Download Astro 7B – Solution Set 7 1 A Star is Born

Astro 7B – Solution Set 7
1
A Star is Born
Consider the gravitational collapse of a cloud of gas and dust in the interstellar
medium. Assume the temperature of the cloud is such that the cloud is initially
marginally Jeans-unstable. The gas consists predominantly of molecular hydrogen.
As the cloud collapses inward, its density increases. In the absence of cooling
mechanisms, the gas will heat adiabatically (gravitational potential energy is converted into kinetic energy and thence into heat). If the temperature rises too
quickly compared to the increase in density, the cloud will become Jeans-stable,
cease to collapse, and fail to form a star.
Fortunately, efficient cooling mechanisms are present in today’s interstellar medium.
Most notable among these is cooling by dust grains. Hot gas molecules can collide
with and transfer their energy to dust grains. The dust grains can then radiate
away this energy by thermal emission of infrared radiation. The radiation escapes
immediately, assuming the cloud is optically thin in the infrared.
The cloud can stay Jeans-unstable if the timescale for cooling is shorter than
the timescale for gravitational contraction (the latter is the same as the heating
timescale). That is, the collapse can proceed nearly isothermally rather than
adiabatically, if cooling mechanisms are efficient.
(a) Consider a spherical cloud of mass density ρ. Derive, to order-of-magnitude,
the timescale for this cloud to gravitationally contract and halve its radius. This is the gravitational contraction timescale, a.k.a. the free-fall
time.
One way to do this is to assume that the gravitational acceleration on the surface
of the cloud is constant as the cloud shrinks from its original radius to half its
radius.
Express your answer for the free-fall time, tff , in terms of G (the gravitational constant) and ρ. For simplicity, set all numerical coefficients of order
unity equal to 1 in your answer.
1
[3 points] If the cloud collapses by gravitational contraction, then its acceleration is g ≈
GM/R2 . As the radius shrinks by a factor of 2, g changes by a factor of 4 (since g ∝ R−2 ),
which is an “order-unity” change so for this problem we can say that g is approximately
constant. Thus, we can write
1 2
R
gtff ≈
2
2
1 GM 2
R
tff ≈
2
2 R
2
where R is the radius of the cloud, M is the mass of the cloud, and tff is the free-fall (aka
gravitational collapse) time. If we solve for tff we get
t2ff ≈
R3
1
≈
GM
Gρ
where we have used the fact that ρ ≈ M/R3 . Finally, we get
1
tff ≈ √
Gρ
As long as your answer doesn’t have a numerical coefficient that differs from 1 by more than
a factor of 10, you should receive full credit.
(b) What is the timescale for cooling? This is the time for a hot gas
molecule to collide with a dust grain.
Assume canonical parameters for the interstellar medium today: (i) dust is in
the form of spheres of radius r = 0.1 µm; (ii) a single sphere has an internal
mass density ρp ≈ 1 g cm−3 ; and (iii) the average density of dust in the cloud is
ρd ≈ Zρ, where Z ≈ 10−2 (also called the metallicity of the gas).
Assume the velocity of dust grains is small compared to the velocity of hydrogen
molecules. Assume also that an individual hydrogen molecule has negligible size
compared to the size of a dust grain.
Derive a symbolic expression for the cooling time, tcool , in terms of r, ρd ,
ρp , T (the temperature of the gas molecules), mH , and k (Boltzmann’s
constant). Again, for simplicity (and ease of grading), drop all factors of order
unity from your answer.
2
[5 points] The average time between collisions of a gas molecule and a dust grain is
tcool = tcollide =
λmfp
1
=
v
nσv
2
2
where n is the number density
p σ = πr ∼ r is the cross-section for collision
p of dust grains,
with a dust grain, and v = 3kT /(2mH ) ∼ kT /mH is the (rms) thermal speed of hydrogen
molecules.
Note that we are neglecting the speed of the dust grain (which we are told is small compared
to the speed of the hydrogen molecules) and we are neglecting the cross-sectional area of a
gas molecule (which is tiny—[3 Angstroms]2 —compared to the dust grain).
We can write
n=
ρd
ρd
3
= ρd
∼
3
md
4πρp r
ρp r 3
Combining these three expressions we can write the cooling time as
tcool
ρp
=r
ρd
r
mH
kT
(c) Initially, the cloud is marginally Jeans-unstable; that is, its mass is just
slightly above the Jeans mass. At this moment, which is shorter, the cooling time or the contraction (heating) time? Evaluate both times in
years. Answers good to a factor of 10 will receive full credit, but you must get
the sign of the inequality correct.
Assume the cloud has a total mass of 1 M and a radius of 0.1 pc.
Note that the temperature T is not given. Use the fact that the cloud is marginally
Jeans-unstable to solve for T .
√
[5 points] The free-fall (or collapse) time is tff = 1/ Gρ and ρ is approximately M/(4πR3 /3) ≈
2 × 10−20 g cm−3 . Thus, tff ≈ 106 yrs .
To calculate cooling time, we need a T . As stated above, the cloud was marginally Jeansunstable, which means M ≈ MJeans or R ≈ RJeans . So, from lecture,
s
kT
R≈
mH Gρ
3
and therefore
r
ρ p mH
tcool = r
ρd kT
ρp 1
√
tcool = r
ρd R Gρ
tcool =
ρp r 1
√
ρd R Gρ
Plugging in the numbers, ρd = Zρ = ZM/R3 ≈ 2 × 10−22 g cm−3 , ρp = 1 g cm−3 , r = 0.1 µm
and R = 0.1 pc, we get tcool ≈ 2 × 105 yrs.
Thus, the cooling time is shorter than the contraction time, tcool < tff , or more
precisely, tcool ≈ 0.2tff . If you dropped the 4π/3 then your answers could be off by a factor
of 8 or so. Again, full credit will be awarded for answers within a factor of 10 of ours.
(d) Assume that if tcool < tff that the collapse proceeds isothermally. Explain
why you might or might not expect to maintain the inequality of
timescales found in (c) as the cloud collapses. Is this calculation
promising for the formation of stars?
[5 points] The question is asking if we expect the cooling time to remain shorter than the
free-fall time as the cloud continues to collapse. If we divide our expression for the cooling
time by our expression for the free-fall time, we get
p H
r ρρpd m
tcool
kT
≈
1
√
tff
Gρ
r
tcool
ρp mH Gρ
≈r
tff
ρd
kT
r
tcool
ρp mH Gρ
≈r
tff
Zρ
kT
s
tcool
ρp mH G
≈r
tff
Z
kT ρ
tcool
1
∝√
tff
Tρ
4
where the proportionality only depends on parameters that might change during the cloud’s
collapse. ρ is increasing as the cloud collapses (since M is staying constant and R is decreasing). We have proven that initially, tcool < tff , which means that the cloud starts to collapse
nearly isothermally (T = constant). Therefore,
tcool
1
∝√
tff
ρ
and since the right hand side continues to decrease as the cloud collapses, tcool will become
smaller and smaller relative to tff . In other words, yes, the inequality is maintained. This is
promising for star formation because it means that the cloud becomes increasingly Jeansunstable and thus collapses under its own gravity unimpeded by thermal pressure, thus
forming a baby star (technically called a “protostar”).
2
To Collide or Not to Collide
(a) GAS: The HI (neutral atomic hydrogen) disk of our Galaxy has a total mass
of roughly 5 × 109 M and is distributed in a cylinder of radius 25 kpc and height
100 pc (a very thin pancake).
The gas is thin because it is collisional: gas atoms collide many times over the
age of the Galaxy. The collisions dissipate energy. The gas cools, loses pressure
support in the direction perpendicular to the disk (the “vertical” direction), and
settles toward the disk midplane.1
Estimate the time for a gas atom to collide with another gas atom,
in years. Your answer should be good to a factor of 3. Take each
hydrogen atom to be a hard sphere having a Bohr radius (= 0.5 Angstrom). Take
the temperature of the gas to be 100 K. Verify that the gas collision time
is shorter than the age of the Galaxy, roughly 1010 yr. Thus gas is
collisional.2
[4 points] The timescale over which gas atoms collide with each other is given by the mean
free path traveled between collisions divided by the average (thermal) speed of the atoms
tcol ∼
1
λmfp
,
cs
The gas resists drifting radially toward the Galactic Center because it has orbital angular momentum
which helps it maintain near-circular orbits; there is, however, a slow radial drift inward because of energy
dissipation; the gas disk is, after all, just another (very big) accretion disk.
2
And can be modeled as a fluid.
5
where cs ∼
p
kT /mH ∼ 1 km/s is the thermal speed of hydrogen atoms of mass mH ≈
1.7×10−24 g at temperature T = 100 K. We estimate the mean free path λmfp ∼ (nσ)−1 of the
2
hgal mH ) ∼ 1
gas as follows. The average number density of gas molecules is n ∼ Mgas /(πRgal
−3
cm (with Rgal and hgal the dimensions of the “cylindrical” galaxy, and Mgas its mass in
gas). The collisional cross section of hydrogen gas is σ ∼ π(2rB )2 , where rB ∼ 0.5 Angstrom
is the Bohr radius. Note the factor of 2, which arises when considering collisions of hard
spheres of equal radius. Numerical evaluation yields
tcol ∼
λmfp
∼ 3 × 1010 s ∼ 1000 yr.
cs
which indeed is much shorter than the 1010 yr age of the Galaxy. Any answer within a
factor of 3 of ours will be given full credit (if you missed the factor of 22 in the collision cross
section, then 1 point will be deducted).
(b) STARS: Consider an elliptical galaxy: a giant quasi-spherical (read: approximately spherical) collection of stars moving on near-radial (read: non-circular)
orbits. If you google-image ‘elliptical galaxy’, you will get plenty of pretty pictures.
By contrast to gas, stars in galaxies are essentially collisionless.
There are different measures of collisionality. One way to “collide” is to literally
have two stars smash together. Another way to “collide” is for a star to gravitationally deflect the trajectory of another star — to change its direction by an
angle on the order of, say, unity (i.e., ∼1 rad).
We will work out the latter case. The timescale for a star to gravitationally
deflect the trajectory of another star is called the “relaxation time”. We will
show that the relaxation time of a present-day elliptical galaxy is much longer
than the galaxy’s age (and if stars don’t relax, they definitely don’t smash into
each other).
Model an elliptical galaxy as a uniformly dense sphere of radius R composed of N
stars each of mass m? . The stars move around randomly with typical velocities
v.
Derive an approximate formula for v in terms of the variables given
and fundamental constants. You can do this problem any way you like. One
way is to recognize that the elliptical galaxy is in dynamical equilibrium such that
the typical stellar velocity is comparable to the escape velocity from the galaxy (ask
6
yourself what would happen if one velocity were very different from the other).
Another way to proceed is to use the virial theorem (if you know what that is) —
elliptical galaxies are said to be in virial equilibrium.
[3 points] The virial theorem states that the time-averaged total kinetic energy < T > and
the time-averaged total potential energy < V > in a self-gravitating system are related by
< T >= −
<V >
2
In our model for the elliptical galaxy
1
< T >= N m∗ v 2
2
and
3 G(N m∗ )2
,
5
R
where the factor of 3/5 arrises from the assumption that the galaxy is a sphere of uniform
< V >= −
(stellar) density. We solve for v using the viral theorem and the expressions above to obtain
r
v=
3 G(N m∗ )
.
5
R
Any answer that gets the scalings right will be awarded full credit (who cares about the
3/5).
(c) Consider two stars passing in the night. They start infinitely far apart with
relative velocity v, and approach one another to a minimum distance of b. In
other words, the impact parameter of the encounter is b.
Estimate, to order-of-magnitude, the value of b required for the stars to
deflect their trajectories by an angle on the order of unity. Express in
terms of v, m∗ , and fundamental constants. Because we are only interested
in an order-of-magnitude answer, you may use the impulse approximation (recall
gravitational lensing), and assume that one of the stars is fixed while the other
star moves.
[3 points] The impulse approximation yields a change in velocity orthogonal to the (initial)
relative direction of motion between the stars of
∆v ∼
2Gm∗
,
bv
7
(see lecture notes for derivation). A order-unity deflection will occur when ∆v/v ∼ 1, i.e.,
2Gm∗ /(bv 2 ) ∼ 1. Solving for b, we obtain
b∼
2Gm∗
.
v2
(d) Estimate the relaxation time in terms of N , R, and v. You will need
to use your answers in (b) and (c).
[3 points] The relaxation time is
λmfp
,
v
∼ (nσ)−1 , with n ∼ 3N/(4πR3 ) and σ ∼ πb2 . Results from (b) and (c) can
trelax ∼
where λmfp
be combined to yield (to order-of-magnitude) b ∼ R/N . Substituting the expressions above
and dropping numerical coefficients, we obtain
trelax ∼
RN
.
v
The time R/v is called, for reasons which I hope are clear, the “crossing time” of the system.
The relaxation time is N times longer than the crossing time. If you go on in physics,
you will learn that this estimate of the relaxation time is an overestimate because it only
considers “close” encounters that change trajectories by order unity in a single encounter.
What we should also consider are “distant” encounters that are less effective per encounter,
but which occur more frequently than close encounters. The net result when we include
distant encounters is to reduce the relaxation time by of order the logarithm of the number
of stars (this is called the “Coulomb logarithm” because the same physics applies when we
consider relaxation between charged particles in a plasma — it’s the same physics because
both gravity and the Coulomb force are inverse square law forces).
(e) Numerically evaluate the relaxation time for N = 1011 , R = 30 kpc,
and v = 200 km/s, and show that this time is the age of the oldest
ellipticals, ∼1010 yr. Thus stars in elliptical galaxies (and in disk galaxies) are
collisionless.
[1 point] Evaluation of trelax yields
trelax ∼
RN
∼ 5 × 1026 s ∼ 1019 yr 1010 yr,
v
thus stars in galaxies are collisionless.
8
3
Hot Halo Gas
Why is the gas in the hot galactic halo so hot?
It is thought that hydrogen falls from the intergalactic medium into the gravitational wells established by dark matter halos. That is, dark matter halos attract
gas. The gas takes the form of filaments that accelerate as they stream into the
dark matter halo. When the fast-moving streams collide and shock, gas is heated
to high temperatures. We say that the halo gas “shocks and virializes”.
(a) Consider a dark matter halo of mass ∼1012 M and radius ∼200 kpc. Estimate the speed with which a hydrogen atom falls from infinity (assuming it is initially at rest) onto the halo (say its surface). Express
in km/s. Your answer should be good to a factor of 3.
[2 points] The energy of the particle consists of its kinetic energy and its gravitational
potential energy. Initially, the gravitational potential energy is zero because the particle is
infinitely far from the halo. And we are told it is at rest, so the kinetic energy is also zero.
Thus the initial energy is zero.
Energy is conserved in this problem, so the final energy, when the particle has fallen
to the surface of the halo, is also zero. But now the kinetic energy and the gravitational
potential energy are both not zero. The kinetic energy is mH v 2 /2 and the gravitational
potential energy is −GMhalo mH /Rhalo . The sum equals zero; solve for v:
v=
2GMhalo
Rhalo
1/2
(1)
Plugging in the numbers, you should get v = 211 km s−1 . This is also the escape velocity
from the halo.
(b) Now assume that a shock converts the kinetic energy of the infalling hydrogen
atom into heat. Estimate the resultant temperature, in K. Your answer
should be good to a factor of 3.
[2 points] Assuming the gas to have reached thermal equilibrium,
1
3
mH v 2 =
kT
2
2
mH v 2
T =
3k
(2)
Plugging in the numbers, you should get T = 1.8 × 106 K . This matches to within a
factor of 2 the temperature given in class.
9
4
Galactic Coordinates 10-0-11-0-0-by-0-2
From Galactic Zero Centre
(a) The distance between the Sun and the Galactic Center is 8 kpc. We know
this using measurements of trigonometric parallax (i.e., ordinary parallax) of the
Sagittarius B2 cloud (a point source of radiation emitted near the vicinity of,
although technically not quite on, the central supermassive black hole).
Draw the orbit of the Earth around the Sun, assuming the orbit is circular. Draw
Sagittarius B2 a distance d away from the Sun. Note that Sagittarius B2 DOES
NOT have to be located in the plane of the Earth’s orbit (and indeed it is not).
Say that the line between the Sun and Sgr B2 makes an angle i relative to the
Earth’s orbit plane.
What is the maximum apparent shift in the angular position of Sgr
B2 over the course of an Earth year? Take d = 8 kpc and express in
milliarcseconds as a function of i. This is called the “parallactic angle” or
simply the “parallax”.3
Assume for this problem that there are enough telescopes scattered all around the
Earth (and even in orbit around the Earth) that Sgr B2 is always visible by some
astronomer somewhere.
[4 points] Your drawing should look similar to the top panel of Figure 1.
Recall that the parallax angle p is 2r/d where r is the radius of the orbit and d is the
distance to the source. In this case, r = 1 AU and d = 8 kpc. Note that some texts
(including Caroll & Ostlie) define parallax angle as r/d, one-half the maximum change in
apparent angular position of the object of interest. Since the question asks for the maximum
shift, use 2r/d.
From Figure 1, it’s apparent that regardless of the inclination angle, there is always a
baseline of length 2r that is perpendicular to the line connecting the Sun and Sgr B2. The
maximum apparent shift p occurs along this baseline. Any other baseline of length 2r will
give a smaller angular shift which is shown in Figure 2.
Therefore, for Earth orbit of any given inclination, the maximum apparent angular shift
is p = 2 AU / 8 kpc = 1.25 × 10−9 radians = 0.26mas .
(b) Another way to measure the distance to the Galactic Center is to exploit the
proper motions in Sgr B2. In reality, Sgr B2 is a collection of individual water
3
For the answer, see Reid et al. 2009, ApJ, 705, 1548, in particular the right panel of their Figure 2, or
their Table 4.
10
i = 90 deg
i = arbitrary
Earth orbit
2r
{
i = 0 deg
d
p
i = 90 deg
d
d
2r
Figure 1: Top: sketch of the orientation of Earth’s orbit and the Sgr B2 for different inclinations. Bottom: a side view of i = 90 deg case. Note that the angle p is the parallax.
maser clouds. The clouds are observed to occupy the surface of an EXPANDING
spherical shell: presumably the gaseous outflow from a cluster of young stars
located near the Galactic Center (imagine a polka-dotted expanding balloon, where
each of the polka dots is a water maser cloud). Each of the clouds produces
an emission line from a rotational transition in the water molecule. The rest
frequency (not the observed frequency) of the water transition is 22.235 GHz (a
radio frequency).
Consider the cloud located at the apparent center of the shell as seen on the sky.
This cloud is observed to produce a water emission line whose frequency is 22.238
GHz.
11
Figure 2: A representation of relative orientation of Sgr B2 and different baselines. At the
incident baseline (baseline aligns with the line connecting the Sun and Sgr B2), the angular
shift is 0. At the perpendicular baseline (baseline is perpendicular to the line connecting the
Sun and Sgr B2), the angular shift is p as defined in Figure 1. At the in-between baseline
(somewhere between incident and perpendicular), the angular shift is some value smaller
than p.
Consider a cloud located at the limb (edge) of the shell as seen on the sky. This
cloud is observed to have a proper motion of 1 milliarcsec / yr.
Deduce from this information the distance to Sgr B2. Express in kpc.
This is a classic example of combining a radial velocity and an angular proper
motion to deduce a distance; we applied a variation of this method to find the
distance to NGC 4258.
[4 points] Since the clouds are on the surface of an expanding shell, the line-of-sight
velocity vr (the radial velocity) observed from the apparent centre of the shell is the same
as the total expansion velocity vexp . From the Doppler shift formula,
vr
∆ν
= −
c
ν
0.003 MHz
vr = −
c
22.235 MHz
vr = −40.48 km s−1
vexp = |vr | = 40.48 km s−1
(3)
Note that the sign of vr shows that the apparent centre of the shell is moving towards
us (blueshifted). Since we’re only interested in the magnitude and not the direction when
considering the expansion velocity, vexp = |vr |.
12
Note that I used ∆ν instead of ∆λ. You can use ∆λ/λ and convert the two frequencies
to wavelength but here, I’ll show how the Doppler shift formula can be expressed in terms
of frequencies:
c
ν
c
= − 2 dν
ν
cdν
= − 2
ν λ
νdν
= − 2
ν
dν
= −
ν
∆ν
= −
.
ν
λ =
dλ
dλ
λ
dλ
λ
dλ
λ
∆λ
λ
(4)
Therefore, from the Doppler shift formula you’re more used to
vr
∆λ
=
c
λ
∆ν
.
= −
ν
(5)
Note that the opposite sign between ∆λ and ∆ν reflects the inverse proportionality
between wavelength and frequency: higher wavelength corresponds to lower frequency.
At the edge of the shell, the proper motion is due to the total expansion velocity. Using
the small angle approximation:
vexp
d
vexp
d =
µ
µ =
(6)
Plugging in the right numbers (don’t forget to convert mas to radian or vice versa), you
should get d ' 8.8 kpc .
(c) The apparent proper motion of Sagittarius A* (the central supermassive black
hole sitting at the Galactic Center) has been measured by radio astronomers to
be 6.2 milliarcseconds / yr.
Assume this proper motion is due entirely to the Sun’s motion around the Galactic
Center. Assume further that the Sun moves on a perfectly circular orbit about
13
the Galactic Center.4 Deduce from this information the angular velocity
ω0 of the Sun’s motion around the Galactic Center. Express ω0 in units
of km/s/kpc.
[2 points] If the proper motion of Sagittarius A* is entirely due to the Sun’s motion
around the Galactic centre, the proper motion of the Sun with respect to the Galactic centre
is 6.2 mas / yr. Changing this to the appropriate unit:
6.2mas
radian
yr
3 × 1021 cm
km
=
yr
2.1 × 103 mas
3.2 × 107 sec
kpc
1 × 105 cm
= 28.57 km s−1 kpc−1
(7)
ω0
(d) From your answers in (b) and (c), calculate the rotational speed Θ0 of
the Sun around the Galactic Center in km/s, and the duration of the
Galactic Year in Earth years.
[2 points] Using the distance from (b), I get Θ0 = ω0 d = 251.46 km s−1 . The duration
of Galactic year is 2πd/Θ0 . Plugging in the numbers, I get 2.1 × 108 years .
4
In reality, the Sun’s motion deviates by a few percent relative to a perfectly circular orbit. We know this
by taking an average over all the motions of the stars (relative to the Sun) in the local solar neighborhood.
So actually, the full apparent motion of Sgr A* is 6.4 milliarcsec / yr.
14