# Stellar evolution

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### Stellar Evolution

\The hydrogen burning phase, or main sequence lifetime of a star, can be found by dividing the energy available in fusion of hydrogen by the luminosity of the star, with

${\displaystyle t_{MS}={\frac {E_{\rm {nuc}}}{L}}\propto {\frac {M_{\rm {core}}}{L}}\approx 10^{10}{\rm {\;years\;}}\left({\frac {M}{M_{\odot }}}\right)^{-2}.\,\!}$

This tells us that more massive stars are shorter lived, and stars only a bit less massive than the Sun have lifetimes longer than the age of the universe. Stellar evolution is due to changes in composition at the centers of stars where fusion is taking place. As hydrogen is consumed, and helium is produced, eventually hydrogen runs out, which drives the structure of a star to change. Even while a star is on the main sequence, its structure will change as a result of changes in composition. If we look back at the radiative transfer equation, remembering how we used this to find how luminosity scales with mass (assuming, say, Thomson scattering), we can see the compositional dependence.

${\displaystyle F_{r}={\frac {L_{r}}{4\pi r^{2}}}=-{\frac {4}{3}}\ell {\frac {d}{dr}}\left(\sigma T^{4}\right),\,\!}$
${\displaystyle kT\sim {\frac {GM\mu m_{p}}{R}},\,\!}$
${\displaystyle L\propto M^{3}\mu ^{4}\mu _{e}.\,\!}$

The mean molecular weight of electrons comes in through the fact that the electrons are doing the scattering, while the total mean molecular weight comes from the virial theorem substitution. Physically, what happens is that the number of particles in the center per unit mass is changing, meaning the temperature in the center must adjust slightly in order to maintain hydrostatic equilibrium, and the rate at which energy leaks out also changes slightly. We can write the opacity in terms of the mass fractions of hydrogen and helium as

${\displaystyle \kappa ={\frac {n_{e}\sigma _{T}}{\rho }}={\frac {\sigma _{T}}{\rho }}(n_{p}+2n_{\alpha })={\frac {\sigma _{T}}{m_{p}}}\left(X+{\frac {Y}{2}}\right),\,\!}$

where ${\displaystyle X}$ is the mass fraction of hydrogen and ${\displaystyle Y}$ is the mass fraction of helium. For the simple case of only hydrogen and helium, this simplifies to

${\displaystyle \kappa ={\frac {\sigma _{T}}{2m_{p}}}\left(1+X\right).\,\!}$

Then the mean molecular weight of electrons is

${\displaystyle \mu ={\frac {2}{1+X}},\,\!}$

and

${\displaystyle \kappa ={\frac {\sigma _{T}}{\mu _{e}m_{p}}}.\,\!}$

At the start of a star’s lifetime, it is roughly 75% hydrogen by mass, and the rest is helium, so ${\displaystyle \mu =0.6}$. For a pure helium star, which is approximately the case for a very massive star at the end of its lifetime, ${\displaystyle \mu =4/3}$. Plugging this in to the luminosity scaling above, we find that

${\displaystyle {\frac {L(t_{MS})}{L(0)}}=40.\,\!}$

For the Sun, only about 10% of the hydrogen fuses, and at the end of its lifetime, ${\displaystyle \mu =0.64}$. This gives

${\displaystyle {\frac {L(t_{MS})}{L(0)}}=1.37.\,\!}$

If we were interested, we could follow this changing composition through all the previous equations of stellar structure we have used, to see how the central temperature, effective temperature, and radius must also change as the composition changes. At the end of a star’s main sequence lifetime, we are left with a situation where nearly the entire core of the star is composed of helium. Outside the core, there is a hydrogen envelope, which has a composition that is more or less the same as the star at the beginning of its lifetime. At this point, we can give a schematic overview of the stellar evolution post main sequence. \\When hydrogen fusion ceases, the star loses its source of luminosity, and, as we saw for pre-main sequence stars, the core of the star must undergo Kelvin-Helmholtz contraction. As it contracts, it heats up, and it becomes possible to fuse heavier elements than was possible before, and the fusion of heavier elements will become large enough to halt contraction. Eventually, the material will run out, and with no energy source, the core must contract, and the cylce stars anew. This cycle can be halted if, during contraction, the density increases enough to the point that electron degeneracy pressure takes over and halts contraction. \\The overall fate of the star thus depends on two facts. One is that a star supported by degeneracy pressure has a maximum mass, known as the Chandrasekhar mass. If this mass is exceeded, electron degeneracy pressure is incapable of halting the contraction. The other is that there is a maximum central temperature, which depends on mass, for an object that is supported by gas pressure and degeneracy pressure. This means that the end state of fusion is set by the mass of the star. If we set the mass of our object to the Chandrasekhar mass, we find that the maximum temperature of an object that can ever be electron degeneracy pressure supported is a few times ${\displaystyle 10^{8}}$ K. Stars cannot fuse to arbitrarily highly charged elements. Some stars that become white dwarfs are able to fuse helium to carbon and oxygen, but it cannot fuse beyond that. Another way of saying this is that the ultimate fate of a star depends on its initial mass. \\One fate is the become a white dwarf. For stars with core masses less than 1.4 ${\displaystyle M_{\odot }}$, which corresponds to stars with initial masses of 8 ${\displaystyle M_{\odot }}$, the end state is an electron degeneracy pressure supported object that has no need for fusion. Why the difference between core mass and initial mass? Without going in to detail, mass loss. This is an uncertain area of stellar evolution, as we do not understand in detail the mass loss history of stars. While the 1.4 ${\displaystyle M_{\odot }}$ limit is theoretical (with plenty of observational support), the 8 ${\displaystyle M_{\odot }}$ limit is empirical. The second fate, for stars with initial masses larger than 8 ${\displaystyle M_{\odot }}$ (meaning no white dwarf), is for fusion to continue all the way up to iron. At this point, the star cannot produce more energy from fusion, since iron is the most tightly bound nucleus. This means that there is no way to keep the star hot and maintain pressure support. This means that the iron core will collapse under its own gravity. When the collapse starts, the core will have a radius of about that of Earth. At the end of the collapse, the radius will be of order 10 km, and the star becomes either a neutron star or a black hole, and produces a supernova explosion. Some sources will say that neutron stars will be produced for stars less massive than 30 ${\displaystyle M_{\odot }}$, and black holes for stars more massive. This is wrong. Definitely wrong. For this reason, we’ll just say neutron star or black hole. \\Our overview finished, we can now talk about evolution in a bit more detail. After the main sequence fusion ends, and the core begins contracting, the density and temperature increase. In addition to the core that contracts, there is also a hydrogen envelope around the core that contracts. This envelope becomes denser and hotter, and before any fusion in the helium core begins, the hydrogen in envelope gets hot enough to fuse. Even so, the core is still contracting, which drives energy generation in the envelope up. The outer part of the star can only carry so much energy out though through radiative diffusion, and eventually the energy generated by shell fusion is larger than the energy the star can carry out. The result is that the star expands, even while the core is contracting. Eventually, the star expands until the effective temperature is around 3000–4000 K, and convection can effectively carry the energy out, and the star will actually go up the Hayashi line. \\This is a key point to understanding stellar evolution. During post main sequence evolution, the core and the outer parts of the star become decoupled. Even while the core contracts, the outer part of the star expands. For stars on the main sequence, the properties and the core and the properties outside the core were not independent. \\For stars more massive than 2 ${\displaystyle M_{\odot }}$, the core continues contracting and the star continues moving up the Hayashi line until helium ignites, and the helium main sequence begins. For stars less massive than ${\displaystyle M_{\odot }}$, the core actually becomes degenerate. The star still moves up the Hayashi line though. The reason for this is that the radius of the core is proportional to ${\displaystyle M^{-1/3}}$. This matters because shell fusion dumps helium on the core, increasing the mass, and causing the core to become smaller and the central temperature to increase. Eventually, helium fusion starts in a slightly different fashion, which we’ll discuss later. Thus, both mass regimes lead to giants, but the details are slightly different.

### Helium fusion physics

\The physics of helium fusion is very different from the fusion of hydrogen. If we have a core composed of hydrogen and helium, and try to fuse helium with helium or a proton with helium, we do not get anything stable. Both beryllium and lithium are unstable. What happens instead is what is called the “triple ${\displaystyle \alpha }$” reaction, where three helium fuse to form carbon. This is not quite an actual triple reaction, but involves two helium fusing and then on a timescale faster than beryllium decay, having another helium come along and fuse. At the same temperature at which this happens, helium and carbon can fuse to form oxygen. As such, this leads to a mixture of carbon and oxygen. \\The first step of the triple-${\displaystyle \alpha }$ process, the fusion of helium to beryllium, is actually endothermic, requiring an energy of 92 keV.

${\displaystyle {}^{4}{\rm {He}}+{}^{4}{\rm {He}}\rightarrow {}^{8}{\rm {Be}}\,\!}$

Thus, the energy of the fusing helium must be greater than 92 keV. We can write down the same sort of ${\displaystyle E_{0}}$ as we did before to find the energy that dominates the fusion reaction, with

${\displaystyle E_{0}=\left({\frac {E_{G}(kT)^{2}}{4}}\right)^{1/3}.\,\!}$

This is a bit different than what we saw before, since helium has to overcome the Coulomb barrier as well as overcome the fact that the reaction is endothermic. Using ${\displaystyle E_{G}}$ from before, we find that

${\displaystyle E_{0}=84T_{8}^{2/3}{\rm {\;keV}}.\,\!}$

Thus temperatures greater than ${\displaystyle 10^{8}}$ K are required for helium fusion. The half-life of beryllium is

${\displaystyle \tau =2.6\times 10^{-16}{\rm {\;s}}.\,\!}$

Since both fusion and decay can happen on short timescales, the number of beryllium will come in to equilibrium, through the arguments we used for the Saha equatoin. In thermal equilibrium,

${\displaystyle 2\mu ({}^{4}{\rm {He}})=\mu ({}^{8}{\rm {Be}}).\,\!}$

Though any one beryllium will not be around for long, there is a population of beryllium that will be maintained, and can then fuse with another helium to form a carbon atom,

${\displaystyle {}^{8}{\rm {Be}}+{}^{4}{\rm {He}}\rightarrow {}^{12}{\rm {C}},\,\!}$

with Q = 7.367 MeV. If this were all there was to it, the amount of carbon produced would not be enough to explain observed abundances. This led to one of the most brilliant predictions in astrophysics history, when Fred Hoyle predicted that the carbon atom had a resonant state at an energy close to the energy released by the fusion reaction. Such a resonant excited state would make fusion easier, and produce carbon. This prediction was verified, and a resonant state in carbon was observed with an energy E = 7.65 MeV above the ground state. Thus, reactions with energies of 290 keV would lead to producing an excited carbon, and we have

${\displaystyle E_{0}=290{\rm {\;keV.}}\,\!}$

Since ${\displaystyle E_{0}}$ that we would find from the typical equation is

${\displaystyle E_{0}=150T_{8}^{1/3}{\rm {\;keV}},\,\!}$

the required temperature is ${\displaystyle 2\times 10^{8}}$ K for the reaction. \\The excited state of carbon (denoted with an asterisk) is itself unstable, and will break apart to form beryllium and helium, with

${\displaystyle {}^{12}{\rm {C}}^{*}\leftrightarrow {}^{8}{\rm {Be}}+{}^{4}{\rm {He}}.\,\!}$

Since this reaction goes both ways, it too will come in to chemical equilibrium. Since the same was true for the first step,

${\displaystyle {}^{4}{\rm {He}}+{}^{4}{\rm {He}}\leftrightarrow {}^{8}{\rm {Be}},\,\!}$

we have a total reaction with

${\displaystyle {}^{4}{\rm {He}}+{}^{4}{\rm {He}}+{}^{4}{\rm {He}}\leftrightarrow {}^{12}{\rm {C}}^{*}.\,\!}$

Then we can write down the chemical equilibrium,

${\displaystyle 3\mu ({}^{4}{\rm {He}})=\mu ({}^{12}{\rm {C}}^{*}).\,\!}$

If this were all, this would still not be that helpful, since we do not observe excited carbon, and any carbon that leaks out of the center of a star will just decay and not leave us with any carbon. What saves us is that there is one other way for excited carbon to decay, with

${\displaystyle {}^{12}{\rm {C}}^{*}\rightarrow {}^{12}{\rm {C}}+\gamma .\,\!}$

This gives us energy and ground state carbon atoms. All of this happens by reactions that are in thermodynamic equilibrium, and only occasionally does an excited carbon make it out of the equilibrium and form a ground state carbon atom. Since it all happens in equilibrium, we know how much carbon there is in the excited state. Then, we can use the decay timescale of the excited carbon to the ground state, and the Saha equation, to find the energy produced by helium fusion. The chemical equilibriums are

${\displaystyle \mu _{12}^{*}=m_{12}^{*}c^{2}-kT\ln \left({\frac {gn_{Q,12}}{n_{12}}}\right),\,\!}$

and

${\displaystyle \mu _{4}=m_{4}c^{2}-kT\ln \left({\frac {gn_{Q,4}}{n_{4}}}\right),\,\!}$

The difference in energy between the masses is

${\displaystyle \Delta E=280{\rm {\;keV}}=m_{12}^{*}c^{2}-3m_{4}c^{2}.\,\!}$

This is positive, unlike our usual reactions, where the energy is negative. This is a result of going to an excited state. Energy is released in the reaction when the excited carbon decays to the ground state. Then, chemical potential equilibrium gives

${\displaystyle -kT\ln \left({\frac {gn_{Q,4}}{n_{4}}}\right)^{3}=\Delta E-kT\ln \left({\frac {gn_{Q,12}}{n_{12}}}\right).\,\!}$

Solving this equation for the number density of excited carbon, and skipping some simplification,

${\displaystyle {\frac {n_{12}^{*}}{n_{4}^{3}}}=3^{3/2}\left({\frac {h^{2}}{8\pi m_{p}kT}}\right)^{3}e^{-\Delta E/kT}.\,\!}$

The number density of hydrogen is

${\displaystyle n_{4}={\frac {\rho }{4m_{p}}}Y,\,\!}$

where ${\displaystyle Y}$ is the mass fraction of hydrogen. The energy difference term is

${\displaystyle {\frac {\Delta E}{kT}}={\frac {44}{T_{8}}}.\,\!}$

Then our final expression for the number density in the excited state is

${\displaystyle n_{12}^{*}={\frac {Y^{3}\rho ^{3}}{(4m_{p})^{3}}}3^{3/2}\left({\frac {h^{2}}{8\pi m_{p}kT}}\right)^{3}e^{-44/T_{8}}.\,\!}$

To get the number in the ground state, we can write the rate of ground state production as

${\displaystyle {\frac {dn_{12}}{dt}}={\frac {n_{12}^{*}}{\tau }}.\,\!}$

This is the step that gives off energy, with an energy generation rate

${\displaystyle \epsilon ={\frac {dn_{12}}{dt}}{\frac {Q}{\rho }},\,\!}$

where ${\displaystyle Q=7.65}$ MeV. A more useful final expression is

${\displaystyle \epsilon =5.4\times 10^{11}{\frac {\rho ^{2}Y^{3}}{T_{8}^{3}}}e^{-44/T_{8}}{\rm {\;erg/s/g}}.\,\!}$

We can turn this in to a simple scaling relationship as before, with

${\displaystyle \epsilon \propto \rho ^{\alpha }T^{\beta }.\,\!}$

At a temperature of ${\displaystyle 10^{8}}$ K, ${\displaystyle \beta sim40}$. At a temperature of ${\displaystyle 2\times 10^{8}}$ K, ${\displaystyle \beta sim20}$. So this is an even stronger function of temperature than CNO is when looking at the temperatures at which the reactions actually happen in each case. \\

### Helium fusion in stars

\Now that we understand how helium fusion operates, we can talk about the helium fusing lifetime of a star. A star on the helium main sequence (horizontal branch), will have helium fusion in the core and hydrogen fusion in a shell around the core. This phase will last for a few percent of the main sequence lifetime for a star, and the luminosities are something like a factor of a few tens larger than the main sequence luminosities. \\Once the helium main sequence finishes, the star has a carbon/oxygen core that, for low mass stars, is almost a white dwarf. The core will contract somewhat, and if the initial mass of the star is less than 8 ${\displaystyle M_{\odot }}$, then the carbon/oxygen core will be halted by electron degeneracy pressure. The rest of the material somehow leaves the star, but this is not particularly well understood. There are helium and hydrogen shells, both of which will fuse material, and produce luminosity in excess of that which can be carried out. The outer layers of the star will then puff out again, and lead to the star becoming a giant again, but this time it moves up the asymptotic giant branch. This line is more or less at the same spot as the red giant branch, the only real difference between the two being that the red giant branch stars have helium cores, and the asymptotic giant branch stars have carbon/oxygen cores. What we think happens is the outer layers basically just keep expanding until they are unbound from the carbon/oxygen core. Radiation pressure on dust probably plays a role. There are also instabilities associated with fusion in thin shells, which will produce bursts of luminosity that will help push off material. As material is blown off, the star moves across the HR diagram to hotter temperatures. This is called the planetary nebula phase. It is this process that is responsible for the fact that stars born with masses less than 8 ${\displaystyle M_{\odot }}$ end their lives as white dwarfs with masses less than ${\displaystyle 1.4\;M_{\odot }}$.

### Evolution of Massive Stars

\An initial stellar mass above 8 ${\displaystyle M_{\odot }}$ implies that the final core mass will be larger than 1.4 ${\displaystyle M_{\odot }}$, and the star cannot be supported against gravity be electron degeneracy pressure. One important consequence of this is that there is no maximum temperature for an object that cannot be supported by electron degeneracy pressure. The existence of a maximum temperature comes about because of the competition between degeneracy pressure and thermal gas pressure, but once we are above the Chandrasekhar mass, degeneracy pressure is irrelevant. Now, as the star contracts, the temperature and density both increase, but ${\displaystyle P_{\rm {degen}}, and the core can get to arbitrarily high temperatures. This means the star can freely go through the cycle we described before, where the core contracts, temperature goes up, the core fuses heavier elements, the core runs out of the heavy element, and then the core contracts again. The difference this time is that this cycle cannot be stopped, and will proceed all the way to iron since it is the most tightly bound nuclei. Once the core is iron, the cycle is no longer useful, as fusing iron does not produce any energy. \\The timescales of each step in this cycle are shorter and shorter as the cycle proceeds. This occurs primarily because of the role that neutrino radiation plays in taking energy out of the core of a star. Once the central temperature of a star is above ${\displaystyle 10^{9}}$ K, the dominant energy loss mechanism is thermal neutrinos, meaning neutrinos that are produced because the core is so hot, not neutrinos produced by fusion. The energy loss by neutrinos drives the fusion cycling to happen very quickly, and cause the final fusion stages to last for only days or months. \\Normally, we have thought about the energy from fusion balancing the energy radiated away by photons, or

${\displaystyle L_{\rm {fusion}}=L_{\rm {rad}}.\,\!}$

Now, we all of the energy is carried out by neutrinos, so

${\displaystyle L_{\rm {fusion}}=L_{\nu }.\,\!}$

Since the mean free path of neutrinos is large compared to the size of the star, this energy easily escapes. \\One means of produces neutrinos is from an electron and positron annihilating, via

${\displaystyle e^{-}+e^{+}\rightarrow \nu _{e}+{\bar {\nu _{e}}}.\,\!}$

Typically, what happens in annihilation is photon production, since the electromagnetic force is so much stronger than the weak force, with

${\displaystyle e^{-}+e^{+}\rightarrow \gamma +\gamma .\,\!}$

Since neutrinos get out easily and photons do not, neutrino production is still important even though it happens more rarely. Where do the positrons come from? The inverse of the above step:

${\displaystyle \gamma +\gamma \rightarrow e^{-}+e^{+}.\,\!}$

The energy required for this reaction is the rest mass energy of the electron and positron. This limits this process to temperatures with ${\displaystyle kT\sim 1}$ MeV, corresponding to ${\displaystyle T\sim 10^{10}}$ K. This sort of process can happen with any particle, as long as things like charge and lepton number are conserved, and the energy of the photons is sufficient. It happens primarily for electrons and positrons because they have lower rest mass energies. If we could get to ${\displaystyle T\sim 10^{13}}$ K, then protons and antiprotons could be produced in this way. \\These reactions are in equilibrium, meaning we can use the Saha equation. Since the chemical potential of photons is zero, we conclude that

${\displaystyle \mu (e^{+})=-\mu (e^{-}).\,\!}$

If they are both non-degenerate and non-relativistic, then

${\displaystyle m_{e}c^{2}-kT\ln \left({\frac {gn_{Q}}{n_{e^{-}}}}\right)=-m_{e}c^{2}+kT\ln \left({\frac {gn_{Q}}{n_{e^{+}}}}\right).\,\!}$

Rearraging, we get

${\displaystyle 2m_{e}c^{2}=kT\ln \left({\frac {g^{2}n_{Q}^{2}}{n_{e^{-}}n_{e^{+}}}}\right).\,\!}$

Solving for number densities gives

${\displaystyle n_{e^{-}}n_{e^{+}}=g^{2}n_{Q}^{2}e^{-2m_{e}c^{2}/kT}.\,\!}$

There are two potential cases here. One is that electrons primarily come from ionized ions, and the other is that they primarily come from this process. Let’s look at the first case. The number density of electrons from ions is

${\displaystyle n_{e^{-}}={\frac {\rho }{\mu _{e}m_{p}}}\approx 3\times 10^{29}\rho _{6}{\rm {cm^{-3}}},\,\!}$

where ${\displaystyle \rho _{6}}$ is the density in units of ${\displaystyle 10^{6}}$ in cgs. In this case, we can substitute this number density for the number density in our Saha solution, and

${\displaystyle n_{e^{+}}={\frac {8\times 10^{28}}{\rho _{6}}}T_{9}^{3}e^{-11.9/T_{9}}{\rm {cm^{-3}}}.\,\!}$

This gives

${\displaystyle {\frac {n_{e^{+}}}{n_{e^{-}}}}=10^{-6}{\rm {\;at\;10^{9}\;K}},\,\!}$
${\displaystyle {\frac {n_{e^{+}}}{n_{e^{-}}}}=10^{-2}{\rm {\;at\;2\times 10^{9}\;K}},\,\!}$
${\displaystyle {\frac {n_{e^{+}}}{n_{e^{-}}}}=1{\rm {\;at\;4\times 10^{9}\;K}}.\,\!}$

Of course, this last result violates our initial assumption that most of the electrons are coming from ions. So for temperatures above a few times ${\displaystyle 10^{9}}$ K, we just set the number density of positrons equal to the number density of electrons, and get

${\displaystyle n_{e^{+}}=10^{29}T_{9}^{3/2}e^{-6/T_{9}}{\rm {\;cm^{-3}}}.\,\!}$

Now, we can use this to find the production rate of neutrinos. The neutrino generation rate is

${\displaystyle r={\frac {n_{e^{+}}}{\tau }},\,\!}$

where ${\displaystyle \tau }$ is the timescale for collisions,

${\displaystyle \tau ={\frac {\ell }{v}}={\frac {1}{n_{e^{-}}\sigma v}}.\,\!}$

Then the rate is

${\displaystyle r=n_{e^{+}}n_{e^{-}}\sigma v.\,\!}$

In order to get an energy generation rate, we need to multiply by the energy taken away by neutrinos, which is of order the rest mass energy of the positron and electron. The energy per time per mass is then

${\displaystyle \epsilon _{\nu }={\frac {n_{e^{+}}n_{e^{-}}}{\rho }}\sigma v(2m_{e}c^{2}).\,\!}$

Without worrying to much about weak physics, we’ll just assume the cross section is of order ${\displaystyle 10^{-20}\sigma _{T}}$, since the weak interaction is weaker than the electromagnetic interaction by roughly that factor. Plugging in our constants and number density expressions, we find the neutrino energy generation rate is

${\displaystyle \epsilon _{\nu }=4\times 10^{18}{\frac {T_{9}^{3}}{\rho }}e^{-11.9/T_{9}}{\rm {\;erg/s/g}}.\,\!}$

To estimate the total luminosity, we can write

${\displaystyle L_{\nu }=\int \epsilon _{\nu }dM.\,\!}$

Since we are concerned with the core, which has a size of order the size of a white dwarf, this is

${\displaystyle L_{\nu }={\frac {4}{3}}\pi R_{WD}^{3}\rho \epsilon _{\nu }.\,\!}$

Plugging in numbers, we find

${\displaystyle L_{\nu }10^{12}T_{9}^{3}e^{-11.9/T_{9}}\left({\frac {R_{\rm {core}}}{R_{WD}}}\right)^{3}L_{\odot }.\,\!}$

How large is this? At ${\displaystyle 5\times 10^{8}}$ K, the luminosity from neutrinos is 10 ${\displaystyle L_{\odot }}$. This is fairly negligible compared to the photon luminosity of a massive star. At ${\displaystyle 10^{9}}$ K, this becomes ${\displaystyle 10^{7}L_{\odot }}$. This is considerable. Lots of luminosity is being radiated by neutrinos now, much more than from photons. At ${\displaystyle 2\times 10^{9}}$ K, ${\displaystyle L_{\nu }\sim 10^{10}L_{\odot }}$. This is huge, and is what drives the short time scale of fusion in massive stars. As stars fuse more massive elements, the Coulomb barrier becomes larger, requiring higher temperatures for fusion. The higher the temperature, the larger the neutrino losses. The larger neutrino losses drive the temperature even higher so that ${\displaystyle L_{\nu }=L_{\rm {fusion}}}$. \\We can roughly estimate this analytically. The timescale is, as always, the energy divided by the luminosity,

${\displaystyle t_{\rm {nuc}}={\frac {E_{\rm {fusion}}}{L_{\nu }}}.\,\!}$

We can rewrite this as

${\displaystyle t_{\rm {nuc}}={\frac {n_{i}Q}{\rho \epsilon _{\nu }}}.\,\!}$

For carbon fusion, ${\displaystyle Q\sim 14}$ MeV, and ${\displaystyle T=10^{9}}$ K. This gives a timescale of ${\displaystyle 10^{3}}$ years. If the temperature is increased by a factor of two, this timescale decreases by a factor of 1000, to 1 year. \\For a 25 ${\displaystyle M_{\odot }}$ star, the timescales are summarized in the table below.

 Fusion Stage ${\displaystyle t_{\rm {nuc}}}$ ${\displaystyle L_{\nu }/L_{\gamma }}$ C ${\displaystyle 10^{3}}$ years 10 Ne 1 year 6000 O 1 year ${\displaystyle 2\times 10^{4}}$ Si 1 day ${\displaystyle 3\times 10^{6}}$

All of the reactions towards iron at the very end happen in what is called nuclear statistical equilibrium, which is basically just Saha for nuclei. All of the reactions leading to iron are happening in both directions since the temperatures are so high. The reason then that iron is produced is a consequence of the most bound nucleus being favored at “low” temperatures, where low means lower than ${\displaystyle 6\times 10^{9}}$ K. For equal numbers of protons and neutrons, the most bound nucleus is ${\displaystyle {}^{56}}$Ni, but in the cores of stars, there are slightly more neutrons than protons from electron capture, and the result is a core with ${\displaystyle {}^{56}}$Fe.