PROBLEM SET #6 AST142 Due in class Tuesday Mar 17, 2015 First

PROBLEM SET #6
AST142
Due in class Tuesday Mar 17, 2015
First, some useful formulas:
Scaling on the main sequence:
Mean free path
T 3/5 ρ¯2
l ∝ −1
ρ¯
for
M . M
M & M
Luminosity
L∝
M 5.5 R−0.5
M3
(low mass)
(high mass)
And also
M 4
L
M
M
R
M
M 1/2
Te,
M
Te 8
L
Te,
L ≈
R ≈
Te ≈
L ≈
τ
∝ M −3
Where M, R, L, Te , τ are mass, radius, luminosity, effective temperature and main sequence
lifetime.
The mean molecular weight µ
mp
3
1
= 2X + Y + Z
µ
4
2
where X, Y , and Z are the fractions of the stars mass in the form of hydrogen, helium,
and all other elements, respectively.
The initial mass function (IMF) describes the numbers of stars born as a function of stellar
mass.
N (M )dm
1
2
PROBLEM SET #6
AST142
represents the total number of stars point with mass between M and M + dm. A Saltpeter
initial mass function has
N (M ) ∝ M −2.3
.
Problems
1. Initial mass function
Star formation regions contain both low and high mass stars but they are not born
with equal probability. The mass distribution at birth is described with a function
called the initial mass function (IMF). Consider the initial mass function with number
of stars in a mass bin dM equal to N (M ) where M is the stellar mass. A power law,
N (M ) = AM −α
with constant A and index α is an approximate description for the initial mass function.
Recent work has used more complex functions to describe it and found that the initial
mass function may vary with location in the Galaxy (and other galaxies). An early
measurement for the exponent in this function was due to Saltpeter. For a Saltpeter
initial mass function the exponent or index is α = 2.3.
(a) Masses of stars in star clusters or star forming regions are not directly observed.
Instead an observer may compile a luminosity distribution by counting the number
of stars as in different luminosity bins. Using a mass-luminosity relation for stars on
the main sequence and the Saltpeter Initial Mass function, compute the luminosity
function N (L) numbers of stars in a luminosity bin of size dL. You can ignore the
constant of proportionality, finding an exponent, γ with N (L) ∝ L−γ .
(b) Observers are more likely to use magnitudes than luminosities. Compute the magnitude distribution N (m) or the number of stars in a magnitude bin dm with
magnitude m. This may vary over orders of magnitude so it is convenient for
an observer to compute log10 N (m). Ignore color and bolometric corrections so
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AST142
3
m = −2.5 log10 L+ constant. On a histogram plot of log10 N (m) versus m you
expect a line. What is the slope of the expected line?
(c) High mass stars have shorter lifetimes on the main sequence. Consider the distribution of main sequence stars in the Solar neighborhood on the main sequence
assuming a constant star formation rate. In this case the number of stars in a mass
bin is Nτ (M ) ∝ N (M )τ (M ) with τ (M ) the lifetime of a star with mass M . How
do you expect the fraction of the main sequence field stars in a mass bin dM to
depend on M in a volume limited and complete sample in the Solar neighborhood?
Use a mass-lifetime relation for stars on the main sequence and the Saltpeter initial
mass function.
2. Numbers of massive stars
Assume a minimum birth mass of Mmin = 0.1M and a maximum mass of Mmax =
100M .
¯ (in solar masses) born using a Saltpeter IMF? What is
(a) What is the average mass, M
the average mass taking into account the lifetime on the main sequence? Evaluate
R Mmax
¯ =
M
Mmin dM M N (M )
R Mmax
Mmin dM N (M )
and
R Mmax
¯ =
M
Mmin dM M Nτ (M )
R Mmax
Mmin dM Nτ (M )
¯ more sensitive
Do do these computations you can assume that Mmin Mmax . Is M
to the high or low mass limit?
(b) For every 1000 stars born how many have masses above 8 M (and could become
a type II supernovae?)
3. Extinction vectors
The extinction as a function of wavelength in the optical and near-infrared wavelengths is approximately
4
PROBLEM SET #6
AST142
Band Wavelength A(λ)/AV
B
0.45 µm
1.29
V
0.55 µm
1.00
R
0.66 µm
0.75
J
1.2 µm
0.28
H
1.6 µm
0.19
K
2.2 µm
0.112
On a R − J vs H − K magnitude color plot determine the length and orientation of a
vector corresponding to visual extinction AV = 1. This vector gives the change on the
color plot between intrinsic and observed colors of an object seen through extinction
corresponding to AV = 1.
4. Extinction and how it affects distance estimates
An F0 main sequence star has an absolute magnitude of MV = 2.7 and intrinsic color
(B − V )0 = 0.30. You observe an F0 main sequence star (with spectral type identified
spectroscopically) finding apparent magnitudes mV = 12.00 and mB = 12.56.
(a) What is the extinction AV for this star?
(b) What is the distance to this star?
(c) What distance would you have computed ignoring extinction?