Disclaimer: I am including a fair bit of the physics background, but I believe this can be solved by someone with a strong calculus understanding

I am following the derivation of the critical electric field strength (Dreicer Field) that leads towards runaway electrons in a plasma exposed to a strong external electric field.

Source: https://journals-aps-org.stanford.idm.oclc.org/pr/pdf/10.1103/PhysRev.115.238

As part of the process they define the dynamical friction between two species (ions (i) and electrons (e)) using Fokker-Planck with Rosenbluth potentials.

The following definitions of the Rosenbluth potentials are given:

$H_{e,i} = \frac{m_e+m_i}{m_i}\Gamma_e\int{\frac{F_i(r,c',t)}{w}d^3c'}$

$w = |c-c'|$

$\Gamma_e= 4\pi\left(\frac{c^2}{4\pi \epsilon_0 m_e}\right)^2\log{\frac{\lambda}{\rho_0}}$


  • r: position space

  • c: electron velocity space

  • c': ion velocity space

  • $F_i(r,c',t)$: 7 dimensional ion distribution function

everything else listed is assumed to be constant including $\log{\frac{\lambda}{\rho_0}}$

At a later point in the paper they define $F(r,c,t)$ in the general sense as a displaced Maxwellian distribution given as:

$F_\alpha(r,c,v(t)) = n(r)\left(\frac{\beta_\alpha(r)}{\pi}\right)^{3/2}\exp{\left(-\beta_\alpha(r)|c-v_\alpha(t)|^2\right)}$

$\beta_\alpha(r) = \frac{m_\alpha}{2kT_\alpha(r)}$

where $v_\alpha(t)$ is the bulk species velocity, and $T_\alpha(r)$ is the species temperature.

With this general definition of the species distribution function the author claims to take this definition and apply it to the previously listed definition of the Rosenbluth potentials to generate the outcome:

$H_{e,i} = \frac{m_e+m_i}{m_i}\Gamma_e\frac{\xi(\beta_i^{1/2}q)}{q}$

$\xi(x) = \frac{2}{\sqrt(\pi)}\int_{0}^{x}\exp{(-t^2)}dt$

$q = |c-v_i(t)|$

My question is how do the previous two definitions combine to produce this output? In my mind substituting and integrating does not produce this result, but my Math could be wrong. Does anyone with a better mind for Calculus have any thoughts on this?

  • $\begingroup$ The $\xi(x)$ term is just an error function. The reason it comes in is that only electrons up to that normalized speed (i.e., drift speed of electrons over the ion thermal speed) play a role in contributing to the potential. You will notice that they don't actually integrate anything, they just replace the error function with a symbolic expression. The error function is integrated numerically. The $q$ factor is just accounting for the frame transformation (i.e., evaluate the potential in the ion rest frame). $\endgroup$ Commented Apr 4, 2018 at 14:51

1 Answer 1


You will not be able to see the result just by inspection because the Rosenbluth potentials are triple integrals, and the final result is just a single integral. You must actually perform the integrals over two velocity coordinates in order to arrive at the quoted result. Below are the first several steps to do so.

First change variables to $C = c'-c$. \begin{align} \int_{\mathbb R^3}d^3c' \frac{F(\vec r, c', t)}{w} &= n_i\left(\frac{\beta_i}{\pi}\right)^{3/2}\int_{\mathbb{R}^3} d^3c' \frac{1}{|c'- c|} e^{-\beta_i|c-v_i|^2}\\ &= n_i\left(\frac{\beta_i}{\pi}\right)^{3/2} \int_{\mathbb R^3} d^3C \frac{1}{|C|}e^{-\beta_i|C+c-v_i|^2} \end{align} Next, expand the exponent using the Law of Cosines, $$ |C+c-v_i|^2 = |C|^2 + |c-v_i|^2 + 2|c-v_i||C|\cos\theta = w^2 + q^2 + 2qw\cos\theta , $$ where $\theta$ is the angle between $C$ and $c-v_i$. Now express the integral in a spherical coordinate system $(C_x,C_y,C_z)\to(w,\theta,\phi)$, \begin{align} &n_i\left(\frac{\beta_i}{\pi}\right)^{3/2} \int_{\mathbb R^3} d^3C \frac{1}{|C|}e^{-\beta_i(w^2+q^2)}e^{-2\beta_i qw\cos\theta} \\ &= n_i\left(\frac{\beta_i}{\pi}\right)^{3/2} \int_0^\infty dw \int_0^{\pi} d\theta\int_0^{2\pi} d\phi ~w^2 \sin\theta \frac{1}{w}e^{-\beta_i (w^2+q^2)}e^{-2\beta_i qw\cos\theta} \end{align} The integrand is independent of $\phi$, so the integral over $\phi$ just contributes an overall factor of $2\pi$. The $\theta$ integral is easiest to do with the substitution $u = \cos\theta$, $du = -\sin\theta d\theta$ \begin{align} & n_i\left(\frac{\beta_i}{\pi}\right)^{3/2} \int_0^\infty dw \int_0^{\pi} d\theta\int_0^{2\pi} d\phi ~w^2 \sin\theta \frac{1}{w}e^{-\beta_i (w^2+q^2)}e^{-2\beta_i qw\cos\theta} \\ &= 2\pi n_i(\beta_i/\pi)^{3/2}\int_0^\infty dw~w e^{-\beta_i(w^2+q^2)}\int_{-1}^1 du ~e^{-2\beta_i q w u} \\ &= n_i\beta_i^{3/2}\frac{2}{\sqrt{\pi}} \int_0^\infty dw~w e^{-\beta_i (w^2+q^2)}~\frac{1}{2\beta_i q w}\left[ e^{2qw} - e^{-2qw} \right] \\ &= \frac{n_i\sqrt{\beta_i}}{q\sqrt{\pi}} \left[ \int_0^\infty dw~e^{-\beta_i(w^2 -2qw + q^2)} - \int_0^\infty dw~e^{-\beta_i(w^2 + 2qw +q^2)} \right] \end{align} Thus the three-dimensional integral has been reduced to just two one-dimensional integrals. I will leave off the final steps of expressing these in terms of $\xi(\sqrt{\beta_i}q)$. It can be done with very routine changes of variables.


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