Homework # 8 Solutions 1. Show that a simple closed-box model ...

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 = Z 0 + 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 Z 0 = 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−Z 0)/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/τ f
)
.
Assume an infall timescale of τ f = 24 Gyr and initial densities Σ g (0) + Σ ∗ (0) =
Σ)T (0). Determine the value of A from the observed surface density Σ T (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, dΣ T /dt = f. Integrating from t = 0 to t = T , we find
Σ T (T ) = AT (1 − T/(2τ f ), so A = 50/9M ⊙ pc −2 Gyr −1 .
Assuming the Schmidt SFR Ψ = νΣ g , determine the value of ν that will
give the present gas fraction Σ g (T )/Σ T (T ) = 0.24 at the present time. Hint:
Note that
df
dt + αf = d ( e−αt fe
αt )
dt
if α is a constant.
Setting q = (1 − R)ν, we have
Ae qt [ 1 − t/τ f
]
dt = d
[
Σg e qt] .
After integrating, we find
(A/q)
[(1 + ( ) ) −1 (1
qτ f − e
−qt ) ]
− t/τ f = Σ g (t)
where the constant of integration is found from Σ g (0) = 0. Since Σ g (T ) =
0.24Σ T (T ), we can eliminate A from the above equation. Then we find
qT [ 1 + (Σ g (T ) /Σ T (T )) q ( τ f − 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
√
r = a (1 − cos η) , t = a 3 / [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 × 10 12 M ⊙ .
The two galaxies will again come close together when η = 2π or
√
t = 2π a 3 / [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 × 10 10 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.