<p>The ratio of neutrons to protons (and electrons, since the fluid is neutral) <em>does</em> depend on the overall density. In an ideal n,p,e fluid, the ratio is of order 100 to 1 at average neutron star densities, but decreases towards 8 to 1 as the density becomes very large.</p>
<p>To understand this note that there will be an equilibrium set up, where neutrons can decay and place new electrons and protons at the tops of their respective Fermi seas, but the most energetic electrons and protons can combine to produce new neutrons. A simple argument based on minimising the total energy density of the fluid, whilst conserving charge and baryon number then leads to the following relationship between the respective Fermi energies:
<span class="math-container">$$E_{F,n}= E_{F,p} + E _{F,e}\ .$$</span>
NB: This assumes the species are completely degenerate. At neutron star densities this should be ok if interior temperatures are <span class="math-container">$\ll 10^{10}$</span> K, which should be true for neutron stars more than a day old.</p>
<p>The electrons are always ultrarelativistic, but at low densities, the neutrons and protons are non-relativistic. And, since the number densities of protons and electrons are equal, so are their Fermi momenta (<span class="math-container">$p_{F,p}=p_{F,e}$</span>). Thus:
<span class="math-container">$$( p_{F,n}^2c^2 + m_n^2c^4)^{1/2} \simeq (p_{F,p}^2 c^2 + m_p^2 c^4)^{1/2}+ p_{F,p}c\ . $$</span></p>
<p>Given that <span class="math-container">$p_F = h(3n/8\pi)^{1/3}$</span>, then the above equation can be solved to give <span class="math-container">$n_n/n_p$</span> and then the total mass density obtained by summing the energy densities of the neutrons, protons and electrons. The plot below shows this calculation; for typical neutron star densities of a few <span class="math-container">$10^{17}$</span> kg/m<span class="math-container">$^3$</span> it is about 100-1000.</p>
<p><a href="https://i.sstatic.net/l5hJt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/l5hJt.png" alt="Neutron to proton ratio" /></a></p>
<p><em>Neutron to proton ratio as a function of mass density for an ideal npe gas.</em></p>
<p>If you allow the density to increase drastically at the core of a neutron star, <em>and</em> you assume that only neutrons, protons and electrons can exist, then <em>all</em> the Fermi energies become ultrarelativistic,
<span class="math-container">$$p_{F,n}=p_{F,p} + p_{F,e} = 2p_{F,p}$$</span>
and thus <span class="math-container">$n_n = 8 n_p$</span>.</p>
<p>In practice, as the neutron Fermi energy becomes large, other species are expected to appear, starting with muons at around <span class="math-container">$8\times 10^{17}$</span> kg/m<span class="math-container">$^3$</span> and then perhaps heavy hadrons, pions/kaons or even a quark plasma. At these densities, the neutron to proton ratio in an ideal n,p,e, gas should be around 10-20, but these non-ideal effects, which are still poorly understood, lead to considerable uncertainty and therefore considerable uncertainty in the structure and cooling rates, particularly of massive neutron stars which are expected to have the highest central densities.</p>

CC BY-SA 4.0
2021-08-22T12:29:17.767Z
Interesting! What have you assumed about the temperature in this calculation? Is it nonzero? Or are you considering a microcanonical ensemble?
Daddy Kropotkin
CC BY-SA 4.0
2021-08-22T13:03:05.407Z
So it sounds like the answer to my question is yes, and the relationship is that as the neutron star's mass (and by proxy, density) increases, more of the neutrons turn into proton-electron pairs.
zucculent
CC BY-SA 4.0
2021-08-23T07:39:57.223Z
@DaddyKropotkin I am assuming the neutron star is cold enough to be considered completely degenerate, so $<10^{10}$ K.
ProfRob
CC BY-SA 4.0
2021-08-23T07:41:24.393Z
@zucculent yes, that is the result.
ProfRob
CC BY-SA 4.0
2021-08-22T08:35:50.070Z
2021-08-23T07:46:29.043Z
2021-08-23T07:46:29.043Z
ProfRob
<p>My understanding is that although electron degeneracy prevents beta decay, there are still a few protons and electrons hanging around. I also understand that, at least in white dwarfs, as the mass increases, more electrons are forced into higher energy/momentum states. Does this carry over to neutron stars, and if so, does it affect the density of protons and electrons?</p>

CC BY-SA 4.0
2021-08-22T05:03:36.150Z
Related posts from our sister site: https://physics.stackexchange.com/a/149656/123208 https://physics.stackexchange.com/q/63383/123208 https://physics.stackexchange.com/a/133139/123208 https://physics.stackexchange.com/q/6533/123208
PM 2Ring
<neutron-star><degenerate-matter>
Does the average proton/electron density in a neutron star change with mass?
zucculent
# Does the average proton/electron density in a neutron star change with mass?

(Asked by: username_0 on 2021-08-22)

My understanding is that although electron degeneracy prevents beta decay, there are still a few protons and electrons hanging around. I also understand that, at least in white dwarfs, as the mass increases, more electrons are forced into higher energy/momentum states. Does this carry over to neutron stars, and if so, does it affect the density of protons and electrons?

**Tags:** neutron-star, degenerate-matter
> **Comments:**
> **username_1**: Related posts from our sister site: https://physics.stackexchange.com/a/149656/123208 https://physics.stackexchange.com/q/63383/123208 https://physics.stackexchange.com/a/133139/123208 https://physics.stackexchange.com/q/6533/123208  

---

## 1 answers

### Answer by username_2 on 2021-08-22. Score: 15 | Accepted answer

The ratio of neutrons to protons (and electrons, since the fluid is neutral) *does* depend on the overall density. In an ideal n,p,e fluid, the ratio is of order 100 to 1 at average neutron star densities, but decreases towards 8 to 1 as the density becomes very large.

To understand this note that there will be an equilibrium set up, where neutrons can decay and place new electrons and protons at the tops of their respective Fermi seas, but the most energetic electrons and protons can combine to produce new neutrons. A simple argument based on minimising the total energy density of the fluid, whilst conserving charge and baryon number then leads to the following relationship between the respective Fermi energies:
$$E\_{F,n}= E\_{F,p} + E \_{F,e}\ .$$
NB: This assumes the species are completely degenerate. At neutron star densities this should be ok if interior temperatures are $\ll 10^{10}$ K, which should be true for neutron stars more than a day old.

The electrons are always ultrarelativistic, but at low densities, the neutrons and protons are non-relativistic. And, since the number densities of protons and electrons are equal, so are their Fermi momenta ($p\_{F,p}=p\_{F,e}$). Thus:
$$( p\_{F,n}^2c^2 + m\_n^2c^4)^{1/2} \simeq (p\_{F,p}^2 c^2 + m\_p^2 c^4)^{1/2}+ p\_{F,p}c\ . $$

Given that $p\_F = h(3n/8\pi)^{1/3}$, then the above equation can be solved to give $n\_n/n\_p$ and then the total mass density obtained by summing the energy densities of the neutrons, protons and electrons. The plot below shows this calculation; for typical neutron star densities of a few $10^{17}$ kg/m$^3$ it is about 100-1000.

[![Neutron to proton ratio](https://i.sstatic.net/l5hJt.png)](https://i.sstatic.net/l5hJt.png)

*Neutron to proton ratio as a function of mass density for an ideal npe gas.*

If you allow the density to increase drastically at the core of a neutron star, *and* you assume that only neutrons, protons and electrons can exist, then *all* the Fermi energies become ultrarelativistic,
$$p\_{F,n}=p\_{F,p} + p\_{F,e} = 2p\_{F,p}$$
and thus $n\_n = 8 n\_p$.

In practice, as the neutron Fermi energy becomes large, other species are expected to appear, starting with muons at around $8\times 10^{17}$ kg/m$^3$ and then perhaps heavy hadrons, pions/kaons or even a quark plasma. At these densities, the neutron to proton ratio in an ideal n,p,e, gas should be around 10-20, but these non-ideal effects, which are still poorly understood, lead to considerable uncertainty and therefore considerable uncertainty in the structure and cooling rates, particularly of massive neutron stars which are expected to have the highest central densities.

> **Comments:**
> **username_3**: Interesting! What have you assumed about the temperature in this calculation? Is it nonzero? Or are you considering a microcanonical ensemble?  
> **username_0**: So it sounds like the answer to my question is yes, and the relationship is that as the neutron star's mass (and by proxy, density) increases, more of the neutrons turn into proton-electron pairs.  
> **username_2**: @DaddyKropotkin I am assuming the neutron star is cold enough to be considered completely degenerate, so $<10^{10}$ K.  
> **username_2**: @zucculent yes, that is the result.  

---

