
H i = (∇ × →
w )i = εijk ∂j wk
(11)
Plug in equations (10) and (11) into equation (9) we are left with the following
dpi
1
−
= E[Gi + (→
v × H)i − 2(∂0 hij )v j − (∂(j hk)i − ∂i hjk )v j v k ]
dt
2
(12)
We can see the first 2 terms in equation (12) resembles E+M. Now lets look at the Einstein
Tensor
G00 = 2∇2 Ψ + ∂k ∂l S kl
(13)
Equation (13) is a constraint on the system. (There is no ∂0 )
1
1
G0j = − ∇2 wj + ∂j ∂k wk + 2∂0 ∂j Ψ + ∂0 ∂k S0k
2
2
(14)
Equation (14) is a constraint. Notice each term deals with initial data. (There is 1 ∂0 )
k
Gij = (δij − ∂i ∂j )(Φ − Ψ) + δij ∂0 ∂k wk − ∂0 ∂(i wj) + 2δij ∂02 Ψ − Sij + 2∂k ∂(i Sj)
− δij ∂k ∂l S kl
(15)
This is not a constraint on the system. (There are 2 ∂02 ) Now lets look at some Gauge
transforms
Φ → Φ + ∂0 ζ 0
(16)
wi → wi + ∂ 0 ζ i − ∂ i ζ 0
1
Ψ → Ψ − ∂i ζ i
3
1
Sij → Sij + ∂i ζi − ∂k ζ k δij
3
(17)
(18)
(19)
1) Transverse
∂i w i = 0
(20)
∂i S ij = 0
(21)
This leads to the following differential equations for ζ
1
∇2 ζ i + ∂i ∂j ζ j = −2∂i S ij
3
(22)
We can think of equation (22) like a matrix equation (M ζ k ). In this sense DetM> 0 and
thus invertible and therefor solvable.
∇2 ζ 0 = ∂i w i + ∂0 ∂i ζ i
2
(23)
Equation (23) is clearly solvable. With this gauge we can see that equations (13)-(15)
simply a lot. If we use these equations to solve for T = Tii 6= 0 then we can say ∂02 Ψ is fed
by the source. 2) Synchronous Gauge
Φ=0
(24)
wi = 0
(25)
Using equations (24) and (25) we can write equation (5) in the following way
ds2 = −dt2 + (δij + 2Sij )dxi dxj
(26)
This leads to the following equations for ζ
∂0 ζ 0 = −Φ
Z
ζ0 = − Φdt
(28)
∂0 ζi = −wi + ∂i ζ0
(29)
(27)
3) Lorenz (Harmonic) Gauge
1
(30)
∂µ hµν − ∂ν h = 0
2
Equation (30) corresponds to something we have already talked about earlier in the class
µ
(∂µ hν = 0). We will break up the wi vector into a transverse and longitudinal piece
i
+ wki
w i = w⊥
(31)
i
=0
∂ i w⊥
(32)
k
εijk ∂j wk
(33)
i
= εijk ∂j ζk
w⊥
(34)
wik = ∂i λ
(35)
ij
+ Ssij + Skij
S ij = S⊥
(36)
ij
∂i S⊥
=0
(37)
∂i ∂j Ssij = 0
(38)
This leads to the following equations
ε
ijk
l
∂j ∂l Ski
=0
1
Skij = (∂i ∂j − ∇2 δij )θ
3
Ssij = ∂(i ζj)
ij
Such that ∂i ζ i = 0. Notice that S⊥
is Spin 2, ζ i is Spin 1, and θ is Spin 0.
3
(39)
(40)
(41)