<p>Proton degeneracy is not important, because its effect is much smaller -- much like nuclear particles in theory also are dictated by gravity, but the electromagnetic and nuclear forces are dominating, since they are much stronger. Proton degeneracy is weaker than electron degeneracy due to the far greater mass of the proton compared to the electron. <a href="https://en.wikipedia.org/wiki/Degenerate_matter">The Wikipedia article on degenerate matter</a> explains this very well;</p>

<blockquote>
  <p>Because protons are much more massive than electrons, the same momentum represents a much smaller velocity for protons than for electrons. As a result, in matter with approximately equal numbers of protons and electrons, proton degeneracy pressure is much smaller than electron degeneracy pressure, and proton degeneracy is usually modeled as a correction to the equations of state of electron-degenerate matter.</p>
</blockquote>

CC BY-SA 3.0
2016-06-28T12:27:53.500Z
2016-06-28T12:27:53.500Z
nataliaeire
<p>Straightforwardly no.</p>

<p>For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic).
There are a few protons near the surface, but not in sufficient numbers to be degenerate.</p>

<p>Let us assume though that you were able to build a hydrogen white dwarf that had equal numbers of free protons and electrons.</p>

<p>The density at which the electrons become degenerate is set by the requirement that their Fermi (kinetic) energy exceeds $kT$. The Fermi energy is given by
$$E_F = \frac{p_F^{2}}{2m} = \left(\frac{3}{8\pi}\right)^{2/3} n^{2/3} \left(\frac{h^2}{2m}\right),$$
where $n$ is the number density (which would be the same for protons and electrons), but $m$ is the mass of a proton or electron, which is different by a factor 1800.</p>

<p>Thus for a given white dwarf temperature, the electrons become degenerate at number densities a factor of $(m_p/m_e)^{3/2} = 78,600$ times lower than do protons.</p>

<p>Even if we were to compress a hypothetical hydrogen white dwarf to the point where the protons were also degenerate (which for a typical white dwarf interior temperature of $10^{7}$ K, would need mass densities considerably in excess of $10^{12}$ kg/m$^{3}$), the (ideal) degeneracy pressures would then be given by
$$ P = \frac{h^2}{20m}\left(\frac{3}{\pi}\right)^{2/3} n^{5/3}$$
and so we immediately see that the degeneracy pressure due to protons would be $\sim 1800$ times less than that due to the same number density of electrons.</p>

<p>If we push up to higher densities then both the protons and electrons will become <em>relativistically</em> degenerate. In this case the pressure becomes <em>independent of the particle mass</em>. However, the Fermi energies of the particles would now be high enough ($&gt;1$ GeV !) to block beta decay and instigate neutronisation. The protons and electrons start combining to form neutrons and an n,p,e fluid is formed where the neutrons vastly outnumber the protons and electrons. This in fact prevents the protons ever becoming relativistic, even at neutron star densities, and the contributions of electrons to the pressure is <em>always</em> orders of magnitude higher than for the protons.</p>

<p>A <strong>handwaving</strong> way of understanding this is that the degeneracy pressure depends on the <em>product</em> of the momentum and velocity of the particles. In turn, the momentum of the fermions depends on how much they are compressed through the uncertainty principle. For a given particle number density, the separation $\Delta x$ is the same for protons and electrons and therefore the uncertainty principle says that the $\Delta p \sim \hbar/\Delta x$ of the momentum is also the same. This is another way of saying that the Fermi momentum does <em>not</em> depend on the mass of the fermion; however, for a given fermion momentum, the velocity clearly does! Therefore the degeneracy pressure must be lower by approximately the mass ratio of the fermions</p>

CC BY-SA 3.0
2017-02-12T05:06:52.240Z
Quite late, but I was thinking. You mentioned that the number of protons and electrons are not equal in a white dwarf. Does that mean that the original star (and most stars, for that matter) have significantly more electrons than protons?
Sir Cumference
CC BY-SA 3.0
2017-02-12T22:56:55.407Z
@SirCumference I have not said that. I said there are very few free protons. White dwarfs are electrically neutral.
ProfRob
CC BY-SA 3.0
2016-06-28T17:22:24.277Z
2017-02-13T07:36:35.043Z
2017-02-13T07:36:35.043Z
ProfRob
<p>In general, fermions form a degenerate gas under high density or extremely low temperature. It's clear that white dwarf stars are supported by electron degeneracy pressure. However, there are still a significant number of protons in a white dwarf. Under those high densities, do the protons form a degenerate gas?</p>

<p>In addition, are white dwarfs supported by proton degeneracy pressure as much as they are by electron degeneracy pressure?</p>

<gravity><white-dwarf><density><degenerate-matter><quantum-mechanics>
Are white dwarf stars supported by proton degeneracy as well?
Sir Cumference
# Are white dwarf stars supported by proton degeneracy as well?

(Asked by: username_0 on 2016-06-28)

In general, fermions form a degenerate gas under high density or extremely low temperature. It's clear that white dwarf stars are supported by electron degeneracy pressure. However, there are still a significant number of protons in a white dwarf. Under those high densities, do the protons form a degenerate gas?

In addition, are white dwarfs supported by proton degeneracy pressure as much as they are by electron degeneracy pressure?

**Tags:** gravity, white-dwarf, density, degenerate-matter, quantum-mechanics


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## 2 answers

### Answer by username_2 on 2016-06-28. Score: 10 | Accepted answer

Straightforwardly no.

For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic).
There are a few protons near the surface, but not in sufficient numbers to be degenerate.

Let us assume though that you were able to build a hydrogen white dwarf that had equal numbers of free protons and electrons.

The density at which the electrons become degenerate is set by the requirement that their Fermi (kinetic) energy exceeds $kT$. The Fermi energy is given by
$$E\_F = \frac{p\_F^{2}}{2m} = \left(\frac{3}{8\pi}\right)^{2/3} n^{2/3} \left(\frac{h^2}{2m}\right),$$
where $n$ is the number density (which would be the same for protons and electrons), but $m$ is the mass of a proton or electron, which is different by a factor 1800.

Thus for a given white dwarf temperature, the electrons become degenerate at number densities a factor of $(m\_p/m\_e)^{3/2} = 78,600$ times lower than do protons.

Even if we were to compress a hypothetical hydrogen white dwarf to the point where the protons were also degenerate (which for a typical white dwarf interior temperature of $10^{7}$ K, would need mass densities considerably in excess of $10^{12}$ kg/m$^{3}$), the (ideal) degeneracy pressures would then be given by
$$ P = \frac{h^2}{20m}\left(\frac{3}{\pi}\right)^{2/3} n^{5/3}$$
and so we immediately see that the degeneracy pressure due to protons would be $\sim 1800$ times less than that due to the same number density of electrons.

If we push up to higher densities then both the protons and electrons will become *relativistically* degenerate. In this case the pressure becomes *independent of the particle mass*. However, the Fermi energies of the particles would now be high enough ($>1$ GeV !) to block beta decay and instigate neutronisation. The protons and electrons start combining to form neutrons and an n,p,e fluid is formed where the neutrons vastly outnumber the protons and electrons. This in fact prevents the protons ever becoming relativistic, even at neutron star densities, and the contributions of electrons to the pressure is *always* orders of magnitude higher than for the protons.

A **handwaving** way of understanding this is that the degeneracy pressure depends on the *product* of the momentum and velocity of the particles. In turn, the momentum of the fermions depends on how much they are compressed through the uncertainty principle. For a given particle number density, the separation $\Delta x$ is the same for protons and electrons and therefore the uncertainty principle says that the $\Delta p \sim \hbar/\Delta x$ of the momentum is also the same. This is another way of saying that the Fermi momentum does *not* depend on the mass of the fermion; however, for a given fermion momentum, the velocity clearly does! Therefore the degeneracy pressure must be lower by approximately the mass ratio of the fermions

> **Comments:**
> **username_0**: Quite late, but I was thinking. You mentioned that the number of protons and electrons are not equal in a white dwarf. Does that mean that the original star (and most stars, for that matter) have significantly more electrons than protons?  
> **username_2**: @username_0 I have not said that. I said there are very few free protons. White dwarfs are electrically neutral.  

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### Answer by username_1 on 2016-06-28. Score: 7

Proton degeneracy is not important, because its effect is much smaller -- much like nuclear particles in theory also are dictated by gravity, but the electromagnetic and nuclear forces are dominating, since they are much stronger. Proton degeneracy is weaker than electron degeneracy due to the far greater mass of the proton compared to the electron. [The Wikipedia article on degenerate matter](https://en.wikipedia.org/wiki/Degenerate_matter) explains this very well;

> Because protons are much more massive than electrons, the same momentum represents a much smaller velocity for protons than for electrons. As a result, in matter with approximately equal numbers of protons and electrons, proton degeneracy pressure is much smaller than electron degeneracy pressure, and proton degeneracy is usually modeled as a correction to the equations of state of electron-degenerate matter.


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