GR Notes for Dec. 1st
Yaqi Han
Look at the metric gµν =ηµν + hµν under the condition that the source is a
far away purterbation from the observer(detector), i.e., |hµν | 1.
To the linear order the variation of the Einstein tensor would be
1
δGµν = − h̄µν = 8πGTµν
2
(1)
where h̄µν = hµν − 12 hηµν .
We can choose a gauge ∂µ h̄µν = 0(Lorentz gauge).The equation we need to
solve now is
h̄µν = −16πGN Tµν
(2)
where ”” is the d’Alembertian for flat Minkowski space.
To solve this equation, choose the retarded Green’s function to be
Gr (xσ − y σ ) =
−1
δ(|~x − ~y | − (x0 − y 0 ))θ(x0 − y 0 ).
4π|~x − ~y |
(3)
Thus the solution to Eq.(2) is
σ
h̄µν (x ) = 4GN
Z
d3 y
1
Tµν (t − |~x − ~y |, ~y ).
|~x − ~y |
(4)
In principle, there are 10-4=6 variables to solve here due to the constraint
from the gauge we chose. Now consider another constraint
∂µ T µν = 0
∂0 T 0µ = −∂i T iµ .
=⇒
(5)
Take the Fourier transformation of Eq.(4) and Eq.(5),
1 Z
h̄µν (ω, ~x) =
dteiωt h̄µν (t, ~x)
2π
(6)
iω Te 0µ (ω, ~x) = −∂i Te iµ (ω, ~x)
(7)
e
1
If we have Teij (ω, ~x), we can obtain Te0i from Eq.(5) and thus Te00 .
Write Eq.(6) explicitly, with tr = t − |~x − ~y |
Z
4GN Z
eiωt
dt d3 y
Tµν (tr , ~y )
2π
|~x − ~y |
Z
iω(tr +|~
x−~
y |)
4GN Z
3 e
=
dtr d y
Tµν (tr , ~y )
2π
|~x − ~y |
Z
eiω|~x−~y| e
= 4GN d3 y
Tµν (ω, ~y ).
|~x − ~y |
h̄µν (ω, ~x) =
e
(8)
When the source is far away,
Figure 1: A source that’s far away from the detector
to the leading order |~x − ~y | ' r = constant.
h̄µν (ω, ~x) = 4GN
e
eiωr Z 3 e
d y Tµν (ω, ~y )
r
(9)
Now examine the integral,
Z
3
d
y Te ij (ω, ~y )
=
Z
d3 y[∂k (y i Te jk ) − y i ∂k Te jk ]
(10)
The first term can be converted into a surface integral which would vanish
at the boundary. Using Eq.(5),
Z
d3 y Te ij (ω, ~y ) =
Z
d3 y[iωy i Te 0,j (ω, ~y )].
(11)
1
d3 y[ iω(y i Te 0,j + y j Te 0,i )].
2
(12)
We can symmetrize it so that
Z
d
3
y Te ij (ω, ~y )
=
Z
Use Eq.(5) again, we get
2
Z
d
3
y Te ij (ω, ~y )
iω Z 3
d y[∂k (y i y j Te 0k ) − y i y j ∂k Te 0k ] (first term vanishes at boundary)
=
2
ω 2 Z 3 i j e 00
=−
d y[y y T ] (Te 00 → density)
2
ω2
' − Ieij (ω)
2
(13)
where Ieij (ω) is the Fourier transformed quadupole tensor of the source. Thus,
h̄ij (ω, ~x) = −
e
2GN ω 2 eiωr eij
I (ω).
r
(14)
And inverse Fourier transformation gives
2GN d2
Iij (tr )
r dt2
2GN ¨
Iij (tr )
=
r
h̄ij (t, ~x) =
It depends on postion through r and tr = t − r.
For a compact binary system with equal masses,
Figure 2: A compact binary with equal masses
3
(15)
we know from the flat space solution that the angular velocity
dφ
2π
GM
=
= ( 3 )1/2 = Ω.
dt
T
4R
(16)
Since this motion is restricted to a plane, we can write

i


x
y
i


 i
z
= (−)R cos Ωt
= (−)R sin Ωt
=0
(17)
where i=1,2 and the ”-” is for the second body.
Iij =
Z
d3 y[y i y j T 00 ]
cos2 Ωt
cos Ωt sin Ωt 0
2
sin2 Ωt
0 
= 2M R  cos Ωt sin Ωt

0
0
0


= MR
2

(18)

1 + cos 2Ωt
sin 2Ωt
0
sin 2Ωt
1 − cos 2Ωt 0 
.
0
0
0


Iij is not traceless but I¨ij is. Thus h̄ij (t, ~x) is transverse and traceless.


− cos 2Ωt sin 2Ωt 0
8GN M R2 Ω2 

h̄ij (t, ~x) =
 − sin 2Ωt cos 2Ωt 0  .
r
0
0
0
The dimension of GM ∼ Rc2 , adding in the missing dimension factor
(19)
1
c4
GN M R3 c2 Ω2
)
∼≤ 10−21 .
2
4
Rc
rc
(20)
dE
2 c5 GN M 5
=−
(
).
dt
5 GN Rc2 )
(21)
h̄ij ∼ 8(
The power radiated is
P =
The fourier transformed results would be
√
e
h̄(ω) ≈ 10−21 / Hz.
(22)
Given the frequency band, we can obtain the total energy emitted
Eemitted =
Z
dω(h̄(ω))2
e
4
(23)