- What makes a neutron star burn, and what kind of fusion/decay is happening there?
- What is supposed to happen with a neutron star in the long run? What if it cools, then what do the degenerated matter looks like after it cools? Will the gravitational equilibrium be ruined after some burn time? How does it explode if it can explode at all?
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14$\begingroup$ It doesn't burn. $\endgroup$– Cedric H.Commented Nov 11, 2010 at 13:02
5 Answers
With normal stars, the "burning", that is to say the fusion reaction, produces a pressure that counteracts the pull of gravity to keep the star from collapsing. But with neutron stars, the protons and electrons in the star have combined into neutrons*.
The Pauli exclusion principle causes the neutrons to resist further compression. That is, the neutrons, being identical fermions, can't all be put in the same state. So to get them closer and closer together you have to go into higher and higher energy states. Thus, there is an energy cost in compressing the star, and this results in a sort of pressure called "degeneracy pressure".
It is this pressure that stabilizes the neutron star against collapse (assuming it doesn't have enough mass to overcome this pressure and become a black hole). So they don't need to "burn" to maintain their stability, and so far as I know, they don't. At least not in the sense of a normal star where you have atomic nuclei fusing.
- Note: Neutrons aren't made of protons and elections, but this transformation can happen by means of the weak nuclear force. Normally neutrons aren't stable outside of the atomic nucleus -- instead the transformation would go the other way and a free neutron would decay into a proton and electron (there's also an anti-electron neutrino produced). But under the intense gravitational pressure in a collapsed star, the neutrons are stable, which allows us to end up with neutron stars.
Edit: This is of course a very approximate picture. The link posted by Thomas Thernel has much more detail. One good point to emphasize is that, as you might expect, the density is greater at the center of a neutron star than at its outskirts, so the star won't really be all neutrons... you'll have more neutrons closer to the center, and more ordinary atomic nuclei further out. Apparently some interesting sorts of structures can form from the remaining nuclei, even at the point where it's 90-95% neutrons.
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1$\begingroup$ Neutron degeneracy pressure can only be a fraction of the pressure required to support neutron stars. The Tolman-Oppenheimer-Volkoff limit for ideal neutron degeneracy pressure in GR is 0.7 solar masses. All neutron stars with measured masses are more massive than this. $\endgroup$– ProfRobCommented Nov 30, 2014 at 19:08
Tim Goodman's answer is right, and I just want to add a couple of things:
Though Neutron stars do not "burn" anything, for the most part, they DO shine, thanks to the fact that they are the remnants of the core of a star, and thus, when they are born, are roughly as hot as the center of a star. They will thus shine in the same way that a hot poker shines when you stick it in a fire.
Second, neutron stars have a maximum possible mass, dictated by the rules of general relativity and the mass of a neutron. If a neutron star has extra mass beyond this dumped on its surface (say, because it is orbiting an ordinary star, and that star accretes mass onto the NS), they can explode in a violent way. To my understanding, this has not been observed, but it has been known to happen in White Dwarfs, which are held together by a very similar principle to neutron stars. The end result is a supernova, and then a black hole.
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$\begingroup$ Thank you. I assume they stay cold forever, when nothing happens $\endgroup$– user299Commented Nov 14, 2010 at 20:13
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1$\begingroup$ Yes, that's right. They just cool off and reach a final, static state at the same temperature as the space around them. $\endgroup$ Commented Nov 14, 2010 at 21:41
The correct comparison for a neutron star is with a cinder spat out of the fire. The cinder will glow brightly for a short period of time and then fade rapidly. Such is the fate of neutron stars, because although born at $10^{11}$ K in the heart of a supernova, they have an extremely low heat capacity.
Contrary to common belief - neutron stars are not supported by neutron degeneracy pressure. Yes, this contributes, but a star supported solely by NDP would be limited to masses less than 0.7$M_{\odot}$. All known neutron stars are more massive than this. It is the strong nuclear force in incompressible, asymmetric nuclear matter that provides most of the support.
Neutron degeneracy is though largely responsible for the thermal properties of neutrons stars. Degenerate matter has an extremely low heat capacity, because there are no empty energy states for fermions to cool and fall into. That means that even though they are born with extremely high temperatures, they will be nearly isothermal (degenerate fermions also have long mean free paths and high thermal conductivity), with very little thermal energy.
Neutrino processes (URCA process, followed by modified URCA process - see What allows the modified Urca process to work at lower density than direct Urca in neutron star cooling?) can effectively and rapidly cool the star on timescales of initially seconds and later tens or hundreds of years. After about ten thousand years cooling from the surface by photons dominates.
As the neutron star cools, what distinguishes it from a normal star, is that its internal pressure is independent of temperature. That means that if it is in equlibrium once it cools to say a billion degrees (maybe after a year), it is already "effectively cold" and further cooling will not change its radius and density.
If we assume that most supernvovae from stars with initial masses between 8 and 20 solar masses produce neutron stars then there are probably about a billion of these dead cinders floating around in our Galaxy.
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$\begingroup$ +1 For mentioning that neutron stars are not (solely) supported by neutron degeneracy pressure but (largely) by strong nucleon interactions. $\endgroup$ Commented May 7, 2015 at 22:29
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$\begingroup$ I think this was the best of the answers. One pet peeve though-- why do so many people treat it as important that the pressure doesn't depend on temperature for fully degenerate matter? That's not actually important at all, what matters is that the pressure is very high when the temperature is low (even zero). But that has nothing to do with the slope, dP/dT, at T=0, it has to do with P at T=0, which is something completely different. The slope could be steep there, and as long as P was high when T was low, it would still have the same effect. $\endgroup$– Ken GCommented Mar 23, 2017 at 1:27
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$\begingroup$ @KenG I take your point that the change in pressure simply depends on the change of internal kinetic energy. However, it is much easier to explain why the size of a neutron star does not change even though it changes its temperature by orders of magnitude, by simply saying that $P \neq f(T)$. If $P$ was a function of $T$ that would affect the thermal history of a neutron star, since the release of GPE would keep it warm. $\endgroup$– ProfRobCommented Mar 23, 2017 at 10:09
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$\begingroup$ But that's just it, it's actually easy to explain that by simply saying that P = f(T), where f(T) is a function that has a large value at T=0. That's all that matters. The common language "P is independent of T" is not clear enough. If it is taken as a statement about dP/dT being 0, it is of no importance, and not even true in white dwarfs. It's much better to just say what P(0) is. The other language is common, yet puzzling, since it sounds like it is claiming that any function that takes a nonzero value at zero argument is therefore independent of that argument. $\endgroup$– Ken GCommented Mar 23, 2017 at 12:44
I think this is a good almost-popular introduction to neutron stars, the processes expected to occur therein and their evolution.
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$\begingroup$ Thanks for the link. I had a rough idea of how neutron stars work (as given in my answer), but this business of "spaghetti" and "lasagna" like structures is really fascinating. $\endgroup$ Commented Nov 11, 2010 at 19:00
Neutron stars can actually "burn" and leave quark matter as an ash, according to some hypotheses.
The Bodmer-Witten hypothesis states that quark matter made of up, down, and strange quarks might be the most stable form of matter. If one could find a way to melt neutrons and protons into stable, strange quark matter, this could unleash a very exothermic event, which can be described as a combustion process. Some argue that the core of a neutron star, due to its high density, might be an ideal place for this phase transition to happen. So in short, the neutron star could "burn" into a quark star and produce an explosion. The Bodmer-Witten hypothesis is a bit controversial but has very interesting astrophysical imiclations.
This is the theoretical basis of the quark nova model:
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$\begingroup$ +1 Just for curiosity, the "nuclear" version of the BW hypothesis is this one: physics.stackexchange.com/q/803817/226902 $\endgroup$– QuilloCommented Feb 25 at 11:03