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<p>
<b>Materials</b>: Chapters 3 and 5 of Onno Pols' lecture notes, chapter 5 of
Kippenhan &amp; Weigert's book.
</p>

<div id="outline-container-orge907fb6" class="outline-2">
<h2 id="orge907fb6"><a href="#orge907fb6">Energy transport in stars</a></h2>
<div class="outline-text-2" id="text-orge907fb6">
</div>
<div id="outline-container-org062787d" class="outline-3">
<h3 id="org062787d"><a href="#org062787d">Summary of where we are</a></h3>
<div class="outline-text-3" id="text-org062787d">
</div>
<div id="outline-container-orgbfff812" class="outline-4">
<h4 id="orgbfff812"><a href="#orgbfff812">Mass conservation:</a></h4>
<div class="outline-text-4" id="text-orgbfff812">
<div class="latex" id="org65df13a">
\begin{equation}\label{eq:mass_cont}
\frac{dm}{dr} = 4\pi \rho r^{2} \ \ .
\end{equation}

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

<div id="outline-container-orgf96cd65" class="outline-4">
<h4 id="orgf96cd65"><a href="#orgf96cd65">Hydrostatic equilibrium:</a></h4>
<div class="outline-text-4" id="text-orgf96cd65">
<div class="latex" id="orga9b6c00">
\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>
where we use Eq. \ref{eq:mass_cont} to use a Lagrangian mass
coordinate as independent functional variable.
</p>

<p>
<b>N.B.:</b> This is the particular form that momentum conservation takes
since we can assume accelerations are &sim;0 because stars do not evolve
on timescales comparable to the dynamical timescale
</p>
</div>
</div>

<div id="outline-container-orgc460380" class="outline-4">
<h4 id="orgc460380"><a href="#orgc460380">Equation of state:</a></h4>
<div class="outline-text-4" id="text-orgc460380">
<div class="latex" id="orgda56c65">
\begin{equation}\label{eq:EOS}
P \equiv P(\rho, \mu, T) = P_\mathrm{rad} + P_\mathrm{gas} = \frac{1}{3}aT^{4} + \frac{\rho}{\mu m_{u}}k_{B}T + P_\mathrm{QM}\ \ .
\end{equation}

</div>

<p>
These three equations have as variables m &equiv; m(r) (or equivalently r&equiv;
r(m)), P, &rho;, and T. By adding an EOS to relate P and &rho; (and the
composition of the star which enters in the mean molecular weight &mu;),
in the general case, we have also added a new variable T, so the
system is not close again, and we need to look for one more equation
to be able to solve it! In today's lecture we will actually derive
two: energy transport and energy conservation. However, today for the
former we will not cover yet all possible physical situations in
stars.
</p>
</div>
</div>
</div>

<div id="outline-container-orgf8e5ecd" class="outline-3">
<h3 id="orgf8e5ecd"><a href="#orgf8e5ecd">Energy flow in stars</a></h3>
<div class="outline-text-3" id="text-orgf8e5ecd">
<p>
To find another equation, we can consider how energy flows in a star.
We have already seen that the "surface" temperature (e.g., of the Sun),
and the average temperature (estimated using the <a href="./notes-lecture-VirTheo.html">virial theorem</a>) are
<i>not</i> the same, so we know that the star is <i>not</i> in global thermal
equilibrium. We also know, from similar arguments, that the average
temperature is higher than the surface temperature, so there should be
an energy flow from the interior outwards.
</p>

<p>
This energy flow can occur in three forms in a star:
</p>
<ol class="org-ol">
<li><b>Diffusion</b>: thermal energy can be moved by the random motion of
particles. This is often the dominant energy transport mechanism in
a star &#x2013; unless for some reason it is insufficient. The diffusion
of energy can be:
<ul class="org-ul">
<li><b>Radiative</b>, if the particles carrying the energy through their
random motion are photons</li>
<li><b>Conductive</b>: if it is the thermal motion of gas particles
(typically electrons, unless you are in a neutron star) that
carries the energy</li>
</ul></li>
<li><b>Convection</b>: in this case thermal energy is transported by bulk
motion of matter. This occurs as an instability if the other
means of energy transport are insufficient to carry the energy flux
required by the local stellar structure, and it will be the topic
of a <a href="./notes-lecture-convection.html">future lecture</a>.</li>
<li><b>Neutrino losses</b>: except for collapsing iron cores and neutron
stars, the stellar gas is transparent to neutrinos. As soon as one
is produced, it can free stream out of the star carrying away
energy effectively instantaneously (or more precisely on a
timescale &sim; R/v<sub>&nu;</sub> &cong; R/c comparable to the light crossing time
if neutrinos are non interacting and we neglect their mass, so they
are ultra-relativistic particles with v<sub>&nu;</sub>&cong; c).</li>
</ol>

<p>
The next <i>two</i> stellar structure equations to add to our system will
come from combining all of these together and applying conservation of
energy.
</p>
</div>
</div>

<div id="outline-container-org42b0e8f" class="outline-3">
<h3 id="org42b0e8f"><a href="#org42b0e8f">Diffusion processes</a></h3>
<div class="outline-text-3" id="text-org42b0e8f">
<p>
In general, the "diffusion approximation" is useful to describe the
net flux of "something" when the average path of the carrier of said
"something" is small compared to the lengthscale over which the
"something" is transported, that is the mean free path \(\ell\) is much smaller
than the size \(L\) of the region \(\ell \ll L\).
</p>

<p>
In this approximation, the net flux of this "something" is related to
the density of "something" by Fick's law:
</p>

<div class="latex" id="orgd07ea0a">
\begin{equation}\label{eq:diff}
\mathbf{J} = - D\nabla n \ \ ,
\end{equation}

</div>

<p>
where \(\mathbf{J}\) is the flux (which is in general a vector
quantity), \(D\) is the so called "diffusion coefficient", and \(\nabla n\) is
the gradient of the volumetric density of the "something" we are
considering. The negative sign appears because the flux is opposite to
the gradient ("something" flows from where there is a lot to where
there is less!). By dimensional analysis, regardless of what
"something" is (and what are its dimensions), you can see that:
</p>
<ul class="org-ul">
<li>[\(J\)] = [something]/(L<sup>2</sup>t) with L length dimension and t time;</li>
<li>[\(\nabla\)] &equiv; [\(d/dx\)] = 1/L;</li>
<li>[\(n\)] = [something]/L<sup>3</sup></li>
</ul>

<p>
Therefore, the diffusion coefficient has the dimensions of [\(D\)] =
L<sup>2</sup>/t &equiv; v &times; L, so we can expect \(D\) to be proportional to the average
velocity of the "something" that is diffusing times the length scale
for the motion of this something (which by hypothesis is small
compared to the length scale over which we want to study the motion of
this "something"). In fact this is almost all there is to know!
</p>

<p>
To get the correct pre-factor, let's say "something" are particles
moving in 3 dimensions and focus on one direction, say \(z\). If all
these particles have a mean free path \(\ell\) and an average velocity \(v\),
then the number that crosses a given coordinate \(z\) with a positive \(\delta
z\) as a function of time is:
</p>
<div class="latex" id="orgbafd9bc">
\begin{equation}
\frac{dN}{dt}(z) =\frac{1}{2} n \frac{1}{3}v \ \ ,
\end{equation}

</div>
<p>
where the first 1/2 factor comes from the fact that their motion is
random, so half the particles have negative \(\delta z\) and half have a
positive one, and we just want to count the latter, and the factor of
1/3 in front of their average velocity comes from the fact that the
motion is isotropic (see <a href="notes-lecture-EOS1.html#org36a9235">derivation of the pressure</a>).
</p>

<p>
If there is a gradient \(\partial n /\partial z\) of particles along the \(z\) direction,
the particles moving with positive \(\delta z\)  come from \(n(z-\ell)\)
while those moving with the negative \(\delta z\)  come from \(n(z+\ell)\), so the net
flux across \(z\) is
</p>

<div class="latex" id="org2975911">
\begin{equation}
J = \frac{dN}{dt}(z-\ell) -\frac{dN}{dt}(z+\ell) = \frac{1}{6}v\left(n(z-\ell)-n(z+\ell)\right)\simeq\frac{1}{6}v\left(-2\ell\frac{\partial n}{\partial z}\right) = -\frac{1}{3}v\ell\frac{\partial n}{\partial z} \ \ ,
\end{equation}

</div>
<p>
where the only approximation we use is a first order Taylor expansion
of the density \(n(z)\) assuming that the mean free path &ell; is small
compared to the scale of interest! Eq. \ref{eq:diff} is the
generalization in 3 dimensions of the above, where \(\nabla = (\partial_{x}, \partial_{y},
\partial_{z})\) for Cartesian coordinates, and \(D=v\ell/3\).
</p>
</div>

<div id="outline-container-org6890f80" class="outline-4">
<h4 id="org6890f80"><a href="#org6890f80">Energy transport by radiative diffusion</a></h4>
<div class="outline-text-4" id="text-org6890f80">
<p>
We have already calculated that <a href="notes-lecture-EOS1.html#org75d15c1">mean free path for photons</a>
\(\ell_{\gamma}\) and estimated that it is very small compared to the typical
size of stars (and the typical size of a resolution element in a
numerical simulation of a star!). Therefore, we can treat the energy
transport by photons in the diffusion approximation.
</p>

<p>
<b>N.B.</b>: if the star were a perfect black body, there would not be any
transport of energy by photons, because by definition the radiation
field would be isotropic, and the gradient of photon energy density
would be zero! In reality, we have already seen that stars are <i>not</i>
black bodies at the surface (in the atmospheric layers where \(\ell_{\gamma}\)
is not small) and neither they are in the interior because there is a
small deviation from LTE of the order of \(\ell_{\gamma} dT/dr\sim10^{-4}\) K.
While this is a small enough deviation that we can assume LTE to write
down an EOS, it is also big enough to introduce a non-negligible flux
of energy in the stars!
</p>

<p>
If the "something" that we are considering in our diffusion equation
is energy, then in Eq. \ref{eq:diff} \(J\rightarrow F_\mathrm{rad}\) is a energy flux of
radiative energy, and \(n\rightarrow u\) is the energy density. Moreover, in the
diffusion coefficient \(D\) the mean velocity of photons is \(v\rightarrow c\), and we
have already written \(\ell_{\gamma} = 1/\kappa_\mathrm{rad}\rho\) as a function of \(\rho\).
</p>

<p>
<b>N.B.:</b> today we will introduce different kinds of opacity \(\kappa\), \(\kappa_\mathrm{rad}\) is
the one impeding the diffusion of photons.
</p>

<p>
Thus, the radiative diffusion equation is
</p>
<div class="latex" id="orgea46165">
\begin{equation}
F_\mathrm{rad} = - \frac{1}{3}\frac{c}{\kappa_\mathrm{rad}\rho}\frac{du}{dr} \ \ ,
\end{equation}

</div>
<p>
where we use the spherical symmetry of the problem to explicit the
gradient and turn it into a total derivative. The radiation energy
density is \(u=aT^{4}\). We can then explicit these into our equation
obtaining:
</p>

<div class="latex" id="org6b0489f">
\begin{equation}
F_\mathrm{rad} = -\frac{4ac}{3\rho} \frac{T^{3}}{\kappa_\mathrm{rad}}\frac{dT}{dr} \ \ ,
\end{equation}

</div>

<p>
which can be turned into an equation for the temperature gradient.
This is a <i>local</i> quantity and it is valid in a region of the star where
the dominant energy transport is radiative diffusion only:
</p>

<div class="latex" id="orgd84585e">
\begin{equation}
\frac{dT}{dr} = -\frac{3}{4ac}\frac{\rho}{T^{3}}\kappa_\mathrm{rad} F_\mathrm{rad} \propto \kappa_\mathrm{rad} F_\mathrm{rad}.\ \ .
\end{equation}

</div>
<p>
<b>In a radiative region the temperature is proportional to the opacity
\(\kappa_\mathrm{rad}\) times the radiative energy flux!</b>
</p>

<p>
We can further rewrite the flux \(F_\mathrm{rad} = L_\mathrm{rad}/(4\pi r^{2})\). This introduces
the <i>local</i> luminosity \(L_\mathrm{rad}\equiv L_\mathrm{rad}(r)\) which is the rate (that is
per unit time) at which radiation transports energy through a surface
of radius \(r\) within the star (or in other words, the "power" that is
in the photon field at the location r):
</p>

<div class="latex" id="orgd4e3ab8">
\begin{equation}
\frac{dT}{dr} = -\frac{3}{16 \pi ac}\frac{\rho\kappa_\mathrm{rad}}{r^{2}}\frac{L_\mathrm{rad}}{T^{3}} \ \ .
\end{equation}

</div>

<p>
This is, for the case of radiative energy transport only, the extra
differential equation relating T and &rho;, but unfortunately it also
brings in a new variable, the local radiative luminosity \(L_\mathrm{rad}\).
</p>

<p>
<b>N.B.:</b> If radiative energy transport is the <b>only</b> energy transport
mechanism at radius \(r\) (that is no convection, no neutrino emission
and no electron conduction), then \(L_\mathrm{rad}(r)\equiv L(r)\) where
\(L(r)\) is the <i>total</i> luminosity at that location in the star. This is
in general not true because of the contribution by neutrinos and
convection: \(L(r) = L_\mathrm{rad} (r)+ L_\mathrm{cond} (r) + L_{\nu}
(r)+L_\mathrm{conv}(r)\).
</p>

<p>
<b>N.B.:</b> Yes, we are introducing yet two other variables, \(L_\mathrm{rad}\) and
\(\kappa_\mathrm{rad}\) here. We will have to write an equation for the former, but
fortunately for us \(\kappa_\mathrm{rad}\) is determined by atomic and particle
physics, as <a href="notes-lecture-kappa.html">we will see in the next lecture</a>. While this is an active
topic of research (including classified research for military
purposes&#x2026;), for stellar physics application we have lookup tables
for \(\kappa_\mathrm{rad}\equiv\kappa_\mathrm{rad}(T,\rho)\), and thus we will not count it as a
new variable after discussing the physics it represents.
</p>

<p>
Because of the assumption underpinning the diffusion approximation,
this is <i>not</i> the right equation whenever \(\ell_{\gamma}\) is not negligible
compared to the scale over which one wants to consider the gradient:
in the stellar atmosphere we need a more detailed approach requiring
to treat the radiative transfer.
</p>

<p>
Now, before looking at the equation for \(L_\mathrm{rad}\), it is useful to consider
next the case where energy is carried not by photons, but by the local
motion of particles, that is <b>conduction</b>.
</p>
</div>
</div>

<div id="outline-container-org5a5cf4b" class="outline-4">
<h4 id="org5a5cf4b"><a href="#org5a5cf4b">Energy transport by conduction</a></h4>
<div class="outline-text-4" id="text-org5a5cf4b">
<p>
Energy transport by diffusion, and especially conduction that is
diffusion of energy through particle motion, is not limited to stars.
For example, in a piece of metal left half in the Sun and half in the
shade, the thermal motion of particles (atoms, electrons, ions)
carries energy from the hotter parts to the colder ones, and the
transfer occurs through collisions between the loose electrons in the
metallic energy bands.
</p>

<p>
Conduction, although always present, is important only in certain kind
of stars. To demonstrate this, we can consider the diffusion
coefficient \(D\simeq v\ell/3\) and compare it to the radiative diffusion
coefficient \(D_\mathrm{rad} = c\ell_{\gamma}/3 =c/(3\kappa\rho)\). In the diffusion
coefficient \(D\), the velocity that appears is the thermal velocity of
the particles (\(v^{2}\simeq 2k_{B}T/m\) for a non-relativistic gas): at a
given temperature \(T\), the least massive particles are faster, and
will contribute more to the conduction of thermal energy. In a star,
this means the electrons are going to dominate conduction whenever
there is some.
</p>

<p>
The other thing to consider is the mean free path &ell;, but since the
collisional (Coulomb-scattering) \(\sigma \leq 10^{-18}\) cm<sup>2</sup>, the mean free path
\(\ell = 1/(n_{e}\sigma) \ll \ell_{\gamma}\) . Thus, since \(v\le c\) and
\(\ell\ll\ell_{\gamma}\) for most stars energy conduction by particle (electron)
collisions is sub-dominant compared to radiative energy.
</p>

<p>
Things are different though for degenerate electron gas (so inside
white dwarfs and neutron stars, but also evolved stellar cores that
are dense enough for degeneracy to occur). In the case of degeneracy,
the thermal velocities increase (up to \(v \simeq c\) for an degenerate gas
of ultra-relativistic electrons!), and the mean-free path for
electron-electron scattering also increases, because for such a
scattering to be possible the final state must be available for an
electron to populate it, but in the case of (partial) degeneracy
(most) states in the "Fermi sea", with \(\varepsilon\le \varepsilon_\mathrm{Fermi}\) are <b>not</b>
available.
</p>

<p>
In general though, in (partially) degenerate layers of the star we
cannot neglect conduction, and it can dominate over radiative
diffusion even! To consider it, we can follow the same reasoning as
above and write an equation for the conductive flux
</p>
<div class="latex" id="orgf7fb92c">
\begin{equation}
F_\mathrm{cond} = - \frac{1}{3}\frac{c}{\kappa_\mathrm{cond}\rho}\frac{d T}{dr} \ \ ,
\end{equation}

</div>
<p>
where we are implicitly defining a "conductive opacity"
\(\kappa_\mathrm{cond}\) and assuming that the energy density of the gas is
proportional to the temperature \(T\) (not a big assumption, since we
know we are very close to LTE, so we can define a local \(T\)). With
this implicit definition of \(\kappa_\mathrm{cond}\) then we can just sum the
contribution to the energy flux from radiative diffusion and
conduction: \(F = F_\mathrm{rad} + F_\mathrm{cond}\) and
</p>

<div class="latex" id="orgd5d7b6e">
\begin{equation}
F = - \frac{1}{3}\frac{c}{\kappa\rho}\frac{d T}{dr} \ \ ,
\end{equation}

</div>
<p>
where now
</p>
<div class="latex" id="org296d108">
\begin{equation}\label{eq:kappas}
\frac{1}{\kappa} = \frac{1}{\kappa_\mathrm{rad}} + \frac{1}{\kappa_\mathrm{cond}} \ \ .
\end{equation}

</div>

<p>
In the absence of convection (which we will treat <a href="https://www.as.arizona.edu/~mrenzo/materials/Convection.pdf">later</a>) and neutrinos
(which leave the star instantaneously without further interaction,
unless it's a neutron star), this \(F_\mathrm{rad} + F_\mathrm{conv}\)
is the total energy flux.
</p>

<p>
From Eq. \ref{eq:kappas} we can infer an interpretation of these
radiative and conductive opacities, which is corroborating the
definition of \(\kappa\): the equation corresponds to the combination of two
resistances in parallel! \(\kappa_{i}\) is the "resistance" to the flow of
energy carried by radiation (\(i=\mathrm{rad}\)) or particle collisions
(\(i=\mathrm{cond}\)). The lowest resistance allows for the largest
energy flux, and the star will use that mechanism as the dominant
energy transport.
</p>

<p>
Moreover, since we have <i>defined</i> \(\kappa_\mathrm{cond}\) so that the
conductive flux has the same form as the radiative flux, we can (using
Eq. \ref{eq:kappas} and \(L(r) = L_\mathrm{rad} + L_\mathrm{cond}\))
continue the analogy and write down:
</p>

<div class="latex" id="orgec51222">
\begin{equation}
\frac{dT}{dr} = -\frac{3}{16 \pi ac}\frac{\rho\kappa}{r^{2}}\frac{L}{T^{3}} \ \ ,
\end{equation}

</div>

<p>
which is the radiative+conductive energy transport equation that
related \(T\), \(\rho\), and the new variable \(L\equiv L(r)\equiv L(m)\) we introduced and
depends on the opacity (radiative and conductive combined in parallel)
\(\kappa\), which we treat as a parameter dependent on atomic and condensed
matter physics (\(\kappa\equiv\kappa(T,\rho)\)).
</p>

<p>
<b>N.B.:</b> For conduction, we have considered the motion of electrons as
the ions are "frozen in place" since \(v_{e} \gg v_\mathrm{ions}\).
However, this will quickly lead to a <i>local</i> charge imbalance! In stars
where conduction is important (typically at least partially
degenerate) there will be a small but non-zero electric field created
by this charge imbalance that slows down the electrons, until their
motion is such that there is a net transfer of their thermal energy
without any net motion of electrons!
</p>
</div>
</div>
</div>

<div id="outline-container-orgd3aee21" class="outline-3">
<h3 id="orgd3aee21"><a href="#orgd3aee21">Local energy conservation</a></h3>
<div class="outline-text-3" id="text-orgd3aee21">
<p>
Let's finally write an equation for the <i>local</i> luminosity in a star \(L\)
that we have introduced above. Since the luminosity is just the local
"power", it makes sense to look into the <i>local</i> energy conservation to
derive such equation. For a unit mass, the "first law of
thermodynamics" states that the change \(d u\) in internal energy (the
specific internal energy) is equal to the heat added/extracted \(d q\)
plus the work done on the unit mass \(Pd v\) with \(v=1/\rho\) the specific
volume:
</p>

<div class="latex" id="org5a78b3c">
\begin{equation}
d u = d q + Pd v \equiv d q - \frac{P}{\rho^{2}}d\rho \ \ ,
\end{equation}

</div>
<p>
where we express things as a function of the density &rho; which already
appears in the other equations.
</p>

<ul class="org-ul">
<li><b>Q</b>: if we compress the gas (\(d\rho > 0\) because \(\rho\) increases), without
adding/extracting heat (\(d q = 0\)) what happens to the internal
energy?</li>
</ul>

<p>
The heat term in a star can only be due to:
</p>
<ul class="org-ul">
<li>energy generation by an internal source (nuclear fusion!), which can
release per unit mass and time energy equal to \(\varepsilon_\mathrm{nuc}\) ([\(\varepsilon_\mathrm{nuc}\)] =[E]/([t][M])).</li>
<li>energy loss by some particle escaping, this can be for example
neutrinos &nu;. Neutrinos in a star can come from nuclear reactions and
they effectively just reduce \(\varepsilon_\mathrm{nuc} \rightarrow \varepsilon_\mathrm{nuc} -
  \varepsilon_{\nu, \mathrm{nuc}}\), or they can come from so-called <b>cooling
processes</b>, for example \(e^{-} +\gamma \rightarrow e^{-} + \nu + \bar{\nu}\),
which really decrease the energy by extracting internal energy,
since as soon as they are produced neutrinos will leave the star
with no further interaction (with the exception of neutron stars).
The neutrino energy cooling rate per unit mass is indicated by \(\varepsilon_{\nu}\)
and it has always a <b>negative</b> contribution to the heat (it's a loss
term for the star)</li>
<li>energy can flow in and out from the boundary of a thin shell of
matter. Above, we have defined: \(L = 4\pi r^{2} F\) (where now both \(L\)
and \(F\) include the contribution from conduction and radiation).
Therefore, the energy per unit time coming from below is \(L\equiv L(m)\)
and the energy per unit time leaking from above is \(L(m+dm)\), to
obtain the specific energy per unit mass we just divide by \(dm\).</li>
</ul>

<p>
Putting all these together we have, at a given mass location m
</p>
<div class="latex" id="orgdf42590">
\begin{equation}
dq(m) = \varepsilon_\mathrm{nuc}(m) dt -\varepsilon_{\nu}(m)dt+(L(m)-L(m+dm))dt/dm \simeq \varepsilon_\mathrm{nuc}(m) dt -\varepsilon_{\nu}(m)dt-\frac{dL}{dm}dt \ \ .
\end{equation}

</div>
<p>
Thus, substituting in the local energy conservation we obtain:
</p>
<div class="latex" id="org0fcfbfb">
\begin{equation}
\frac{dL}{dm} = \varepsilon_\mathrm{nuc}(m) -\varepsilon_{\nu}(m) - \frac{du}{dt} +\frac{P}{\rho^{2}}\frac{d\rho}{dt} \ \ .
\end{equation}

</div>
<p>
Often the last two terms are combined together to define:
</p>
<div class="latex" id="orgab57a72">
\begin{equation}
\varepsilon_\mathrm{grav} = - \frac{du}{dt} +\frac{P}{\rho^{2}}\frac{d\rho}{dt} = -T\frac{ds}{dt} \ \ .
\end{equation}

</div>
<p>
which being a term dependent on \(dt\) it is usually small for a star in
a static (\(\partial_{t} \simeq 0\)) configuration. However, a star may
occasionally be out of thermal equilibrium (\(du/dt \neq 0\)) and/or
expanding or contracting (\(d\rho/dt\neq0\)). This will change the internal
state of the gas, and that is why it is often convenient to write
things in terms of the (specific) entropy \(s\). Moreover, since most
often this occurs because of contraction/expansion of a star,
historically this has been called \(\varepsilon\) "grav", although it really has
more to do with the internal energy of the gas. With this definition,
the next equation of stellar structure becomes
</p>

<div class="latex" id="orge3374c8">
\begin{equation}
\frac{dL}{dm} = \varepsilon_\mathrm{nuc} -\varepsilon_{\nu} + \varepsilon_\mathrm{grav} \ \ .
\end{equation}

</div>

<p>
<b>N.B.:</b> In regions where no energy is produced (\(\varepsilon_\mathrm{nuc} = 0\)), there are no
neutrino losses (\(\varepsilon_{\nu}=0\)) and in thermal equilibrium (\(\varepsilon_\mathrm{grav} =
T\partial s/\partial t = 0\)), the luminosity is constant as one moves inward or
outward in mass coordinate: \(dL/dm = 0 \Rightarrow L = constant\).
</p>

<ul class="org-ul">
<li><b>Q</b>: can you think of a region of the Sun with constant \(L(m)\)?</li>
</ul>


<p>
<b>N.B.:</b> Once again, we found another equation but it comes with new
unknowns. \(\varepsilon_\mathrm{grav}\) is fortunately only dependent on the thermodynamics
of the gas, so with the EOS we can calculate that (the specific
entropy is also a function of \(\rho\) and \(T\)). The other two terms
instead are input physics for the star. We will <a href="https://www.as.arizona.edu/~mrenzo/materials/nuclear_reaction_rates.pdf">later</a> unpack more
\(\varepsilon_\mathrm{nuc}\) by discussing nuclear energy generation &#x2013; but
ultimately it will depend on cross sections for nuclear interactions
which in stellar physics are taken as known input physics (again
coming often from military research). Similarly, \(\varepsilon_{\nu}\) depends on
neutrino physics and contains many neutrino loss terms. We will
discuss also these a bit more later on, but effectively in stellar
physics \(\varepsilon_{\nu}\) is also a quantity that we assume to know as a function
of \(T\) and \(\rho\), borrowing the work of neutrino and particle physicists.
</p>

<p>
So, in total at this point, we have \(\kappa\), \(\varepsilon_{\nu}\),
\(\varepsilon_\mathrm{nuc}\) assumed to be known input physics, and we have an
equation for the local conservation of energy, and the energy
transport in the case of diffusion (mediated by photons or particles,
i.e. conduction).
</p>

<p>
We still need an equation for the convective energy transport, and
while unpacking \(\varepsilon_\mathrm{nuc}\) we will write a set of equations for
the chemical evolution due to nuclear burning, but we are getting
close!
</p>
</div>
</div>
</div>

<div id="outline-container-org6841b72" class="outline-2">
<h2 id="org6841b72"><a href="#org6841b72">Homework</a></h2>
<div class="outline-text-2" id="text-org6841b72">
</div>
<div id="outline-container-orgf824451" class="outline-3">
<h3 id="orgf824451"><a href="#orgf824451">Eddington Luminosity</a></h3>
<div class="outline-text-3" id="text-orgf824451">
<p>
Consider an optically thick, hot, and stratified gas: this could be
(some layers of) a star, or a sufficiently dense accretion or
decretion flow to/from a compact object. Because of the assumption of
optical thickness, we can assume that the layer is in LTE and the
radiation field is well approximated by a black body. If the gas is
sufficiently hot, radiation pressure is the dominant term
(\(P_\mathrm{rad} \gg P_\mathrm{gas}\)).
</p>
<ol class="org-ol">
<li>Write \(dP_\mathrm{rad}/dr\) as a function of \(L\), \(\kappa\), and \(\rho\) and
\(r\) expressing \(dT/dr\) assuming energy is transported throughout
our layer of gas by radiative diffusion.</li>
<li>Impose hydrostatic equilibrium for this gas, and derive the
functional form for the luminosity (call it \(L_\mathrm{Edd}\))
required for radiation pressure in an optically thick gas to
balance out gravity.</li>
</ol>

<p>
The expression that you found was first derived by <a href="https://en.wikipedia.org/wiki/Arthur_Eddington">Arthur Eddington</a>,
assuming that \(\kappa \equiv \kappa_{es} = 0.2(1+X)\) g cm<sup>-2</sup>. In this derivation you did
not need to assume anything for \(\kappa\): the expression you derived is
sometimes referred to as "modified Eddington Luminosity". Because of
its dependence on \(\kappa\), which we will see can vary throughout the star,
it can occasionally occur that \(L_\mathrm{Edd} <L\): in this case
radiative energy transport and hydrostatic equilibrium cannot be
simultaneously satisfied - and this occurs in the envelope of massive
stars for example. For an (optically thick) accretion flow, this
luminosity corresponds to the limit when the in-falling material
liberates gravitational potential in the form of heat to the point
that the photons produced balance out the gravitational pull that
brings in the in-falling material in the first place.
</p>

<p>
<b>N.B.:</b> the only central hypothesis necessary to derive the Eddington
luminosity here is that the photons are a black body, that is an
optically thick environment is necessary.
</p>
</div>
</div>

<div id="outline-container-org3ffadc3" class="outline-3">
<h3 id="org3ffadc3"><a href="#org3ffadc3">Exercise 5.3 in Onno Pols' lecture notes</a></h3>
<div class="outline-text-3" id="text-org3ffadc3">
<p>
Without solving the stellar structure equations, we can already derive
useful scaling relations. In this question you will use the equation
for radiative energy transport with the equation for hydrostatic
equilibrium to derive a scaling relation between the mass and the
luminosity of a star.
</p>
<ol class="org-ol">
<li>Derive how the central temperature, \(T_\mathrm{center}\), scales
with the total mass \(M\), outer radius \(R\), and luminosity \(L\), for
a star in which the energy transport is by radiation. To do this,
use the stellar structure equation for the temperature gradient in
radiative equilibrium (<b>hint:</b> use the \(dT/dr\) form).</li>
<li>Assume that \(r \sim R\) and that the temperature is proportional to
\(T_\mathrm{center}\), \(L(m) \sim L\) and estimating \(dT/dr ∼
   −T_\mathrm{center} /R\).</li>
<li>Derive how \(T_\mathrm{center}\) scales with \(M\) and \(R\), using the hydrostatic
equilibrium equation, and assuming that the ideal gas EOS holds.</li>
<li>Combine the results obtained in 1. and 2., to derive how \(L\) scales
with \(M\) and \(R\) for a star whose energy transport is radiative throughout.</li>
</ol>

<p>
You have arrived at a mass-luminosity relation <i>without assuming
anything about how the energy is produced</i>, only about how it is
transported (by radiation). This shows that the luminosity of a star
is not determined by the rate of energy production in the centre, but
by how fast it can be transported to the surface: <b>stars don't shine
because they burn, conversely they burn because they shine!</b>
</p>
</div>
</div>
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