# Where does the 10 km of neutrons stars comes from? [duplicate]

Neutrons stars are known to have a radius of about 10 km to 15 km. Where that size comes from ? How to theoretically derive that number, using Newton's theory of gravity ?

I know the magnetic flux conservation : $\Phi = \pi R_0^2 \, B_0 \approx \pi R^2 \, B$, but this requires that we know the initial and final magnetic field at the surface of the star, and the star's size before the supernova. So this isn't satisfying.

Conservation of angular momentum alone isn't telling the radius neither : $S = \frac{2}{5} \, M R_0^2 \, \omega_0 = \frac{2}{5} \, M R^2 \, \omega$, since this requires that we know the star size and angular velocity before the supernova.

Using Newton's theory of gravitation and conservation of energy (or another method ?), how can we derive the theoretical size of NS ? The only input number that I could accept in the derivation is the NS angular velocity $\omega$ (and mass $M \approx 1.44 \, M_{\odot}$), since this can be found from conservation of angular momentum and simple models of supernova (if we already know the NS radius !).

Currently, the only rough argument that I know is the following : assuming an uniform sphere rotating at its maximal value to support gravity, we should have a balance relation like this : $$\tag{1} F_{\text{centripetal}} \approx M \, \omega^2 R = F_{\text{grav}} \approx \frac{G M^2}{R^2}.$$ Isolating $R$ gives $$\tag{2} R \approx \Big( \frac{G M}{\omega^2} \Big)^{\frac{1}{3}}.$$ Inserting mass $M \approx 1.44 M_{\odot}$ and period $T \approx 1 \text{ms}$ give something interesting : $$\tag{3} R_{\text{NS}} \approx 16.9 \text{km}.$$ I strongly suspect this could be improved or made more rigorous (even if it's less precise), using the conservation of energy.

EDIT : Another argument comes from the density. A neutron star has a density comparable to a nucleus or a neutron, so $$\tag{4} \rho = \frac{3 M}{4 \pi R^3} \approx \rho_{\text{neutron}} = \frac{3 m_N}{4 \pi r_N^3}.$$ This gives the star's radius (using $M \approx 1.44 \, M_{\odot}$ and $r_N \approx 10^{-15} \, \mathrm{m}$) : $$\tag{5} R \approx \Big( \frac{M}{m_N} \Big)^{\frac{1}{3}} \, r_N \approx 12 \mathrm{km}.$$

## marked as duplicate by John Rennie, peterh, Kyle Kanos, Void, David HammenJun 17 '17 at 15:29

• Hi Cham. The answer to the question I've linked cover this in some detail. Also see this search for more related questions. The size is determined by the phase change to degenerate matter not the properties of the original plasma in the collapsing star. – John Rennie Jun 17 '17 at 5:21
• @JohnRennie, all these answers are more about the mass than the radius itself. The Chandrasekhar limit gives the mass, not the radius. I'm accepting mass and angular velocity as known inputs, then would like to deduce the radius from classical mechancis. – Cham Jun 17 '17 at 10:48
• Hi Cham. You cannot deduce the radius from classical mechanics because you need to consider the phase change from normal matter to degenerate matter. It's like trying to understand the behaviour of a compressed gas without taking into account liquifaction under pressure. – John Rennie Jun 17 '17 at 10:51
• @JohnRennie, I think we can deduce it if we accept a few simple inputs, like the mass and the angular velocity. – Cham Jun 17 '17 at 10:57
• Hi Cham. The phase change to degenerate matter is an essential part of the neutron star formation. If you ignore this phase change you are doing a calculation that doesn't describe what is actually happening. The number you get will be physically meaningless. – John Rennie Jun 17 '17 at 11:32

Using Newton's theory of gravitation and conservation of energy

Neutrons are quantum mechanical entities, as are also electrons and protons so Newton and angular momentum conservation cannot take you too the limits where the degeneracy of fermions stops collapses:

In any case, the basic idea is that when the central part of the star fuses its way to iron, it can't go any farther because at low pressures iron 56 has the highest binding energy per nucleon of any element, so fusion or fission of iron 56 requires an energy input. Thus, the iron core just accumulates until it gets to about 1.4 solar masses (the "Chandrasekhar mass"), at which point the electron degeneracy pressure that had been supporting it against gravity gives up the ghost and collapses inward.

At the very high pressures involved in this collapse, it is energetically favorable to combine protons and electrons to form neutrons plus neutrinos. The neutrinos escape after scattering a bit and helping the supernova happen, and the neutrons settle down to become a neutron star, with neutron degeneracy managing to oppose gravity.

Italics mine

Electron degeneracy is a stellar application of the Pauli Exclusion Principle, as is neutron degeneracy. No two electrons can occupy identical states, even under the pressure of a collapsing star of several solar masses.

.....

Above 1.44 solar masses, enough energy is available from the gravitational collapse to force the combination of electrons and protons to form neutrons. As the star contracts further, all the lowest neutron energy levels are filled and the neutrons are forced into higher and higher energy levels, filling the lowest unoccupied energy levels. This creates an effective pressure which prevents further gravitational collapse, forming a neutron star. However, for masses greater than 2 to 3 solar masses, even neutron degeneracy can't prevent further collapse and it continues toward the black hole state.

All these are models that fit observations using not only newtonian mechanics but equally quantum mechanics.

The actual modelling of neutron star formation and properties are a matter of ongoing research, for example:

In this work, we determine the neutron star mass-radius relation and, based on recent observations of both transiently accreting and bursting sources, we show that the radius of a 1.4 solar mass neutron star lies between 10.4 and 12.9 km,