<p>Another way to think of the answer is that the pressure of any nonrelativistic gas (degenerate or ideal) is always 2/3 the kinetic energy per unit volume.  So we can say that when the electrons go degenerate, and contribute almost all of the pressure, it must be because they also contain almost all of the kinetic energy.  This is true, the reason they contribute most of the pressure is because they hold most of the kinetic energy, but we can then ask, why do they contain most of the kinetic energy?</p>

<p>Framing the question this way focuses our attention on what really matters here: the thermodynamics of degeneracy.  A purely degenerate gas has zero temperature, that is what degeneracy means in the thermodynamic sense.  However, the kinetic energy is quite large.  So it is this <em>thermodynamic</em> disconnect between the kinetic energy and the temperature that is the cause of the importance of the electron pressure.  Since we can assume the electrons and protons equilibrate to the same temperature (that assumption is always in place when you see these kinds of analyses), the protons must also have essentially zero temperature when the electrons are highly degenerate.  But protons are not degenerate, due to their higher mass (as explained above), so if they have low temperature, they also have low kinetic energy.  It's due to their different thermodynamic connection between kinetic energy and temperature (which is more like an ideal gas rather than a degenerate one).</p>

<p>Put this way, we see what degeneracy is doing-- it is causing the electrons to hog all the kinetic energy, because degenerate electrons have a very high ratio of kinetic energy to kT, whereas protons have a more classical such ratio.  That's why they produce the pressure, it's all about kinetic energy.  You don't need to think about any weird forces, you only need the quantum mechanical thermodynamics, which drives the temperature of electrons down, relative to their kinetic energy, owing to the fact that degenerate electrons are very stingy about giving up heat (and temperature is the statement of which objects will give up heat in equilibrium, as per the zeroth law of thermodynamics).</p>

CC BY-SA 4.0
2019-03-29T22:28:19.007Z
Interesting. However, doesn't really answer my question. Still, +1.
Max0815
CC BY-SA 4.0
2019-03-30T02:45:46.323Z
Thanks.  The direct answer to your question is there is no force of either electron degeneracy or proton degeneracy.  Degeneracy is not a force-- it is a thermodynamic effect that alters the partitioning of kinetic energy between electrons and protons, and also among the electrons.  The electrons dominate the pressure because the thermodynamics says they get almost all the kinetic energy.
Ken G
CC BY-SA 4.0
2019-03-29T16:38:46.600Z
2019-03-29T16:38:46.600Z
Ken G
<p>Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.</p>

<p>Since the Fermi momentum <span class="math-container">$p_F$</span> of a species is dependent only on number density<span class="math-container">$^1$</span>, the Fermi momentum of the electrons and protons are the same. Pressure however<span class="math-container">$^1$</span>, is an increasing function of <span class="math-container">$p_F/mc$</span> and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.</p>

<p>Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.</p>

<p>Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as <span class="math-container">$p_F/mc \gg 1$</span> (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.</p>

<p>(1) For those interested: The Fermi momentum is
<span class="math-container">$$p_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$</span>
where <span class="math-container">$n$</span> is the number density of fermions.</p>

<p>Whilst ideal fermion degeneracy pressure is given by
<span class="math-container">$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3
  \sinh^{-1}(x) \right]\ , $$</span>
where <span class="math-container">$x = p_F/mc$</span>.</p>

CC BY-SA 4.0
2019-03-29T07:17:08.237Z
2019-03-29T14:19:05.537Z
2019-03-29T14:19:05.537Z
ProfRob
<p>Come to think of it, the only things we hear about degeneracy is neutron and electron.<br>
I made a generalization:</p>

<ul>
<li><strong>White Dwarf</strong>: Supported by electron degeneracy.</li>
<li><strong>Neutron Star</strong>: Supported by neutron degeneracy and nuclear forces, and sometimes centrifugal force.</li>
<li><strong>Quark Star</strong>: Most likely supported by quark degeneracy.</li>
</ul>

<p>Now, we can see that there is no mention of proton degeneracy. </p>

<p>A quick search lead to the Wikipedia subsection <a href="https://en.wikipedia.org/wiki/Degenerate_matter#Proton_degeneracy" rel="nofollow noreferrer">Degenerate matter; Proton degeneracy</a>, and the answer that yes, there is proton degeneracy, but is much less effective than other forms of degenerate matter.</p>

<p><strong>My question:</strong> What is the force exerted by degenerate protons, and what equation is it modeled by?</p>

CC BY-SA 4.0
2019-03-29T05:55:35.337Z
Possible duplicate of [Are white dwarf stars supported by proton degeneracy as well?](https://astronomy.stackexchange.com/questions/16402/are-white-dwarf-stars-supported-by-proton-degeneracy-as-well)
PM 2Ring
CC BY-SA 4.0
2019-03-29T10:36:07.143Z
Not a duplicate, since that question, and my answer, are restricted to white dwarfs.
ProfRob
CC BY-SA 4.0
2019-03-29T15:31:25.283Z
@RobJeffries Ah, ok. I figured it was a reasonable match, but I'm glad you've written a specific answer for this new question.
PM 2Ring
CC BY-SA 4.0
2019-03-29T19:19:13.240Z
@uhoh, okay. I searched on google which brings me to wikipedia. I clarrified in the question.
Max0815
CC BY-SA 4.0
2019-03-29T22:15:57.267Z
@Max0815 looks good, thanks! I've adjusted the format a bit and taken out the non-essential bits. I don't think there's any chance of it being closed as duplicate now.
uhoh
<degenerate-matter><proton>
Measures for the degeneracy from protons?
Max0815
# Measures for the degeneracy from protons?

(Asked by: username_0 on 2019-03-29)

Come to think of it, the only things we hear about degeneracy is neutron and electron.  
I made a generalization:

* **White Dwarf**: Supported by electron degeneracy.
* **Neutron Star**: Supported by neutron degeneracy and nuclear forces, and sometimes centrifugal force.
* **Quark Star**: Most likely supported by quark degeneracy.

Now, we can see that there is no mention of proton degeneracy.

A quick search lead to the Wikipedia subsection [Degenerate matter; Proton degeneracy](https://en.wikipedia.org/wiki/Degenerate_matter#Proton_degeneracy), and the answer that yes, there is proton degeneracy, but is much less effective than other forms of degenerate matter.

**My question:** What is the force exerted by degenerate protons, and what equation is it modeled by?

**Tags:** degenerate-matter, proton
> **Comments:**
> **username_1**: Possible duplicate of [Are white dwarf stars supported by proton degeneracy as well?](https://astronomy.stackexchange.com/questions/16402/are-white-dwarf-stars-supported-by-proton-degeneracy-as-well)  
> **username_2**: Not a duplicate, since that question, and my answer, are restricted to white dwarfs.  
> **username_1**: @RobJeffries Ah, ok. I figured it was a reasonable match, but I'm glad you've written a specific answer for this new question.  
> **username_0**: @uhoh, okay. I searched on google which brings me to wikipedia. I clarrified in the question.  
> **username_3**: @Max0815 looks good, thanks! I've adjusted the format a bit and taken out the non-essential bits. I don't think there's any chance of it being closed as duplicate now.  

---

## 2 answers

### Answer by username_2 on 2019-03-29. Score: 10 | Accepted answer

Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.

Since the Fermi momentum $p\_F$ of a species is dependent only on number density$^1$, the Fermi momentum of the electrons and protons are the same. Pressure however$^1$, is an increasing function of $p\_F/mc$ and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.

Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.

Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as $p\_F/mc \gg 1$ (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.

(1) For those interested: The Fermi momentum is
$$p\_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$
where $n$ is the number density of fermions.

Whilst ideal fermion degeneracy pressure is given by
$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3
\sinh^{-1}(x) \right]\ , $$
where $x = p\_F/mc$.


---

### Answer by username_4 on 2019-03-29. Score: 2

Another way to think of the answer is that the pressure of any nonrelativistic gas (degenerate or ideal) is always 2/3 the kinetic energy per unit volume. So we can say that when the electrons go degenerate, and contribute almost all of the pressure, it must be because they also contain almost all of the kinetic energy. This is true, the reason they contribute most of the pressure is because they hold most of the kinetic energy, but we can then ask, why do they contain most of the kinetic energy?

Framing the question this way focuses our attention on what really matters here: the thermodynamics of degeneracy. A purely degenerate gas has zero temperature, that is what degeneracy means in the thermodynamic sense. However, the kinetic energy is quite large. So it is this *thermodynamic* disconnect between the kinetic energy and the temperature that is the cause of the importance of the electron pressure. Since we can assume the electrons and protons equilibrate to the same temperature (that assumption is always in place when you see these kinds of analyses), the protons must also have essentially zero temperature when the electrons are highly degenerate. But protons are not degenerate, due to their higher mass (as explained above), so if they have low temperature, they also have low kinetic energy. It's due to their different thermodynamic connection between kinetic energy and temperature (which is more like an ideal gas rather than a degenerate one).

Put this way, we see what degeneracy is doing-- it is causing the electrons to hog all the kinetic energy, because degenerate electrons have a very high ratio of kinetic energy to kT, whereas protons have a more classical such ratio. That's why they produce the pressure, it's all about kinetic energy. You don't need to think about any weird forces, you only need the quantum mechanical thermodynamics, which drives the temperature of electrons down, relative to their kinetic energy, owing to the fact that degenerate electrons are very stingy about giving up heat (and temperature is the statement of which objects will give up heat in equilibrium, as per the zeroth law of thermodynamics).

> **Comments:**
> **username_0**: Interesting. However, doesn't really answer my question. Still, +1.  
> **username_4**: Thanks.  The direct answer to your question is there is no force of either electron degeneracy or proton degeneracy.  Degeneracy is not a force-- it is a thermodynamic effect that alters the partitioning of kinetic energy between electrons and protons, and also among the electrons.  The electrons dominate the pressure because the thermodynamics says they get almost all the kinetic energy.  

---

