Homework # 3 Solutions
1. Calculate the size of a Stromgren sphere in a gas of density 103 cm−3 surrounding an O3 star. This is the volume of gas nearly completely ionized by
the star. Assume the recombination rate of protons and electrons to form
neutral hydrogen is
−3/4
dne
T
−13
= −2 × 10 ne np
cm−3 s−1 .
dt
104 K
A Stromgren sphere represents a balance between the rate of production
of ionizaing photons and recombinations. The ionization rate doesn’t matter
once equilibrium is established since it is generally faster than recombination.
The flux of ionizing photons is
Z ∞
Z
L∗ (ν)
R∗2 ∞ 2ν 2
1
Fν =
dν = 2
dν
2
2
hν/kT
r ν1 hc e
−1
ν1 4πr hν
The spectral distribution peaks near hνp ∼ 2kT . When ν1 ≥ νp the lower limit
of the integral can be taken to be 0 and most of the star’s photons ionize, but
when ν1 < νp only a fraction of the star’s photons can ionize. For these limits
Z ∞ 2
−a
x dx
2
→
a
+
2a
+
2
e
a > 0.9
x
a e −1
→ 2ζ (3) − a2 /2 ' 2.4 − a2 /2
a < 0.9
For an O3 star, T = 44, 500 K, a = hν1 /kT = 3.6. For a G3 star, T = 5780 K,
a = 27.3. Thus, the fraction of photons available is 25% for the O3 star, but
only 1.6 × 10−11 for the G3 star.
From the notes, the quantity NLyα ∼ 1050 s−1 is the production rate for
an O3 star, roughly. Equating this to the recombinations per second when
completely ionized gives
2/3
1/3
Rs = 1.2 103 cm−3 /n
NLyα /5 × 1049 s−1
pc.
For a G3 star, the production rate of photons will be smaller by
2 4
RG
TG
× 1.6 × 10−11 ∗ 4 ∼ 2 × 10−16 ,
RO
TO
so the Stromgren sphere will be smaller by 1.4 × 10−8 , or about 10−2 AU. This
would be almost at the surface of the Sun! We expect H in space near the
Earth to be neutral.
If the star suddenly turned on, estimate how long it would take for the
Stromgren sphere to come into radiative equilibrium.
The recombination is the slower rate, so computing ne /(dne /dt) gives about
5 × 1019 s for np ∼ 103 cm−3 and T ∼ 104 K. This is not much longer than the
light-travel time throught the Stromgren sphere.
2. Suppose the number density of stars normal to the Galactic plane varies as
n(z) = n0 exp(−z/zh ). Suppose you do a survey of G2 stars in the direction of
the North Galactic Pole complete to a magnitude mv = 30.
What value would you obtain for V/V max ?
First, calculate zmax , the largest distance you could see a G2 star in the
survey. Use M = 5 for a G2 star; then
zmax = 101+(m−M )/5 pc ∼ 106 pc.
This is practically infinity. We find
R zmax
3 R ∞ 5 −x
3
5 dz
n
(z)
z
x e dx
z
V
zh
h
0R
0
R∞
= 3
=
= 60
<< 1
z
2 −x
Vmax
zmax
zmax
zmax 0 max n (z) z 2 dz
0 x e dx
which shows that the objects are very non-uniformly distributed, as expected.
3. Compute expressions for the potential and kinetic energies of the singular
isothermal sphere.
Assume the temperature is 20 K. Suppose the sphere’s boundary ro is where
it’s pressure is the same as that of the warm phase in the ISM.
The mass of the sphere, if we terminate it at ro , is
Z ro
Z ro
µ
µ
2
2
M = 4π
ρ (r) r dr = 4π no ro
dr = 4π no r03 .
No
No
0
0
If the cloud satisfies hydrostatic equilibrium dp/dr = −GM(r)ρ(r)/r2 , then
kT = 2πG(µ/No )2 no ro2 . The matching pressure is po , and
r
po
No kT
1
po = no kT,
no =
,
ro =
.
kT
µ
2πGpo
Eliminating no and ro , and solving for po , we find
4
2
No kT
po =
.
πG3 M2
µ
Therefore, the mass and temperature of a cloud will adjust to force pressure
equilibrium at the boundary.
If we take the gravitational potential of the sphere to be zero at this boundary, ro , the potential and kinetic energies are
Z
No kT
No kT
KE =
dM =
M
µ
µ
Z
M (r)
M2
PE = −G
dM = −G
.
r
ro
What is the radius of the sphere if its mass is 10 M ? We cannot specify po ,
M and T independently and maintain hydrostatic equilibrium. So to obtain
a certain radius requires that the cloud’s temperature or the value of po be
varied.
What will happen if the sphere cools uniformly?
If constant pressure is applied at the cloud’s edge and hydrostatic equilibrium is maintained, the mass and radius of the cloud will both decrease.
Before cooling occurs, the condition 2 KE + PE = 4πpo ro3 applies. Thus,
2 KE > |PE| since the right-hand side is positive; the cloud is stable against
collapse. But if cooling occurs rapidly enough, collapse can be triggered since
the kinetic energy is decreased and the magnitude of the potential energy is
increased by the quasi-static contraction.