I know that after a star in the main sequence runs out of H in the core, it will start burning H in the shells surrounding the (now) He core.

(1) Why now the Hydrogen shells are hot enough for burning H and not before? What made them hotter?

I know too that the burning of H in the shells surrounding the He core will produce new He which will be deposited in the He core and therefore the He core will increase its mass and it will be compressed as a result (since we are talking about a degenerate matter core then its size/radius will follow the relation $R\propto M^{-1/3})$

(2) Does it shrinkage heats up either the core itself or the surrounding H shells? If it is so, Why?

  • I know that when it comes to an ideal gas and gravitational collapse, then the Virial Theorem tells us that half of the released gravitational energy will be used for increasing the internal energy of the system, that is, its temperature. However, I am not quite sure if the same holds for a degenerate core. (If I am not wrong, the Virial Theorem derives from the condition of hydrostatic equilibrium).

I also know that the degenerated He core is isothermal so it will have the same temperature than the immediate surrounding H burning shell, so, if you can answer me why the surrounding H shells becoming hotter and hotter, it may be the same as answer me why the degenerate He core becomes hotter and hotter.

Thanks! :)

  • 1
    $\begingroup$ When you say hotter and hotter, I guess you mean only for a limited time. An energy sink is provided by neutron emission. Nuclear burning in degenerate material produces a runaway effect, the material becomes thermally unstable. Eventually, the helium flash occurs when the He ignition threshold is reached. This process in mass independent, after a relatively short time, the increase in core temperature removes the degeneracy, expansion of the core occurs and the He burning archives stability. I am sure you are aware of this, sorry for being pedantic on hotter and hotter $\endgroup$
    – user154420
    Commented Jul 6, 2017 at 0:01

2 Answers 2


The answer lies in something called the virial theorem.

A He core that has ceased nuclear burning and that is in quasi-static equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) that twice the summed internal energy of particles ($U$) in the gas plus the (negative) gravitational potential energy ($\Omega$) equals zero. $$ 2U + \Omega = 0$$

This assumes that the pressure of gas outside the core is negligible (which is a reasonable assumption given the fall in temperature and density with radius.)

Now you can write down the total energy of the core as $$ E_{tot} = U + \Omega$$ and hence from the virial theorem that $$E_{tot} = \frac{\Omega}{2},$$ which is negative.

If we now remove energy from the core, by allowing the gas to radiate away energy (or maybe even some neutrino losses), such that $\Delta E_{tot}$ is negative (we can ignore energy generation since fusion has ceased), then we see that $$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$

So $\Omega$ becomes more negative - which is another way of saying that the core is attaining a more collapsed configuration. This process will occur unless the gas is completely degenerate, which in practice the core never attains (the assumption of complete degeneracy is essentially saying that the gas has a negligible temperature). The temperatures are always high enough for the degeneracy to be only partial.

Oddly, at the same time, we can use the virial theorem to see that $$ \Delta U = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$ is positive. i.e. the internal energies of particles in the gas (and hence their temperatures) actually become hotter. In other words, the gas has a negative heat capacity. Because the temperatures and densities are becoming higher, the interior pressure increases and can support a more condensed configuration. However, if the radiative losses continue, then so does the collapse.

This process is ultimately arrested in the core by the onset of He fusion.

Another way of thinking about this is that gravitational compression is doing work on the gas, thus raising its internal energy and hence temperature. The pressure does not rise significantly because most of the energy goes into raising the temperature of the non-degenerate ions which have a much larger heat capacity.

As the core contracts, the H-shell immediately around the core will also contract in response and achieve higher temperatures. As a result the shell burns even more fiercely than hydrogen burned in the core during the main sequence lifetime and the luminosity of the star increases as it ascends the red giant branch.

  • $\begingroup$ Am I right about thinking that the key in your explanation is to acknowledge that the He core is not completely degenerated and so it can react as a "normal" (say, ideal) gas and therefore increase its temperature? Is that the way to overcome the sentence "an increase in pressure does not lead to an increase in temperature for a degenerated gas"? $\endgroup$
    – Stefano
    Commented Sep 25, 2017 at 17:47
  • 1
    $\begingroup$ @stefano where are you getting the sentence you have put in quotation marks? The He core is not completely degenerate, but even when it is highly degenerate, it can continue to contract if it's mass increases. $\endgroup$
    – ProfRob
    Commented Sep 25, 2017 at 20:04
  • $\begingroup$ Thank for clarifying it to me @Rob Jeffries; I had to ask it. The intention of the quotation marks was not to quote something that you said but something that is commonly said during a class of astrophysics. I apologize for the misunderstood. $\endgroup$
    – Stefano
    Commented Sep 26, 2017 at 0:51

Both the core and the shell become hotter and more dense.
As He builds up in the core, since it is not yet hot enough to burn it does not support itself at the old volume. As volume decreases, temperature and density increase. The shell being immediately outside the core is supported by the core (even though at a decreasing radius), so as the core experiences a reduction in volume, the shell follows.
As the shell falls inward it experiences an increase in density and temperature.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.