When stars go supernova, a neutron star may be left over if there isn't enough mass to create a black hole. Based on the mass present, will a neutron star have a varying density or is the radius simply larger for heavier neutron stars?

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    $\begingroup$ The central density will be a function of the star's radius, but it cannot depend on the original composition of the star that shrunk to a neutron star as it aged. The ratio of neutrons, protons, and electrons in neutron star matter at low temperature is universal, fully determined by principles of chemical equilibrium. $\endgroup$ Apr 15, 2019 at 11:19

1 Answer 1


First of all, let me state in mathematical terms the the assumptions I think you are considering, based on the content of your question.

(a) To relate the mass and radius of the star with density, you employ

$$ M_{\text{star}} = \frac{4}{3}\pi\times \rho\times R^{3}_{\text{star}}\ \ \ \ \ \ \ \ \ \ (1)$$

(b) $M_{\text{star}}$ is assumed as fixed.

Let me also introduce the following considerations:

  1. Eq.(1) should be considered as an $\textit{approximate}$ relation among the mass and radius of the star and its $\textit{average}$ density $\langle \rho\rangle$. Moreover, as mass and radius can be inferred from observations this expression is more useful to constraint $\langle \rho\rangle$ than the radius itself. Although $\langle\rho\rangle$ does not necessarily agrees with the central density of the object, it is good for a quick order-of-magnitude estimation, so I suggest to consider

$$ M_{\text{star}} \approx \frac{4}{3}\pi\times \rho_{\text{central}}\times R^{3}_{\text{star}}\ \ \ \ \ \ \ \ \ \ (2)$$

  1. The structure of any standard neutron star is determined by:

$$ (\rho_{\text{center}}, P_{\text{center}}) + \text{Equations of structure} + \text{Condition for radius} \Rightarrow (M_{\text{star}}, R_{\text{star}})\ .$$

where $R_{\text{star}}$ is defined as the point where pressure vanishes, and the equations of structure come from General Relativity. We also expect density, pressure and composition, at all points inside the star, to be related by an Equation of State (EOS) $P(\rho)$, and $\textit{different central densities leading to different}\ M_{\text{star}}, R_{\text{star}}$.

  1. A priori, we postulate $\textit{all}$ neutron stars will obey the same EOS. Currently there exist many different candidates to be the "correct" EOS, but for simplicity in the argument we shall consider it as unique.

Now, the answer. Since $4\pi/3$ is a constant, notice that by fixing $M_{\text{star}}$ in Eq.(2), you still have two free parameters. Without more information, in principle there exists many combinations of numbers whose product is equal to this fixed value you impose. Consequently, $\textit{a priori}$ it is not valid to assume massive stars will occupy a large volume, since we can also set $R_{\text{star}}$ to be small and $\rho_{\text{central}}$ to be the actual big value from the two, and yet we would obtain the same $M_{\text{star}}$.

But, from considerations 1-3 we deduce that if $M_{\text{star}}$ is fixed, and we know which central density $\rho_{\text{central}}$ produces exactly this value, then $R_{\text{star}}$ is not free anymore and is uniquely determined. Therefore, we can affirm that once the EOS is known there exist unique values for $M_{\text{star}}$, $\rho_{\text{central}}$ and $R_{\text{star}}$.

As a final comment: contrary to your expectation, from a theoretical point of view, and from what is inferred from observations, it has been found the more massive a neutron star, the shorter its radius is. This implies that, in Eq.(2), it is $\rho_{\text{central}}$ and not $R_{\text{ast}}$ the big term for massive stars. I suggest to see, for instance, Figs. 4 and 7 from


or the figures from the more recent


  • $\begingroup$ Does general relativity enter into this or can we still approximate in a flat spacetime? It seems to me that the surface gravity of a neutron star should require general relativity to deal with questions of geometry. Maybe a better measure than radius would be surface area, then? Is my intuition wrong? $\endgroup$ Jun 6, 2023 at 18:47

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