# How small would a neutron star be to see the entirety of it?

How small in Schwarzschild radii would a neutron star need to be for its gravity to be strong enough to deflect light emitted from one side toward an observer on the opposite side? I know the figure is above $$1.5R_s$$.

• do you mean Schwarzschild_radius ?en.wikipedia.org/wiki/Schwarzschild_radius . What do you mean by "see"? Commented Nov 28, 2021 at 18:16
• Commented Nov 28, 2021 at 18:21

For a Schwarzschild spacetime outside the neutron star (i.e. spherically symmetric and non-rotating), the neutron star surface would need to be at a radial coordinate $$\leq 1.76 r_s$$ (e.g. Pechenik et al. 1983), where $$r_s$$ is the Schwarzschild radius. This corresponds to an apparent radius at infinity of $$\leq 2.68r_s$$.
There are possible equations of state being considered for neutron stars that would allow them to be as small as this for large neutron star masses. This effect - that you can see more than $$2\pi r^2$$ of a neutron star surface - must be taken account of when interpreting the flux of radiation from a neutron star.
EDIT: I attach a plot from a script I've written to calculate the total deflection angle either as a function of the impact parameter of light ($$b$$) or the closest approach ($$r_{tp}$$). The plot shows that a total deflection angle of 180 degrees (corresponding to a 90 degree bend for light emitted tangentially from the centre of the opposite side of the neutron star) occurs for $$r_{tp}=1.76r_s$$ and for an impact parameter of $$2.68r_s$$. In other words, the apparent radius of the neutron star at infinity would be $$2.68r_s$$, with the centre of the rear-side of the neutron star forming the outermost ring of the apparent image.