quantum mechanics - 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

Answer

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.