<p>Gravity is only important insofar that it is capable of compressing the material to high densities. Whether that material is capable of solidifying depends on the competition between Coulombic potential energy and the thermal energy of the particles. The former increases with density, the latter increases with temperature. A dense plasma can still be a gas if it is hot enough.</p>

<p>A rough formula for the exponential scale height of the atmosphere is
$$ h = \frac{kT}{\mu m_u g},$$
where $T$ is the temperature of the gas, $m_u$ is an atomic mass unit, $\mu$ is the number of atomic mass units per particle and $g$ is the surface gravity, with $g = GM/R^2$.</p>

<p>For a typical neutron star with $R=10$ km, $M= 1.4M_{\odot}$, we have $g=1.86\times 10^{12}$ m/s$^2$. The atmosphere could be a mixture of ionised helium ($\mu=4/3$) or perhaps iron ($\mu = 56/27$), so let's say $\mu=2$ for simplicity. The temperature at the surface of the neutron star will change with time; typically for a young pulsar, the surface temperature might be $10^{6}$ K.</p>

<p>This gives $h = 2$ mm.</p>

<p>Why is this not a "solid"? Because the thermal energy of the particles is larger than the coulombic binding energy in any solid lattice that the ions could make. That is not the case in the solid surface below the atmosphere because the density grows very rapidly (from $10^{6}$ kg/m$^{3}$ to more than $10^{10}$ kg/m$^{3}$ (where solidification takes place) only a few cm in, because the scale height is so small. Of course the temperature increases too, but not by more than a factor of about 100. After that, the density is high enough for electron degeneracy, and the material becomes approximately isothermal and at a small depth the "freezing temperature" falls below the isothermal temperature.</p>

CC BY-SA 3.0
2016-05-24T15:55:00.873Z
I am confused about your usage of $\mu$ and $m_u$.
geometrian
CC BY-SA 3.0
2016-05-24T15:59:33.130Z
@imallet Atomic mass unit $m_u = 1.67\times 10^{-27}$ kg. $\mu$ - the number of mass units per particle. ionised Helium 3 particles, 4 mass units (whoops, I made a mistake).
ProfRob
CC BY-SA 3.0
2016-05-24T16:14:53.777Z
Still confused. I'm parsing e.g. $56/27$ as "$56$ amu per $27$ atoms". Maybe, you can provide the source for the formula?
geometrian
CC BY-SA 3.0
2016-05-24T16:51:03.650Z
So in laysman terms... That stuff is too hot to hold as a solid or even a liquid. Cool.
Geeky Guy
CC BY-SA 3.0
2016-05-24T19:18:19.563Z
@imallet per PARTICLE. Electrons plus nuclei.
ProfRob
CC BY-SA 3.0
2016-05-24T19:43:50.027Z
Technically speaking, plasma is a distinct phase of matter from gas.  So we can apparently expect a thin, dense "plasma atmosphere", but is there a meaningful "gaseous atmosphere" a little further out?
zibadawa timmy
CC BY-SA 3.0
2016-05-24T19:46:17.903Z
@zibadawatimmy if you wish to go with that definition, then there is no gas. It's all ionised.
ProfRob
CC BY-SA 3.0
2016-05-24T19:50:27.773Z
"*The atmosphere could be a mixture of ionised helium or perhaps iron...*" Aren't neutron stars made of neutronium?
RBarryYoung
CC BY-SA 3.0
2016-05-24T20:05:22.110Z
@RBarryYoung The equation of state for a neutron star has yet to be determined, but there is expected to be a thin crust of normal nuclei, possibly iron (due to the high binding energy per nucleon) or helium (if heavier elements would "sink" deeper into the star).  You have to go a bit further in before nuclei cease to exist.
zibadawa timmy
CC BY-SA 3.0
2016-05-24T22:08:42.933Z
@RBarryYoung Neutronium is a made-up SciFi word. Neutron stars have crusts of neutron-rich nuclei accompanied by degenerate electrons. The outer cm or so is a non-degenerate gas of uncertain composition, but one thing it isn't is free neutrons.
ProfRob
CC BY-SA 3.0
2016-05-26T12:40:13.113Z
Good answer to a hard question! While it was published in 1980, the hard Sci-Fi book [Dragon's Egg](https://en.wikipedia.org/wiki/Dragon%27s_Egg) by Robert L. Forward does help one *imagine* what the surface *might be like* - including neutron-rich nuclei on the surface as well as the atmosphere discussed here.
uhoh
CC BY-SA 3.0
2016-05-26T20:43:33.570Z
@RobJeffries . . . so the arithmetic is "$56$ amu / ($26$ electrons + $1$ nucleus)"?
geometrian
CC BY-SA 3.0
2016-05-26T21:34:36.840Z
@imallett Correct.
ProfRob
CC BY-SA 3.0
2016-05-24T13:29:53.467Z
2016-05-24T19:30:45.247Z
2016-05-24T19:30:45.247Z
ProfRob
<p>Neutron stars can have small atmospheres. However, they also have extremely strong gravitationally pulls. Shouldn't the all the gas molecules be drawn to the star's surface, and become solids under the immense pressure?</p>

<p>Maybe I'm thinking about this the wrong way, but I don't see how it could be possible.</p>

CC BY-SA 3.0
2016-05-24T13:15:18.273Z
4 inch thick atmospheres.   :-)
userLTK
CC BY-SA 3.0
2016-05-24T13:15:51.643Z
@userLTK It still seems absurd that matter so close to the star will be gaseous.
Sir Cumference
CC BY-SA 3.0
2016-05-24T13:16:10.420Z
What do you mean by large atmospheres? If you mean the magnetospheres, well the clue is in the name. Gravity is not the only force acting.
ProfRob
CC BY-SA 3.0
2016-05-24T13:16:37.817Z
@RobJeffries Yeah, made a mistake by saying "large". I meant the small gaseous atmospheres surrounding neutron stars.
Sir Cumference
CC BY-SA 3.0
2016-05-24T13:20:40.910Z
As a source:  http://chandra.harvard.edu/press/09_releases/press_110409.html  Hydrogen and Helium fuse on the surface to make Carbon.  "Atmosphere" might be a bit vague, it's probably more of a dense, nearly solid plasma. . . . but I'm guessing.
userLTK
CC BY-SA 3.0
2016-05-24T18:30:14.420Z
So it seems *less* absurd that there should be a completely abrupt transition from a degenerate fluid to vacuum with nothing in between? No transition region at all? Really?
dmckee --- ex-moderator kitten
CC BY-SA 3.0
2016-05-24T19:02:59.077Z
@dmckee Considering the immense gravity of a neutron star, I simply didn't see how a gas could possibly surround it.
Sir Cumference
<star><gravity><neutron-star><stellar-atmospheres>
How can neutron stars have gaseous atmospheres?
Sir Cumference
# How can neutron stars have gaseous atmospheres?

(Asked by: username_0 on 2016-05-24)

Neutron stars can have small atmospheres. However, they also have extremely strong gravitationally pulls. Shouldn't the all the gas molecules be drawn to the star's surface, and become solids under the immense pressure?

Maybe I'm thinking about this the wrong way, but I don't see how it could be possible.

**Tags:** star, gravity, neutron-star, stellar-atmospheres
> **Comments:**
> **username_1**: 4 inch thick atmospheres.   :-)  
> **username_0**: @userLTK It still seems absurd that matter so close to the star will be gaseous.  
> **username_2**: What do you mean by large atmospheres? If you mean the magnetospheres, well the clue is in the name. Gravity is not the only force acting.  
> **username_0**: @RobJeffries Yeah, made a mistake by saying "large". I meant the small gaseous atmospheres surrounding neutron stars.  
> **username_1**: As a source:  http://chandra.harvard.edu/press/09_releases/press_110409.html  Hydrogen and Helium fuse on the surface to make Carbon.  "Atmosphere" might be a bit vague, it's probably more of a dense, nearly solid plasma. . . . but I'm guessing.  
> **username_3**: So it seems *less* absurd that there should be a completely abrupt transition from a degenerate fluid to vacuum with nothing in between? No transition region at all? Really?  
> **username_0**: @dmckee Considering the immense gravity of a neutron star, I simply didn't see how a gas could possibly surround it.  

---

## 1 answers

### Answer by username_2 on 2016-05-24. Score: 12 | Accepted answer

Gravity is only important insofar that it is capable of compressing the material to high densities. Whether that material is capable of solidifying depends on the competition between Coulombic potential energy and the thermal energy of the particles. The former increases with density, the latter increases with temperature. A dense plasma can still be a gas if it is hot enough.

A rough formula for the exponential scale height of the atmosphere is
$$ h = \frac{kT}{\mu m\_u g},$$
where $T$ is the temperature of the gas, $m\_u$ is an atomic mass unit, $\mu$ is the number of atomic mass units per particle and $g$ is the surface gravity, with $g = GM/R^2$.

For a typical neutron star with $R=10$ km, $M= 1.4M\_{\odot}$, we have $g=1.86\times 10^{12}$ m/s$^2$. The atmosphere could be a mixture of ionised helium ($\mu=4/3$) or perhaps iron ($\mu = 56/27$), so let's say $\mu=2$ for simplicity. The temperature at the surface of the neutron star will change with time; typically for a young pulsar, the surface temperature might be $10^{6}$ K.

This gives $h = 2$ mm.

Why is this not a "solid"? Because the thermal energy of the particles is larger than the coulombic binding energy in any solid lattice that the ions could make. That is not the case in the solid surface below the atmosphere because the density grows very rapidly (from $10^{6}$ kg/m$^{3}$ to more than $10^{10}$ kg/m$^{3}$ (where solidification takes place) only a few cm in, because the scale height is so small. Of course the temperature increases too, but not by more than a factor of about 100. After that, the density is high enough for electron degeneracy, and the material becomes approximately isothermal and at a small depth the "freezing temperature" falls below the isothermal temperature.

> **Comments:**
> **username_4**: I am confused about your usage of $\mu$ and $m_u$.  
> **username_2**: @imallet Atomic mass unit $m_u = 1.67\times 10^{-27}$ kg. $\mu$ - the number of mass units per particle. ionised Helium 3 particles, 4 mass units (whoops, I made a mistake).  
> **username_4**: Still confused. I'm parsing e.g. $56/27$ as "$56$ amu per $27$ atoms". Maybe, you can provide the source for the formula?  
> **username_5**: So in laysman terms... That stuff is too hot to hold as a solid or even a liquid. Cool.  
> **username_2**: @imallet per PARTICLE. Electrons plus nuclei.  
> **username_6**: Technically speaking, plasma is a distinct phase of matter from gas.  So we can apparently expect a thin, dense "plasma atmosphere", but is there a meaningful "gaseous atmosphere" a little further out?  
> **username_2**: @zibadawatimmy if you wish to go with that definition, then there is no gas. It's all ionised.  
> **username_7**: "*The atmosphere could be a mixture of ionised helium or perhaps iron...*" Aren't neutron stars made of neutronium?  
> **username_6**: @RBarryYoung The equation of state for a neutron star has yet to be determined, but there is expected to be a thin crust of normal nuclei, possibly iron (due to the high binding energy per nucleon) or helium (if heavier elements would "sink" deeper into the star).  You have to go a bit further in before nuclei cease to exist.  
> **username_2**: @RBarryYoung Neutronium is a made-up SciFi word. Neutron stars have crusts of neutron-rich nuclei accompanied by degenerate electrons. The outer cm or so is a non-degenerate gas of uncertain composition, but one thing it isn't is free neutrons.  
> **username_8**: Good answer to a hard question! While it was published in 1980, the hard Sci-Fi book [Dragon's Egg](https://en.wikipedia.org/wiki/Dragon%27s_Egg) by Robert L. Forward does help one *imagine* what the surface *might be like* - including neutron-rich nuclei on the surface as well as the atmosphere discussed here.  
> **username_4**: @RobJeffries . . . so the arithmetic is "$56$ amu / ($26$ electrons + $1$ nucleus)"?  
> **username_2**: @imallett Correct.  

---

