Physics – Astronomy
PHYS 1902 Assignment 2
Due: Thursday, March 23, 2017
AGAIN, DON’T PANIC
Once again, this assignment’s bark is worse than its bite. Please ask for help if you need it. And
please don’t delay trying the assignment, term is ending faster than it seems!
20 marks
1. The Galactic Mass Fraction of Interstellar Dust: Interstellar gas and dust mix to
form giant clouds and complexes in the galaxy. Dust affects the way wee see the
galaxy because of general obscuration (interstellar extinction) and by selective blue
light extinction (interstellar reddening); yet dust makes up a tiny fraction of the total
mass in the galaxy. We can estimate the mass fraction of dust at our location in the
galaxy using some basic inputs.
(a) Let’s approximate dust grains as spheres, each with a radius of R = 10-5
cm.
Each grain of dust presents a cross-sectional shadow of pR
2
. Recall that a photon’s
mean-free-path, l, is the typical distance a photon flies before bumping
into matter. If the number density of dust grains (number of grains per cm3
) is
n, the mean-free-path length relationship to the cross-sectional shadow is:
l =
1
npR2
(1)
Given that interstellar extinction observations indicate that l = 3 × 103
lightyears,
calculate the number density, n, of dust grains in units of cm-3
. How
many dust grains would you expect to find in a volume equal to the Rogers
Centre in Toronto (a cube about 100 metres in all directions)?
(b) Assuming that dust grains have a typical matter density of about 2 gm/cm3
, estimate
the mass, mgrain, of a dust grain. A volume V = 300 (lty)
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in the galaxy
typically contains one solar mass of stars. What is the mass of all the dust in
this volume, Mdust = nVmgrain? What is the mass fraction of dust compared to
stars at our position in the galaxy?
The interstellar material with its interactions with embedded stars and dilute
radiation fields produces extremely rarefied matter that is far from thermodynamic
equilibrium. The birth and death of stars intimately link to the interstellar
material as old stars fill space with heavy elements and new stars form
when the gas and dust become dense and cold enough to gravitationally collapse.
Gravity and the second law of thermodynamics shape the material between
the stars. The enriched material forms a new generation of stars and
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planets—and new life who asks where it comes from. As we look out into the
Universe from our small stage, the interstellar material reminds us that we are
all made of star dust, fashioned by gravity and thermodynamics.
20 marks
2. White Dwarf Stars and the Chandrasekhar Mass Limit: In the last assignment we
discussed pressure as force per unit area. Let’s revisit that idea from the microscopic
perspective. Imagine a cloud of small pellets all moving in random directions in a
box. Now imagine one of the walls of the box as pellets bounce against it. Over
the course of time t, as the pellets move to the wall of area A, they sweep out a
volume of volume = Avt, where v is the velocity of the pellets heading toward the
wall. The pellets reflecting off the wall generates pressure. The force of the reflection
comes from the change in momentum (mass times velocity p = mPv, mP is the pellet’s
mass) of each pellet. In time t, half of the pellets are moving towards the wall
and the other half are moving away after reflection and the change in momentum
is twice p since the velocity switches sign (the collision with the wall is perfectly
elastic, the particles don’t lose energy in the collision). If the number of particles per
unity volume is n, then the pressure must be:
P = [number of particles per unit volume]×[volume swept out by particles heading
towards and bouncing off the wall]×[change in momentum per unit time]÷[the
area of the wall] = (n)(Avt/2)(2p/t)/A = nvp.
In three dimensions, we would adjust this result by a factor of 3, but we won’t
worry about that here. We have the formula for pressure in terms of microscopic
quantitites: P = nvp, where p is the particle momentum, p = mv.
Now imagine an electron gas with number density ne
. This means that on average,
the distance between electrons is ?x = (1/ne)
1/3. Let us suppose that the electron
gas is so dense that it’s degenerate meaning that the rules of quantum mechanics
become important. In particular, the gas obeys Pauli’s exclusion principle, which
prevents two fermions (electrons are fermions) from occupying the same quantum
state, and Heisenberg’s uncertainty principle, which tells us that it is impossible to
define the position and the momentum of a particle through their product to an accuracy
better than Planck’s constant: (?x)(?p) > h. For our degenerate electron gas,
these principles mean that p = h/x = hn1/3
e
. If the electrons in the gas are moving
much less than the speed of light, then v = p/me
. These conditions are met inside
a typical white dwarf star—the ions, which are the nuclei of the atoms that make
up the white dwarf, are relatively stationary, but the electrons form a degenerate gas
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inside the solid body. Unlike a main sequence star, in which pressure generated
from thermal effects prevents gravitational collapse, it’s the quantum mechanical
degeneracy pressure of the electron gas that supports a white dwarf.
(a) Based the relationship Pe = nevp = nep
2/me
, and p = hn1/3
e
, show that the electron
degeneracy pressure is Pe = h
2n
5/3
e /me
. In reality, this result is modified
by a factor of 0.0485—not too bad for our rough estimate!
(b) Recall that the atomic number, Z, counts the number of protons in the nucleus
of the atom and that the atomic weight, A, counts both the number of protons
and the number of neutrons. For example carbon has atomic number 6, and
atomic weight 12; oxygen has atomic number 8, and atomic weight 16. The
white dwarf has overall charge neutrality meaning that the ions of atomic number
Z that make up the star have the number density relationship Zn+ = ne
,
where n+ is the ion number density. A white dwarf is typically composed of
carbon-oxygen and so the ratio of the atomic number to the atomic weight,
Z/A, is about 0.5. Since the proton and the neutron have similar mass, we
will approximate the neutron mass with the proton mass, mp. The proton is
significantly more massive than the electron, so the density of the white dwarf
is approximately ? = Ampn+ and so ne = Z?/(Amp). Show that we can write
the electron degeneracy pressure as
Pe =
h
2
me
Z
A
5/3
?
5/3
m
5/3
p
(2)
(c) Recall from the first assignment that the central pressure required to hold up a
self-gravitating sphere of mass M and radius R is approximately Pc = GM2/R
4
.
If we take the density to be ? = M/R
3
in our expression for Pe
, and if we set
Pe = Pc, show that the mass-radius relationship of a white dwarf is:
R =
h
2
Gmem
5/3
p
Z
A
5/3
M-1/3 (3)
The actual relationship is modified by a factor of 0.114—again, not bad for our
rough calculation. Notice that the radius of a white dwarf shrinks as the cube
of the mass. Unlike a main sequence star, the more massive a white dwarf
becomes, the smaller its radius. Using the 0.114 correction factor, and setting
Z/A = 0.5, calculate the radius of a white dwarf that has the mass of the Sun;
Msun = 1.99 × 1033 gm, G = 6.67 × 10-8 gm-1
cm3
s
-2
, h = 6.63 × 10-27 erg s,
me = 9.11 × 10-28 gm, mp = 1.67 × 10-24 gm.
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White dwarfs are like stellar embers, slowly cooling as they reach thermal equilibrium
with the Universe. Since the degeneracy pressure holds up the white
dwarf, it can cool without shrinking. Eventually, the white dwarf will end up
with crystallized ions and a degenerate electron gas moving around inside the
lattice structure.
(d) So far we have been dealing with a white dwarf with a degenerate electron gas
in which the electrons are whizzing around at speeds much less than the speed
of light. As we increase the mass of a white dwarf, the radius shrinks and
eventually the degenerate electron gas becomes relativistic, that is, the electrons
move around at near light speed, c. Under such extreme conditions, we need
to modify our approach. The pressure for the relativistic degenerate electron
gas is,
Pe =
hc
m
4/3
p
Z
A
4/3
?
4/3. (4)
If we again use ? = M/R
3
, and we set Pe = Pc, show that the radius R drops
out of the calculation and we end up with the result,
M* =
Z
A
2
hc
Gm2
p
!3/2
mp. (5)
In actuality M* is modified by a factor of 0.2. This result is the Chandrasekhar
mass limit of a white dwarf star, named for Subrahmanyan Chandrasekhar, the
Nobel prize winning Indian-American physicist. The Chandrasekhar limit is
1.4 solar masses (you can verify the result numerically, if you’d like). Above
the Chandrasekhar mass, even the quantum mechanical degeneracy pressure
of the electron gas cannot stop the white dwarf from gravitationally collapsing.
If the white dwarf has a mass greater than 1.4 solar masses, the white dwarf
will “drive” the electrons into the protons to form a neutron star. If the neutron
star exceeds about three solar masses, even the quantum mechanical properties
of the neutron star cannot halt the gravitational collapse—the stellar remnant
will collapse forever, forming a black hole—the topic of our next question.
15 marks
3. Hawking’s Evaporating Black Holes, and the Scale of Quantum Gravity: The escape
velocity of a body is the minimum speed required to leave its gravitational
clutch. The formula for the escape velocity is vesc =
v
2GM/r, where r is the radius
of the body, and M is its mass.
(a) Suppose the gravitation field of a body with mass M is so strong that you
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would need to reach the speed of light, c, to escape. The radius of this object is
called the Schwarzschild radius, named after Karl Schwarzschild, the physicist
who provided the first exact solution to Einstein’s field equations of general
relativity—and he did it while serving in the German army on the Russian front
during the first world war. Using the escape velocity formula above, show that
RSch = 2GM/c
2
. What is the Schwarzschild radius of a 3 solar mass black hole
(3Msun = 5.97 × 1033 gm, G = 6.67 × 10-8 gm-1
cm3
s
-2
, c = 3.00 × 1010 cm/s
)?
Think about our collapsing star with a mass larger than three solar masses. As
we learned in the last question, quantum mechanical degeneracy pressure is
powerless to stop the collapse. Once the radius of the collapsing star reaches
the Schwarzschild radius, light itself cannot escape. But the star continues to collapse
forever. In truth, the situation is a bit more complicated. Einstein’s theory
of General Relativity becomes important for such strong gravity fields and
since spacetime becomes highly distorted, it doesn’t make much sense to talk
about a “radius” as such. Nevertheless, we can think about the point where an
outwardly traveling photon could just barely escape—this is the event horizon.
If we imagine tracing out these barely-escape points by going around the collapsing
star, we will find that the circumference is 2pRSch and that the surface
area is 4pR
2
Sch = 16pG
2M2/c
4
. That is, we can interpret the Schwarzschild radius
as the “radius” of the black hole. We will not consider rotation, as rotating
black holes make the situation even more complicated.
(b) In part (a), notice that the surface area of the event horizon, A = 16pG
2M2/c
4
,
depends on the square of the mass. This observation lead Stephen Hawking
to suppose that in any natural process, the surface area of the event horizon
must always increase, or at best, stay the same. In 1972, Jacob Bekenstein noticed
that the statement looks very much like a thermodynamic law—it looks a
bit like the second law of thermal dynamics except instead of entropy increasing,
it’s the surface area of the event horizon. Jacob Bekenstein made the bold
suggestion that the surface area of the event horizon is a measure of the black
hole’s entropy. But if a black hole behaves like a thermodynamic body, what
would it mean for the black hole to have temperature? Thermal bodies emit radiation,
so how can a black hole have thermal properties if nothing can escape?
In quantum mechanics the vacuum is not a quiet place. Virtual particles of all
types are constantly coming into “existence” and annihilating. Since the uncertainty
principle states that energy differences and time differences have the
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Figure 1: Hawking radiation near the event horizon of a black hole. The creation and
annihilation of virtual particles usually ends with complete destruction unless the tidal
forces cause one of the particles to fall into the black hole. In such cases, quantum mechanics
allows the other particle to escape by “tunneling”. The radiating particle reduces
the mass of the black hole. Image Credit: Northern Arizona University
relationship (?E)(?t) > h, particle-antiparticle states can pop out of the vacuum
with energy ?E = mc2 as long as it returns the energy to the Universe
inside the time ?t ˜ h/?E. We have lots of evidence for these vacuum processes
in laboratory experiments. Stephen Hawking wondered about particleantiparticle
states emerging near the event horizon of a black hole. Normally,
the particle-antiparticle state would quickly return the energy back to the Universe
through annihilation, but suppose that one of the particles falls into the
black hole before it gets a chance to annihilate with its partner. See figure 1.
In that case, the event horizon will become a source of radiation. This particle
emission from the event horizon, which carries away energy, steals mass
from the black hole by the relationship E = mc2
. Stephen Hawking realized
that black holes evaporate by thermal emission—a process we know today as
Hawking radiation. Quantum mechanical processes1
shrink the surface area of
the event horizon and evaporate the black hole’s mass!
Suppose that virtual particle-antiparticle “energy borrowing” from the uncertainty
principle is governed by ?E?t = h/(4p). Let us further suppose that
1For stellar mass black holes, the evaporation process proceeds extremely slowly, taking many of orders
of magnitude longer than the current age of the Universe for complete evaporation.
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if the virtual pair separates by half the circumference of the event horizon,
c?t/2 = 2pGM/c
2
, then one of the pair has a reasonable chance of falling
into the black hole while the other one escapes. If the energy of the escaping
particle has a thermal distribution, ?E = kT, where k is Boltzmann’s constant,
show that the temperature of the black hole is,
T =
hc3
16p2kGM. (6)
This expression is Stephen Hawking’s famous black hole radiation result. The
event horizon is a source of radiation, and thus black holes evaporate with a
temperature inversely proportional to the black hole’s mass—small black holes
are hotter than large ones. Compute the temperature of a 3 solar mass black
hole, (3Msun = 5.97×1033 gm, G = 6.67×10-8 gm-1
cm3
s
-2
, h = 6.63×10-27
erg s, c = 3.00 × 1010 cm/s, k = 1.38 × 10-16 erg/K).
Given the violent stages of the early universe, it is possible that micro black
holes, with masses of a typical asteroid or terrestrial mountain (1016 gm), might
have formed at that epoch. What would be the temperature of such a low
mass black hole? These primordial micro black holes would have a Hawking
radiation lifetime of about the current age of the Universe. Some astrophysicists
have suggested that micro black holes could form the dark matter and
we would be able to detect them from their Hawking radiation as they evaporate.
The Fermi Gamma-ray Space Telescope is currently looking for gammaray
burst signatures from evaporating primordial micro black holes.
(c) Imagine if we make the black hole quantum mechanically small so that twice
its Compton wavelength, 2h/(mc), is equal to its Schwarzchild radius, 2Gm/c
2
.
Show that m = (hc/G)
1/2. This mass, mPl, is called the Planck mass and
on this scale gravity and quantum mechanics are both equally relevant. We
don’t know the laws of physics in this regime—it is the scale of quantum gravity.
Compute the Planck mass in grams (G = 6.67 × 10-8 gm-1
cm3
s
-2
,
h = 6.63 × 10-27 erg s, c = 3.00 × 1010 cm/s) and compare it to the mass of
the recently discovered Higgs Boson, MHiggs = 2.2 × 10-22 gm. What does the
scale difference tell you about how much more powerful our particle accelerators
need to become before we can directly explore this physics?
In this course, we have seen how astrophysical processes involve the competition
between gravitation and the second law of thermodynamics. A black hole
is the most pure gravitational stellar end state, but as we have seen from Hawking
radiation, quantum mechanics implies that even black holes are thermody-
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namic bodies. What is the ultimate end state for matter? If grand unification
ideas in particle physics are correct, all atomic nuclei are unstable, eventually
decaying into photons and leptons. If those ideas are correct, even white
dwarfs and neutron stars will decay away. The observation that the Universe’s
expansion is accelerating suggests that the second law of thermodynamics will
win out. The Universe will expand forever, eventually reaching thermodynamic
equilibrium with no black holes, no stars, no galaxies, no planets—all
material entities being just way-stations on Nature’s journey of turning all matter
into photons and leptons.
55 marks
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