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<p>
<b>Materials</b>: Chapter 3 of Onno Pols' lecture notes, Sec. 2.2 and Chapter
5 and 13 of Kippenhahn's book, Sec. 2-4 of Clayton's book.
</p>

<div id="outline-container-org1670270" class="outline-2">
<h2 id="org1670270"><a href="#org1670270">Equation of state 1/2</a></h2>
<div class="outline-text-2" id="text-org1670270">
</div>
<div id="outline-container-orgb2a96f3" class="outline-3">
<h3 id="orgb2a96f3"><a href="#orgb2a96f3">Summary of where we are</a></h3>
<div class="outline-text-3" id="text-orgb2a96f3">
<p>
We have introduced now two equations of the stellar structure (<b>N.B.:</b>
at the end of the course we'll need 5 to close the system and get a
solvable model), and we know we can write them using the Lagrangian
mass coordinate, that is the mass enclosed in a certain radius, as
independent variable. But to reason conceptually about the problem, we
can stick to writing them as a function of r: the Lagrangian formalism
is important when doing numerical work, but less so for analytic
reasoning, when we can easily switch back and forth between r, m, or
any other variable as independent.
</p>

<p>
<b>Mass continuity:</b>
</p>
<div class="latex" id="org4c60c82">
\begin{equation}\label{eq:mass_cont}
\frac{dm}{dr} = 4\pi r^{2}\rho \Leftrightarrow  \frac{dr}{dm} = \frac{1}{4\pi r^{2} \rho}\ \ ,
\end{equation}

</div>
<p>
where we implicitly are assuming static equilibrium (&part;<sub>t</sub> = 0).
</p>

<p>
<b>Hydrostatic equilibrium:</b>
</p>
<div class="latex" id="org063e7f7">
\begin{equation}\label{eq:HSE}
\frac{dP}{dr} = -\frac{Gm}{r^{2}}\rho \Leftrightarrow \frac{dP}{dm}=-\frac{Gm}{4\pi r^{4}}\ \ .
\end{equation}

</div>

<p>
These two equations express the conservation of mass and momentum,
respectively (<b>N.B.:</b> it is often useful to write all equations in the
form of conservation laws to solve them numerically), and have as
variables the function \(m \equiv m(r)\) (or equivalently \(r\equiv r(m)\)), \(P\) and
\(\rho\). So clearly we cannot yet solve them, we need some relation
between the pressure and the density - a so-called <i>equation of state</i>
(EOS).
</p>
</div>
</div>

<div id="outline-container-org19af375" class="outline-3">
<h3 id="org19af375"><a href="#org19af375">General consideration on the EOS</a></h3>
<div class="outline-text-3" id="text-org19af375">
<p>
In general, this is a relation between the <i>local</i> value of pressure \(P\),
density \(\rho\), and temperature \(T\), which may also depend on the
composition, that is the ensemble of fraction by mass of elements (\(X_{i}\)
= mass of $i$-th element/total mass):
</p>

<div class="latex" id="org1b1c3df">
\begin{equation}\label{eq:general_EOS}
P\equiv P(\rho, T, \{X_{i}\})
\end{equation}

</div>

<p>
For a given composition \({X_{i}}\), given two out of three between \(P, \rho,
T\), one can obtain the third using the EOS.
</p>

<p>
The aim of this lecture is to introduce first simple EOS which do not
introduce any other variables (so independent of T), and then go over
a more appropriate EOS which is a good approximation for most stars,
the ideal gas EOS. We will see other EOS (for degenerate or
radiation-dominated gas) in a <a href="./notes-lecture-EOS2">future lecture</a>.
</p>
</div>
</div>

<div id="outline-container-org75d15c1" class="outline-3">
<h3 id="org75d15c1"><a href="#org75d15c1">Global and local thermal equilibrium</a></h3>
<div class="outline-text-3" id="text-org75d15c1">
<p>
Before we get into the EOS though, we have to ask ourselves whether we
can use thermodynamics/statistical mechanics to describe the
<i>microscopic</i> properties of the stellar gas and find some relationship
between \(P\), \(\rho\) (and other thermodynamic quantities such as \(T\), entropy,
etc). Are these even well defined?
</p>

<p>
Thermal equilibrium means by definition that all the particles in a
system (including photons, if there is radiation) have the same
temperature \(T\), and their velocity distribution are described by a
Maxwell-Boltzmann distribution (for particles) and the black-body
spectrum (for radiation). This situation can only be achieved if there
are sufficient interactions so each particle can exchange energy with
the others (collisions among particles, and scattering/absorption for
photon-particles interactions).
</p>

<ul class="org-ul">
<li><b>Q</b>: are stars in global thermal equilibrium? If they were, they would
have a single temperature throughout the star, does this sound
realistic?</li>
</ul>

<p>
<b>Stars are not in global thermal equilibrium</b>. They are in "<i>gravothermal
equilibrium</i>": the hydrostatic equilibrium requirement and the
self-gravity determine the thermal state of each layer so that the
pressure gradient can balance gravity.
</p>

<p>
However, <i>locally</i>, we can still assume that there are stellar layers
with a thickness \(dr\ll R\) that are in thermal equilibrium (with \(R\) outer
radius of the star, which we have seen to be an idealization, since in
reality \(R\equiv R(\lambda)\) depends on the wavelength &lambda; at which one observes).
For this to happen, this layer needs still to be much larger than the
mean-free-path for particle-particle and particle-photon interactions.
This allows for many interactions between the particles to occur and
establish <i>local</i> thermal equilibrium within the layer of thickness dr.
Let's consider the mean-free-path for particle-photon interactions first:
</p>

<div class="latex" id="org641be86">
\begin{equation}
\ell_{\gamma} = \frac{1}{n\sigma} \equiv \frac{1}{\kappa\rho} \ \ ,
\end{equation}

</div>

<p>
where n is the number density of particles ([\(n\)] = cm<sup>-3</sup>) and &sigma; is the
interaction cross section ([&sigma;] = cm<sup>2</sup>), or equivalently, &rho; is the mass
density (appearing in the two equations we already have), and &kappa; is the
"opacity", which is a way to measure the probability of a photon to
interact with a certain amount of mass.
</p>

<ul class="org-ul">
<li><b>Q</b>: what are the units of &kappa;? How can we interpret it?</li>
</ul>

<p>
Let's for simplicity assume that a star is all made of hydrogen (a
result first derived &sim;100 years ago by Cecilia Payne in her PhD thesis
in 1925). Let's also assume that it is fully ionized (i.e., protons
and electrons are separated from each other). The very minimum opacity
in this case is given by scattering of photons onto the electrons and
is: \(\kappa = 0.2(1+X)\) cm<sup>2</sup> g<sup>-1</sup> (see also <a href="./notes-lecture-kappa.html">lecture on opacities</a>), where \(X\)
is the mass fraction hydrogen, so for our rough approximation, &kappa;&cong;0.4.
The average density of the Sun is &rho;<sub>☉, avg</sub>&cong;1.4 g cm<sup>-3</sup>,
therefore &ell;<sub>&gamma;</sub> &cong; 2 cm. The mean-free-path for
particle-particle interactions \(\ell_\mathrm{b}\) is much smaller (because
the cross section &sigma; is larger): the interaction is electromagnetic and
has infinite range, in principle all charged particles feel all other
(we will return on this at the end of this lecture). Therefore, we can
conservatively ask what is the temperature variation across &ell;<sub>&gamma;</sub>:
</p>

<div class="latex" id="org56e5862">
\begin{equation}
\Delta T \simeq \ell_{\gamma} \frac{dT}{dr} \le \ell_{\gamma} \frac{T_\mathrm{center} - T_\mathrm{surface}}{R} \simeq \ell_{\gamma} \frac{T_\mathrm{center}}{R}
\end{equation}

</div>

<p>
(where we neglect \(T_\mathrm{surface}\) because we expect it much lower
than \(T_\mathrm\){center}$, since we already know \(P_\mathrm{center} \gg
P_\mathrm{surface}\)). For the Sun we will see that we can estimate
\(T_\mathrm{center}\simeq 10^{7}\) K, and we know \(R_{☉}\simeq 10^{11}\) cm, so
\(\Delta T\simeq 10^{-4} K \ll T_{surf} \ll T_{center}\).
</p>

<p>
Thus, we can define layers of the star with \(\ell_\mathrm{b} \ll \ell_{\gamma}
\ll dr \ll R\) where locally the temperature variation across the layer is
negligible, and the assumption of <i>local thermal equilibrium</i> (LTE)
holds well. (<b>N.B.:</b> you can revise this a posteriori once we have an
EOS and a way to estimate of \(T_\mathrm{center}\)!)
</p>

<p>
This may not be true close to the surface, where &ell;<sub>&gamma;</sub> progressively
increases (because &rho; decreases). There we may need to care about the
wavelength (or frequency) dependence of the cross section &sigma;&equiv;&sigma;(&lambda;)
and thus the mean free path: a star has a different "surface" for
different photon frequencies! In those layers, the assumption of LTE
does not work.
</p>

<p>
As long as we discuss regions of the star with a thickness large
w.r.t. &ell;<sub>&gamma;</sub> but small enough that the T variation across them are
negligible, we can use thermodynamics!
</p>

<p>
Also, as we have seen before, the stars are usually in equilibrium and
change very slowly. Any <i>local</i> thermal imbalance will be restored on
the <i>local</i> thermal timescale, which can be estimate in multiple ways,
but it is usually very short compared to the star evolutionary
timescale (we will see this in more detail later). Thus the assumption
of LTE holds locally (as the "L" indicates) at any time! Therefore,
while we cannot really define a physically meaningful \(T\) for the
entire star (beyond some rough averaged quantity), we can meaningfully
define a <i>local</i> \(T(r)\) or \(T(m)\) temperature that characterizes the gas
<i>and</i> the radiation at that radius/mass coordinate. The local validity
of the LTE assumption allows us to define thermodynamic quantities
from statistical mechanics and look for a relation between them, that
is an EOS.
</p>
</div>
</div>

<div id="outline-container-org79fc55c" class="outline-3">
<h3 id="org79fc55c"><a href="#org79fc55c">Polytropic EOS</a></h3>
<div class="outline-text-3" id="text-org79fc55c">
<p>
Polytropic EOS as a special case of <i>barotropic</i> EOS, which are all the
EOSs for which <i>the density depends only on the pressure and not on
the temperature or composition</i>: \(\rho\equiv\rho(P)\). These have applications
beyond stellar physics (e.g., for atmospheric physics). A polytropic
EOS is thus a barotropic EOS where the \(\rho\equiv\rho(P)\) is in the form of a
powerlaw, although usually written (following the general form of Eq.
\ref{eq:general_EOS}):
</p>

<div class="latex" id="org15f01d6">
\begin{equation}\label{eq:polytrope}
P = C\rho^{\Gamma} \equiv C\rho^{1+1/n}\ \ ,
\end{equation}

</div>

<p>
where C is a constant, and by definition \(\Gamma=\partial \ln P /\partial \ln \rho = 1+1/n\).
The form of the EOS in Eq. \ref{eq:polytrope} may seem particularly
artificial, but it is helpful because it does not introduce any new
variable, and thus "closes" (together with Eq. \ref{eq:mass_cont} and
Eq. \ref{eq:HSE}) the system of equations describing a star! This is
the system of equations used to create, for example, analytic stellar
models (e.g., the Lane-Emden equation), which is one of the topics for
the projects. These can be useful to initialize roughly correct
stellar structures in multi-dimensional hydrodynamic codes for
particular applications (e.g., a common envelope evolution
simulation!), but they are usually insufficient for modern stellar
physics application.
</p>

<p>
More importantly, there are various <i>physical</i> situations (as we will
return on during <a href="./notes-lecture-EOS2.html">the second lecture on EOSs</a>) in which EOS of this form
occur, and are useful to describe real observed stars.
</p>

<ul class="org-ul">
<li>fully convective stars (see <a href="./notes-lecture-conv.html">relative lecture</a>)</li>
<li>stars supported by quantum mechanical effects such as white dwarfs
(WDs)</li>
</ul>

<p>
Often, different polytropes in the form of Eq. \ref{eq:polytrope} can
be used for different layers of the stars (piece-wise polytropes) as
useful approximation. In this cases, it is important to ensure the
<i>continuity</i> of \(P\) at the points where different polytropes are
stitched together.
</p>

<ul class="org-ul">
<li><b>Q</b>: why do we want \(P\) to be a continuous function in stars?</li>
</ul>
</div>
</div>

<div id="outline-container-org36a9235" class="outline-3">
<h3 id="org36a9235"><a href="#org36a9235">Ideal gas</a></h3>
<div class="outline-text-3" id="text-org36a9235">
<p>
In general, it is not possible to have an EOS independent of \(T\)
and/or the composition \({X_{i}}\). Let's consider a <i>perfect gas</i>, that is a
gas of particles that do interact with each other (otherwise they
could never exchange energy and be in thermal equilibrium!), but such
as that the energy exchanged in the particle-particle interactions is
much small than their kinetic energy (due to their thermal motion)
</p>

<div class="latex" id="org642a127">
\begin{equation}
\Delta E_\mathrm{interaction} \ll k_{B} T \ \ ,
\end{equation}

</div>

<p>
and other than these interaction between particles, they are free to
move (no external potential).
</p>

<p>
To obtain the pressure of such gas, we need to consider the
distribution in velocity space of these particles. Let's for a moment
consider particles that all have the same mass m, we can then
equivalently consider the distribution in momentum p=mv of the
particles &#x2013; this is convenient to generalize later to relativistic
particles, and we will see how to deal with mixtures of gases (each
with particles m<sub>i</sub>) further down.
</p>

<p>
Since the particles of the ideal gas move isotropically within their
volume, the momentum distribution of particles is a Maxwell-Boltzmann
distribution. The number of particles with momentum between \(p\) and
\(p+dp\) is
</p>

<div class="latex" id="orgfd86851">
\begin{equation}\label{eq:Maxwellian}
n(p)dp = \frac{n}{(2\pi m k_{B} T)^{3/2}} \exp\left(-\frac{p^{2}}{2mk_{B}T}\right)4\pi p^{2}dp \ \ ,
\end{equation}

</div>

<p>
where on the r.h.s., \(n\) is the total number density, the prefactor of
the exponential is the normalization constant, and the exponential
comes from assuming a Gaussian distribution of kinetic energies for
each momentum component (you can demonstrate that a Maxwellian
distribution is equivalent to a Gaussian in each direction \(p_{x}\), \(p_{y}\),
\(p_{z}\), by going to spherical-polar coordinates in momentum space, thus
introducing a Jacobian for the change of variables, and integrating
over the \(p_{\theta}\), \(p_{\varphi}\) components).
</p>

<p>
From this, we can calculate the pressure of the gas. By definition
this will be isotropic, and so we can imagine to put a "box" with unit
linear size through our gas (the orientation of the walls does not
matter). To determine the gas pressure we want the force exerted by
the gas on the walls. By Newton's third law this is equal to the
change in momentum of each gas particles as they hit the walls. We
will first consider the momentum change for a generic single particle,
and then integrate over the distribution in Eq. \ref{eq:Maxwellian} to
get the whole pressure.
</p>

<p>
Let's call the imaginary wall the xy plane and assume the collisions
to be elastic (because we are considering an ideal gas, by definition
any exchange of energy is negligible). For a generic particle of the
gas, it will collide with the wall at an angle \(\theta \in [0, \pi/2]\) and the
collision will change its momentum from (\(p_{x}\), \(p_{y}\), \(p_{z}\)) to (\(p_{x}\),
\(p_{y}\), \(-p_{z}\)): only the component perpendicular to the wall changes
sign. Thus
</p>

<div class="latex" id="orgd4bf657">
\begin{equation}
\Delta p = 2 p \cos(\theta)\  \mathrm{with}\  p=\sqrt{p_{x}^{2} + p_{y}^{2} +p_{z}^{2}}
\end{equation}

</div>

<p>
The time between two collisions of a particle on the same wall is
</p>

<div class="latex" id="org48305b6">
\begin{equation}
\Delta t = \frac{2L}{v\cos(\theta)} = \frac{2}{v\cos(\theta)} \ \ ,
\end{equation}

</div>

<p>
where we used the \(L=1\) assumption. Thus the force exerted on this
imaginary wall of a unit box is \(F=\Delta p/\Delta t = vp cos^{2}(\theta)\), dividing
by \(L^{2} = 1\) we obtain the contribution to the pressure from one particle
coming from one specific direction &theta; (w.r.t. the wall of the box), and
introducing the distribution of particles in angle and momentum we
have
</p>


<div class="latex" id="org0a22d9e">
\begin{equation}
dP = vp\cos^{2}(\theta)n(\theta,p)d\theta dp \ \ ,
\end{equation}

</div>

<p>
But we can assume that the motion of the particles is isotropic
(spherical symmetry), so \(n(\theta,p)d\theta = n(p)sin(\theta)d\theta\), and thus
</p>

<div class="latex" id="org9fd506a">
\begin{equation}\label{eq:P_statistic}
dP = vp\cos^{2}(\theta)n(p)\sin(\theta)d\theta dp \Rightarrow P = \frac{1}{3}\int_{0}^{+\infty} n(p)p v dp \ \ ,
\end{equation}

</div>
<p>
where we integrated over \(\theta\). Combining Eq. \ref{eq:P_statistic} with Eq. \ref{eq:Maxwellian} gives the pressure.
</p>
</div>

<div id="outline-container-org8f9afa6" class="outline-4">
<h4 id="org8f9afa6"><a href="#org8f9afa6">Non-relativistic, classical gas</a></h4>
<div class="outline-text-4" id="text-org8f9afa6">
<p>
Let's consider a non-relativistic gas of classical particles (no
quantum effects). Then \(p= mv \Leftrightarrow v=p/m\) (<b>N.B.:</b> we are considering an
ideal gas, so each particle is freely moving, no potential of
interaction!). Carrying out the integral above using the
Maxwell-Boltzmann distribution for \(n(p)\) and \(v=p/m\) gives
</p>

<div class="latex" id="orgac7b5e5">
\begin{equation}
P = n k_{B} T \ \ .
\end{equation}

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

<div id="outline-container-org1964f41" class="outline-4">
<h4 id="org1964f41"><a href="#org1964f41">Mixture of (non-relativistic, classical) gases</a></h4>
<div class="outline-text-4" id="text-org1964f41">
<p>
Let's now say that we have multiple gas mixed, for example, a gas of
ions of various species and electrons. Each gas will contribute to the
pressure: \(P_{tot} = P_{ion, tot} + P_{e} = \sum_{i} P_{ion, i} + P_{e} = (\sum_{i}n_{i} +n_{e}) k_{B}T\),
where \(n_{i}\) is the number density of the ions \(i\), which have mass \(m_{u} A_{i}\)
with \(m_{u}\) the <i>atomic mass unit</i>:
</p>

<div class="latex" id="org6f3acf3">
\begin{equation}
m_{u} = 1 / N_{A} \,\mathrm{g} \simeq 1.66 \times 10^{-24} \,\mathrm{g} \ \ .
\end{equation}

</div>

<p>
Thus, we can relate the number density of the ions of species i with
the mass density that already appears in the equations we already have
with \(n_{i} = X_{i}\rho/(A_{i} m_{u})\). Note that we are implicitly using the fact
that everything has the same T because of the assumption of LTE!
</p>

<p>
We can re-write the total contribution of the ions defining the
ion <i>mean molecular weight</i> such that \(\mu_\mathrm{ion} \times m_{u}\) = "average mass of
ions", that is \(\mu_\mathrm{ion} n_\mathrm{ion} = \rho/m_{u}\) or \(n_\mathrm{ion} = \sum_{i} n_{i} = \sum_{i} X_{i}\rho/(A_{i}m_{u}) \equiv
\rho/(\mu_\mathrm{ion}m_{u})\) and:
</p>

<div class="latex" id="orgbdc00cc">
\begin{equation}
\frac{1}{\mu_\mathrm{ion}} = \sum_{i}\frac{X_{i}\rho}{A_{i}} \ \ .
\end{equation}

</div>

<p>
Similarly, we can define an electron mean molecular weight noticing
that to maintain a total charge of zero per unit volume, since each
ion carries charge \(+Z_{i}e\) and each electron as charge \(-e\): \((\sum_{i} Z_{i}n_{i} -
n_{e})e = 0\). Thus
</p>

<div class="latex" id="orgd3daa96">
\begin{equation}
\frac{1}{\mu_\mathrm{e}} = \sum_{i}\frac{Z_{i}X_{i}\rho}{A_{i}} \ \ .
\end{equation}

</div>

<p>
and we can define a combined mean molecular weight:
</p>

<div class="latex" id="orgc90eb12">
\begin{equation}
\frac{1}{\mu} = \frac{1}{\mu_\mathrm{ion}}+\frac{1}{\mu_{e}} \ \ ,
\end{equation}

</div>
<p>
So that the total pressure of a mixture of ideal, classical and
non-relativistic gas is
</p>

<div class="latex" id="orga02f419">
\begin{equation}\label{eq:ideal_gas_EOS}
P = \frac{\rho}{\mu m_{u}}k_{B}T
\end{equation}

</div>

<p>
The introduction of the <i>mean molecular weight</i> allows us to treat a
mixture of gases (assumed to be in LTE) as a single gas!
</p>

<p>
<b>N.B.:</b> this holds as long as every species satisfies our assumption of
ideal, non-relativistic, classical gas.
</p>
</div>
</div>

<div id="outline-container-orge2420ca" class="outline-4">
<h4 id="orge2420ca"><a href="#orge2420ca">Physical interpretation of \(\mu\)</a></h4>
<div class="outline-text-4" id="text-orge2420ca">
<p>
The <i>mean molecular weight</i> we have introduced above may seem a bit
arbitrary, but it has a clear physical interpretation: it is the
average number of particles per unit atomic mass \(m_{u}\).
</p>

<p>
For a fully ionized gas (i.e., where every ion is stripped of <i>all</i> its
electrons):
</p>

<div class="latex" id="orgec9f38b">
\begin{equation}
\mu = \frac{1}{\sum_{i}X_{i}\left(\frac{Z_{i}+1}{A_{i}}\right)} \ \ ,
\end{equation}

</div>
<p>
In fact, for every $i$-th species that corresponds to a fraction \(X_{i}\)
of the total mass, that is \(X_{i}/A_{i}\) in number density, we have \(Z_{i}\)
electrons plus one nucleus that contribute. This approximation is
usually good in the stellar interior, but as one moves outwards in the
star, \(P\) decreases, therefore by Eq. \ref{eq:ideal_gas_EOS}, \(T\)
decreases, and elements recombine, meaning the term \(Z_{i}+1\) is reduced.
This can have important consequences in certain layers of the stars
("partial ionization layers").
</p>

<p>
We can further simplify the expression for \(\mu\) by noting that for
hydrogen \((Z_{i} +1)/A_{i} = 2\) (i.e., hydrogen contributes 2 particles per
unit atomic mass, one proton and one electron), for helium \((Z_{i} +1)/A_{i}
= 3/4\) (we are actually considering only the stable isotope of helium
that contributes 3 particles every 4 atomic mass units, one nucleus
and 2 electrons), and for the vast majority of stable metals \(Z_{i} \gg 1\)
and \(Z_{i} \simeq A_{i}/2\) (i.e., most metals contribute per each \(A_{i}\) atomic
mass units roughly \(A_{i}/2\) particles which are the electrons that are
typically half as many as the nucleons). Therefore, for fully ionized
gas, we can simplify the mean molecular weight to:
</p>

<div class="latex" id="org95da4e2">
\begin{equation}
\mu \simeq \frac{1}{2X + \frac{3Y}{4} + \frac{Z}{2}} \ \ ,
\end{equation}

</div>
<p>
where \(X\), \(Y\), \(Z\) are the mass fraction of hydrogen, helium, and the
metallicity, respectively.
</p>
</div>
</div>
</div>

<div id="outline-container-orgb9e7853" class="outline-3">
<h3 id="orgb9e7853"><a href="#orgb9e7853">Relation between pressure and internal density</a></h3>
<div class="outline-text-3" id="text-orgb9e7853">
<p>
Eq. \ref{eq:P_statistic} can be used to relate \(P\) to the internal
energy density of a gas. This can be defined as:
</p>

<div class="latex" id="org470b991">
\begin{equation}\label{eq:E_statistic}
du_\mathrm{int} = \varepsilon(p)n(p)dp  \ \ ,
\end{equation}

</div>
<p>
with &epsilon;(p) the energy per particle.
</p>

<p>
If the particles in the (ideal) gas are non-relativistic, then
\(\varepsilon=p^{2}/2m\), thus in the term \(pv\) in Eq. \ref{eq:P_statistic} is \(pv = 2\varepsilon\),
and thus:
</p>

<div class="latex" id="org25a6de6">
\begin{equation}
P = \frac{2}{3} u_\mathrm{int} \ \ .
\end{equation}

</div>

<p>
If instead the gas is ultra-relativistic, then \(\varepsilon = pc\) (neglecting the
mass term in the energy since \(pc \gg mc^{2}\)), and thus:
</p>

<div class="latex" id="org02c4280">
\begin{equation}
P=\frac{u_\mathrm{int}}{3} \ \ .
\end{equation}

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

<div id="outline-container-org5bab33f" class="outline-3">
<h3 id="org5bab33f"><a href="#org5bab33f">Can we really use an ideal gas EOS in a plasma?</a></h3>
<div class="outline-text-3" id="text-org5bab33f">
<p>
This is legitimate as long as the interaction energy between the
particles are small compared to their kinetic energy. The dominant
interaction between the particles (ions and electrons) is going to be
through the Coulomb force, leading to energy exchanges of the order of:
</p>

<div class="latex" id="orgd931881">
\begin{equation}
\varepsilon_\mathrm{Coulomb} \simeq \frac{Z^{2}e^{2}}{d} \ \ ,
\end{equation}

</div>

<p>
for particles of charge \(Ze\) (the ions) and average distance \(d \simeq (4\pi
n/3)^{-1/3}\) with \(n\simeq\rho/Am_{u}\) number density. We want to compare this with the
kinetic energy, which for point-like particles is \(\varepsilon_\mathrm{thermal} = 3k_{b}T/2\).
The ratio of these two is often called the Coulomb parameter
(neglecting constants of order unity):
</p>

<div class="latex" id="orgb8eac94">
\begin{equation}
\Gamma_{C} = \frac{\varepsilon_\mathrm{Coulomb}}{\varepsilon_\mathrm{thermal}} \simeq \frac{Z^{2}e^{2}}{dk_{B}T} \simeq \frac{(Ze)^{2}}{k_{B}T}\left(\frac{\rho}{Am_{u}}\right)^{1/3} \simeq 2.275\times 10^{5} Z^{2} A^{-1/3}\left(\frac{\rho}{\mathrm{g\ cm^{-3}}}\right)^{1/3 }\left(\frac{T}{\mathrm{K}}\right)^{-1}\ \ .
\end{equation}

</div>
<p>
We can assume the ideal gas situation if \(\Gamma_{C} \ll 1\), which is the case
for the average \(T\) and &rho; of the Sun (we will estimate the average
temperature of the Sun in the <a href="notes-lecture-VirTheo.html#org6c2f766">next lecture</a>). We also see that at
progressively lower temperature the Coulomb interaction start to
matter (this is important for the crystallization of WDs for
instance), or at increasingly high densities.
</p>
</div>
</div>
</div>

<div id="outline-container-org4e4689e" class="outline-2">
<h2 id="org4e4689e"><a href="#org4e4689e">Homework</a></h2>
<div class="outline-text-2" id="text-org4e4689e">
<ul class="org-ul">
<li>We have discussed that the internal layers of the star are in LTE
using an argument based on the photons mean free path &ell;<sub>&gamma;</sub>.
Assuming a star of constant density (use the mean density for
this!), pure hydrogen composition, and that electron scattering is
the dominant interaction of the photons in the stellar interior so
&kappa;&cong;&kappa;<sub>es</sub>=0.2(1+X), using one-dimensional random-walk arguments,
estimate:
<ol class="org-ol">
<li>how many scatterings a photon created in the center of the Sun
will experience before coming out at the surface?</li>
<li>Knowing that photons travel at the speed of light c and assuming
scatterings to be instantaneous, what is the photon diffusion
timescale throughout the star? How does it compare to the
dynamical timescale?</li>
</ol></li>
<li>Run with <code>MESA-web</code> a 0.3M<sub>☉</sub> star up to 10<sup>8</sup> yrs, and plot the
P(&rho;) profile of the star at this age (<b>Hint</b>: it may be useful to plot it on
a log-log plot). Do you think it is possible to find a good
approximation of this profile with a polytropic relation? Note that
<code>MESA-web</code> does <b>not</b> assume a poytropic EOS! As usual, the deliverables
are the plot, the code used to make it, and a small paragraph of
text with your answer. <b>Extra</b>: you may even try to fit a polytrope
throughout the star and provide the polytropic index &Gamma;.</li>
</ul>
</div>
</div>
</div>
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