# When a star turns into a red giant what stops its core contracting

I recently asked this question When do stars become red giants? and am now wondering when the star's core is contracting after it has fused $H$ to $He$ what force stops the contraction just before it becomes a red giant?

• Closely related: physics.stackexchange.com/q/91716 Commented Apr 19, 2014 at 19:49
• Do you mean what force slows the contraction? The star is already a red giant before the core even comes close to stopping its contraction. Commented Oct 31, 2014 at 21:33

The answer depends upon the mass of the star.

For stars of less than 2 solar masses, electron degeneracy pressure stops the collapse.

For more massive stars, helium fusion begins which stops collapse, without a degenerate state being reached.

• I believe even low-mass stars continue to contract until He is ignited. So the answer is He ignition in both cases. Commented Oct 31, 2014 at 21:32
• @RobJeffries, for a solar mass size star, the inert helium core is sustained from collapse by electron degenercy pressure for over a billion years until the helium flash. Electron degeneracy pressure is the force that opposes gravity, allowing the shell hydrogen to fuse for over a billion years. Commented Nov 5, 2014 at 12:32
• I don't disagree. I was interpreting the question as what stops the core getting smaller? The core continues to get smaller until He burning is initiated. Commented Nov 5, 2014 at 12:45
• does the core really keep getting smaller once it is degenerate, or does it just get more dense by adding more helium as the shell hydrogen fuses to helium? Commented Nov 5, 2014 at 14:03
• Good question. I think it must because as you know the radius of objects supported by degeneracy gets smaller as they increase in mass. I might look for chapter and verse, but I can't see how the density can increase by a couple of orders without shrinkage. Commented Nov 5, 2014 at 20:49

$$\frac{dP} {dr} = -\frac{GM\rho}{r^2}$$Where $P$, is electron degeneracy pressure, $G$ is a constant, $M$ is the mass enclosed in a shell of radius $r$ with density $\rho$.

In short, electron degeneracy pressure ( left hand side ) generated by electron degeneracy is opposite and equal to gravity ( Right hand side ). Where electron degeneracy is a result of the Pauli Exclusion Principle. We have so many electrons in such a small volume, and of course per the exclusion principle, none of these can exist in the same quantum state. By trying to force these electrons into this smaller and smaller region, the laws of quantum mechanics fight back by resisting being in such a small volume, this generates the aforementioned electron degeneracy pressure. For more detail see here, where the analytic solution for the pressure part of the above Hydrostatic Equilibrium Equation is defined.

• but in a red giant of <2 solar masses, there is no fusion happening in the core. Only electron degeneracy pressure stops the collapse. abyss.uoregon.edu/~js/ast122/lectures/lec16.html Commented Apr 18, 2014 at 18:06
• Would the triple alpha process not be stopping the contraction? Commented Apr 18, 2014 at 18:09
• Not at first in stars <2 solar masses. Once helium starts fusion, the star is considered a horizontal branch star rather than simply a red giant. For red giants <2 solar masses, there is a distinct period of time where there is no fusion in the core. Later there can be a sudden "helium flash" when helium starts fusing. For stars >2 solar masses there is no helium flash, helium gradually starts fusing without the core becoming degenerate. Commented Apr 18, 2014 at 18:15
• Good to know! I'll modify my answer! Commented Apr 18, 2014 at 18:21
• @DavePhD: Seem about right? Commented Apr 18, 2014 at 18:28

Nothing stops the core contraction until He ignition takes place!

Yes, the core is partly supported by electron degeneracy, this slows the contraction and allows a relatively stable period of shell H-burning, but the central density continues to rise and the core contracts until He is actually ignited. This is especially true where there is a high degree of degeneracy. The mass of the core increases as more He ash accumulates from the H-burning shell above; the core shrinks further because, for an object supported by a degenerate equation of state, the more massive it is, the smaller it has to be. It is the core contraction and relocation of the H-burning shell to hotter temperatures that causes the increase in luminosity characterizing the ascent of the giant branch.

In a lower mass star, the pressure builds up very rapidly until He is ignited in a runaway "flash", the degeneracy is lifted, and the core expands significantly (see diagram below). In higher mass stars ($>2M_{\odot}$), where the core was not very degenerate, the He is ignited more gradually and the core contraction is halted more gradually.

Here is a diagram of the central density vs central temperature for stars of 1.3 and 2.1$M_{\odot}$ leaving the main sequence and igniting He in their cores - at point A for the lower-mass star, which undergoes a helium flash. The main sequence is to the bottom-left of the diagram. The diagram is from a thesis by Gautschy (2012). The two green lines mark first (left) where partial degeneracy is felt in the core, then on the right (only reached by the lower mass stars) where a high level of degeneracy is reached.

Note that the central density monotonically increases from the main sequence until He is ignited (the star is already a red giant at this point).

Edit and postscript:

It can't be true that electron degeneracy halts the core contraction (except in stars with $M<0.5M_{\odot}$). If that were the case, the core could cool and yet still support the star and He burning would never take place - leading to the formation of a He white dwarf.

• Laughlin, Bodenheimer, and Adams, "The end of the main sequence," The Astrophysical Journal 482.1 (1997) seems to contradict that parenthetical remark and final sentence in your edit and postscript. They discuss the life of small ($< 0.25\,M_{\odot}$) stars in detail. It's only in very small stars ($< 0.08\,M_{\odot}$) that a degenerate shell fails to form. Helium burning doesn't occur in these small stars. The larger ones form things that look like red giants, but they never hit the tip of the red giant branch. Commented Nov 9, 2014 at 15:49
• @DavidHammen What does it contradict? This paper confirms what I wrote - (p.431, point 3) that stars below 0.5Msun (or thereabouts) "do not reach the tip of the red giant branch". ie. they do end up being supported by degeneracy pressure in the He core and He doesn't ignite and they end up as He WDs. ? Commented Nov 10, 2014 at 1:00
• @RobJeffries, you wrote, "It is the core contraction and relocation of the H-burning shell to hotter temperatures that causes the increase in luminosity". I have been looking for an explanation precisely for that phenomena. Could you please explain to me: (1) Why the contraction of the core implies an increase in temperature? and (2) Why and how the H-burning shells manage to be relocated to higher temperatures? I would be very grateful if you can answer me. Thanks for your time :) Commented Jul 5, 2017 at 20:20
• @Stefano These are new questions. Contraction leads to an increase in temperature - basic application of the virial theorem. If the core contracts and becomes hotter, then a surrounding shell also moves inwards and gets hotter. Commented Jul 5, 2017 at 23:15
• @RobJeffries. So the Virial theorem is the answer. I understand that the Virial theorem applies to the H surrounding shell, it moves inwards and gets hotter. Does the Virial Theorem also apply to degenerate matter like the He core? (I know that the Virial Theorem derives from the condition of hydrostatical equilibrium). I know that the He core is isothermal so it will have the same temperature as the H surrounding shell, therefore it will get hotter and hotter as the surrounding shell does, however, I would like to know if the shrinkage of the degenerate core alone also contributes to it. Commented Jul 6, 2017 at 9:51