# Neutron star size in different reference frames

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?

• 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. – 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:

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:

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