Homework # 6 Solutions
1. Demonstrate that the temperature applicable to a stellar system is
T =
m < v 2 (x) >
3k
where it is assumed that the stars have equal mass m, v(x) is the velocity of a
star relative to the center-of-mass, and <> represents an average.
On the assumption that the stellar velocities are distributed in the most
probably way, i.e., a Boltzmann distribution, one has
n (x, v) = no exp − mΦ (x) + mv 2 /2 / (kT ) .
The quantity kT is simply a constant representing an average energy. To see
how it is related to the average velocity, we compute the mean square stellar
velocity at x, which is
R∞
4πno v 4 exp − mΦ (x) + mv 2 /2 / (kT ) dv
2
0
hv (x)i = R ∞
2
2
0 4πno v exp [− (mΦ (x) + mv /2) / (kT )] dv
R∞ 4
v exp −mv 2 / (2kT ) dv
Γ (5/2) (2kT /m)5/2
3kT
0
=R∞ 2
=
.
=
3/2
2
m
Γ (3/2) (2kT /m)
0 v exp [−mv / (2kT )] dv
2. Calculate the gravitational potential a distance z from a steet of matter with
uniform surface density Σ, and show that the vertical force is independent of
z.
We use Poisson’s equation and Gauss’ divergence theorem:
Z
Z
Z
3
3
2
4πGM = 4πG ρd x = d x∇ Φ (x) = d2 S · ∇Φ (x) = 2S∇Φ (x) .
We could take ∇Φ out of the integral since it depends, in this geometry, only on
z. The surface mass density is M/S = Σ, so ∇Φ(z) = 2πGΣ which is constant.
This is proportional to the force, which is also constant. Thus Φ(z) = 2πGΣ|z|.
3. In this problem, you are asked to compute the frequencies of the Sun’s orbital
excursions from its mean position perpindicular and parallel to the Galactic
plane, ν and κ, respectively. You are to assume a Plummer model for the
Galctic gravitational potential. Find the Plummer parameter a from the observed values of the Oort A and B constants,
A = 15 km s−1 kpc−1 ,
B = −12 km s−1 kpc−1 ,
assuming the Galactocentric distance of the Sun is R0 = 8.5 kpc and its circular
velocity is V0 = 220 km s−1 .
The Plummer potential and derivatives are:
GM R
∂Φ
,
=
3/2
∂R
(r2 + a2 )
GM a2 − 2R2
GM a2 − 2z 2
∂ 2Φ
∂ 2Φ
=
,
=
.
5/2
5/2
∂R2
∂z 2
(r2 + a2 )
(r2 + a2 )
Φ = −√
One has
GM
,
r 2 + a2
and
2
ν =
κ2 =
∂ 2Φ
∂z 2
∂ 2Φ
∂R2
∂Φ
∂R
= Ω20 R0
R=R0 ,z=0
=
R=R0 ,z=0
R=R0 ,z=0
GM a2
Ω20 a2
=
5/2
R02 + a2
R02 + a2
+ 3Ω20 = Ω20
4a2 + R02
= −4BΩ0 .
R02 + a2
p
Using Ω0 = A − B we find a =p 7/20R0 .
√
−4B(A − B) = 36 km s−1 kpc−1 and ν =
Also, κ = −4BΩ0 =
q
p
κa/ R02 + 4a2 = 7/48κ = 14 km s−1 kpc−1 .
Also determine the total mass of the Galaxy and the fraction which is
interior to the solar circle. Explain the ratio κ/ν that you find.
The total mass M is
3/2
Ω20 R02 + a2
M=
= 1.4 × 1011 M .
G
The fraction within the solar circle is
3/2
Z R0
M (R0 )
R03
20
3a2 r2 dr
=
=
=
' 0.64.
5/2
3/2
M
27
0
(r2 + a2 )
R02 + a2
The ratio of κ/ν > 1 is opposite to what we would have found if we assumed
the Galactic material was concentrated in the Galactic plane, in which case
ν ' 2κ.