 Solution

ID: past-exam-of-the-collisionless-plasma-physics-course-of-the-university-of-oxford/2025/1/a/solution

e^{i k \cdot r - iω t}

, and the signed cyclotron frequency

Ω_{s} = q_{s} B_{0} / (m_{s} c)

. Write

Ω_{e} = e B_{0} / (m_{e} c) > 0

for its electron magnitude, so the signed electron value is

- Ω_{e}

ω_{p s}^{2} = 4 π n_{0 s} q_{s}^{2} / m_{s}

- iω u_{x s} - Ω_{s} u_{ys} = \frac{q _{s}}{m _{s}} E_{x}, Ω_{s} u_{x s} - iω u_{ys} = \frac{q _{s}}{m _{s}} E_{y}, - iω u_{zs} = \frac{q _{s}}{m _{s}} E_{z} .

j_{s} = n_{0 s} q_{s} u_{s}

σ_{s} = \frac{n _{0 s} q _{s}^{2}}{m _{s}} \frac{iω}{ω ^{2} - Ω _{s}^{2}} \frac{Ω _{s}}{ω ^{2} - Ω _{s}^{2}} 0 - \frac{Ω _{s}}{ω ^{2} - Ω _{s}^{2}} \frac{iω}{ω ^{2} - Ω _{s}^{2}} 0 00 \frac{i}{ω} .

ϵ = I + (4 πi / ω) \sum_{s} σ_{s}

. Thus

ϵ = S - i g 0 i g S 0 00 P, S = ϵ_{⊥} = 1 - \sum_{s} \frac{ω _{p s}^{2}}{ω ^{2} - Ω _{s}^{2}}, P = ϵ_{∥} = 1 - \sum_{s} \frac{ω _{p s}^{2}}{ω ^{2}}, g = - \sum_{s} \frac{ω _{p s}^{2} Ω _{s}}{ω ( ω ^{2} - Ω _{s}^{2} )} .

At the stated electron frequency scale, the ion contribution is smaller by

O (Z m_{e} / m_{i})

for

q_{i} = Z e

and quasineutrality. To this accuracy,

S = 1 - \frac{ω _{p e}^{2}}{ω ^{2} - Ω _{e}^{2}}, P = 1 - \frac{ω _{p e}^{2}}{ω ^{2}}, g = \frac{ω _{p e}^{2} Ω _{e}}{ω ( ω ^{2} - Ω _{e}^{2} )} .

These expressions require negligible collisions and the cold ordering

k v_{th, s} ≪ ∣ ω ∣

; sufficiently close to a cyclotron resonance, thermal broadening or collisions must also be retained.

 New to topics? Read the docs here!