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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
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3 Answers 3

up vote 9 down vote accepted

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
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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
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This really nails it! Thanks to you both for a quick and decisive resolution. –  Slaviks May 5 '13 at 15:42
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@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
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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.

This means, every time when a neutron decays, a proton captures (in average) an electron and the star stays stable.

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This doesn't sound right to me. Equilibrium just guarantees that the concentrations stay constant. It doesn't guarantee that they have a certain value. The equilibrium concentration of neutrons in a white dwarf is zero, so you need to explain why the equilibrium concentration would be zero in one case and 100% in the other. –  Ben Crowell May 5 '13 at 15:23
    
@BenCrowell Who says it's 100%? It's just tilted towards neutrons. –  Loren Pechtel May 5 '13 at 19:26
    
I ment "there is beta decay in neutron!! stars", I'm sorry for that. It was the answer to the question " what prevents neutrons from decaying in neutron stars" and my answer is "nothing". There is neutron i.e. beta decay and it is in equilibrium with electron capture. Am I wrong here at this point? –  Noldig May 5 '13 at 21:48
    
I agree this doesn't fully explain the value of the equilibrium concentration, but +1 for emphasizing the equilibrium nature of neutron stars. There is nothing magic preventing them from decaying, and indeed they will go back and forth between neutron and proton states. –  Chris White Jul 19 '13 at 4:05
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In a neutron star there are mostly "free" neutrons and the question then is why they don't all beta decay into electrons and protons?

Well, some of them do, but the point is that when the electron (or proton, there are equal numbers of each) numbers build up then they become degenerate (meaning no more than two electrons can occupy the same energy state and all energy states are filled up to a "Fermi energy" which increases with electron density) and their Fermi-energies increase. At some threshold number density, their Fermi energies will exceed the maximum energy of the particles that can be produced by beta-decaying neutrons. At that point beta decay pretty much stops because there are no available states that can be filled by the decay electron/proton and an equilibrium is set up between occasional beta decays and inverse beta decays such that the Fermi energies of the species are related by

$$ E_{F,n} = E_{F,p} + E_{f,e}$$

It isn't the case that this is just an equilibrium condition where half the neutrons in a neutron star will decay in 10 mins but be replaced by inverse beta decay at the same rate. The beta decay and inverse beta decay reactions are heavily suppressed (at least when the neutron to proton ratio is >8) because it is not possible (in degenerate gases) to simultaneously conserve both energy and momentum in these reactions once the equilibrium state has been achieved, and so other processes involving bystander particles (modified URCA process) have to be invoked, which are much less efficient.

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A postscript to back up the last part of my answer. If the modified URCA process operates, this generates a neutrino luminosity of about 10^33 W in a typical neutron star (Friman and Maxwell 1979, ApJ, 232, 541) at interior temperatures of ~10^9K. Each neutrino has an energy ~kT and there are ~10^57 neutrons in a neutron star. Hence the lifetime of a typical neutron is ~10^10 seconds. –  Rob Jeffries Mar 28 at 13:08
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The site uses the MathJax rendering engine for LaTeX-alike math. In-line between single $s and block equations between $$s. Given the way you wrote the above equation I imagine that you can figure out the rest. –  dmckee Mar 28 at 14:49
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