<p>The most important thing that happens is hydrogen is turned into helium, which essentially means that electrons and protons are getting merged into neutrons (the neutrons inside the helium). That reduces the number of particles.  For example, if you turn a pure hydrogen gas into a pure helium gas, you now have half the electrons you had before.  Also, you have 3 particles (a He nucleus and 2 electrons) where before you would have had 8 (in the case of pure H initially, just as an example).  Losing particles means you lose pressure at a given temperature and volume (it would be proportional to number of particles), so the core must contract.  The temperature does go up a little, but not much because nuclear burning is very temperature sensitive.  But the particle density drops a great deal, so for the density to stay the same, the core pressure must drop quite a bit, perhaps by almost a factor of 1/2.</p>

<p>The reason this might not require the core density to change is that the core pressure scales roughly like the radius to the -4 power (because of the requirements of force balance, which balances the pressure times the surface area, which scales like P*R^2, with the gravity, which scales like 1/R^2).   So that's quite a steep drop in P as R rises a little, and apparently the small rise in R is enough to compensate for the large drop in core P, to keep the density the same (based on your professor's statement that it does stay nearly the same).  It must be essentially a coincidence of how these things work out, it's not obvious to show why the density should be exactly constant but at least we can see that a large drop in P is compensated well by a small rise in R.  The physics of convection is pretty tricky, and there is only an approximate treatment called "mixing length theory" but it's never really been clear how to do this in a formally correct way.</p>

CC BY-SA 4.0
2019-06-24T03:14:30.637Z
2019-06-24T03:29:44.860Z
2019-06-24T03:29:44.860Z
Ken G
<p>I was able to get some insight from my professor, so I figured I'd answer this question. Basically, convection works AGAINST gravity and keeps the density constant. </p>

CC BY-SA 4.0
2019-06-23T16:55:40.853Z
2019-06-23T16:55:40.853Z
Astroturf
<p>We know the CNO cycle produces the majority of energy production in a "higher mass star" of approx. over 2 solar masses and the core is convective due to the large temperature gradient. My professor has noted that radius, luminosity, T_eff, and T_center all increase but rho_center (density) does not. I cannot find anything in my text that explains this, but I assume it is because convection works against gravity. If someone could clear this concept up (as to why rho_center) stays generally constant, I'd appreciate it. Conceptual explanations are helpful!</p>

<star><stellar-evolution><main-sequence>
Does the density in the core of a "higher mass" star stay generally constant while it evolves on the MS?
Astroturf
# Does the density in the core of a "higher mass" star stay generally constant while it evolves on the MS?

(Asked by: username_0 on 2019-06-12)

We know the CNO cycle produces the majority of energy production in a "higher mass star" of approx. over 2 solar masses and the core is convective due to the large temperature gradient. My professor has noted that radius, luminosity, T\_eff, and T\_center all increase but rho\_center (density) does not. I cannot find anything in my text that explains this, but I assume it is because convection works against gravity. If someone could clear this concept up (as to why rho\_center) stays generally constant, I'd appreciate it. Conceptual explanations are helpful!

**Tags:** star, stellar-evolution, main-sequence


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## 2 answers

### Answer by username_1 on 2019-06-24. Score: 1

The most important thing that happens is hydrogen is turned into helium, which essentially means that electrons and protons are getting merged into neutrons (the neutrons inside the helium). That reduces the number of particles. For example, if you turn a pure hydrogen gas into a pure helium gas, you now have half the electrons you had before. Also, you have 3 particles (a He nucleus and 2 electrons) where before you would have had 8 (in the case of pure H initially, just as an example). Losing particles means you lose pressure at a given temperature and volume (it would be proportional to number of particles), so the core must contract. The temperature does go up a little, but not much because nuclear burning is very temperature sensitive. But the particle density drops a great deal, so for the density to stay the same, the core pressure must drop quite a bit, perhaps by almost a factor of 1/2.

The reason this might not require the core density to change is that the core pressure scales roughly like the radius to the -4 power (because of the requirements of force balance, which balances the pressure times the surface area, which scales like P\*R^2, with the gravity, which scales like 1/R^2). So that's quite a steep drop in P as R rises a little, and apparently the small rise in R is enough to compensate for the large drop in core P, to keep the density the same (based on your professor's statement that it does stay nearly the same). It must be essentially a coincidence of how these things work out, it's not obvious to show why the density should be exactly constant but at least we can see that a large drop in P is compensated well by a small rise in R. The physics of convection is pretty tricky, and there is only an approximate treatment called "mixing length theory" but it's never really been clear how to do this in a formally correct way.


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### Answer by username_0 on 2019-06-23. Score: 0

I was able to get some insight from my professor, so I figured I'd answer this question. Basically, convection works AGAINST gravity and keeps the density constant.


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