# How much energy is released per unit mass from depressurizing degenerate matter?

A neutron star remnant consists mostly of neutron degenerate matter. If you happened to suddenly have 1 kg of it in your lap without the pressure necessary to keep it degenerate, I suppose it would "undegenerate" and release alot of energy. How much energy and/or effect compared to, say, 1 kg of hydrogen fusing in the Sun or a bomb? And for the different kinds of degenerate matter such as electron degeneracy, proton degeneracy, quark degeneracy (and maybe more), what would the relative proportions of energy released from them being suddenly depressurized or becoming "undegenerate". How much more powerful would one be compared to another.

• Hi, imo, too many questions, and And what are the proportions for the different kinds of degenerate matter such as electron degeneracy, proton degeneracy, quark degeneracy (and maybe more) I don't know what you mean that that. Could you split them up? There are probably related duplicates for your first question on this site. – user81619 Jul 31 '15 at 11:05
• – userLTK Jul 31 '15 at 11:57
• I also found this to be a fun read. io9.com/5805244/… With Neutron decay, 0.08% of the mass is converted into energy. Hydrogen to Helium, 0.7% so ignoring the kinetic energy, 1 KG of Neutrons would be hotter than you'd want to be around. 1/9th the power of hydrogen fusion into helium. – userLTK Jul 31 '15 at 13:04
• @userLTK "so ignoring the kinetic energy..." - but that's where almost all the energy is. The energy released by beta decay is negligible in comparison. – Rob Jeffries Jul 31 '15 at 15:14
• @Rob Jeffries, I know. I just thought I'd mention that as a sidebar. – userLTK Jul 31 '15 at 19:45

The amount of energy per cubic metre in a degenerate gas depends on the density and composition of that material. I can only give some examples - any specific mixtures or densities would require individual calculation.

The energy density of an ideal, completely degenerate gas of fermions is given by $$u = \frac{\pi m^4 c^5}{h^3} \left[ x(1+2x^2)(1+x^2)^{1/2} - \sinh^{-1}(x) \right],$$ where $m$ is the fermion rest mass, $x = p_F/mc$ and $p_F$ is the Fermi momentum given by $$p_{F} = \left(\frac{3h^{3}}{8\pi}\right)^{\frac{1}{3}}\, n^{\frac{1}{3}},$$ and $n$ is the fermion number density. Note that this energy density includes the rest mass energy of the fermions.

Two useful approximations are those for non-relativistic and ultra-relativistic fermion gases. The kinetic energy density of a non-relativistic gas is $$u_{KNR} = \frac{3p_F^{2}}{10m} n$$ whereas for an ultrarelativistic gas, the kinetic energy density is $$u_{KUR} = \frac{3}{4} p_{F} c n$$

To turn this into an energy per kg you simply divide by the mass density. In the non-relativistic case $\rho = nm$; for the ultrarelativistic case $\rho = u_{KUR}/c^2$.

So let's get the easy one out of the way. If you have a gas of ultra-relativistically degenerate Fermions, they are all travelling at (just less than) $c$ and the energy per kilogram is $c^2$. It is unlikely that neutron star interiors get dense enough to make this happen. In a typical neutron star interior with say $\rho = 10^{18}$ kg/m$^{3}$, the relativity parameter $x \simeq 0.5$, so the neutrons are only mildly relativistic.

The non-relativistic approximation gives a kinetic energy density of $8\times 10^{15}$ J/kg. This then is a little on the low side as the neutrons are mildly relativistic.

If you can be bothered to put the numbers into the correct formula above you would get a little more than this (NB: This is not a simple calculation because you have to work out the right number density of neutrons and $n \ne \rho/m$ in a mildly relativistic gas). This is larger than I calculated here, because I've used a larger density.

How much energy is this compared with "hydrogen fusing in the Sun" - difficult to know what you mean. Hydrogen fusion releases 0.7% of the rest mass as energy - so if we assume all the Hydrogen in the core fuses into Helium, this would release $6 \times 10^{14}$ J/kg - but spread over 10 billion years. A more equivalent calculation is to ask what is the energy density of the gas in the Sun's core. It is more-or-less an ideal gas at a temperature of $1.5\times10^{7}$K. The energy per particle is $3kT/2$, the mean number of mass units per particle (ions and electrons) is about 0.6, so the kinetic energy per unit mass is $3\times 10^{11}$ J/kg.

A typical yield for a nuclear weapon can be as high as 5 kilotonnes of TNT per kg which is about $2\times 10^{13}$ J/kg.

Thus, even when judged by energy per unit mass (as opposed to per unit volume) degenerate neutron star material would be $\sim 100-1000$ times more explosive than a nuclear weapon and have more than 10 times the specific energy of hydrogen in the core of the Sun.

I'll deal with one more example - electron degeneracy. Degenerate charged fermions do not exist in isolation (actually there are a small fraction of protons, electrons and maybe muons in neutron star material too). For example a massive white dwarf could have an interior density as high as $10^{13}$ kg/m$^3$, but this is mostly made up of non-degenerate carbon ions. The degenerate electrons will be ultra-relativistic and have a number density of $n_e = 3\times 10^{39}$ m$^{-3}$. The kinetic energy density of the ions can be neglected; the energy density of the electrons is given by the ultra-relativistic approximation and is $3\times 10^{27}$ J/m$^3$. Hence the energy per kg of this material is $3\times10^{14}$ J/kg - so somewhere in between neutron-star material and a nuclear weapon and similar to the amount of energy per kg released by fusion in the Sun over its whole lifetime.

If Rob Jeffries numbers are right (Here) and it seems as good a baseline as any, you get about 3*10^15th Joules per KG of Neutron Star matter. The Hiroshima bomb was about 6*10^13th Joules, so just 1 KG of degenerate matter, you'd be looking at about 50 Hiroshima bombs released in kinetic energy, in the form of (I think) mostly Neutrons.

1 KG of hydrogen fuel for fusion, a little googling tells me would be about 6.3x10^14 Joules. source

So, nearly 5 times more energy than 1 KG of hydrogen fusion into Helium, and 1 KG of Neutron Star material would be so small you couldn't see it.

As for the other questions . . . . not my strong suit, but the first one was easy to extrapolate from the answer already given.

• An invisible amount of neutron degeneracy is like a standard sized fusion bomb in today's arsenals. Then I understand why astrophysicists are so interested in stellar mass neutron stars colliding with each other. And they can't hide, so observers must love them too. – LocalFluff Jul 31 '15 at 12:48