Solutions for Homework #2

that, we get equation (7):
dr d 2 r
dt = −4πG
dt dt2 3r 2 r3 0 ρ 0dr (6)
( ) 2
1 dr
= 4πG
2 dt 3r r3 0ρ 0 + C 1 (7)
At r = r 0 (t = 0) we know that dr/dt = 0, so the constant of integration is
C 1 = − 4πG
3 r2 0 ρ 0 (8)
so this gives:
√
( )
dr 8πG
dt = 3 ρ 0r0
2 r0
r − 1
(9)
Let us use new parameters to make the derivation more simple. Let K be:
K =
√
8πG
3 ρ 0 (10)
and let θ be
θ = r r 0
(11)
This way
so dθr 0 = dr. The equation of motion then becomes:
Now let θ = cos 2 ξ. This way
dθ
dr = 1 r 0
(12)
dθ
dt = K √
1
θ − 1 (13)
dθ
dξ
The equation of motion then becomes:
= 2 sin ξ cos ξ (14)
2 sin ξ cos ξ dξ
dt
cos 2 ξ dξ
dt
√
1
= K
cos 2 ξ − 1 (15)
= K sin ξ
(16)
cos ξ
= K 2
(17)
2