What is the estimated size of neutron stars observed in their reference frame and in our reference frame?

That is, how bent is space-time around neutron stars?

  • 2
    $\begingroup$ This is actually a quite non-trivial question to answer, due to the so-called Ehrenfest paradox regarding Lorentz contraction of a rotating body, but since the involved velocities are, after all, small compared to the speed of light (at most ~10%), the difference which involves the factor $\sqrt{1-v^2/c^2}$ will be on the the <1% scale. $\endgroup$
    – pela
    Nov 1 '19 at 12:42

If we ignore the rotation issue (as Pela pointed out, the correction is <1%), then we can approximate the spacetime curvature inside the neutron star using the interior Schwarzschild metric and use the exterior one for the outside. The $r$ coordinate is not the radial distance as one would expect, but defined in terms of $r=$ constant circles having circumference $2\pi r$.

A neutron star of radius $R$ in these metrics has a circumference measured on its surface as $2\pi R$ as expected. But the distance to the core (that is, the length of a hole from the surface to the core) is: $$d(R) = \int_0^{R} \frac{dr}{1-Kr^2} = \frac{1}{\sqrt{K}}\tanh^{-1}\left(\sqrt{K}R\right).$$ where $K=r_s/R^3$ and $r_s$ is the Schwarzschild radius of the star (all of this assumes constant density in the interior). If we plot this for stars of different $R$ (but same mass, say one solar mass, producing a constant $r_s$) we get the following:

enter image description here

As the neutron star approaches becoming a black hole it gets "deeper": there is more volume than one would expect. Less dense neutron stars have depths that scale linearly with their radius... except that it is $r_s/3$ larger! This odd result comes from the assumption that we look at stars of the same mass even though we make them much larger. A big constant mass object will curve spacetime too, and this produces this effect. (Yes, this means that the sun is $r_s/3 = 984.73$ meters deeper than it looks!)

If we instead decide that we use a constant density (say nuclear density) and plot the depth, the collapse to a black hole instead happens to the right as the neutronium sphere becomes too large. Here I still use the solar Schwarzschild radius as a length scale to keep things comparable. Now, for small spheres the coordinate and measured depths converge:

enter image description here

In actual neutron stars things are complicated by rotation, the core pressure diverging as one approaches $(9/8)r_s$, and of course that it is hard to dig a hole through superfluid neutrons.


The inferred radius by a distant observer is given by $$ R_{\infty} = \frac{R}{\sqrt{1- R_s/R}},$$ where $R_s = 2GM/c^2$, $M$ is the gravitational mass and $R$ is the radius measured at the surface.

The fact that $R_{\infty} > R$ is because an observer can see more than 50 per cent of the neutron star surface. See https://physics.stackexchange.com/questions/350805/seeing-something-from-only-one-angle-means-you-have-only-seen-what-of-its-su/350814#350814


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