Cross section for photon-electron interaction in high temperature, high density plasma

Question

I need a plot for the cross section of the photon-electron scattering at high density and high temperature. I am working on a project for which I have to calculate the opacity of a hot, dense plasma (similar that in the core of stars). The opacity is the inverse of the mean free path of the photon-electron scattering, so I need a plot for the cross section of this process.

Answer 1

I need a plot for the cross section of the photon-electron scattering at high density and high temperature.

Elastic scattering
I think you are looking for something called Compton scattering, where a photon gives energy to an electron. Suppose a photon of energy, $\epsilon_{i} = h \nu_{i}$ (where $h$ is the Planck constant and $\nu_{i}$ is the incident photon frequency), is incident on an electron, if the interaction is perfectly elastic then the scattered photon will have a final energy, $\epsilon_{f} = h \ \nu_{f} = \epsilon_{i}$. In this limit, and for low energies (i.e., $h \nu_{i} \ll m_{e} c^{2}$, where $m_{e}$ is the electron rest mass and $c$ is the speed of light in vacuum), the interaction obeys something called Thomson scattering. The Thomson scattering cross section is given by: $$ \sigma_{T} = \frac{ 8 \ \pi }{ 3 } r_{e}^{2} = \frac{ 8 \ \pi }{ 3 } \left( \frac{ e^{2} }{ 4 \pi \ \varepsilon_{o} \ m_{e} c^{2} } \right)^{2} \tag{0} $$ where $r_{e}$ is the classical electron radius, $\varepsilon_{o}$ is the permittivity of free space, and $e$ is the fundamental charge.

The differential cross-section is then given by: $$ \frac{ d \sigma_{T} }{ d \Omega } = \frac{ r_{e}^{2} }{ 2 } \left( 1 + \cos^{2}{\theta} \right) \tag{1} $$ where $\theta$ is the scattering angle of the photon relative to its incident direction and $d \Omega$ is the differential solid angle.

Inelastic scattering
Suppose we cannot assume that $\epsilon_{f} = \epsilon_{i}$, then we find the Klein-Nishina formula given by: $$ \frac{ d \sigma }{ d \Omega } = \frac{ r_{e}^{2} }{ 2 } \left( \frac{ \epsilon_{f} }{ \epsilon_{i} } \right)^{2} \left[ \frac{ \epsilon_{f} }{ \epsilon_{i} } + \frac{ \epsilon_{i} }{ \epsilon_{f} } - \sin^{2}{\theta} \right] \tag{2} $$

Note that in the limit that $\epsilon_{f} \rightarrow \epsilon_{i}$, Equation 2 will reduce to Equation 1 above.

To simplify matters, let us define $\zeta_{j} = \tfrac{ h \nu_{j} }{ m_{e} c^{2} }$. Then we can write down the full Compton scattering cross-section as: $$ \sigma = \frac{ 3 \ \sigma_{T} }{ 4 } \left\{ \frac{ 1 + \zeta }{ \zeta^{3} } \left[ \frac{ 2 \zeta \left( 1 + \zeta \right) }{ 1 + 2 \zeta } - \ln{\lvert 1 + 2 \zeta \rvert} \right] + \frac{ 1 }{ 2 \zeta } \ln{\lvert 1 + 2 \zeta \rvert} - \frac{ 1 + 3 \zeta }{ \left( 1 + 2 \zeta \right)^{2} } \right\} \tag{3} $$

So we can see in the ultra-relativistic regime (i.e., $\zeta \gg 1$) Equation 3 reduces to: $$ \sigma \approx \frac{ 3 \ \sigma_{T} }{ 8 \ \zeta } \left( \frac{ 1 }{ 2 } + \ln{\lvert 2 \zeta \rvert} \right) \tag{4} $$

We can also see that Compton scattering becomes less efficient as the photon energy increases.

You can add extra complexities (e.g., inverse Compton scattering) onto this as necessary but this should give you a decent start.