Homework 3 Due class time Oct 18
1) Assume fully convective protostars have and effective
T of 3000 K. Consider when the young Sun had a radius
100 times what it is now.
a) Estimate its surface pressure, using P = g/kappa,
where g is the acceleration of gravity and kappa is the
cross section per gram (just assume ~1 cm^2/gram as a
very rough approximation for simplicity). Compare your
answer to the atmospheric pressure around you, which is
10^5 Newtons/m^2, or 10^6 in cgs.
answer: g=GM/R^2 = 2.7 cm/s^2 so
P ~ 2.7 gram/cm s^2 so a bit over a millionth of an atm.
b) Estimate the core P of this young Sun, using force
balance.
answer: P ~ GM^2 / R^4 ~ 10^8 in cgs, so about 8 orders of
magnitude larger than at the surface.
c) The rule of adiabatic convection produces T
proportional to P to the 2/5 power, so use that (and
the other information in this problem) to estimate
the core T of this young Sun.
answer: T_c ~ T_s * (P_c / P_s)^(2/5) ~ 3000 K * 10^(16/5) ~ 5 x 10^6 K
(note: the original question contained a typo and said T is proportional
to P to the 5/2 not 2/5, so count that correct but it would give a core
T that was way too hot-- it would fuse everything!)
d) Estimate how much time was required for this young
Sun to contract its radius by the factor 1/2. (This
is essentially the "Kelvin-Helmhotz time" of this
young Sun). Show all work.
answer: t ~ E/L ~ GM^2/RL where L ~ sigma*T_s^4*4pi*R^2 ~ 3 x 10^36 erg/s,
so t ~ 10^10 s (so only about 10^3 years, it forms quickly at first as
it transitions from free-fall).
2) Using the same information as in (1), estimate the
Sun's radius, relative to what it is today, when it
transitioned from primarily convective to primarily
radiative heat transport. (You may use the M-L
relation for radiative heat transport to justify that
the solar luminosity hasn't changed much since then,
and the fact that the surface T today is nearly
6000 K.)
answer: L is proportional to T_s^4, and T_s has risen from
about 3000 K to about 6000 K, so that would be a factor of ~16
rise in L, but L has not risen much at all over the Henyey track
(of radiative diffusion), so the R^2 must make up the difference
by dropping by a factor of 16. So R was 4 times larger then.
3) The amount of heat per particle that must be added
to an ideal gas to get a T increase of 1 Kelvin is
nk/2, where k is the Boltzmann constant and n is the
number of degrees of freedom to put energy (so n=3 is
for a monatomic gas in 3D, n=5 is for a diatomic gas
that adds two degrees of freedom for rigid rotation.)
a) If equal numbers of particles of two such ideal
gases combined (i.e., an equal number of monatomic
and diatomic molecules), and total net heat Q is
added to the system, what fraction of that Q goes
into each ideal gas? (Use the fact that the two
gases are in thermal contact so stay at the same T.)
answer: dQ = nk/2 * dT , so dQ_1 / dQ_2 = n_1 / n_2 * (dT_1 / dT_2),
but dT_1 = dT_2 to keep T the same, so the dQ ratio equals the n ratio
for the two gases (with equal number of particles). So the gas with
n=3 gets 3/8 of the heat, and the gas with n=5 gets 5/8.
b) What does your result tell you about the
connection between the specific heat per particle
(essentially its "n" value) and the fraction of
added heat that each particle receives?
answer: added heat is partitioned among the species in proportion
to their specific heat (when the number of particles is the same--
in general, we would say it is partitioned in proportion to the
"heat capacities").
c) Consider that a highly degenerate gas acts as
if n is much less than 1 (because most of the
particles cannot receive heat, so a little added
heat goes a long way in raising T). So that when
He fusion initiates in a sea of highly degenerate
electrons, where does the heat released by fusion
go? What does this fact have to do with the
"helium flash," and should that perhaps be
mentioned in textbook descriptions of this
phenomenon?
answer: When He fusion adds heat to a gas of He ions bathed in
degenerate electrons, the low heat capacity of the electrons means
they get very little of that heat-- it mostly all goes to the He ions.
The expansion work, which is not any different for a degenerate gas
from an ideal gas (that's what the textbooks get wrong), is done by
the electrons, because they have the kinetic energy (remember that
at high degeneracy, the average kinetic energy per particle is way
larger than kT, yet kT is what is the same as the He ions).
The combination of expansion work not needing to be done by the
He ions, with the He ions getting the lion's share of the fusion heat,
means that more fusion leads to more kinetic energy in the He ions,
and that leads to fusion runaway. Fusion always runs away when it
initiates in a gas containing lots of degenerate electrons, so this
can never be maintained-- either the star explodes (type Ia supernova,
which requires a binary so will not happen to our Sun),
or the degeneracy is removed before that happens (the "helium flash"
in our Sun's future). Yes, it would be nice if this would start
appearing in textbooks, it is much more insteresting than the patently
false claim that heat can be added to any kind of gas without it
experiencing an expansion from the unavoidable pressure rise!
4) You cannot believe everything you read,
especially when it comes to verbal descriptions
of the physics going on in stars. If you google
"red giant," among the first few hits you will
get a seemingly authoritative site called
"Universe Today: Why Do Red Giants Expand?".
Not to bash the heroic efforts of scientific
communicators to bring this fascinating physics
to the public, we should still perhaps hold them
to a standard that what they say should be at
least to something like the truth! To test our
own understanding of these phenomena, give a
critique of these two sentences, and assess how
close they are to the truth. Speaking about
shell fusion:
"As this outer layer contains a bigger volume
than the original core of the Sun, it heats up
significantly, releasing far more energy. This
increase in light pressure from the core pushes
much harder against gravity, and expands the
volume of the Sun."
answer: The volume in the fusing shell of
a red giant is not bigger, it is much smaller
(about 10000 times smaller, as the length scale
of the core is about 10 times less than before).
Also, being bigger would not make its T higher,
in fact the reason the T is much higher is because
the very contracted core has a very strong gravity,
and the shell particles must be moving fast to
avoid being sucked into the core. This high T
does indeed lead to a high fusion rate, but it is
not the light pressure that pushes out the envelope,
it is the simple fact that fusion releases heat
that enters the envelope and forces (via the virial
theorem) the reverse of the contraction that happened
to the envelope when it was a protostar. If we would
track the forces, they would be gas pressure forces,
not light pressure forces (which only matter in much
more massive stars, see next semester). Some students
might say there is also a kind of implication that the
expansion of the envelope is not in force balance, even
though it actually is in a good force balance, so that
can be accepted also (although in fairness this particular
description does not claim that as clearly as we often see
in other places).