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Is there a critical size that an hydrogen bomb detonation needs to have in order to produce neutron-degenerated matter?

Does anyone knows if matter in this state would be stable at ambient temperature or pressures? is there any way we can obtain stable "nuggets" of nuclear matter?

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2 Answers 2

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According to Wikipedia the pressure inside an atomic explosion maxes out around $10^{15}$Pa. By comparison the pressure in a neutron star is around $10^{34}$Pa. The 19 orders of magnitude difference is too large to be achievable just by making the bomb larger i.e. you will never be able to make neutronium in an atomic bomb.

As Ron explains in his answer, neutronium is unstable under normal conditions. If you extend your definition of degenrate matter to include strange matter then there have been suggestions that this could be stable. However I don't know of anyone who takes the prospect very seriously.

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It is not possible to make a gigantic neutron ball because the neutrons will beta decay to protons, and emit electrons (and neutrinos) and the protons will sit on the surface, and blow the nugget apart through their electrostatic repulsion. The only way to resist this fragmentation is to have a star-massed neutron cluster, so that the cluster is held together in the compact configuration by gravity. There is no other attractive force which can counteract the electrostatic repulsion.

There is a large gap in size between the largest stable nucleus and the smallest stable neutron star essnetially caused by the weakness of gravity. If you made a neutron ball, it wouldn't even be stable for 8 minutes, since the only reason the neutron is long-lived is that the proton and neutron mass are so close, but in nuclear matter empty of protons, a proton's binding energy to the surrounding neutrons reduce the lifetime to nothing. Heavy neutron rich nuclei tend to beta-decay much more quickly than a free neutron.

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isn't a wide held belief (for nuclear physicists anyway) that islands of stability of nuclear matter will grow with the logarithm of the net neutron number? how can you be so sure that the nuclear surface will be unstable? –  lurscher Jul 13 '12 at 3:00
beta decay of neutrons does not play a role in nuclear stability in general, so it should be even less important with a big chunk of matter with larger volume-to-area ratio –  lurscher Jul 13 '12 at 3:10
"beta decay is only observable in free neutrons" Er...simply not true. There are many isotopes which undergo beta decay. They don't do so on the free-neutron life-time because, as you say, nucleons in a nuclear context are different from their free counterparts but it is still beta decay. It is still weak mediated and it still results in the conversion of one nucleon to one of the other isospin. –  dmckee Jul 13 '12 at 3:20
beta decay of neutrons is only observable in free neutrons. Presumably, when neutrons are bound in a nucleus, there is no well-defined notion of neutron-proton identity so the eigenstates half-lifes of the nucleus as a whole cannot be explained away by the half-lifes of the eigenstates of the parts –  lurscher Jul 13 '12 at 3:41
@dmckee, i deleted the comment and re-added with your provided correction. thanks! –  lurscher Jul 13 '12 at 3:42

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