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Can't really put 2-2 and together as to how having an inert degenerate He core translates to a He-flash. Also, at which points exactly do degeneracy and the He-flash start occurring?

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Short answer: it is a combination of (1) the ignition occurring in an electron-degenerate, isothermal core in which the equation of state is independent of temperature; and (2) the extreme temperature dependence of the triple alpha He fusion reaction.

Details:

The helium flash occurs at the tip of the first ascent red giant branch in stars with masses between 0.5 and about 2 solar masses. At this point the star consists of a helium core surrounded by a vigorously burning shell of hydrogen, surrounded by a very large convective envelope.

The core is the left over from core hydrogen burning, supplemented by helium produced by the hydrogen shell burning that takes over once the core hydrogen is exhausted. The inert core shrinks in radius from its main sequence size because it has more mass per particle, so must increase in density to maintain pressure. As it does so, the virial theorem demands that it also gets hotter. The shell burning drops more and more He into the core, the core shrinks further and gets hotter.

In stars > 2 solar masses, the core gets hot enough to ignite helium in the triple alpha process. This raises the core temperature, but not massively, because at the same time, the pressure increases, the core expands vigorously and the hydrogen burning shell is pushed outwards and extinguished.

In a lower mass star it is different. The He core density rises to the point that the core electrons become degenerate. Electron degeneracy pressure (EDP) dominates the total pressure of the gas and arises because at high enough densities, the electrons fill all low energy quantum states.EDP only depends on density, not temperature.

A core supported by EDP gets smaller and denser, the more massive it is, so becomes more degenerate as He ash is dropped onto it. But it is also sitting inside a hugely luminous H burning shell which heats it. In stars more massive than 0.5 solar masses, eventually the core gets hot enough ($\sim 10^{8}$ K) to ignite the He. Because degenerate electrons are extremely conductive, the core is almost isothermal, so ignition spreads rapidly through the core. This raises the core temperature, but crucially, not the core pressure (EDP is independent of temperature and, as Ken G explains in his answer, most of the heat gets absorbed by the non-degenerate ions that hardly contribute to the pressure). It just gets hotter and He fusion increases massively because it is very temperature sensitive (roughly proportional to $T^{40}$ !!). This runaway process is termed the "helium flash".

Eventually the temperature rises enough (to about $3\times10^{8}$ K) to break the electron degeneracy, the core expands rapidly, the H shell is extinguished and the core luminosity falls.

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Helium is chemically inert, but in the conditions present in the core of a star or on the surface of an accreting white dwarf helium is prone to fusion. The helium is degenerate, which means that the structure of the Helium core/white dwarf is not being supported by temperature, which means the energy produced during fusion does not cause the core to expand as it would if it were supported by thermal energy. This lack of expansion means the fusion is able to runaway and consume the entire core, producing tremendous amounts of energy very quickly.

As to when it happens, there must be a core of degenerate matter. If the star's mass is greater than about $2.5M_{\odot}$, the helium core will not become degenerate before it begins fusing, so there will be no helium flash. In the case of white dwarfs, the helium is already degenerate, so as long as accretion continues the pressure will build until the He flash occurs.

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    $\begingroup$ "This lack of expansion means the fusion is able to runaway and consume the entire core, producing tremendous amounts of energy very quickly." Is this a consequence of the Virial Theorem and the star needing to produce more energy to oppose gravitational collapse? PS: I think you submitted before finishing the last sentence. $\endgroup$ – Tetradic Apr 7 '15 at 23:44
  • $\begingroup$ I fixed the last sentence. Also, I would say that this is not a consequence of the Virial Theorem, since the pressure on the core is coming from the rest of the star, and not from self gravity. What you say is correct, though, and is the reason why the core collapses once fusion stops producing energy at Fe. $\endgroup$ – Blake Hartley Apr 9 '15 at 0:51
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This might not be a good explanation, but the gist of it is, as stars get hotter, even though they get less dense over time, the extra heat speeds up the fusion process. The helium fusion can only happen at about 100 million degrees. The temperature at the core of our sun is 15 million degrees - new energy is created all the time by hydrogen and deuterium fusion, but energy also radiates out, so it heats up slowly and the core will never reach 100 million degrees, until it collapses and helium fusion starts.

Once helium fusion starts, the heat increases and that helps keep the fusion process going.

2 things to keep in mind. 1, when the sun has a degenerate helium core it will still have hydrogen around the core burning around the edge, so when it enters the helium flash and expands to a red giant, it's a combination of helium and hydrogen fusion at that time.

2, hydrogen fusion is rare. something like 99.9999% of the time, when 2 protons meet, they just split up again. Proton-proton fusion is like 2 shy people at a bar - most of the time, even if they bump into each other, often no connection is made. It's only 1 in a million interactions where the 2 protons become a deuterium and a positron - the positron quickly meets an electron, but the proton-proton interaction is more reluctant. Helium-helium, as I understand it, is less reluctant to undergo the fusion transition in the CNO cycle, once the temperature is high enough. So, once it starts, it can happen more quickly.

If that's wrong, let me know. That's my understanding of the process in layman's terms.

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  • $\begingroup$ Stars don't get "less dense over time" - it is a lot more complicated than that. It is the core conditions that matter, and the core gets more dense over the time relevant to this question, leading to the He flash. The helium flash terminates the red giant phase, it is not a precursor. First ascent red giants are burning hydrogen in a shell, that's all. $\endgroup$ – Rob Jeffries Apr 8 '15 at 13:02
  • $\begingroup$ Thanks for the fix. I didn't know that about red giants, I'd assumed it was helium burning. I guess it's the fusion closer to the surface that makes the star expand so much then. And, I see what you're saying about density too - He is much denser than H, I was thinking the hotter core would expand due to heat, which is basic physics, but that might be a smaller factor than the reduction of hydrogen and increase in helium - so that makes sense. $\endgroup$ – userLTK Apr 9 '15 at 10:48
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There is a very widespread misconception that a plasma supported by the pressure of degenerate electrons will not expand when heat is added to it. This is incorrect, it would represent a basic violation of the virial theorem. When heat is added to either an ideal gas or a degenerate gas, the expansion is exactly the same if they are both nonrelativistic, because the expansion (and the pressure, by the way) depends only on the increase in internal kinetic energy. This is just as true for degenerate gas as ideal gas, the distinction there involves the temperature behavior, not the pressure behavior. This is such a widely held misconception, I almost despair of trying to fix it!

By a process of remarkable elegance that is sadly overlooked, what actually happens in a helium flash is that when heat is added to a gas supported by the pressure of degenerate electrons, all that heat does go into raising the temperature of the ions, just as is normally claimed. However, it is completely untrue that the gas does not expand-- instead, the added heat causes a reduction in the degeneracy of the electrons that produces precisely the same expansion, and expansion work, that adding that same heat to an ideal gas would do. That's the virial theorem. So it means that the gas does expand, but all the work to support that expansion comes from the kinetic energy of the electrons, the ions are left scot free to pick up all the heat that is being added. The reason the ions get the heat is that they have a much higher specific heat because they are not degenerate (their temperature rises much more slowly when heat is added), and the species with the higher specific heat always gets the proportionately higher share of the added heat! So that's what causes the thermonuclear runaway, it is not that expansion is absent, it is that expansion is irrelevant to the temperature, if the electrons are degenerate. The expansion is the same as for an ideal gas, for a given heat added.

The way to tell that the usual explanation of the helium flash, which you will find everywhere, is wrong is that in the usual explanation, you should expect that a thermonuclear runaway would occur even if the number and mass of the ions was the same as the electrons. Clearly, if you think it happens because there's no expansion of degenerate gas, there's no problem if the ions are just as degenerate as the electrons. But in fact, no thermonuclear runaway would occur in that case, because then the expansion would rob kinetic energy from the ions just as it already does from the electrons. If the ions are not less degenerate, they do not have the higher specific heat that allows them to gobble up the lion's share of the added heat. The internal energy of the gas would behave just as it does for ideal gases, and fusion would be self-regulated in the usual way.

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  • $\begingroup$ "When heat is added... to a degenerate gas". Degenerate gases contain no heat - if you add "significant" heat to a degenerate gas, then it is not degenerate any more. The only reason there is any expansion is because the gas is partially degenerate. $\endgroup$ – Rob Jeffries Sep 13 '16 at 13:40
  • $\begingroup$ I'm also struggling with the last paragraph. You say consider no expansion but with the ions and electrons as degenerate species, but then go on to say "the expansion would rob KE from the ions..." ? In fact if the ions were degenerate, the runaway would be more extreme. The same added heat would result in even higher temperatures and even more rapid nuclear reaction rates, but with little increase in pressure. The internal energy of a perfect gas depends on its temperature, that of a degenerate gas does not. $\endgroup$ – Rob Jeffries Sep 13 '16 at 13:55
  • $\begingroup$ No. For all types of nonrelativistic gas, including degenerate gas, the pressure is simply 2/3 of the internal kinetic energy. So anything that adds internal kinetic energy is always going to have exactly the same effect on the pressure, it makes no difference whatsoever if the gas is degenerate or not. What degeneracy affects is the temperature, the pressure doesn't care if the gas is degenerate or not in any situation where you can track the history of the internal kinetic energy, such as whenever you add heat and apply the first law of thermodynamics. $\endgroup$ – Ken G Sep 13 '16 at 13:58
  • $\begingroup$ Your second comment demonstrates exactly the misconception I'm talking about. This is a very widespread misconception, it is found in otherwise reliable sources, so it's not surprising you have fallen victim to it. The usual wrong explanation suggests that if both electrons and ions were degenerate, the runaway would be even worse, but in fact it would not happen at all-- the logic for that is given above. $\endgroup$ – Ken G Sep 13 '16 at 14:00
  • $\begingroup$ OK, I see what you are saying, and the second part of my second comment is incorrect. But you need to reword something because you start by saying consider a model where there is no expansion and then say that the expansion would rob KE from the ions. As far as I can see my answer above is fine. I am arguing that the pressure of the gas does not increase with temperature - and that is true. The reason is what you describe above - the heat goes into the ions. $\endgroup$ – Rob Jeffries Sep 13 '16 at 14:08

protected by Qmechanic Sep 13 '16 at 17:10

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