Electrostatic ion cyclotron waves
On page 102 of the 3rd edition of <PERSON>'s introduction to plasma physics and controlled fusion, he discusses the phenomenon of an ion acoustic wave propagating at right angles to a magnetic field $\mathbf{B}_0$ as shown in the diagram below:
He mentions that it is unrealistic for $\mathbf{k}$ to be perpendicular to $\mathbf{B}_0$ and it is better to let $\mathbf{k}$ be almost perpendicular to $\mathbf{B}_0$ as shown in the diagram above.
He proceeds to mention that since the angle is small, when analyzing the motion of IONS, it suffices to take $\mathbf{E} = E \mathbf{e}_x$ and $\mathbf{k} = k \mathbf{e}_x$, the reason for this is because $k_z$ is small, and the inertia of the ions pretty much guarantees that motion in the $z$ direction is negligible.
However, this argument does not WORK for ELECTRONS because there is a big difference between $\frac{\pi}{2} - \theta = 0$ and $\frac{\pi}{2} - \theta << 1$
<PERSON> mentions that in the context of electrons, if $\frac{\pi}{2} - \theta = 0$, then:
Statement 1: The electrons have such a small Lamor radii that they cannot move in the $x$ direction to preserve charge neutrality; all that the $\mathbf{E}$ field does is make them drift back and forth in the $y$ direction
Chen also mentions that:
Statement 2: The critical angle $\chi = \frac{\pi}{2} - \theta$ is proportional to the ratio of ion to electron parallel velocities: $$ \chi \approx \sqrt{\frac{m_e}{m_i}} $$
(a) In statement 1, how does the small Larmor radii of the electron restrict the motion of electrons in the $x$ direction to preserve charge neutrality? It seems to me that it would be better to argue that because the Larmor radii is small, we can ignore the Larmor gyration in the $x$ direction and only focus on the $y$ direction due to the presence of an $\mathbf{E} \times \mathbf{B_0}$ drift.
(b) In statement 2, <PERSON> does not provide any justification as to how he arrives at this conclusion regarding the angle being proportional to that ratio, how does he arrive at this? Even if I take it for granted, to reach the desired expression, I require: $$ \frac{1}{2}m_e v_{e \parallel}^2 = E_\parallel = \frac{1}{2}m_i v_{i \parallel}^2 $$
In order for: $$ \chi \approx \frac{v_{e \parallel}}{v_{i \parallel}} \approx \sqrt{\frac{m_e}{m_i}} $$
Again, the fact that the two kinetic energies are equal are not justified 649