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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.

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  • $\begingroup$ 4 inch thick atmospheres. :-) $\endgroup$ – userLTK May 24 '16 at 13:15
  • $\begingroup$ @userLTK It still seems absurd that matter so close to the star will be gaseous. $\endgroup$ – Sir Cumference May 24 '16 at 13:15
  • $\begingroup$ 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. $\endgroup$ – Rob Jeffries May 24 '16 at 13:16
  • $\begingroup$ @RobJeffries Yeah, made a mistake by saying "large". I meant the small gaseous atmospheres surrounding neutron stars. $\endgroup$ – Sir Cumference May 24 '16 at 13:16
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    $\begingroup$ As a source: 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. $\endgroup$ – userLTK May 24 '16 at 13:20
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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.

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  • $\begingroup$ I am confused about your usage of $\mu$ and $m_u$. $\endgroup$ – imallett May 24 '16 at 15:55
  • $\begingroup$ @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). $\endgroup$ – Rob Jeffries May 24 '16 at 15:59
  • $\begingroup$ Still confused. I'm parsing e.g. $56/27$ as "$56$ amu per $27$ atoms". Maybe, you can provide the source for the formula? $\endgroup$ – imallett May 24 '16 at 16:14
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    $\begingroup$ So in laysman terms... That stuff is too hot to hold as a solid or even a liquid. Cool. $\endgroup$ – Renan May 24 '16 at 16:51
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    $\begingroup$ @zibadawatimmy if you wish to go with that definition, then there is no gas. It's all ionised. $\endgroup$ – Rob Jeffries May 24 '16 at 19:46

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