Categories: Physics, Plasma physics.

# Spitzer resistivity

If an electric field with magnitude $E$ is applied to the plasma, the electrons experience a Lorentz force $q_e E$ (we neglect the ions due to their mass), where $q_e$ is the electron charge.

However, collisions slow them down while they travel through the plasma., This can be modelled as a drag force $f_{ei} m_e v_e$, where $f_{ei}$ is the electron-ion collision frequency (we neglect $f_{ee}$ since all electrons are moving together), $m_e$ is their mass, and $v_e$ their typical velocity relative to the ions in the background. Balancing the two forces yields the following relation:

\begin{aligned} q_e E = f_{ei} m_e v_e \end{aligned}

Using that the current density $J = q_e n_e v_e$, we can rearrange this like so:

\begin{aligned} E = f_{ei} m_e \frac{J}{n_e q_e^2} = \frac{m_e f_{ei}}{n_e q_e^2} J = \eta J \end{aligned}

This is Ohm’s law, where $\eta$ is the resistivity. From our derivation of the Coulomb logarithm $\ln(\Lambda)$, we estimate $f_{ei}$ to be as follows, where $n_i$ is the ion density, $\sigma$ is the collision cross-section, and $\mu$ is the reduced mass of the electron-ion system:

\begin{aligned} f_{ei} = n_i \sigma v_e = \frac{1}{2 \pi} \Big( \frac{q_e q_i}{\varepsilon_0 \mu} \Big)^2 \frac{n_i}{v_e^3} \ln(\Lambda) \approx \frac{1}{2 \pi} \frac{Z q_e^4}{\varepsilon_0^2 m_e^2} \frac{n_e}{v_e^3} \ln(\Lambda) \end{aligned}

Where we used that $\mu \approx m_e$, and $q_i = -Z q_e$ for some ionization $Z$, and as a result $n_e \approx Z n_i$ due to the plasma’s quasi-neutrality. Beware: authors disagree about the constant factors in $f_{ei}$; recall that it was derived from fairly rough estimates. This article follows Bellan.

Inserting this expression for $f_{ei}$ into the so-called Spitzer resistivity $\eta$ then yields:

\begin{aligned} \boxed{ \eta = \frac{m_e f_{ei}}{n_e q_e^2} = \frac{1}{2 \pi} \frac{Z q_e^2}{\varepsilon_0^2 m_e} \frac{1}{v_e^3} \ln(\Lambda) } \end{aligned}

A reasonable estimate for the typical velocity $v_e$ at thermal equilibrium is as follows, where $k_B$ is Boltzmann’s constant, and $T_e$ is the electron temperature:

\begin{aligned} \frac{1}{2} m_e v_e^2 = \frac{3}{2} k_B T_e \quad \implies \quad v_e = \sqrt{\frac{3 k_B T_e}{m_e}} \end{aligned}

Other choices exist, see e.g. the Maxwell-Boltzmann distribution, but always $v_e \propto \sqrt{T_e/m_e}$. Inserting this $v_e$ into $\eta$ then gives:

\begin{aligned} \eta = \frac{1}{6 \pi \sqrt{3}} \frac{Z q_e^2 \sqrt{m_e}}{\varepsilon_0^2 (k_B T_e)^{3/2}} \ln(\Lambda) \end{aligned}

## References

1. P.M. Bellan, Fundamentals of plasma physics, 1st edition, Cambridge.
2. M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.