# Why do neutron stars with more mass have smaller volume?

I know about Heisenberg uncertainty which makes more localized neutrons have a wider range of undefined momentum, and Pauli exclusion principle which prohibit neutrons from getting too close or "occupying the same quantum state" so as to say. But they only explain how neutron stars don't increase in volume while increasing in mass, yet it doesn't (at least from my understanding) explain how it gets smaller. I understand this degeneracy pressure only accounts for part of the opposing forces inside the neutron stars, with the others being a variety of stuff including strong force repulsion. I also know there are equations for calculating this, but I want to know if there's a more intuitive way of understanding how this phenomenon is created by these rules and forces or if there is a good explanation as to how the equations was formed and what they meant

• Related (the same concept as in David's answer): physics.stackexchange.com/q/366468 – safesphere Feb 11 at 3:55
• Note that what you're asking about would apply to (hypothetical) neutron stars governed by neutron degeneracy pressure. But since they're not, real neutron stars have a harder equation of state and are likely about the same size regardless of mass. – Rob Jeffries Feb 11 at 7:28

Try this argument.

To be in hydrostatic equilibrium, the pressure gradient inside a star must equal (minus) the density multiplied by the gravitational field $$\frac{dP}{dr} = - \rho g$$

If we take the average pressure gradient to be $$-P_c/R$$, where $$R$$ is the stellar radius and $$P_c$$ is the central pressure, then this will roughly be equal to average density multiplied by the average gravitational field. So in proportionality terms $$\frac{P_c}{R} \sim \frac{M}{R^3}\frac{M}{R^2}\, ,$$ where $$M$$ is the stellar mass and thus $$P_c \sim M^2 R^{-4} \tag*{(1)}$$

The relationship between mass and radius will depend on what provides the pressure.

If it is perfect gas pressure (in a non-degenerate star), then $$P_c \propto \rho T \sim MT/R^3$$. But in a main sequence star, the core temperature is roughly fixed, because hydrogen burning has a strong temperature dependence and hardly changes as the mass changes. Thus we have from eqn (1): $$P_c \sim MR^{-3} \propto M^2 R^{-4}$$ $$\rightarrow \ \ \ \ R \propto M$$

Now consider (non-relativistic) degeneracy pressure. This scales as $$\rho^{5/3}$$ and is independent of temperature. Thus $$P_c \propto M^{5/3} R^{-5}$$. Putting this into eqn (1): $$P_c \sim M^{5/3} R^{-5} \propto M^2 R^{-4}$$ $$\rightarrow \ \ \ \ R \propto M^{-1/3}$$

Note that real neutron stars are not governed by the ideal equation of state for degenerate neutrons. Neutrons in fact are strongly intercting particles when compressed to separations of $$\sim 10^{-15}$$ m. The interaction is repulsive and leads to a "hardening" of the equation of state, such that $$P_c \sim \rho^2 \propto M^2 R^{-6}$$. If we put this into equation (1) we find that there is only one value of $$R$$ that will satisfy the equation. i.e. That the radius does not depend on mass. If you look at many examples of theoretical mass-radius relations for neutron stars you will see that there usually is a range of masses for which the radius is nearly constant.

• may I ask where does rho^(5/3) come from in degeneracy? – never took courses but why Feb 11 at 17:47
• @nevertookcoursesbutwhy The Fermi momentum of a degenerate gas is proportional the number density $n^{1/3}$. The kinetic energy density is the number density, multiplied by the average kinetic energy per particle $\sim p_f^2/2m$. Thus kinetic energy density is proportional to $n^1 n^{2/3}$ and hence to mass density $\rho^{5/3}$. For any non-relativistic gas, pressure is just 2/3 of the kinetic energy density. Hence $P \propto \rho^{5/3}$. – Rob Jeffries Feb 11 at 20:02

The key point is that the Heisenberg principle restricts the number of particles in a (hyper)-volume of phase space rather than a volume of plain 'ole physical space.

That is, there can be at most two neutrons in a (hyper-)box defined by a volume in physical space times a volume in momentum space.

But the kinetic energy of a particle depends on it's momentum.

Neutron stars stop collapsing when, in order to fit all the particles into less physical space you would have to give some of them enough more momentum that the change costs more added kinetic energy that you get back in gravitational energy.

When you add more mass the amount of gravitational energy you get from a particular change in radius increases, so you can afford to move neutrons to higher momentum in order to fit them into less physical space.

You can actually see a similar phenomena at work in the sizes of some atoms: higher Z atoms (with more electrons) can take up less space than lower Z atoms (with fewer). Because more protons in the nucleus means you get more energy from moving the electrons inward and can therefore afford to have them at higher momentum.

• ok so if I interpreted it correctly, so u add mass. that increased gravitational energy, some of which is then consumed to give neutrons higher KE or wtv. Basically gravitational energy is increased then decreased Then how do I know what the net change of gravitational energy will be? – never took courses but why Feb 11 at 17:54
• For that you have to calculate. Rob's answer is the starting point. – dmckee Feb 11 at 18:14

More mass ==> stronger gravity ==> more force pushing the star inwards ==> smaller volume before the process stabilizes.

This is why stars with huge amounts of mass can end up as tiny black holes!

• But the first three points apply to normal stars too. So you haven't explained anything. – Rob Jeffries Feb 11 at 7:24
• An explanation for a superset doesn't apply to a subset within it? – Inertial Ignorance Feb 11 at 15:34
• no but then u failed to distinguish this case from the others which produced different results. Essentially I'm asking providing everything u said applies to all stars, then why does only neutron star have such a peculiar effect while eg white dwarf produce the complete opposite effect which is increase in size? – never took courses but why Feb 11 at 18:16
• If I recall correctly, don't white dwarfs have a rebound effect after shrinking? – Inertial Ignorance Feb 11 at 18:19