Download Problem Set 2 for Astro 322 Read chapter 24.2. (Some of this

Problem Set 2 for Astro 322
Read chapter 24.2. (Some of this material isn’t in the textbook.)
Problem 1: Assume the Salpeter equation describes stars formed in a cluster with masses
between Ml and Mu >> Ml . • Write down and solve the integrals that give (a) the number of
stars, (b) their total mass, and (c) the total luminosity, assuming L = L (M/M )3.5 . • Explain why the number and mass of stars depend mainly on the mass Ml of the smallest stars,
while the luminosity depends on Mu , the mass of the largest stars. • Taking Ml = 0.3M
and Mu >> 5M , show that only 2.2% of all stars have M > 5M , while these account for
37% of the mass. • The Pleiades cluster has M ∼ 800M ; show that it has about 700 stars.
• Taking Mu = 10M , show that the few stars with M > 5M contribute nearly 80% of the
light.
R
RM
a) number: ξ(M )dM = Mlu ξ0 M −2.35 dM ,
R
−1.35
u
ξ(M )dM = −(ξ0 M −1.35 /1.35) |M
− Ml−1.35 )/1.35
Ml = −ξ0 (Mu
b) mass: We’ll need to multiply the integrated function by the mass of the star, to convert our expression for the number of stars into an expression for the sum of their masses.
R
RM
RM
ξ(M )M dM = Mlu ξ0 M −2.35 M dM = Mlu ξ0 M −1.35 dM ,
R
ξ(M )M dM = −ξ0 (Mu−0.35 − Ml−0.35 )/0.35
R
RM
c) luminosity: We use L ∝ L (M/M )3.5 , so ξ(M )L(M )dM = Mlu ξ0 M −2.35 L M 3.5 dM =
R Mu
ξ0 M 1.15 L dM .
Ml
R
2.15
u
− Ml2.15 )/2.15.
Solving this, ξ(M )L(M )dM = (ξ0 L /2.15)M 2.15 |M
Ml = ξ0 L (Mu
The expressions for the number and mass have negative exponents on the mass limits, so
the result will be dominated by the *smaller* number, Ml , rather than Mu . However, the
luminosity expression has positive exponents on the masses, so the larger Mu dominates.
Taking Mu as 100,000 (it needs to be rather large to asymptotically approach the right answer for the mass), the total number of stars is −ξ0 (100000−1.35 − 0.3−1.35 )/1.35 = 3.76ξ0 .
The number of stars above 5 M is then −ξ0 (100000−1.35 − 5−1.35 )/1.35 = 0.083ξ0 . Thus the
number of stars above 5 M is only 0.022=2.2% of the total.
The mass of these stars is −ξ0 (100000−0.35 − 5−0.35 )/0.35 = 1.576ξ0 , while the total mass is
−ξ0 (100000−0.35 − 0.3−0.35 )/0.35 = 4.304ξ0 . Thus the total mass of the high-mass stars is
∼0.37 of the total cluster stellar mass.
If the cluster mass is 800 M , then we can find ξ0 ; 800M = −ξ0 (Mu−0.35 − Ml−0.35 )/0.35, so
ξ0 = 800 ∗ 0.35M /(0.3−0.35 − 100000−0.35 ) = 185.9. Then # stars= −185.9(100000−1.35 −
0.3−1.35 )/1.35 = 699.6, about 700 stars.
If Mu = 10M , then the total light from all stars is ξ0 L (102.15 − 0.32.15 )/2.15 = 65.7ξ0 L ,
while the light from above 5 M is ξ0 L (102.15 − 52.15 )/2.15 = 50.9ξ0 L , about 78% of the
total.
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Problem 2: Thin disk stars make up 90% of the total stellar density in the midplane, while
10% belong to the thick disk. However, hz in the thin disk is roughly three times smaller
than for the thick disk. Show that the surface density of stars per square parsec follows Σ(R,
thin disk)∼ 3Σ(R, thick disk).
The surface density Σ is just the volume density integrated over all z. The volume density
will go as n = n0 exp(−|z|/hz ). So we can integrate this over z (piecewise, cut at z=0;
R∞
or else integrate from z=0 to ∞ and double it), getting Σ(top)= 0 n0 exp(−z/hz )dz =
−n0 hz (exp(−∞) − exp(0)) = no hz for the top half, or Σ = 2n0 hz for the total surface
density.
The midplane volume density is simply n(0) = n0 exp(−0) = n0 , so the ratio n0,thin /n0,thick =
9. Then the ratio of Σthin /Σthick = (n0,thin /n0,thick )(hz,thin /hz,thick ) = 9 ∗ (1/3) = 3.
Problem 3: • By integrating the expression n(R, z, S) = n(0, 0, S) exp[−R/hR (S)] exp[−|z|/hz (S)],
show that at radius R the number of stars per unit area (surface density) of type S is
Σ(R, S) = 2n(0, 0, S)hz (S) exp[−R/hR (S)]. If each has luminosity L(S), the surface brightness I(R, S) = L(S)Σ(R, S). • Assuming that hR and hz are the same for all types of star,
show that the disk’s total luminosity LD = 2πI(R = 0)h2R . • For the Milky Way, taking
LD = 1.5 × 1010 L in the V band and hR = 4 kpc, show that the disk’s surface brightness
at the Sun’s position 8 kpc from the center is ∼ 20L pc−2 . The mass density of stars in the
disk is 40-60 M pc−2 , so we have M/LV ∼2-3. • Why could this be larger than M/LV for
stars within 100 pc of the Sun?
We want Σ(R, S), the z-integrated volume density. Only one part has a z-dependence,
R∞
so we integrate that over z as in question 2 ( 0 exp[−z/hz (S)]dz = −hz (S)(exp[−∞] −
exp[0]) = hz (S); adding the bottom side of the disk doubles it, to 2hz (S)), getting Σ(R, S) =
2n(0, 0, S) exp[−R/hR (S)]hz (S).
To get the disk’s total luminosity, first integrate Σ(R, S) over the area of the galaxy, in
polar coordinates (deriving the total number of stars). We’ll drop the S dependence on scale
heights, as indicated. Integrating in polar coordinates needs the integrating factor R dR dθ
R 2π R ∞
(equivalent to dx dy in Cartesian coords). So LD = 0 0 2n(0, 0, S)hz exp[−R/hR ]R dR dθ
Treat exp[−R/hR ]R dR as x exp[cx] dx, with antiderivative exp[cx](cx−1)/c2 , for c = −1/hR ,
x = R. Then
R
R 2π
Σ = 2n(0, 0, S)hz h2R (−R/hR − 1)(exp[−R/hR ]) |∞
dθ
0
0
R
R 2π
2
Σ = 2n(0, 0, S)hz hR (exp[−∞](−∞/hR − 1) − exp[0](0/hR − 1)) 0 dθ
R
Σ = 2n(0, 0, S)hz h2R (0 − (−1))(θ) |2π
0
R
Σ = 2n(0, 0, S)hz h2R (2π) = 4πn(0, 0, S)hz h2R
To get LD , we need to multiply by L(S) and integrate this expression over S; LD =
R
R
L(S)4πn(0, 0, S)hz h2R dS = 4πhz h2R allstars L(S)n(0, 0, S)dS.
allstars
2
Ok, next we look at surface brightness, I(R, S) = L(S)Σ(R, S). Integrating over S gives
R
the total surface brightness function I(R) = allstars L(S)Σ(R, S)dS. For R = 0, I(R = 0) =
R
R
L(S)2n(0, 0, S)hz exp[−0/hR ]dS I(R = 0) = 2hz allstars L(S)n(0, 0, S)dS.
allstars
Comparing this with the expression for LD above, we see they differ by only 2πh2R , and thus
that LD = 2πh2R I(R = 0).
For the Milky Way, 1.5 × 1010 L = 2πh2R I(R = 0), and I(R) = I(R = 0) exp[−R/hR ].
We find I(R = 0) = LD /(2πh2R ) = 1.5 × 1010 L /(2π(4000pc)2 ) = 149 L pc−2 . Then
I(R = 8kpc) = I(R = 0)exp(−8/4) = 20.2 L pc−2 .
We note that this M/LV ratio is integrated over all z, through the thick disk as well as the
thin. The thick disk (due to its age) has no massive stars, so its M/LV ratio is rather higher
than the ratio for stars near the Sun at the midplane, which include younger stars.
Problem 4: See Figure 1 below, which plots the numbers of stars at each B − V color
with apparent V magnitude 19 < mV < 20, per square degree near the North Galactic Pole.
Use Fig. 2 to represent the stars in the local disk, and the metal-poor globular M92 to
represent stars in the halo (Fig. 3). (a) What is the absolute magnitude MV of a disk star
at B − V = 0.4? How far away must it be to have mV = 20? In M92, the bluest stars still
on the main sequence have B − V ∼ 0.4. Show that, if such a star has apparent magnitude
mv = 20, it must be at d ∼ 20 kpc. (b) What absolute magnitudes MV could a disk star
have, if it has B − V = 1.5? How far away would that star be at mV = 20? In M92, what is
MV for the reddest stars, with B − V = 1.2? How distant must these stars be if mV = 20?
(c) Explain why the reddest stars in Fig. 1 are likely to belong to the disk, while the bluest
stars belong to the halo.
a) Reading from fig. 2, MV = 2.5 (plus or minus 1 are acceptable). Since we’re looking out
of the disk, there’s little dust and extinction. For mV = 20, d = 10(mV −MV +5)/5 = 31622 pc
or ∼32 kpc. (Choices of MV ± 1 from 2.5 give d between 20 and 50 kpc.) For M92, from fig.
3, MV ∼ 4 for B − V = 0.4. As above, d = 10(20−4+5)/5 = 15849 or ∼20 kpc.
b) A disk star could have MV of either 10-11 (red dwarfs) or ∼0 (giants). For mV = 20, the
red dwarfs are at distances d = 1000 to 631 pc, while the red giants would be at ∼100,000
pc. For M92, MV of the reddest stars is ∼ −3, at B − V = 1.2. For mV = 20, their distance
would be 400,000 pc.
c) Consider the distances calculated above. A distance of 630-1000 pc is consistent with
nearby stars in the disk (maybe 1 kpc thick), so the reddest stars in Fig. 1 can be red
dwarfs. A distance of 100 kpc is too far for the disk, so these red stars are not red giants.
A distance of 400 kpc is 50 times farther than our distance to the Galactic Center, so even
the halo runs out of stars. For the blue stars in part a, 32 kpc is too far for the disk. ∼ 20
kpc is in the far reaches of the halo (sim3-4 times our distance from the Galactic Center),
so the bluest stars are in the halo.
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Fig. 1.— Numbers of stars at each B −V color
for 19 < mv < 20 toward north Galactic pole.
The solid line shows the prediction of a model:
thin-disk stars (triangles) are red, halo stars
(stars) are blue, thick-disk stars (squares) have
intermediate colors.
Fig. 2.— CMD from Hipparcos.
Fig. 3.— CMDs for the metal-poor globular cluster M92, and metal-rich globular 47 Tuc.
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