Homework # 8 Solutions
1. Show that a simple closed-box model with instantaneous recycling results in
the gas-phase metallicity increasing linearly with time. If this is true, what
cosmic time does [Fe/H]=-3 correspond to?
From the notes, we see that Z = Z0 + pν(1 − R)t, but only if the Schmidt
law with N = 1 is assumed. Since Z(T − 4.5 Gyr)) = Z , where T =today,
Z(T ) = Z + (4.5 Gyr)pν(1 − R). If Z0 = 0, Z(T ) = Z T /(T − 4.5 Gyr) '
1.6Z . Then t = 7.5 Myr.
Also show in this model that the distribution of stars in metallicity dN/d ln Z
has a maximum where Z = p, the effective yield of heavy elements defined in
the lecture notes.
Since dN/d ln Z = MZe−(Z−Z0 )/p , where N is the number of stars in an
interval of Z, the maximum will be at Z = p.
2. Explore the evolution of the solar neighborhood with a galactic chemical evolution model with instaneous recycling and infall which obeys
f (t) = A 1 − t/τf .
Assume an infall timescale of τf = 24 Gyr and initial densities Σg (0) + Σ∗ (0) =
Σ)T (0). Determine the value of A from the observed surface density ΣT (T ) =
50 M pc−2 of stars and gas at the present time T = 12 Gyr in the Galactic
plane in the solar neighborhood.
In this model, dΣT /dt = f . Integrating from t = 0 to t = T , we find
ΣT (T ) = AT (1 − T /(2τf ), so A = 50/9M pc−2 Gyr−1 .
Assuming the Schmidt SFR Ψ = νΣg , determine the value of ν that will
give the present gas fraction Σg (T )/ΣT (T ) = 0.24 at the present time. Hint:
Note that
df
d
+ αf = e−αt
f eαt
dt
dt
if α is a constant.
Setting q = (1 − R)ν, we have
Aeqt 1 − t/τf dt = d Σg eqt .
After integrating, we find
h
i
−1 −qt
(A/q) 1 + qτf
1−e
− t/τf = Σg (t)
where the constant of integration is found from Σg (0) = 0. Since Σg (T ) =
0.24ΣT (T ), we can eliminate A from the above equation. Then we find
qT 1 + (Σg (T ) /ΣT (T )) q τf − T /2 = 1 − e−qT 1 + qτf
for which the solution is q = ν(1 − R) ' 0.28 Gyr−1 .
3. Explain why the value of [O/Fe] observed in stars is both greater than 0 and
does not vary with [Fe/H] for metal-deficient stars for [Fe/H]< −1.5, but decreases with increasing metallicity for more metal-rich stars. When [Fe/H]=0,
what value should [O/Fe] have?
At low metallicities, all O and Fe is produced by type II supernovae and is
therefore roughly constant. However, at later times than [Fe/H]=-1.5, Type I
supernovae contribute Fe but no O, so O/Fe decreases. When the iron abundance is solar, the oxygen abundance should also be solar.
4. Do problem 4.13 in the text.
The equations described, setting e = 1, are
q
r = a (1 − cos η) , t = a3 / [G (M + m)] (η − sin η) ,
dr/dt = (r/t) sin η (η − sin η) / (1 − cos η)2 .
Using r = 770 kpc, t = 12.8 Gyr and v = −120 km s−1 , one finds η ∼ 4.2, a =
520 kpc and M + m ∼ 5 × 1012 M .
The two galaxies will again come close together when η = 2π or
q
t = 2π a3 / [G (M + m)] ' 16 Gyr.
Since now was assumed to be 12.8 Gyr, this is 3.2 Gyr from now.
One has to add up the luminosities for the Local Group galaxies, about
5.1 × 1010 L . Then the mass-to-light ratio about 100.
With the new value of η and given values of r and dr/dt, one can compute
a new value of t which turns out to be larger. With the new eta and given r,
you recompute the value of a which also becomes larger. Then use the equation
for t to compute the new value of the total mass, which turns out to become
smaller.