Notes for Tuesday, November 25th, 2014
C. L. Wang
1
Orbiting Clock
Suppose there is a clock in a circular orbit in the Schwarzschild geometry,
let’s figure out how much time elapses while moving along the orbit. First
we know the invariant line element in Schwarzschild is,
rs 2 rs −1 2
2
ds = − 1 −
dt + 1 −
dr + r 2 dθ2 + r 2 sin2 θdφ2 ≡ −dτ 2 . (1)
r
r
It’s usually convenient to parameterize the e.o.m by the proper time τ for
time like particles. So dividing both side equation (1) by dτ 2 , we can write,
2
2
r dt 2 r −1 dt 2
dθ
dφ
s
s
2
2
2
− 1−
−r
−r sin θ
= 1 . (2)
1−
r
dτ
r
dr
dτ
dτ
Treat the orbiting clock as a point like test particle whose Lagrangian is given
by :
dxµ dxν
1
1
≡ mgµν ẋµ ẋν .
(3)
L = mgµν
2
dτ dτ
2
Note that this Lagrangian has no explicit dependence on either t or φ while
considering the Schwarzschild metric, so the conjugated momenta are conserved, i.e.
rs dt
pt ≡ gtµ ẋµ = −m 1 −
= −E ,
(4)
r dτ
dφ
= Lz .
(5)
pφ ≡ gφµ ẋµ = mr 2 sin2 θ
dτ
Picking the coordinates system such that the motion takes place at θ = π/2,
we find,
2
dr
L2z
rs E2 1+ 2 2 .
(6)
= 2 − 1−
dτ
m
r
mr
Differentiate equation (6) with respect to τ , we get,
rs
L2z
3rs L2z
d2 r
=
−
+
−
.
dτ 2
2r 2 m2 r 2 2m2 r 4
1
(7)
For a stable circular orbit (r =constant), equation (7) implies that the angular momentum L2z along this orbit should satisfy,
L2z =
m2 rs r
.
2 − 3rrs
(8)
Together with equation (6) for the case of a stable circular orbit, the energy
can be written as,
2
m2 1 − rrs
2
.
E =
(9)
s
1 − 3r
2r
Dividing equation (4) by equation (5) and using equation (8) and (9), we
find,
s
2r 3
dt
=
.
(10)
dφ
rs
To figure how time elapses along the orbit, we consider the invariant line
element,
rs 2
ds2 = − 1 −
dt + r 2 dφ2 ≡ −dτ 2 ,
(11)
r
where we have taken into account the fact of circular orbit with the angle θ
held fixed at π/2. Dividing equation (11) by dφ2 , we find,
1/2
2
rs dt
dτ
(12)
= 1−
− r2 .
dφ r
dφ
2
Detection of Gravitational Wave
Take the problem 7.7 in Carroll as an example. From the plane wave form,
h
i
h
i
ds2 = −dt2 + 1 + A cos[ω(t − z)] dx2 + 1 − A cos[ω(t − z)] dy 2 + dz 2 , (13)
we see that the gravitational wave is polarized along x and y directions.
Thus, to get the largest response we’d better orient the detector along these
two directions so as to measure the difference in arrival time. Suppose these
two masses are lying in the x − y plane (z = 0).
If the detector is oriented along x direction, we can write equation (13)
as the null condition for either the outgoing or returning laser beam,
h
i1/2
dt = ± 1 + A cos(ωt)
dx ,
(14)
2
where “+” denotes the outgoing beam, and “−” denotes returning beam.
Then we can estimate the time pass for either outgoing or returning process
by inverting the relation in equation (14) up to the error of O(A2 ),
A
1 − cos(ωt) dt ≈ ±dx .
(15)
2
For outgoing, we integrate the “+” case,
Z Lx
Z t′ A
dt 1 − cos(ωt) ≈
dx ,
2
0
0
(16)
and find halfway time pass, to the leading order of A which is assumed to be
small,
A
t′ ≈ Lx +
sin(ωLx ) .
(17)
2ω
Then for returning, we integrate the “−” case,
Z Tx Z 0
A
dx ,
(18)
dt 1 − cos(ωt) ≈ −
2
Lx
t′
and use the halfway time given by equation (17) so as to find the arrival time
of the pulse traveling along x direction,
2Lx
A
2ωLx
Tx ≈
.
(19)
+
sin
c
2ω
c
Similarly, we can obtain the arrival time of the pulse if the detector is oriented
along the y direction, i.e., Ty . Then we will be able to see how the difference
in arrival time of pulse in these two different directions is related to the
observable signals corresponding to the gravitational wave.
3
Linearized Gravity and Gauge Transformations
Suppose we add a small perturbation hµν to the flat space to form the full
metric,
gµν = ηµν + hµν , (|hµν | ≪ 1) .
(20)
3
Using the inverse metric,
g µν = η µν − hµν + hµρ hνρ + O(h3 ) ,
(21)
to work out the affine connection, and then the Riemann tensor, we find, to
the leading order of perturbation,
1
Rµνρσ =
∂ρ ∂ν hµσ + ∂σ ∂µ hνσ − ∂σ ∂ν hµρ − ∂ρ ∂µ hνσ + O(h2 ) ,
(22)
2
and the Einstein tensor,
1
ρ
ρ
ρσ
Gµν =
∂ρ ∂ν hµ + ∂ρ ∂µ hν − hµν − ηµν ∂ρ ∂σ h + ηµν h + O(h2 ) , (23)
2
where,
hµν ≡ η µρ hρν ,
≡ ∇µ ∇µ ,
h ≡ η µν hµν ,
(24)
(25)
(26)
At this point it is worth mentioning that under the following coordinate
transformation,
x′µ = xµ − ξ µ (x) ,
(27)
the perturbation will change as,
h′µν = hµν − 2∂(µ ξν) .
(28)
This change is called a gauge transformation, an analog to the gauge transformations of electromagnetism.
Now if we define the so called “tace reverse” of hµν ,
1
hµν ≡ hµν − ηµν h ,
2
(29)
which has this name because,
h ≡ η µ hµν = −h .
(30)
With this “trace reverse” one can show, starting with equation (22), that the
Einstein tensor can be written as,
1
ρσ
ρ
ρ
(31)
Gµν = − hµν + ηµν ∂ρ ∂σ h − ∂ρ ∂ν hmu − ∂ρ ∂µ hnu + O(h2 ) .
2
4
It is clear that equation (31) would get vastly simplified if we require,
∂µ h
µν
=0.
(32)
These are four equations, but recall that we also have four free gauge functions ξ µ . It turns out always possible to choose a gauge so that the transverse
conditions given by equation (32) can be satisfied. To see this, let’s consider
the gauge change of the “trace reverse”. Using the equation (28) and (30)
one can show that,
h′µν = hµν − 2∂(µ ξν) + ηµν ∂ρ ξ ρ .
(33)
Then the divergence is,
∂ µ h′µν = ∂ µ hµν − ξν .
(34)
So, if we insist ∂ µ h′µν = 0, the gauge functions ξ µ can be determined by the
following equation,
µν
ξ µ = ∂ν h .
(35)
Under such gauge, we can write the Einstein tensor in a very simple form,
1
Gµν = − hµν .
2
5
(36)