Chong-Sun Chu
Solution
Mid-Term Examination
Introduction to Relativity II (PHYS432000)
1. (Schwarzschild blackhole)
(a) Consider a blackhole described by the Schwarzschild metric. Show that
Ẽ := −
is conserved.
Soln:
EOM is
m
p0
m
(1)
dpα
1
= gµν,α pµ pν ,
dτ
2
(2)
where pα := mdxα /dτ . Since the metric is independent of t, so p0 and hence Ẽ is conserved.
[10 pts]
(b) Derive the equation of motion for a particle falling radially in the Schwarzschild metric.
Soln:
p0 = g 00 p0 = m(1 − 2M/r)−1 Ẽ,
pr = m
dr
,
dt
pθ = 0
pφ = g φφ pφ = mL̃/r2 ,
(3)
dr 2
)
dt
(4)
where L̃ := pφ /m. For radial motion, L̃ = 0. Subsitute this into
−m2 = p2 = −m2 (1 − 2M/r)−1 Ẽ 2 + m2 (1 − 2M/r)−1 (
giving
(
dr 2
) = Ẽ 2 − (1 − 2M/r)
dt
(5)
[10 pts]
(c) Consider a particle of mass m falls into the blackhole in the radial direction, initially at a distance
R > 2GM from the center of the blackhole. Calculate, for the case that Ẽ = 1, the proper time
elapsed for the particle to reach the horizon.
Soln:
For Ẽ = 1, we have
dr
4M r 3/2 R
dτ = −
=
(
) .
(6)
3 2M
2M
(2M/r)1/2
[10 pts]
(d) Show that once the particle reaches the horizon of the Schwarzschild blackhole, it will reach r = 0
in a proper time
τ ≤ πM,
(7)
no matter how the engines are fired.
Soln:
The equation of motion reads
−1 = p2 = −(1 − 2M/r)−1 Ẽ 2 + (1 − 2M/r)−1 (
dφ
dr 2
) + r2 sin2 θ( )2
dt
dτ
(8)
Inside the horizon
(1 − 2M/r) < 0
1
(9)
and so thefirst and the 3rd terms on the RHS of the above equation are positive. Therefore
dr 2
) > 1,
dt
(10)
dτ < −(2M/r − 1)1/2 dr.
(11)
(2M/r − 1)−1 (
or
Integrating
Z
τ < τmax
=
0
(2M/r − 1)1/2 dr
−
2M
(12)
=
i0
h
1
)
r1/2 (2M − r)1/2 + 2M tan−1 ( 2M
2M
r −1
=
πM.
(13)
(14)
[20 pts]
2. (Gravitational radiation)
Consider two stars of mass M1 and M2 separated by a distance R revolve about each other in a circular
orbit. Let ω be the angular speed of rotation.
(a) In the lowest order approximation, you may assume that the motion is non-relativistic. Show that
the energy of the system is given by
1 M1 M2
E=−
(15)
2 R
Hint: use the reduced mass description of the system
Soln:
For Newtonian mechanics, the energy is given by
E
=
1
M1
1
M1 r˙1 2 + M2 r˙2 2 −
R
2
2
M2
=
1 2
µM
µω R −
2
R
=
µM
.
−
2R
(16)
(17)
(18)
Here M = M1 + M2 is the total mass.
[10 pts]
(b) Suppose the rotation is on the xy-plane and the rotation axis is the z axis. Calculate Ixx , Iyy and
Ixy .
Soln:
1 2
R µ cos 2φ + const,
2
1
= R2 µ sin2 φ = − R2 µ cos 2φ + const,
2
1
= Iyx = R2 µ sin φ cos φ = R2 µ sin 2φ,
2
Ixx = R2 µ cos2 φ =
Iyy
Ixy
(19)
(20)
(21)
where φ = ωt.
[10 pts]
2
(c) Caluclate the gravitational energy radiated per unit second.
Soln:
Note that
Ixx + Iyy = const
(22)
and so
Ixx = Ixx + const,
Perform the sum
d3 I jk d3 I jk
dt3
dt3
Iyy = Iyy + const,
Ixy = Ixy .
(23)
and the time average, we find
1 d3 Ijk d3 Ijk
i =
h
5 dt3 dt3
1
1
(2ω)6 ( R2 µ)2 hsin2 2ωt + sin2 2ωt + 2 cos2 2ωti
5
2
(24)
=
32 6 4 2
ω R µ
5
(25)
3 2
=
32 M µ
5 R5
(26)
[15 pts]
(d) As a result of the gravitational radiation, the radius R changes with time. Derive R(t).
Soln:
dE
1 µM dR
32 M 3 µ2
=
=
−
dt
2 R2 dt
5 R5
This get implified to
dR
64
R3
= − M 2µ
dt
5
and integrated to
256 2
R4 =
M µ(t0 − t),
5
where t0 is a constant. It can be fixed in terms of the initial radius R0 as
R04 =
256 2
M µt0 .
5
(27)
(28)
(29)
(30)
[15 pts]
3