What's the deal with plasma physicists setting heat cap. ratio to 3?

[Twigg]

This has persistently bugged me in my intro plasma course. They keep using $\gamma = 3$ aka $N_d = 1$ (where $N_d$ is the number of degrees of freedom in the molecule) as an approximation. See for example, the Bohm-Gross dispersion curve. I can tell you from deriving this that the factor of 3 in front of the $k_b T$ is the heat capacity ratio $\gamma$. In other words, that equation ought to be $$\omega^2 = \omega_{p,e}^2 + \frac{\gamma k_B T_e k^2}{m_e}$$ and they've taken $\gamma = 3$. What the heck man?

To clarify why I find this confusing, $\gamma = \frac{N_d + 2}{N_d} = 3$ implies that $N_d = 1$. What kind of toy molecule has one degree of freedom?? Pure unobtainium vapor?

Edit: should have specified, the Bohm-Gross dispersion curve is for Langmuir waves (longitudinal waves in the electrons with the ions stationary). In other words, the $\gamma$ there is for electrons. Last time I checked electrons could move in three dimensions?? ($N_d = 3$ therefore $\gamma = \frac{5}{3}$).

 

[Twigg]

Welp, I was looking for another source to prove what I claimed about the factor of 3 being $\gamma$ in the Bohm-Gross dispersion curve, and I stumbled on the answer too. We need a "clown" emoji for situations like this.

From this document,

With the assumption that the electron compression occurs one-dimensionally and faster than thermal conduction, we have $p_1 = \gamma T n_1$, with $\gamma = 3$

Am I right to think that this means that there is a breakdown of the equipartition theorem because thermal energy doesn't have time to redistribute from the longitudinal axis of propagation to the transverse axes?

 

[jasonRF]

If you feel a little uncertain about the legitimacy of the result, linearized kinetic theory is another way to derive the dispersion relation. The factor of 3 naturally falls out of the calculation without explicitly making the $\gamma=3$ assumption.

jason