Why do neutrino account for 99% of the energy release for a SN II, while it is not expected to be the case for SN Ia?
Is it because the densities are not high enough to induce inverse beta-decay?
(The question came from reading Particle Astrophysics of Donald Perkins, section 7.14.1 last paragraph (it is on page 178 of the second edition).)
Answer
With Type II supernovae, the photo-disintegration of iron produces helium and neutrons: $$ \gamma+{}^{56}{\rm Fe}\leftrightarrow13\,{}^4{\rm He}+4{\rm n} $$
These helium atoms can then photo-disintegrate, producing protons and more neutrons: $$ \gamma+{}^{4}{\rm He}\leftrightarrow 2{\rm p}+2{\rm n} $$ The collapse of the star initiates the process of neutronization (generating neutrons from protons & electrons): $$ e^-+p\to n+\nu_e $$ (Note also that the electrons can bind to iron atoms to produce manganese and neutrinos: $e^-+{}^{56}{\rm Fe}\to{}^{56}{\rm Mn}+\nu_e$, so there's lots of neutrinos made here). So clearly the driver of the energetic explosion are the neutrinos themselves.
With Type Ia supernovae, however, the driver of the supernova energy is the decay of ${}^{56}{\rm Ni}$: $$ {}^{56}{\rm Ni}\to{}^{56}{\rm Co}\to{}^{56}{\rm Fe} $$ (this reaction was recently confirmed with SN 2014J, which occurred in M82; prior to this discovery this was a purely theoretical result, albeit one that matched observed light curves).
The decay of nickel-56 occurs via electron-capture: $$ {}^{56}{\rm Ni}+e^-\to{}^{56}{\rm Co}^*+\nu_e\to{}^{56}{\rm Co}+\gamma $$ where the photons produced here are $\gamma$-rays. Cobalt-56 then decays to iron-56 via: $$ {}^{56}{\rm Co}\to\begin{cases} {}^{56}{\rm Fe}+\gamma+\nu_e \\ {}^{56}{\rm Fe}+\gamma+e^++\nu_e\end{cases} $$ So clearly there are neutrinos being produced (and we should observe them, but detections of these events are unlikely as Type Ia's are usually very far away from us), but it is not really the driver of the energetics as it is with Type II supernovae.
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