Velocity distribution function for cyclotron motion

Question

Let us be a plasma at temperature $T_e$. Electrons in it have a VDF (velocity distribution function) :

$$ f_e(\textbf{v})\text{d}^3\textbf{v}=\left( \frac{m}{2\pi k_BT_e}\right) ^{3/2} \times \exp\left[-\frac{m\textbf{v}^2}{2k_B T_e} \right]\text{d}^3\textbf{v} $$

Assume now that there is a constant and static magnetic field along the $\textbf{e}_z$ direction. Electrons will have a cyclotron motion due to Lorentz force, at frequency $\omega_c=eB/m$ around magnetic field lines. If we neglect collisions, an electron will have a velocity $$ \textbf{v}=\textbf{v}_\theta+\textbf{v}_z, $$ in cylindrical coordinates. What is the proper way to go from the VDF above to the new one in $f_e(v_\theta,v_z)$ ?

What I already did: I made the variable change in cilyndrical coordinates and got

$$ f_e(\textbf{v})\text{d}^2\textbf{v}=2\pi v_\theta\left( \frac{m}{2\pi k_BT_e}\right) \times \exp\left[-\frac{m\textbf{v}^2}{2k_B T_e} \right]\text{d}^2\textbf{v} $$

If $\textbf{v}$ was $v_r$ I would be okay, but here I have doubts.

Answer 1

The motion is circular in the directions perpendicular to the field-line along $\mathbf{e}_z$. This means,

\begin{eqnarray} v_x & = & v_\perp \cos(\omega_c t) \; , \\ v_y & = & v_\perp \sin(\omega_c t) \; , \\ \end{eqnarray}

with $v_\perp$ and $v_z$ independent of time. Thus, $v^2$ is independent of time

\begin{equation} v^2 = v_x^2 + v_y^2 + v_z^2 = v_\perp^2 + v_z^2 \; . \end{equation}

The volume element is cylindrical, with $v_\perp$ playing the role of the cylindrical radius (it is customary to write $v_\perp$ not $v_\theta$):

\begin{equation} d^3v = 2 \pi v_\perp d v_\perp d v_z \; . \end{equation}

The factor of $2\pi$ arises from the symmetry about the fieldline. The final result is thus:

\begin{equation} f_e \, d^3 v =\left( \frac{m}{2\pi k_BT_e}\right) ^{3/2} \exp\left[-\frac{m}{2k_B T_e} \left( v_\perp^2 + v_z^2 \right) \right] \, 2 \pi v_\perp d v_\perp d v_z \end{equation}

If you prefer to retain the gyroangle $\phi$ (which is the velocity angle swept by the circular motion), then in the volume element, set $2 \pi \rightarrow d\phi$.