14 Jul '20 02:41

This is a back of an envelope calculation and is out by about 3 orders of magnitude, according to the estimate given in the introduction of [1] which is based on a far more sophisticated calculation.

Given the mass of the neutron, proton and electron, and assuming that the mass of neutrinos is zero (which is fine for this) the amount of energy liberated by the decay of a free neutron into a proton, electron and anti-neutrino is 0.782326 MeV (the error caused by the mass of the neutrino at most affects the last digit given). The gravitational binding energy of a neutron at the edge of a neutron star, in the Newtonian approximation is:

E = -GMmₙ/r = -GAmₙ²/r

Where G is Newton's constant, A is the number of neutrons, mₙ the mass of a neutron and r the radius of the neutron star.

A = 4/3 π n r³

n is the number density of nuclear matter which we take to be 1.60E+44 [2]. Substituting this into the gravitational energy equation we have:

E = D M ^ (2/3)

where D = G cuberoot(4/3 π n mₙ⁴) and M is the mass of the neutron star (= A mₙ). This gives an estimate for the mass required to supress beta decay gravitationally. Assuming I haven't goofed up on the spreadsheet the figure is 0.000325 solar masses or 108 times the mass of the Earth. Reference [1] gives the figure of 0.189 solar masses, so I'm out by a barn mile, but it's not a very sophisticated calculation.

What this means is that the object refered to in the OP cannot be made of neutron matter as it can't have a mass much in excess of 5 solar masses.

[1] https://arxiv.org/pdf/astro-ph/9707230.pdf

[2] https://en.wikipedia.org/wiki/Nuclear_density

Given the mass of the neutron, proton and electron, and assuming that the mass of neutrinos is zero (which is fine for this) the amount of energy liberated by the decay of a free neutron into a proton, electron and anti-neutrino is 0.782326 MeV (the error caused by the mass of the neutrino at most affects the last digit given). The gravitational binding energy of a neutron at the edge of a neutron star, in the Newtonian approximation is:

E = -GMmₙ/r = -GAmₙ²/r

Where G is Newton's constant, A is the number of neutrons, mₙ the mass of a neutron and r the radius of the neutron star.

A = 4/3 π n r³

n is the number density of nuclear matter which we take to be 1.60E+44 [2]. Substituting this into the gravitational energy equation we have:

E = D M ^ (2/3)

where D = G cuberoot(4/3 π n mₙ⁴) and M is the mass of the neutron star (= A mₙ). This gives an estimate for the mass required to supress beta decay gravitationally. Assuming I haven't goofed up on the spreadsheet the figure is 0.000325 solar masses or 108 times the mass of the Earth. Reference [1] gives the figure of 0.189 solar masses, so I'm out by a barn mile, but it's not a very sophisticated calculation.

What this means is that the object refered to in the OP cannot be made of neutron matter as it can't have a mass much in excess of 5 solar masses.

[1] https://arxiv.org/pdf/astro-ph/9707230.pdf

[2] https://en.wikipedia.org/wiki/Nuclear_density