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\\The picture to have in mind is that there are lots of particles moving around and colliding, which deflects their motion and exchanges energy with other particles and causes a flow of energy from more energetic areas to less energetic areas. The same basic physics applies for energy transport by both photons and electrons. \\Particles have an energy density $U$ , a typical velocity $v$ , and a mean free path $\ell$ , where the mean free path represents the average distance traveled between exchanging energy. There is flux $F$ in some direction that is just the energy density multiplied by the velocity. Eliot drew a picture as a basis for the argument that the net flux is

$F={\frac {1}{6}}U(x-\ell )v-{\frac {1}{6}}U(x+\ell )v,\,\!$ $F=-{\frac {1}{3}}v\ell {\frac {dU}{dx}},\,\!$ $F=-{\frac {1}{3}}v\ell {\frac {dU}{dT}}{\frac {dT}{dx}},\,\!$ $F=-\kappa {\frac {dT}{dx}},{\rm {\;where}}\,\!$ $\kappa \sim {\frac {1}{3}}v\ell {\frac {dU}{dT}}.\,\!$ Now that we have gone through this, let’s define the mean free path a bit more concretely, specifically in terms of the cross section for collisions, $\sigma$ . For a flux of incoming particles, (an energy/area/time), we define the energy/time absorbed by the particles as the product of the flux and the cross section $\sigma$ :

$P=\sigma F.\,\!$ Once we have defined a cross section, we can determine how far a particle is likely to travel before interacting with another particle. This depends on the number density of particles as well, with

$\ell ={\frac {1}{n\sigma }}.\,\!$ Let’s estimate this for a particle in the Earth’s atmosphere. The density of air is $\rho \sim 10^{-3}$ g/cm$^{3}$ , which gives a number density of roughly $n\sim 10^{20}$ particles/cm$^{3}$ . The cross section for two neutral atoms or molecules is determined simply by the size of the constituents. For a typical molecule, the size is of order an Angstrom, or $10^{-8}$ cm. The mean free path for molecules in the atmosphere is then

$\ell \sim 10^{-4}{\rm {\;cm,\;or\;one\;micron}}.\,\!$ Let’s now focus on an ideal ionized gas. The energy per unit volume is ${\frac {3}{2}}nk_{B}T$ . The conductivity $\kappa$ is then

$\kappa \sim {\frac {1}{2}}v\ell nk_{b}.\,\!$ This is the appropriate conductivity for energy transport by electrons. For electrons in a gas,

$v\sim {\sqrt {\frac {kT}{m}}}.\,\!$ For the cross section, the Coulomb cross section is what is relevant. We define this in terms of the distance between an electron and a proton at which the proton causes a large deflection of the motion of the electron. This depends on the energy of the electron from thermal motion, and the electrostatic energy that the electron feels from a proton, which is

$E_{e}={\frac {e^{2}}{r}}.\,\!$ When this energy is similar to the thermal energy $k_{B}T$ , an electron will scatter. Thus significant deflections occur at

$b={\frac {e^{2}}{kT}},\,\!$ where $b$ is known as the impact parameter. The Coulomb cross section is then

$\sigma _{c}=\pi b^{2}=\pi {\frac {e^{4}}{k_{B}^{2}T^{2}}}.\,\!$ Note that this is independent of the mass of the particle. When this calculation is done more carefully, the right hand side is multiplied by a factor called the Coulomb logarithm, which typically is of order 10. This accounts for the cumulative effect of many small angle deflections. Though this does not depend on mass, the overall conductivity does, with

$\kappa \sim {\frac {k_{B}^{3}vT^{2}}{\pi e^{4}}}\propto {\frac {T^{5/2}}{\sqrt {m}}}\,\!$ However, electrons move more quickly in a gas, so they are able to carry the energy further between scatterings, and are thus more efficient at energy transport. \\We can perform a crude estimate to determine if this is important in the Sun.

$F_{r}=-\kappa {\frac {dT}{dr}}={\frac {L_{r}}{4\pi r^{2}}}.\,\!$ Making some approximations to simplify (like dropping differentials),

$L\sim 4\pi R\kappa T_{c},\,\!$ $L\sim {\frac {k_{b}^{7/2}T_{c}^{7/2}R}{e^{4}{\sqrt {m_{e}}}}},\,\!$ $L\sim 10^{-4}\left({\frac {R}{R_{\odot }}}\right)\left({\frac {T_{c}}{10^{7}K}}\right)^{7/2}L_{\odot }.\,\!$ This is much too small to carry out the energy we observe, despite evaluating this under the most favorable of conditions for electron conduction to be important. We can then confidently say that conduction is not important to carrying out energy for a star on the main sequence. \\While this exercise showed we did not need to consider conduction, we can use a very similar derivation for photons. For photons, what changes is that $v=c$ , $U=aT^{4}$ , and the mean free path is different (smaller, in fact). Then the flux from photons is

$F=-{\frac {4caT^{3}}{3n\sigma }}{dT}{dx}.\,\!$ This is in analogy with the conduction equation from before, so we can define a photon conductivity in analogy to before of

$\kappa _{ph}={\frac {4caT^{3}}{n\sigma }}.\,\!$ This is often written alternatively in terms of the opacity and density rather than cross section and number density, with

$F=-{\frac {4caT^{3}}{3\kappa \rho }}{dT}{dx}.\,\!$ It is important to keep in mind that the density is the density of the scatters, not the photons. It turns out that electrons are what dominate scattering. The cross section for radiation for electrons can be derived in a nice way. We can start by writing down the Lorentz force for an electron

$m_{e}{\frac {dv}{dt}}=-e({\overrightarrow {E}}+{\frac {\overrightarrow {v}}{c}}\times {\overrightarrow {B}}).\,\!$ Since the magnitude of $E$ and $B$ are comparable, we need to only worry about the electric field, and can see that

${\overrightarrow {a}}=-{\frac {e{\overrightarrow {E}}}{m_{e}}}.\,\!$ We now need to see what the power radiation by an electron that is being accelerated by the presence of an electromagnetic wave. This is given by the Larmor formula as

$P={\frac {2e^{2}}{3c^{3}}}|a|^{2}.\,\!$ Substituting the acceleration from above,

$P={\frac {2e^{4}}{3m_{e}^{2}c^{3}}}|E|^{2}.\,\!$ If we go back to the picture before for determing the cross section, we see that the flux and the power absorbed or scattered are related by the cross section, with

$P=\sigma F.\,\!$ The Larmor power represents then reradiated power due to the photon electron interaction. We know the reradiated power, and we also know that the incoming flux is just the Poynting flux, given as

$F={\frac {c|E|^{2}}{4\pi }}.\,\!$ Solving $P=\sigma F$ , we find the Thomson scattering cross section is

$\sigma ={\frac {8\pi }{3}}{\frac {e^{4}}{m_{e}^{2}c^{4}}}.\,\!$ This is independent of the electric field, as it should be since it is a property of the particle. It is also independent of frequency (except at high frequency, where some implicit assumptions break down). This is also sometimes written as

$\sigma \sim \pi r_{c}^{2},\,\!$ where $r_{c}$ is the classical electron radius, which is the radius found from equation the rest mass energy to the electrostatic self energy of an electron.