# Stellar end states

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### End States of Massive Stars

\We left off talking about core-collapse supernova, which are produced by stars with initial masses of greater than 10 ${\displaystyle M_{\odot }}$ at the end of their lives. Some small fraction of these, perhaps 1 in 1000, will behave differently due to being a rapidly rotating star, and results in what we observe as a gamma ray burst. Gamma ray bursts emit most of their energy in gamma rays, as the name implies. The total energy release is of order (slightly larger) the energy released by supernova. They are also directional, with emission emitted along cones outward, and the material that produces it is moving at highly relativistic speeds. Associated with GRBs in most cases are afterglows at longer wavelengths (x-ray, optical, radio). The interpretation of this is that the gamma ray bursts launch material out, producing gamma ray emission, and then the material runs in to surrounding material, slowing down, and producing emission at longer wavelengths. \\We observed supernova as very bright optical sources. This luminosity is produced by fusion during the explosion. Fusion is ignited because the outer layers are heated very quickly by the injection of kinetic energy from the core. In order for fusion in the outer layers to fuse, the timescale for fusion needs to be shorter than the timescale for the explosion of the star. If the timescale for explosion is shorter, then the material will cool before fusion can get going. In practice, this requires temperatures in the outer layers of 2–5 ${\displaystyle \times 10^{9}}$ K, and, when it occurs, is called explosive nucleosynthesis. Even during the explosion, when this is satisfied, the material is in nuclear statistical equilibrium. This state favors the most bound nucleus, given the ratio of nucleons to protons in the material. This ratio is constant during the explosion because the timescale for ${\displaystyle \beta }$-decay (turning neutrons in to protons or protons plus electrons in to neutrons) is longer than the timescale of the explosion. The material in the outer layers is mostly low atomic number, meaning an equal number of protons and neutrons. When this is the case, the most bound nucleus is ${\displaystyle {}^{56}}$Ni. We might have guessed ${\displaystyle {}^{56}}$Fe, but that does not have an equal number of protons and neutrons. Despite being the most bound nucleus, ${\displaystyle {}^{56}{\rm {Ni}}}$ is only stable for 6.1 days. This is fine for the explosion, since this timescale is much longer than the explosion, but means that the ${\displaystyle {}^{56}{\rm {Ni}}}$ will decay not too long after the explosion. \\Before getting in to the details of this explosion, remember that there are two different kinds of supernovae that arise from very different conditions. One, which we have been talking about, are supernovae that arise from the core collapse in a massive star, and are called Type Ibc or Type II supernovae. The other is a supernovae produced by a thermonuclear runaway in a degenerate white dwarf, called Type Ia. That these have confusing names is a historical result from observations, which classified supernovae by the presence of hydrogen in the spectrum.\\Going back to core collapse supernovae, the total kinetic energy of the explosion is roughly ${\displaystyle !0^{51}}$ ergs. The thermal energy is roughly of order the kinetic energy, which implies a temperature on the surface of the neutron star of ${\displaystyle 10^{9}}$ K. This is not observable, of course, because the photons need to diffuse out. The diffusion timescale is much larger than the expansion timescale though, so we only see the photons when the star has expanded enough that it becomes optically thin. As the radius increases, the thermal energy decreases. At some point, the diffusion time will be of order the expansion time, and the radiation leaks out. This point represents the peak of the supernovae emission. (I missed some stuff, so this result is coming out of nowhere) This peak occurs at

${\displaystyle t_{\rm {peak}}=\left({\frac {M\kappa }{4vc}}\right)^{1/2}.\,\!}$

This turns out to be about two weeks, or

${\displaystyle t_{\rm {peak}}=2{\rm {\;weeks}}\left({\frac {M}{M_{\odot }}}\right)^{1/2}\left({\frac {v}{10^{4}{\rm {\;km\;s^{-1}}}}}\right)^{-1/2}.\,\!}$

The luminosity is about the thermal energy when photons can get out divided by the timescale of diffusion at the peak, which is just ${\displaystyle t_{\rm {peak}}}$, so

${\displaystyle L_{\rm {peak}}\approx {\frac {E_{th}}{t_{\rm {peak}}}}.\,\!}$

This is not the whole story though. There are important decays that contribute considerable energy, particularly for stars that are not large when they explode. This starts with the decay of the unstable ${\displaystyle {}^{56}{\rm {Ni}}}$ atom, which releases a couple of MeV per decay. (A lot missing here. Very sneezy day.)

### Neutron Stars

\Neutron stars are the remnants of supernovae. To first order, neutron stars are like white dwarfs, but with the degeneracy pressure provided by neutrons rather than electrons. This means they are supported by non-relativistic degeneracy pressure, and we know that ${\displaystyle P\propto \rho ^{5/3}}$, meaning it is an ${\displaystyle n=3/2}$ polytrope. We could apply our results from that discussion, and would find that the radius of a neutron star is

${\displaystyle R\approx 15\left({\dfrac {M}{M_{\odot }}}\right)^{-1/3}{\rm {\;km}}.\,\!}$

The reason that neutorn stars are so much smaller than white dwarfs is because the degeneracy pressure scales as

${\displaystyle P_{\rm {degen}}\sim {\frac {n^{5/3}}{m}}.\,\!}$

Thus for a given energy, neutrons have a lower Fermi energy, and thus a lower pressure. In order to support the same mass then, a neutron star must be smaller than a white dwarf. The central density of a neutron star, again assuming it is an ${\displaystyle n=3/2}$ polytrope, is

${\displaystyle \rho _{c}=8\times 10^{14}\left({\frac {M}{M_{\odot }}}\right)^{2}{\rm {\;g\;cm^{-3}}}.\,\!}$

Just as in the white dwarf case, at some point, the neutrons will become relativistic, and the equation of state changes, with ${\displaystyle P\propto \rho ^{4/3}}$. The implication of relativistic neutrons is the same as for electrons, with an upper limit placed on the mass of an object that can be supported by neutron degeneracy pressure. This result, written in a slightly different way, is

${\displaystyle M_{\rm {ch}}=5.8\mu ^{-2}M_{\odot }.\,\!}$

For ${\displaystyle \mu _{e}=2}$, this gives ${\displaystyle 1.4M_{\odot }}$, the result we found for white dwarfs. For a neutron star, most of the particles are neutrons, so we can take as a first approximation ${\displaystyle \mu =1}$, giving a neutron star maximum mass of

${\displaystyle M_{\rm {ch}}=5.8M_{\odot }.\,\!}$

This result is not quite right in detail though, for two main reasons. One of these is that we have ignored general relativity. We can ignore GR when

${\displaystyle {\frac {GM}{Rc^{2}}}\sim {\frac {v_{esc}^{2}}{c^{2}}}<<1.\,\!}$

Using the ${\displaystyle R(M)}$ relationship for neutron stars,

${\displaystyle {\frac {GM}{Rc^{2}}}\approx 0.1\left({\frac {M}{M_{\odot }}}\right)^{4/3}.\,\!}$

Clearly this is violated as the mass approaches the limit that we found. The second reason is that the neutrons are not an ideal gas. Neutrons are an ideal gas when

${\displaystyle \rho _{c}<<\rho _{\rm {nuc}}.\,\!}$

The nuclear density is about ${\displaystyle 2\times 10^{14}{\rm {g\;cm^{-3}}}}$. This is fairly comparable to the central density we found, so neutrons in a neutron star are not an ideal gas. \\The first problem may be solved by considering hydrostatic equilibrium in general relativity, where the metric comes in to play and we have to remember that pressure gravitates:

${\displaystyle {\frac {dP}{dr}}=-{\dfrac {\rho {\dfrac {GM_{r}}{r^{2}}}\left(1+{\dfrac {P}{\rho c^{2}}}\right)\left(1+{\dfrac {4\pi Pr^{3}}{c^{2}M_{r}}}\right)}{\left(1-{\dfrac {2GM_{r}}{rc^{2}}}\right)}}\,\!}$

This is the Oppenheimer-Volk (?) approximation. Solving this (but ignoring the neutron ideal gas issue), we find a maximum mass of 0.7 ${\displaystyle M_{\odot }}$, with a radius of 9.6 km. If this were true, there would be no neutron stars, as neutron stars form from the collapse of the iron core when it exceeds the electron degeneracy pressure maximum mass. The neutron stars do exist is because this density is much greater than the nuclear density, and the neutrons start to overlap. \\Dealing with overlapping neutron stars requires nuclear physics. We’ll start on the outside at low densities where things are easiest, and move in. At ${\displaystyle \rho <4\times 10^{11}{\rm {\;g\;cm^{-3}}}}$, there are no free neutrons. The matter is composed of free electrons and heavy nuclei. Electron capture takes place, with nuclei becoming more neutron rich. As the density increases, the lowest energy state of matter cannot accommodate more neutrons in nuclei. This leads to free neutrons. The name “neutron drip” is applied to densities beyond this limit, as neutrons drip away from nuclei. Moving to higher and higher densities, the number of free neutrons increases, the number of neutrons in nuclei decreases, and the number of protons and electrons decreases. At ${\displaystyle \rho <4\times 10^{12}{\rm {\;g\;cm^{-3}}}}$, non-relativistic layers finally overtake relativistic electrons as the dominant source of pressure. The number of protons, neutrons, and electrons also reach an equilibrium, with ${\displaystyle n_{n}\approx 8n_{p}}$, and ${\displaystyle n_{p}=n_{e}}$. Finally, at ${\displaystyle \rho <2\times 10^{14}{\rm {\;g\;cm^{-3}}}}$, neutrons interact via the strong force. We don’t know what happens here. Unfortunately, 99% of the matter in a neutron star falls in this density regime. \\Observing neutron stars is hard. We already saw it was difficult in white dwarfs, because they cool so quickly after their birth. Neutron stars are born hotter, at ${\displaystyle 10^{11}}$ K, but cool to ${\displaystyle 10^{9}}$ K in a day from neutrino luminosity. The effective temperature remains at a few times ${\displaystyle 10^{6}}$ K for some ${\displaystyle 10^{5}}$ or ${\displaystyle 10^{6}}$ years, after which it drops like a rock because photons are getting out on that timescale. For ${\displaystyle T_{\rm {eff}}\sim 10^{6}}$ K and a radius of 10 km, the luminosity is roughly ${\displaystyle L_{\odot }}$. The bulk of this luminosity comes out in the UV and X-ray region. After this ${\displaystyle 10^{6}}$ year period, the neutron star is almost impossible to see. At the surface of a neutron star, gas pressure dominates, and the scale height is

${\displaystyle h={\frac {kT}{mg}}.\,\!}$

This turns out to be 1 cm, because of the high gravity at the surface, and the density is about ${\displaystyle 1{\rm {\;g\;cm^{-3}}}}$. The conditions in the atmosphere are actually pretty comparable to the interior of the Sun then. Despite the exotic nature of neutron star interiors, the part we see is actually not exotic.

### Rotation of Neutron Stars

\We can expect that neutron stars should rotate very rapidly from conservation of angular momentum. If we look at the Sun (which will not become a neutron star), with a period of roughly 24 days and a radius of ${\displaystyle R_{\odot }\approx 7\times 10^{10}}$ cm, we can make a rough estimate of the period of a neutron star, since we know its final size. From ${\displaystyle J\sim MR^{2}\Omega }$,

${\displaystyle P_{\rm {NS}}\approx P_{\odot }\left({\frac {R_{\rm {NS}}}{R_{\odot }}}\right)^{2}.\,\!}$

This corresonds to a period of order a millisecond, which is a little on the short side of what is actually observed. This corresponds to a rotational energy of

${\displaystyle E_{\rm {rot}}\approx 10^{48}\left({\frac {P_{\rm {rot}}}{100{\rm {\;ms}}}}\right)^{-2}{\rm {\;ergs}}.\,\!}$

The neutron star taps in to this energy because it is strongly magnetized, which we may understand by noticing that stars are excellent conductors. The Sun has a magnetic field of 1 Gauss. By flux conservation,

${\displaystyle B_{\rm {NS}}R_{\rm {NS}}^{2}\sim B_{\odot }R_{\odot }^{2}.\,\!}$

Thus we would expect that the neutron star has a magnetic field with strength ${\displaystyle 10^{10}}$ Gauss. The actual range of magnetic fields observed ranges from ${\displaystyle 10^{8}}$${\displaystyle 10^{15}}$ Gauss. For both the rotation and magnetic field then, the values for the neutron stars are not too different from what we would expect from simple conservation of angular momentum and flux from a star like the Sun. \\As a result of this high magnetic field and rapid rotation, in addition to the misalignment of the rotational and magnetic axes, pulsed emission is generated by spinning neutron stars. Or, what appears pulsed to an observer on Earth. The change in rotational energy is what powers the radio emission, sort of (we do not observe the full change in rotational energy). This change in energy is

${\displaystyle {\frac {dE_{\rm {rot}}}{dt}}=-{\frac {B^{2}R^{6}\Omega ^{4}\sin ^{2}(\alpha )}{6c^{3}}}.\,\!}$

For typical numbers, this is

${\displaystyle {\frac {dE_{\rm {rot}}}{dt}}=10^{35}\left({\frac {B}{10^{12}{\rm {\;G}}}}\right)^{2}\left({\frac {P}{100{\rm {\;ms}}}}\right)^{-4}.\,\!}$

This leads to something called the spin down time, the timescale over which the period changes significantly, which is

${\displaystyle t_{\rm {spindown}}={\frac {\Omega }{\dot {\Omega }}}.\,\!}$

By measuring the period, and rate at which the period is changing, we can infer the magnetic field strength of neutron stars, if we assume that the changing period is due to the loss of rotational energy. This simple model is not quite right though, as we have assumed that the area outside of the neutron star is effectively a vaccuum, since it has a scale height of 1 cm. It turns out that the strong magnetic and electric field of the neutron star means that charged matter on the surface has a much different scale height, and that there is a plasma surrounding the neutron star. This is what is thought to actually be happening. A lot of the energy from spinning down actually goes to ejecting material from the neutron star at relativistic speeds. This manifests itself in a bubble of material like that which is observed around the Crab Pulsar. \\Average neutron stars are traveling through the galaxy at velocities of 350 ${\displaystyle {\rm {km\;s^{-1}}}}$ or greater. They do not participate in the rotation of the galaxy like normal stars, and many pulsars are on courses that will take them out of the galaxy. Traveling at this speed requires a significant amount of kinetic energy, something of order ${\displaystyle 10^{48}}$ ergs. This energy probably comes from the explosion that formed the neutron star. If the explosion is not perfectly symmetric, then the neutron star has to be given a kick, so that momentum is conserved.

### Black Holes

\The “surface” of a black hole is described by the event horizon, that defines the region from no information can escape. In order to detect them, indirect means are required. One method is dynamical in nature. Unseen mass may be detected based on its gravitational influence on another body. The other method relies on detecting radiation from just outside of a black hole. Radiation from deep within the potential well of a black hole will show signatures of this (ie, quasars). \\There is strong evidence for the existence of black holes, even though they can only be detected indirectly. One class is supermassive black holes at the centers of galaxies, and the other is the black holes that are produced at the endpoint of stellar evolution for very massive stars. The SMBHs manifest themselves as active galactic nuclei and quasars, while we see most stellar remnant black holes as X-ray binaries, as they accrete gas from a companion star and the accreted gas emits X-rays as it falls in to the black hole. This mass exchange occurs when the companion star overflows its Roche lobe. \\The result is an accretion disk around the black hole. Particles in orbit around the black hole would like to be in stable, circular orbits. However, the particles feel friction from one another, due to turbulence in the disk caused by the presence of magnetic fields. This moves angular momentum around, and as particles lose angular momentum, they fall inward on the black hole. Of course, angular momentum is globally conserved, so some particles must gain angular momentum and move out. \\The energy of material on a circular orbit is

${\displaystyle E=-{\frac {GM\Delta M}{2R}}.\,\!}$

As matter falls in closer to the black hole, it becomes more bound, going to more negative energies, and the change in energy is

${\displaystyle \Delta E=-{\frac {GM\Delta M}{2R_{in}}}--{\frac {GM\Delta M}{2R_{out}}}.\,\!}$

For large changes in radius, this simplifies to

${\displaystyle \Delta E=-{\frac {GM\Delta M}{2R_{in}}}.\,\!}$

If mass moves this distance over a time ${\displaystyle \Delta t}$, then

${\displaystyle {\frac {\Delta E}{\Delta t}}=-{\frac {GM\Delta M}{2R_{in}\Delta t}}.\,\!}$

We can identify the mass accretion rate,

${\displaystyle {\dot {m}}={\frac {\Delta M}{\Delta t}}.\,\!}$

Then

${\displaystyle L={\frac {GM{\dot {M}}}{2R_{in}}}.\,\!}$

This energy is ultimately converted to radition going out, characterized by the efficiency ${\displaystyle \eta }$ as

${\displaystyle L=\eta {\dot {M}}c^{2}.\,\!}$

Since matter falls so deep in to the potential for a black hole, the efficiency is very high, with

${\displaystyle \eta ={\frac {R_{s}}{4R_{in}}}.\,\!}$

This is in the range ${\displaystyle \eta \approx 0.1}$${\displaystyle 0.4}$. Compare this to fusion, where four protons of material (4 GeV) combine to release 27.3 MeV of energy. This corresponds to an ${\displaystyle \eta \approx 0.007}$. Just as for a star, there is a maximum luminosity for accreting black holes, and it is set by the same physics that set the Eddington luminosity. It turns out that most observed accreting black holes are accreting near the Eddington luminosity. This has a corresponding mass accretion rate, given by

${\displaystyle {\dot {M_{Edd}}}=10^{-8}\left({\frac {M}{M_{\odot }}}\right)\left({\frac {\eta }{0.1}}\right)^{-1}{\frac {M_{\odot }}{\rm {yr}}}.\,\!}$

This energy is emitted in X-rays, and neutron stars and black holes may be found by looking for X-ray sources. Stellar mass black holes are identified when the companion can be seen, and from the orbit, the mass of the accreting object can be determined. If the mass is considerably above the maximum mass of the neutron star, we can be confident that the accreting object is a black hole. \\The other class of black holes, of the super massive variety, manifest themselves as point sources of radiation at the centers of galaxies, and are called active galactic nuclei. The luminosity of AGN vary on very short timescales, as short as hours or days. Based on causality arguments, the size of these objects must be the size of the solar system or smaller. Yet the luminosity they produce can be as high as ${\displaystyle 10^{13}L_{\odot }}$. This is interpreted as being due to an accreting black hole with a mass of ${\displaystyle 10^{6}}$ up to ${\displaystyle 10^{10}}$ ${\displaystyle M_{\odot }}$. For a ${\displaystyle 10^{9}M_{\odot }}$ black hole, the Eddington luminosity is about ${\displaystyle 10^{13}L_{\odot }}$, corresponding to an accretion rate of ${\displaystyle 10M_{\odot }}$/year. It also has Schwarzschild radius of about 20 AU. This is all self consistent then with the idea that black holes power AGN. \\In the case of our own galaxy, we can actually track orbits of stars near the center of the Milky Way, and see elliptical orbits. This is only consistent with a one over ${\displaystyle R}$ potential, and the orbit of each star is consistent with an enclosed mass of ${\displaystyle 4\times 10^{6}M_{\odot }}$. The region in which orbits have been seen is about 10 light days in size, about the size of the Solar System. This is again consistent with a black hole. Explaining it in any other way is very difficult (improbable).