As much as 99% of the gravitational energy liberated in a core collapse supernova emerges as kinetic energy of neutrinos. The neutrinos are produces in nuclear reactions (electron capture on nuclei and free protons) and non-nuclear reactions (e.g. pair annihilation $\gamma\to\text{e}^-+\text{e}^+\to\nu+\overline\nu$).

How can we estimate without detailed model calculations that as much as 99% of the gravitational energy (not, say, 50 or 10%) emerges as kinetic energy of neutrinos?

  • $\begingroup$ For the question, "How can the electrons be degenerate at this extremely high temperature?" the answer is pressure. Lots and lots of pressure, enough that the back-pressure from the thermal motion of the electrons is dwarfed by the effective back-pressure from the Heisenberg uncertainty principle (in fact, the electron degeneracy pressure is independent of temperature). $\endgroup$ – probably_someone Oct 25 '19 at 17:48
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    $\begingroup$ In any case, you night want to split this multi-question into several single questions. They're easier to answer that way and won't risk the question being closed. $\endgroup$ – probably_someone Oct 25 '19 at 17:48
  • $\begingroup$ @probably_someone I realise that the degeneracy pressure is almost independent of temperature. But the question whether or not degeneracy occurs, depends on the degeneracy parameter $\psi=E_\text{F}/(k\cdot T)$. How does pressure enter into this? My questions are connected, so I prefer not to split them in different entries. $\endgroup$ – gamma1954 Oct 25 '19 at 18:44
  • $\begingroup$ @Kyle Kanos the Q&A to which you refer, doesn't give me a clue how to estimate the 99% $\endgroup$ – gamma1954 Oct 25 '19 at 18:45
  • $\begingroup$ @probably_someone the pressure is a consequence of the high density. It is the high density that leads to electron degeneracy. There are too many questions here, I am not going to try and write an answer to all of them $\endgroup$ – ProfRob Oct 25 '19 at 19:43

The 99% value is calculated from comparing maths to observations. In particular, you compute the gravitational potential energy of the precursor, $$ U\sim\frac{GM^2}{R}\sim10^{53}\,\text{erg} $$ Then you compare this to the observed energies in supernovae, which is typically $10^{51}$ erg (sometimes called a foe for fifty one ergs). Hence, the need for something that accounts for the unaccounted 99% of the energy.

That neutrinos could be the source of the missing energy wasn't fully confirmed until SN1987A, though it was theorized in the 60s and 70s.

  • $\begingroup$ "though it was theorized in the 60s and 70s." - I think that is what the OP is looking for; a back of the envelope calculation that says this. $\endgroup$ – ProfRob Oct 28 '19 at 13:01
  • $\begingroup$ @RobJeffries that's not exactly how I read it, but should be able to get it into the post this evening (at work currently & SE only on mobile which isn't too conducive for typing answers) $\endgroup$ – Kyle Kanos Oct 28 '19 at 14:09
  • $\begingroup$ The Newtonian $U\sim GM^2/R$ and the assumption of uniform density do not seem valid in the collapsed final state of a (proto) neutron star. So $|\Delta U|=10^{53}\text{erg}$ is a rough estimate. A kinetic energy of $10^{51}\text{erg}$ probably is an average value, depending on the mass and velocity of the ejecta. In 1966 S.A. Colgate and R.H. White assumed that neutrinos are the energy "sink" for the collapsing core, for lack of alternative sinks. 25 neutrinos in a limited energy range from the "peculiar" SN1987A, do not prove 99% is realistic. Maybe we should read 99% as "a large fraction". $\endgroup$ – gamma1954 Oct 29 '19 at 18:22
  • $\begingroup$ For those who struggle with the distinction between "fifty one ergs" and "ten to the power of 51 erg", Steven Weinberg proposed the alternative unit bethe. $1\text{B}=10^{51}\text{erg}=1\text{foe}$ ;-) $\endgroup$ – gamma1954 Oct 29 '19 at 18:25
  • $\begingroup$ @gamma1954 the GPE calc is clearly for the precursor star and not the NS/BH that results. And even if you assume both $\sim100$ foe & $\sim1$ foe are estimates, they are much too far apart to reconcile without $\nu$s. $\endgroup$ – Kyle Kanos Oct 29 '19 at 20:07

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