# Is the speed of sound almost as high as the speed of light in neutron stars?

Have you ever wondered about the elastic properties of neutron stars?

Such stars, being immensely dense, in which neutrons are bound together by the strong nuclear force on top of the strong gravity that “presses” them together, one would think they must have extremely large Young modulus, and the speed of sound could be on a par with the speed of light in the vacuum.

If we let $c_s$ be the speed of sound, and also assume that the neutron star is isotropic, then using the well known equation for the speed for acoustic waves in solids, we can write the following equation for the crust of the neutron star

$c_s=\sqrt{\frac{E}{\rho}}$

For a neutron star of density $\rho =5.9\times 10^{17}$ Kg m$^{-3}$ and Young modulus of about $E=5.3\times 10^{30}$Pa we get a value for $c_s=3.0\times 10^6$ ms$^{-1}$!

The Questions are:

1) How can sound travel at such immense speeds inside a neutron star?

2) Should nuclear interactions, n-n and q-q , dictate the elastic properties of a neutron star, or is it just the gravitational force?

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In the 1980's Robert Forward wrote some hard-science fiction where the story turned on these properties of a neutron star's crust. See en.wikipedia.org/wiki/Dragon%27s_Egg and en.wikipedia.org/wiki/Starquake_%28novel%29 –  mmesser314 Apr 27 at 0:52

The neutron star crust is a solid and there are indeed elastic waves for which the speed of sound is controlled by the shear modulus. I'm not sure where you got your estimate of the shear modulus from (there is some literature on the subject, see for example http://arxiv.org/abs/1104.0173).

Most of the neutron stat is a liquid, and the speed of sound is given by the usual hydrodynamic result $$c_s^2=\left(\frac{\partial P}{\partial\rho}\right)_{s}.$$ In dilute, weakly interacting neutron matter the speed of sound (in units of the speed of light $c$) is $$c_s^2 = \frac{1}{3}\frac{k_F}{\sqrt{k_F^2+m^2}}$$ where the Fermi momentum $k_F$ is determined by the density, $$n = \frac{k_F^3}{3\pi^2}.$$ You can see that in the high density (relativistic) limit the speed of sound approaches $c/\sqrt{3}$. In the center of a neutron star you get quite close to this. Interaction change this result by factors of order one, but as an order of magnitude estimate these simple results are quite good.

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The dominant supporting force (resisting collapse) comes from neutron degeneracy, so yes, it is the nuclear properties dominate. However, the idea of relating sound speed to Young's modulus comes from Hookian notions ($F\propto k x^2$). The equation of state for degenerate neutrons is not well known, so how exactly they might communicate pressure fluctuations is not well-defined. Since degenerate neutrons are essentially frictionless, they may actually transmit such fluctuations quite a bit faster than the $0.01 c$ you have suggested. (Again, I'm not sure if a Young's modulus is even a meaningful quantity within a neutron star; perhaps it would be in an iron shell around the neutronic core, but that doesn't seem to be what you are interested in). More information on the subject of neutron stars and equations of state is here.

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Thank you for the informative answer, and for the link. Yes, I agree the standard equation I have used is only a hand-waving argument, to get some estimate, at least the order of madnitude, of the speed of sound in such magnificent objects. –  JKL Feb 22 '13 at 2:11

Bad question- the speed of light should be rephrased as "Speed of electromagnetic radiation" and then specify what wavelength of radiation we're talking about and which part of the neutron star is being discussed. Nobody has determined the EM properties of the inside of neutron stars so the question is not on target.

Different parts of a neutron star have different properties - see http://en.wikipedia.org/wiki/Neutron_star

Visible light might even travel slower than sound waves at some points in the star -hmmm! NS.

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Pretty sure the question is just asking about the speed of sound, using the regular speed of light ($3\times 10^8 m/s$) as a reference speed. –  Javier May 23 at 13:42
Don't be rude to the OP. –  Jimmy360 May 26 at 19:33
Also, it is a very good question, although (evident from your answer) you did not understand it. –  Jimmy360 May 26 at 19:34