Solutions for Homework #2

part. Since our system also has a central mass (the black hole), the original Maxwellian
distribution function has to be modified, so that
E i = 1 2 v2 + V (44)
where
Our distribution function then is
V = − GM r
(45)
f 0 (v) =
ν 0
GM
e
(2πσ 2 3/2
)
rσ 2 − v2
2σ 2 (46)
The velocity distribution can be derived from
∫ vmax
ν(r) = 4π v 2 f 0 (v)dv (47)
v min
To calculate that, we have to know the boundary velocities. Since we are looking at "unbound"
particles, we know that the smallest allowable velocity is their "free" velocity,
√
which is v min = 2GM/r. The outer velocity boundary has to be ∞. Our starting equation
then becomes
By substituting our Maxwellian distribution, we get
∫ ∞√
ν(r) = 4π v 2 f 0 (v)dv (48)
2GM/r
∫
ν(r)
∞√
= 4π v 2 1 GM
e rσ
ν 0 2GM/r (2πσ 2 3/2 2 − v2
2σ 2 dv (49)
)
This equation is not straightforwardly solvable, but can be done using Mathematica,
which immediately gives the desired solution:
√ [ (√
ν(r) rH
= 2
ν 0 πr + rH er H/r
1 − erf
r
)]
(50)
3.b. To obtain the behavior at r ≪ r H , we use equation (1C-13) from Binney & Tremaine
which says:
e−x2
lim (1 − erf x) = √ . (51)
x→∞ πx
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