# Supernovae

### Core-collapse Explosions (supernovae)

\At the end of the fusing stage of a very massive star, the result is an onion-like structure with iron at the center and layers of less massive nuclei above, all the way out to a hydrogen exterior. After fusion stops, the star initially undergoes Kelvin-Helmholtz contraction, but accelerates and becomes a free-free collapse as pressure support is lost. There are two processes that contribute to the contraction becoming a collapse. \\The first process is photodissentegration of heavy nuclei. The energetic photons in the core will break up massive nuclei into small constituents. For example, photons can break iron up in to 13 helium nuclei and 4 neutrons. At “low” temperatures (${\displaystyle 10^{9}}$ K), bound nuclei are favored, but at “high” temperatures (${\displaystyle 10^{10}}$), nuclear statistical equilibrium favors breaking nuclei down in to smaller parts. This means that as the core contracts and the temperature goes up, matter in the core is dissentegrated in to lighter nuclei. The star spent tens of millions of years fusing hydrogen all the way up to iron, and in this stage, the star undoes all of that work and breaks the nuclei back down. This process takes energy, which causes the core contraction to accelerate, further removing pressure support, and causing the acceleration to increase. \\The second process is related to the Fermi energy of electrons. At the density of a stellar core at the end of its lifetime, the Fermi energy is of order MeV, as

${\displaystyle E_{F}\approx 5\rho _{9}^{1/3}{\rm {\;MeV}}.\,\!}$

For a density of ${\displaystyle 10^{9}}$ g cm${\displaystyle ^{-3}}$, this energy is greater than the difference in mass of the proton and neutron, meaning the reaction

${\displaystyle e^{-}+p\rightarrow n+\nu _{e}\,\!}$

is favorable. The neutrinos produced in this process take energy away, causing the collapse to accelerate. This process also converts the charged material in the star in to neutrons, and is what leads to a neutron star. That material is converted to neutrons is also important in the process of finding pressure support. The Chandrasekhar mass is

${\displaystyle M_{ch}=1.4\left({\frac {\mu _{e}}{2}}\right)^{-2}M_{\odot }.\,\!}$

This decreases as electrons are consumed in the production of neutrons. At some point, the Chandrasekhar mass will drop below the mass of the core, and the star collapses in free fall. The timescale for the collapse is of order 0.1 seconds. \\Once the star stars collapsing, a new source of pressure must become important in order to halt the collapse. Thermal pressure cannot help, as it drives the star to radiate away energy as neutrinos, and electron degeneracy pressure cannot help since the electrons are being removed. From hydrostatic equilibrium, we know that the pressure needed to balance gravity is

${\displaystyle P_{HE}\approx {\frac {GM^{2}}{R^{4}}}\propto M^{2/3}\rho ^{4/3}.\,\!}$

The problem of pressure support actually becomes harder as the star contracts, as the density increase will drive the required pressure up. Eventually, neutron degeneracy pressure becomes important. If the neutrons are non-relativistic, then

${\displaystyle P\sim E_{F}n\sim {\frac {h^{2}n^{5/3}}{m_{n}}}.\,\!}$

Initially, due to the mass of the neutron, they are unimportant compared to electrons. Their mass is useful though in that they can be non-relativistic at higher energies, and the non-relativistic degeneracy pressure increases at a higher rate with density than relativistic degeneracy pressure does. Also imporant is that the NR degeneracy pressure increases faster than the pressure needed to balance gravity. Eventually then,

${\displaystyle P_{n}\approx P_{HE}.\,\!}$

Since the radius of a degeneracy pressure supported object scales as

${\displaystyle R\propto {\frac {1}{m}},\,\!}$

and the radius of a white dwarf is roughly 10000 km, the radius of a neutron star is about 10 km. This corresponding density is then roughly ${\displaystyle 10^{14}}$ g cm${\displaystyle ^{-3}}$. The separation between particles in the neutron star is about ${\displaystyle 10^{-13}}$ cm, which is roughly the size of neutrons. This invalidates our assumption that the neutron star is an ideal quantum gas. The strong force is important, and affects the pressure. We will talk a bit about this a bit more later, but the important result of this for understanding core collapse is that the strong force actually makes the neutron star stiffer, as it resists compression beyond a certain point. The mass of the neutron star is about the mass of the iron core, which is of order ${\displaystyle 1M_{\odot }}$. All of the layers outside barely know what happened. The collapse happens in an inside out manner, since the free-fall time at the center is shorter than the free-fall time in the layers outside. Focusing on the core, its collapse releases a lot of energy from gravitational potential energy, of order

${\displaystyle E_{G}\sim {\frac {GM^{2}}{R_{NS}}}.\,\!}$

This is of order ${\displaystyle 2\times 10^{53}}$ ergs. This is more energy than the Sun releases in its lifetime, and it is released on a timescale of a tenth of a second. Mostly, this energy gets out as neutrinos, since they more easily escape than photons. Since we are concerned with explosions, how much energy would be required to unbind the outer layers? This is of order

${\displaystyle E_{G}\sim {\frac {GM_{core}M_{env}}{R_{core}}},\,\!}$

where the radius of the core is about the radius of a white dwarf. This turns out to be about ${\displaystyle 10^{51}}$ ergs. which is considerably smaller than the energy released by the core’s collapse. Thus the core collapse only needs to deposit about 1% of its energy in the outer layers in order to unbind it from the proto-neutron star. Even though this is a small fraction, it is not clear how the energy actually is deposited. There are a few ideas, none of which totally work. \\The earliest idea was that the core overshoots in its collapse, and it bounces back outward. The bounce would then cause shocks and push outer material away from the core. As it turns out, this cannot actually blow up the star. The basic reason is that as the star starts to push back out, its kinetic energy is radiated away in neutrinos and photons, causing photodissentegration and energy to leave, and ultimately not accomplishing the goal of blowing up the star. \\Another idea relies upon neutrinos. A neutron star is one of the few examples in astrophysics where the mean free path of a neutrino is of order or smaller than the size of the system. This arises because the energy and density in the centers of neutron stars are so high. The optical depth in a neutron star is

${\displaystyle \tau \approx {\frac {R}{\ell _{\nu }}}\approx n\sigma _{\nu }R\approx 3\left({\frac {E_{\nu }}{m_{e}c^{2}}}\right)^{2}.\,\!}$

In the core of the neutron star, the optical depth is about ${\displaystyle 10^{5}}$. This will produce a blackbody spectrum of neutrinos, just like a star produces a blackbody spectrum of photons. The best guess for the cause of the explosion then is that about 10% of the energy of neutrinos is absorbed by the infalling matter, which is enough to blow up the star. Getting a star to explode in a calculation is possible, but hard, and is very sensitive to details. A simulation which succeeds in exploding the star may not succeed if small details are changed. Simulations like this from 15 years ago succeeded, but more detailed ones today do not. As such, this remains an outstanding problem. \\The time it takes for the neutrinos to get out is found in just the same way we found how long it takes for photons to get out of the Sun. This is

${\displaystyle t_{\rm {escape}}\sim {\frac {R^{2}}{\ell _{\nu }c}}.\,\!}$

This turns out to be of order 1–10 seconds. The neutrino luminosity is similar, but not identical, to our typical Stefan-Boltzmann expression, with

${\displaystyle L_{\nu }\approx 4\pi R^{2}6{\frac {7}{16}}\sigma T_{\rm {eff}}^{4}.\,\!}$

The corrections are due to neutrinos being fermions, and there being 6 neutrino species. \\This has been observed once, from SN 1987 A, a supernova in the LMC. There were 19 neutrinos observed in 13 seconds, which is consisent with predictions for the timescale and the luminosity of neutrinos. It is not enough to do anything much more quantitative than that though. With any luck, a supernova in our galaxy will help better answer questions about neutron star (and black hole) formation, since neutrino detectors are much improved in the last twenty years. \\The focus today was on neutron stars, but the formation of black holes basically follows the formation of neutron stars, with the one major difference being that the material does not get exploded away, the mass on the neutron star exceeds the Chandrasekhar mass for neutron stars, and the star will collapse to a black hole. It is also possible that a black hole could form even with a supernova, if material which initially is pushed away is pushed away at less than the escape speed, and eventually falls back on the central object and leaves a black hole. We do not really understand how this happens or what this depends upon.