2
$\begingroup$

From what I understood from this answer to How can a red giant grow so big?

But this is not at all the way the shell fusing in a red giant self-regulates its fusion rate-- it cannot regulate its temperature, because the temperature is handed to it by the gravity of the degenerate core its sits upon. (This sets the temperature via the virial theorem, that is the key way the degenerate core affects the shell-- it sets its temperature.)

For the fusing hydrogen shell, an increase in volume isn't accompanied by a decrease in temperature, which is how cores of main sequence stars stay in equilibrium. Consequently, this is why such a dramatic reduction in pressure is required for a red giant.

Why doesn't thermal dynamics apply to the shell (or so I assume)? Why is the shell qualitatively different from a fusing core of a main sequence star? Is it because in a red giant, the core is massive and dense enough that the shell is degenerate?

$\endgroup$
1
  • $\begingroup$ Thermodynamics apply both cases. The boundary conditions are different: for core burning you inside the fusion zone no mass, no radius, no luminosity. For shell burning the inner boundary is at a certain radius r and you have an enclosed mass $m_{core}$. You need to solve the (different) equation of states for core and shell such that the thermodynamic properties match at the boundary (and for the core ,the viral theorem is means to get a handle on some of those).. $\endgroup$ Commented Aug 27, 2021 at 14:20

0

You must log in to answer this question.

Browse other questions tagged .