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<p>
<b>Materials</b>: Chapter 3 of Onno Pols' lecture notes, Chapter 14, 15, and
16 of Kippenhahn's book.
</p>

<div id="outline-container-org3bf8a56" class="outline-2">
<h2 id="org3bf8a56"><a href="#org3bf8a56">Equation of state 2/2: Quantum-mechanical effects</a></h2>
<div class="outline-text-2" id="text-org3bf8a56">
<p>
In the <a href="./notes-lecture-EOS1.html">previous lecture on EOS</a>, we have discussed polytropic EOS and
classical, non-relativistic, ideal gas. The latter is an accurate
description of the stellar gas in most, but importantly not <i>all</i> the
astrophysical cases.
</p>

<p>
Now, we will consider cases that don't verify the hypothesis we took
previously and discuss briefly some stars for which we need to go
beyond those.
</p>

<p>
Let's start by relaxing the hypothesis that the gas behaves
"classically". A star is "big", why do we need to care about quantum
mechanics (QM)? Besides its central role in allowing for nuclear
burning, <a href="./notes-lecture-nuclear-burning.html">which we will see later</a>, QM effects can matter also at
extremely "low" temperatures and/or "high" densities, which can be
encountered in stellar physics.
</p>

<p>
From the wave-nature of the solutions of Schroedinger's equation, we
know that there is a limit on the precision to which position and
velocity of a particle can be known ("Heisenberg's uncertainty
principle"): \(\Delta x \Delta p \ge h\) with \(h\) Planck's constant. This naturally
translates in three spatial dimension to
</p>

<div class="latex" id="orgd5b67c1">
\begin{equation}
\Delta x \Delta y \Delta z \Delta p_{x} \Delta p_{y} \Delta p_{z} \ge h^{3} \ \,
\end{equation}

</div>

<p>
where \(h^{3}\) is the volume of a quantized cell in phase-space! Thus the
available number of quantized states with momentum between \(p+dp\) within
a volume \(dV = dxdydz \equiv d^{3}x\) is
</p>

<div class="latex" id="orgb7a1f8b">
\begin{equation}
g(p)d^{3}pd^{3}x = g_{s} \frac{4\pi p^{2} dp dV}{h^{3}} \ \mathrm{with}\ p=\sqrt{p_{x}^{2} +p_{y}^{2} +p_{z}^{2}} .
\end{equation}

</div>
<p>
In the equation above, the factor \(g_{s}\) allows to account for "internal
degrees of freedom" of particles, e.g., their spin, polarization, or
isospin. The key point here is \(g(p)\propto p^{2}\) to satisfy QM. <b>N.B.:</b> the \(p^{2}\)
factor in \(g(p)\) comes from the Jacobian from going from Cartesian to
spherical polar coordinates in momentum space and assuming an
isotropic momentum distribution (\(dp_{x}dp_{y}dp_{z} \equiv d^{3}p = 4\pi p^{2} dp\)).
</p>

<p>
Let's start by considering a classical gas in non-extreme regime. Then
the particles are distributed according to a Maxwell-Boltzmann
distribution \(n(p)\), but whenever \(n(p)> g(p)\), we know this is going
to violate QM!
</p>

<p>
From the <a href="notes-lecture-EOS1.html#org753adea">previous lecture on EOS</a> we have already seen the
Maxwell-Boltzmann distribution
</p>

<div class="latex" id="orgb04bd17">
\begin{equation}
n(p)\propto \frac{n}{(mT)^{3/2}} \exp\left(\frac{-p^{2}}{2mk_{B} T}\right)p^{2 }\ \ ,
\end{equation}

</div>

<p>
where \(n\) is the number density, \(T\) the temperature, and \(m\) the mass of
the particles. Therefore:
</p>

<div class="latex" id="org1555656">
\begin{equation}\label{eq:momentum_ratios}
\frac{n(p)}{g(p)}\propto n (mT)^{-3/2}\exp(\frac{-p^{2}}{2mk_{B}T}) \ \ .
\end{equation}

</div>

<p>
We can see that:
</p>

<ul class="org-ul">
<li>at fixed temperature \(T\), for very high number densities \(n\), this ratio
is going to be larger than one in violation of QM</li>
<li>at fixed number density \(n\), for very low temperatures \(T\), this ratio
will be larger than 1 in violation of QM (because every term in the
Taylor expansion of the exponential is proportional to a negative
power of \(T\)).</li>
<li>the smaller mass particle will violate QM earlier than the higher
mass particles</li>
</ul>

<p>
Thus, we can expect that for "very cold" stars (we will define what is
the relevant scale here) or "very dense" stars the ideal gas EOS will not
be appropriate.
</p>

<p>
To account for QM effects, we need to consider the nature of the
particles making up the star, which can be either
</p>

<ul class="org-ul">
<li><p>
<b>Fermions</b> with semi-integer spin, such as electrons, protons,
neutrons (and protons and neutrons can be seen as two different
isospin states of a generic nucleon, this is useful for example to
discuss the interior composition of neutron stars). In this case,
the distribution function that determines the occupation of quantum
states of energy between \(\varepsilon\equiv\varepsilon(p)\) and \(\varepsilon+d\varepsilon\) is the Fermi-Dirac
distribution:
</p>

<div class="latex" id="org4510637">
\begin{equation}\label{eq:Fermi-Dirac}
 f_{FD}(\varepsilon) = \frac{1}{e^{(\varepsilon/k_{B}T - \eta)}+1} \le 1 \ \,
\end{equation}

</div>

<p>
where \(\eta=\mu/k_{B}T\) is the "degeneracy parameter" dependent on the
<i>chemical potential &mu;</i>, that is how much energy you need to "spend" to
create a new particle in an available energy level (<b>N.B.:</b> do not
confuse this with the mean molecular weight which has the same
symbol, the chemical potential has the dimension of an energy, while
the mean molecular weight is a dimensionless number!) and the
temperature. The fact that this is &le; 1 is an expression of Pauli's
exclusion principle, each quantized energy state can be occupied by
at most one fermion.
</p></li>

<li><p>
<b>Bosons</b> with integer spin, such as photons, or &alpha; particles. In this
case the relevant distribution is the Bose-Einstein's distribution:
</p>

<div class="latex" id="org5d79bc6">
\begin{equation}\label{eq:Bose-Einstein}
 f_{BE}(\varepsilon) = \frac{1}{e^{(\varepsilon/k_{B}T-\eta)}-1} \ \,
\end{equation}

</div>

<p>
which can be &ge; 1, meaning more than one boson can occupy the same
energy level (e.g., in the extreme case of a Bose-Einstein
condensate all particles occupy the level with lowest \(\varepsilon\) - this may
be relevant in the interior structure of neutron stars for example).
</p></li>
</ul>

<p>
The total number of particles with momentum between \(p\) and \(p+dp\) is
thus given by \(f(\varepsilon(p))g(p)dp\) for an appropriate choice of \(f\)
depending on the particle considered (\(f=f_\mathrm{FD}\) or
\(f=f_\mathrm{BE}\)). To determine the chemical potential &mu; one can
impose the normalization following from:
</p>
<div class="latex" id="org2e86132">
\begin{equation}
n = \int_{0}^{+\infty} f(\varepsilon(p))g(p)dp \ \ ,
\end{equation}

</div>
<p>
that is integrating the phase space distribution in momentum one
should find the spatial density \(n\) of particles.
</p>

<p>
Let's now consider a gas of electrons. These are the particles in the
ionized gas of the star that will first start feeling QM effects,
since \(m_{e} \ll m_{ion}\) (in fact \(m_{e} \simeq m_\mathrm{proton}/1836.15 \simeq
m_\mathrm{proton}/2000\)), cf. Eq. \ref{eq:momentum_ratios}.
</p>

<p>
<b>N.B.:</b> In practice, the pressure provided by the <i>ions</i> never becomes
affected by QM directly: because of the \(m^{-3/2}\) term in the
Maxwell-Boltzmann distribution, this would require densities so high
that the ions would not exist as ions anymore. Instead the high
density would allow for electron captures onto the protons of the
ions turning them in neutrons (\(e^{-} + p \rightarrow n + \nu_{e}\)). As we will see this
is what happens at the very end of the evolution of massive stars
that end their life exploding and leaving a neutron-star: those
electron capture reactions are where the neutrons come from! The
pressure provided by the neutrons does depend on QM effects and it
is what sustains the structure of the neutron stars!
</p>

<p>
Because the pressure term from electrons (which in the ideal gas EOS
is accounted for thanks to our definition of the mean molecular
weight) changes because of QM effects first, let's now consider a gas
made of electrons only. These particles have spin 1/2, thus they are
fermions, and obey Eq. \ref{eq:Fermi-Dirac} with \(q_{s} = 2\) (each
quantum cell of the phase space 4&pi; p<sup>2</sup>dp dV can be occupied by 2
electrons, one with spin up and one with spin down).
</p>
</div>

<div id="outline-container-orgc893a70" class="outline-3">
<h3 id="orgc893a70"><a href="#orgc893a70">Fully degenerate electron gas</a></h3>
<div class="outline-text-3" id="text-orgc893a70">
<p>
By definition, a fully degenerate gas is one where all the particles
are in the lowest possible energy state, corresponding to the limit
\(T\rightarrow0\). Of course, if \(T\equiv0\) there would be no cooling through
radiation, the object would not be a <i>star</i> anymore (it would be if
one wants a "black dwarf", a theoretical idea the Universe is too
young to have produced, e.g., <a href="https://ui.adsabs.harvard.edu/abs/2020MNRAS.497.4357C/abstract">Caplan 2020</a>). What we really mean by
taking the \(T\rightarrow0\) limit is that the thermal energy of the particles
is very small compared to the Fermi energy, i.e. the maximum energy
level that particles occupy (assuming to set the ground-level to
\(\varepsilon=0\)): \(k_{B}T \ll \varepsilon_{F}\). In this limit <i>the thermal and
mechanical properties of the gas decouple from each other</i>, and we
can assume \(T \simeq 0\) to discuss the mechanical properties, and
consider \(T\) only for the radiative properties.
</p>

<p>
For fermions (like the electrons we are focusing on), this means that
the electrons occupy a sphere in momentum space with a radius p<sub>F</sub>
called the "Fermi" momentum:
</p>

<div class="latex" id="org9d0cffb">
  \begin{equation}
  g_{e}(p)dp = q_{s}\frac{4\pi p^{2}}{h^{3}}dp \equiv \frac{8\pi p^{2}}{h^{3}}dp \ \  \mathrm{for} \ \
p\leq p_{F}  \ \  \mathrm{otherwise} \ \ 0 \ \ ,
  \end{equation}

</div>
<p>
and we used q<sub>s</sub>=2 for electrons. To find the value of p<sub>F</sub> we can use
the normalization coming from the total number density of electrons
</p>

<div class="latex" id="org2a8bf46">
\begin{equation}
n_{e} = \int_{0}^{+\infty} g_{e}(p)dp = \frac{8\pi}{3h^{3}}p_{F}^{3} \Rightarrow p_{F} = h\left(\frac{3}{8\pi}n_{e}\right)^{1/3} \ \ .
\end{equation}

</div>
<p>
Therefore, the <i>Fermi momentum depends only on the density of
electrons for a fully degenerate electron gas</i>.
</p>

<p>
We can now calculate the pressure exactly like we did for the
classical ideal gas (cf. Eq. 12 in <a href="notes-lecture-EOS1.html#org753adea">the Ideal gas section</a>), we just
need the appropriate p&equiv; p(&epsilon;) relaton
</p>
</div>

<div id="outline-container-org742687a" class="outline-4">
<h4 id="org742687a"><a href="#org742687a">Non-relativistic electron gas</a></h4>
<div class="outline-text-4" id="text-org742687a">
<p>
In this case \(\varepsilon = p^{2}/2m\) is the energy of the electrons (still ideal
gas) and \(v=p/m\), thus from the previous lecture on EOS we have:
</p>
<div class="latex" id="orgbc3f138">
\begin{equation}
P_{e} = \frac{1}{3}\int_{0}^{p_{F}} \frac{8\pi}{h^{3}}p^{2}\frac{p}{m_{e}} p  dp = \frac{8\pi}{15 h^{3} m_{e}}p_{F}^{5} \equiv \frac{h^2}{20m_{e}}\left(\frac{3}{\pi}\right)^{2/3} n_{e}^{5/3} \Rightarrow P_{e} \propto \rho^{5/3}\ \ .
\end{equation}

</div>

<p>
Note the functional form \(P_{e}\equiv P_{e}(n_{e})\): it's a powerlaw, like we
arbitrarily assumed would be a decent guess when discussing
polytropes. <i>A fully degenerate classical electron gas has a
polytropic EOS with exponent</i> \(\Gamma=5/3\).
</p>

<p>
<b>N.B.:</b> \(n_{e} \propto \rho\) with the definition of the electrons mean
molecular weight such that \(n_{e} \mu_{e} m_{u} = \rho\).
</p>
</div>
</div>

<div id="outline-container-org69ba73e" class="outline-4">
<h4 id="org69ba73e"><a href="#org69ba73e">Ultra-relativistic electron gas</a></h4>
<div class="outline-text-4" id="text-org69ba73e">
<p>
As the number density of electrons increases, \(p_{F}\) increases, and
thus at some point the \(v=p/m\) we used above will not hold anymore,
because the electrons become relativistic. In the extremely
relativistic limit, we can assume \(v=c\) (i.e. neglect the rest
energy of the electrons in the \(p(\varepsilon)\) relation), and then we lose
one power of \(p\) in the integral above. Thus, in the /fully
degenerate ultra-relativistic gas, the EOS will again be a polytrope
with exponent now/ \(\Gamma=4/3\).
</p>

<p>
Specifically the calculation yields:
</p>
<div class="latex" id="org08c18e0">
\begin{equation}
P_{e} = \frac{1}{3}\int_{0}^{p_{F}} \frac{8\pi}{h^{3}}p^{2}c p  dp = \frac{hc}{8}\left(\frac{3}{\pi}\right)^{1/3} n_{e}^{4/3} \Rightarrow P_{e} \propto \rho^{4/3} \ \ .
\end{equation}

</div>

<p>
In general, we should expect a <b>smooth</b> transition between these two
polytopes as \(n_{e}\) increases. Since the density in a star increases
towards the center, we can expect this transition to occur as we
move inwards in a star where these effects matter. In this case, we
need to use the relativistic formula \(p^{2} = \varepsilon^{2}-m_{e}c^{2}\) to solve the
integral and obtain the pressure.
</p>

<p>
One can estimate the density at the transition with the
condition \(p_{F} \simeq m_{e} c\):
</p>

<div class="latex" id="org919caff">
\begin{equation}
 \rho_{NR \rightarrow UR} \simeq \mu_{e} m_{u} \frac{8\pi}{3} \left(\frac{m_{e}c}{h}\right)^{3} \ \ .
\end{equation}

</div>

<p>
<b>N.B.:</b> the density around which we expect a transition from
non-relativistic to ultra-relativistic gas only depends on \(\mu_{e}\) and
fundamental constants!
</p>
</div>
</div>
</div>
<div id="outline-container-orga50282e" class="outline-3">
<h3 id="orga50282e"><a href="#orga50282e">Partial degeneracy</a></h3>
<div class="outline-text-3" id="text-orga50282e">
<p>
The equations derived above are valid in the strict limit of \(T=0\),
necessary for <b>full</b> degeneracy. In reality it is sufficient to have
\(k_{B}T \ll \varepsilon_{F} = p_{F}^{2}/2m\) (which defines what is "cold" enough
to get QM effects on the pressure contribution for non-relativistic
electrons). This is equivalent to asking \(\eta\gg 1\) with \(\eta\) electron
degeneracy parameter (cf. Eq. \ref{eq:Fermi-Dirac}).
</p>

<p>
The transition between ideal gas and fully degenerate gas goes
through partially degenerate gas, and in that case the degeneracy
pressure is harder to calculate analytically, and one needs to
calculate \(P = 1/3 \times \int n(p)pvdp\) using \(n(p) = g(p)f(\varepsilon(p))dp\) with
the Fermi-Dirac distribution for \(f\) (in the case of electrons).
</p>

<p>
For \(\eta \ll 1\) the Fermi-Dirac distribution can be Taylor expanded and
one recovers the ideal gas equation of state.
</p>

<p>
So, in summary, because electrons are Fermions that need to obey
Pauli's principle at very low \(T\) (comparing their kinetic energy to
the Fermi energy) and/or very high \(\rho\), they can exert a much larger
pressure than predicted by the classical ideal gas. Moreover, in
those situation, the pressure is a polytrope, independent of
temperature T! The decoupling between mechanical (hydrostatic
structure) and radiative (energy transport) properties of the star
afforded by degeneracy of the gas greatly simplifies the problem.
This also means the stars do not need to heat up anymore in order to
sustain themselves against their own gravity (breaking the
conclusion we obtained from the Virial theorem). This is the
situation of a "white dwarf" (WD), which are the remnants for the
vast majority of stars, including the Sun.
</p>

<p>
These compact objects contract and cool until they fully crystallize
(releasing further latent heat), becoming "planet-sized diamond-like
structures"! In the homework you will also see how there is a
maximum mass for a WD - the so-called Chandrasekhar mass, after the
Nobel-prize winning discovery by Subrahmanyan Chandrasekhar.
</p>
</div>
</div>

<div id="outline-container-orgdd8bf22" class="outline-3">
<h3 id="orgdd8bf22"><a href="#orgdd8bf22">Radiation pressure</a></h3>
<div class="outline-text-3" id="text-orgdd8bf22">
<p>
In some stars, the radiation field is so strong that is has a
non-negligible contribution to the pressure. The particles providing
that pressure are photons, which are <b>bosons</b> with 2 possible
polarization states, so \(g_{s}= 2\) (using an analogy with classical
electromagnetic wave language, this is because for a fixed
propagation direction of a wave the electric field can still be in
two directions, the two defining the plane orthogonal to the
propagation direction - this can be formally demonstrated using
relativistic quantum field theory).
</p>

<p>
Moreover, the number of photons does not need to be conserved,
radiative processes will destroy/create photons as needed to achieve
equilibrium: there is no chemical potential to overcome, thus \(\eta=0\).
</p>

<p>
Finally, noting that the photons are ultra-relativistic by
definition, we have \(\varepsilon = pc = h\nu\), and the Bose-Einstein distribution
in Eq. \ref{eq:Bose-Einstein} becomes the Black body distribution!
We can then calculate the internal energy density of the photon gas
as \(u_{int} = a T^{4}\) with a the radiation constant:
</p>

<div class="latex" id="org30670e1">
\begin{equation}
a = \frac{8\pi^{5} k_{B}^4}{15h^{3}c^{3}} = 7.56\times10^{-15} \mathrm{erg} \ \mathrm{cm}^{-3} \ \mathrm{K}^{-4} \ \ ,
\end{equation}

</div>

<p>
which is closely related to the Stefan-Boltzmann constant \(\sigma\): \(a=4\sigma/c\).
</p>

<p>
Relying again on the ultra-relativistic nature of photons, we know
that \(P=u_{int}/3\) and therefore the radiation pressure is:
</p>

<div class="latex" id="orgbedf7e4">
\begin{equation}
P_\mathrm{rad} =\frac{1}{3}aT^{4} \ \ .
\end{equation}

</div>
</div>
</div>

<div id="outline-container-org8746c66" class="outline-3">
<h3 id="org8746c66"><a href="#org8746c66">Partial ionization effects</a></h3>
<div class="outline-text-3" id="text-org8746c66">
<ul class="org-ul">
<li><b>Q</b>: So far we have assumed full ionization of the gas. What do you
think may change if we account for partial ionization? And where may
that be important?</li>
</ul>

<p>
<i>Ionization is the process of removal of an electron from an ion</i>, which
can be <b>collisional</b> (e.g., molecules/atoms bumping into each other in
the atmosphere charging a cloud and preparing it for lightning
discharge) or <b>radiative</b> (e.g., photoionization in the photoelectric
effect that won Einstein the Nobel prize).
</p>

<p>
For an element with \(Z\) electrons there are \(Z+1\) possible ions, from
the neutral atom to the fully stripped nucleus with no electrons
attached to it. For historical reasons, these are often indicated with
the element symbol followed by a roman number from I - for the neutral
atom to \(Z+1\) in roman numbers for the fully ionized ion, e.g., HII for
fully ionized hydrogen (cf. <a href="notes-lecture-CMD-HRD.html#org384d08b">digression on spectra in the CMD/HRD
lecture</a>).
</p>

<p>
So far in our discussion of the EOS, we have considered always this
last case of full ionization. Since the atomic binding energies are of
order of &sim;1-10 eV (think of the Rydberg, \(\chi=13.6\) eV to strip Hydrogen
of its electron from the fundamental state), that is 1 eV/\(k_{B}\) &sim;
10<sup>4</sup> \(K\), and most of the stellar material is hotter than this, this
was probably not a bad approximation: the (thermal) kinetic energy of
the particles flying around in the stellar gas are much larger than
what is needed to separate electrons and ions, so probably this will
happen a lot.
</p>

<p>
However, in the layers where \(T\) decreases, we can have partial
ionization, which <i>will change the number of particles per unit atomic
mass</i>, so you can expect this to <i>impact the mean molecular weight &mu;</i>,
and thus the pressure from the EOS (and we will see <a href="./notes-lecture-ETransport.html">later</a> also the
temperature gradient).
</p>

<p>
By definition the mean molecular weight \(\mu\) is such that \(\rho =
m_{u}\mu(n+n_{e})\). This is what we used in the ideal gas equation to get
\(P=\rho k_{B}T/(\mu m_{u})\) combining the electrons and ions pressure. Similarly we
can define \(\mu_{0}\) as the mean molecular weight per nucleus, and \(\mu_{e}\) as the
mean molecular weight per electron, and thus
</p>

<div class="latex" id="orge7ba8d4">
\begin{equation}
\rho = (n+n_{e})\mu m_{u}\equiv n\mu_{o}m_{u} \equiv n_{e}\mu_{e} m_{u} \ \ .
\end{equation}

</div>

<p>
We can also define the number of free electrons per ion/atom \(E=n_{e}/n\)
(where \(n_{e}\) is the number density of electrons and \(n\) the number
density of massive ions regarless of their ionization state), and thus
rewrite the above as
</p>

<div class="latex" id="org616bd8f">
\begin{equation}
\mu = \frac{\rho}{m_{u}n}\frac{1}{1+E} \equiv \frac{\mu_{0}}{1+E} \equiv \mu_{e}\frac{E}{1+E} \ \ .
\end{equation}

</div>

<p>
which gives the relation between the mean molecular weight(s) and the
number of free electrons. We will see in a <a href="./notes-lecture-radTrans.html">later lecture</a> how to
calculate \(E\) as a function of \(T\), and \(\rho\).
</p>
</div>
</div>
</div>

<div id="outline-container-org7e188d1" class="outline-2">
<h2 id="org7e188d1"><a href="#org7e188d1">Total pressure in a generic star</a></h2>
<div class="outline-text-2" id="text-org7e188d1">
<p>
Putting all things together:
</p>

<div class="latex" id="org2e7c635">
\begin{equation}
P_\mathrm{tot} = P_\mathrm{gas} + P_\mathrm{rad} = \frac{\rho}{\mu m_{u}}k_{B}T +
P_\mathrm{QM} + \frac{1}{3}aT^{4}  \ \ ,
\end{equation}

</div>
<p>
where we have decomposed the gas pressure into a degeneracy term due
to quantum effects and a classical term. \(P_\mathrm{QM}\) then depends
on whether the electron gas is non-relativistic, mildly relativistic,
or ultra-relativistic, while ions do not contribute to \(P_\mathrm{QM}\)
</p>

<p>
Note that in practice, stellar evolution code often rely on <i>tabulated</i>
EOS, which account for many non-ideal effects that we have only
briefly discussed here. EOS are ultimately one of the points of
contact between stellar physics and atomic physics and statistical
mechanics:
</p>


<figure id="org9933016">
<img src="./images/EOS_blend_paxton18.png" alt="EOS_blend_paxton18.png" width="100%">

<figcaption><span class="figure-number">Figure 1: </span>Blend of tabulated EOS on the T(&rho;) plane used in MESA (Fig. 50 in <a href="https://ui.adsabs.harvard.edu/abs/2018ApJS..234...34P/abstract">Paxton et al. 2018</a>), see also <a href="https://ui.adsabs.harvard.edu/abs/2021ApJ...913...72J/abstract">Jermyn et al. 2021</a> for updates relevant to large portions of this plane. The blue and purple tracks correspond to evolved stellar models of the mass labeled.</figcaption>
</figure>


<p>
A typical issue is how to obtain numerically good derivatives from
tabulated EOS, especially at the boundaries between tables coming from
different studies. These can often be a severe limiting factor in the
numerical accuracy of stellar models, and this was one of the
motivation for the development of a new EOS covering large portions of
the T(&rho;) plane (<a href="https://ui.adsabs.harvard.edu/abs/2021ApJ...913...72J/abstract">Jermyn et al. 2021</a>) now used by default in MESA.
</p>
</div>
</div>

<div id="outline-container-orgb678ba7" class="outline-2">
<h2 id="orgb678ba7"><a href="#orgb678ba7">Homework</a></h2>
<div class="outline-text-2" id="text-orgb678ba7">
<ul class="org-ul">
<li>Using the virial theorem, discuss which pressure term is more
important in the total pressure as a function of the mass (and
radius) of stars.</li>
<li>Derive an upper limit for the temperature T as a function of the
density &rho; for a star supported by fully degenerate
(non-relativistic) electrons, and plot this relation on a T(&rho;)
diagram. To explicit the relation between n<sub>e</sub> and &rho;, assume a
composition made of pure carbon (X<sub>i</sub> = 1 if carbon, 0 otherwise, Z<sub>i</sub> =
6, A<sub>i</sub> = 12). Any T much lower than this limit can be considered T&cong;0
for the purpose of the pressure calculation, but that still leaves a
large range of non-zero T from the radiative point of view!</li>
<li>Using the EOS for non-relativistic degenerate gas (and the other
stellar equations you know), determine a mass-radius relation for
stars entirely supported by (non-relativistic) electron degeneracy.
This is a good approximation for a white dwarf, the end point of the
vast majority (&gt;98%) of stars!</li>
<li>Clayton's problem 2-59: Let's now consider the case where electrons
are are ultra-relativistic, show that the central pressure scales as
P<sub>center</sub> &cong; 1.244 &times; 10<sup>15</sup> (&rho;/&mu;<sub>e</sub>)<sup>4/3</sup> dynes cm<sup>-2</sup>. Consider
the case where the electrons are ultra-relativistic <i>throughout</i> the
star, then P&cong; P<sub>center</sub> <i>throughout</i> the star as well. Using the mass
continuity equation and hydrostatic equilibrium, show that this
implies that the only mass that the ultra-relativistic electron gas
can sustain is M<sub>Chandrasekhar</sub> = 5.80 M<sub>☉</sub> &times; &mu;<sub>e</sub><sup>-2</sup> &cong;
1.44 M<sub>☉</sub> for &mu;<sub>e</sub> &cong; 2 (note the &mu;<sub>e</sub><sup>-2</sup> dependence!). What
does this specific value of the mass (for a given composition, i.e.,
&mu;<sub>e</sub>) mean for stars supported by ultra-relativistic electron
degeneracy pressure? What equation of stellar structure (of the ones
we have seen so far) <i>cannot</i> be satisfied for larger values of the
mass?</li>
</ul>
</div>
</div>
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