I am currently trying to understand the history of the development of the equations of state and structure of neutron stars.

In my textbook, I frequently encounter phrases such as "The neutronization at $\rho \approx \rho_{\text{ND}}$ greatly softens the EOS, but at the crust bottom the repulsive short-range component of the neutron-neutron interaction comes into play and introduces a considerable stiffness.

Another section states:

The model EOSs can be subdivided into the soft, moderate, and stiff ones with respect to the compressibility of the matter.

What is meant by a soft or stiff EOS?

  • 2
    $\begingroup$ I don't know in the context of astrophysics, but I would interpret stiff to mean "Small changes in X result in large changes in Y." $\endgroup$
    – tpg2114
    Commented Mar 22, 2015 at 2:04

1 Answer 1


In the degenerate interiors of neutron stars, the equation of state is usually just density (and composition) dependent. You can express the pressure as a polytropic law of the form $P \propto \rho^\alpha$, where $\rho$ is the density.

A stiff (or hard) equation of state is one where the pressure increases a lot for a given increase in density. Such a material would be harder to compress and offers more support against gravity. Conversely, a soft equation of state produces a smaller increase of pressure for a change in density and is easy to compress.

i.e. if $\alpha$ is large, we have a hard equation of state. The (blurred) dividing line between the two is around $\alpha=5/3$. Anything below this is soft (in neutron star work).

In neutron stars, the hardness of the equation of state controls the mass-radius relationship and maximum possible mass of neutron stars. A harder equation of state gives a bigger radius for the same mass and a larger possible maximum neutron star mass.

The outer crust of a neutron star is, ironically, governed by the soft equation of state of ultra-relativistic degenerate electrons (with non-degenerate ions contributing most of the density, $\alpha=4/3$). Actually, $\alpha <4/3$ because, as your text suggests, neutronisation removes free electrons with increasing density and softens the equation of state even further. A stable star cannot be built out of such material. The reason neutron stars exist is that neutrons drip out of nuclei at a few $10^{14}$ kg/m$^3$ and form a non-relativistic degenerate gas with $\alpha=5/3$. However, this on its own will not support the neutron stars we observe, because it is not hard enough. An ideal degenerate neutron gas can only support a neutron star up to $0.7M_{\odot}$. It is a common misconception that neutron stars are supported by neutron degeneracy pressure - they are not; they are supported by the repulsive core of the strong nuclear force, which produces an equation of state with $\alpha > 5/3$ at densities greater than that of nuclear matter. However, at very high densities ($\sim 10^{18}$ kg/m$^3$) it is likely that neutrons transform into new hadronic or bosonic phases, or even into quark matter. A gas consisting of only ideal, ultra-relativistic fermions/quarks would have $\alpha=1$ ($P = \rho c/3$). All of these will have softer equations of state and probably define the upper limit to neutron star masses.


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