3
$\begingroup$

Here :

http://www.newscientist.com/article/dn20084-neutron-star-seen-forming-exotic-new-state-of-matter.html

are news on superfluidity in a neutron star. The necessary bosons they say are pairs of neutrons. So far, so good.

But then they postulate the production of neutrinos in the formation of those neutron pairs.

When neutrons pair up to form a superfluid they release neutrinos which should pass easily through the star,...

What kind of reaction is that? n + n => neutrino + "n-n" + "XYZ" ?

$\endgroup$

1 Answer 1

2
$\begingroup$

The phenomenon was first predicted in this paper:
"Neutrino pair emission from finite-temperature neutron superfluid and the cooling of young neutron stars"
Flowers E. G., Ruderman M., Sutherland P. G., 1976, ApJ, 205,541
PDF here

The process they are describing is actually $n^*n^*\to\nu\overline\nu$, where $n^*$ is a quasiparticle excitation above the superfluid condensate. It is the annihilation of those quasipartilces that produces those neutrinos.

Of course, the "real" neutrons do not annihilate, but the description of the process in terms of the "real" neutrons is way too complex, because there is a whole medium made of constantly interacting neutrons.

Which is once again shows the usefulness of a concept of a quasiparticle.

$\endgroup$
3
  • 1
    $\begingroup$ Please correct, that should be neutrino antineutrino pair, adsabs.harvard.edu/abs/1976ApJ...205..541F otherwise lepton number conservation is violated. The decay will be mediated by an offshell Z. $\endgroup$
    – anna v
    Commented Feb 4, 2011 at 14:58
  • $\begingroup$ Correction made: One $\nu$ in the first line is now a $\overline\nu$. $\endgroup$
    – Ted Bunn
    Commented Feb 4, 2011 at 15:07
  • $\begingroup$ Let me describe what I understand: Two neutrons are bound as a pair (singulett I suppose), via a quasiparticle way. The net energy set free is carries away by a neutrinoand a antineutrino. $\endgroup$
    – Georg
    Commented Feb 4, 2011 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.