# What stabilizes neutorns against beta decay in a neutron star?

Free neutrons are known to undergo beta decay with a half-life of slightly above 10 minutes. Binding with other nucleons stabilizes the neutrons in an atomic nucleus, but only if the fraction of protons is high enough (at least a third or so). But what keeps a neutron star stable against beta decay? Apparently, this is extra pressure due to gravity in contrast to "negative pressure" of proton Coulomb repulsion in a nucleus but how do we know that this is enough to stabilize the degenerate neutronic fluid?

I am aware of a closely related question but not really happy with the answers there. There is lot of dazzling details here, but I am a looking for an answer suitable form a 8-year old with enhanced curiosity towards astrophysics.

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Why were you "not really happy with the answers" to the analogous question for nuclei? Lagerbaer's answer is correct, and it's also the correct general answer for a neutron star. There simply isn't a lower-energy state to decay to. A neutron star is just a big nucleus, albeit one that's big enough for gravity to play an additional stabilizing role. –  Ben Crowell May 5 '13 at 15:20
@Ben Crowell I like dmckee's answer below much better because it identifies the key difference: there no force to keep electrons in a nucleus in contrast to gravity-"degenerated" electron gas component in a neutron star. Astrophysics is neat! –  Slaviks May 5 '13 at 15:39

Conservation of energy and the electron-degenerate pressure.

For the neutron to decay you must have $$n \to p + e^- + \bar{\nu}$$ or $$n + \nu \to p + e^- \quad.$$

In either case that electron is going to stay around, but in addition to the neutrons being in a degenerate gas, the few remaining electrons are also degenerate, which means that adding a new one requires giving it momentum above the Fermi surface and the energy is not available.

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So, you are saying that the chemical potential of electrons rises so quickly with addition of yet another electron that it is enough to keep just a bit of them around to preclude beta decay? And, unlike others, you emphasize electrons rather than protons because e are so much lighter? –  Slaviks May 5 '13 at 15:34
That is a very reasonable way to view it. Though as Ben alluded in his comment Noldig's answer, the situation is setup starting with a electron degenerate gas and converting protons to neutrons as long as you get energy back taking electrons (with high chemical potential) out of the picture. At equilibrium both the electrons and the neutrons are degenerate and you run into a energy barrier going either way. –  dmckee May 5 '13 at 15:40
This really nails it! Thanks to you both for a quick and decisive resolution. –  Slaviks May 5 '13 at 15:42
@Slaviks I don't want you to go away with the impression that neutrons are unchanging in neutron stars. In fact, neutrons are transforming all the time via the (direct and modified) Urca process (see also here), and the subsequent release of neutrinos is one of the main channels for neutron stars to cool off. The neutron/proton ratio is very much set by thermodynamics, not kinetics. –  Chris White Jul 19 '13 at 4:15
There is Beta decay in neutron stars. This is the simple answer. Since a neutron star is electrical neutral, there is the same amount of $\beta^+$ as $\beta^-$ decay, this is called the chemical equilibrium.