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).