 Solution

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

δ B = (c / ω) k \times δ E

and

k \times (k \times δ E) + \frac{ω ^{2}}{c ^{2}} ϵ \cdot δ E = 0.

n = k c / ω

, the linear system is therefore

S - n^{2} cos^{2} θ - i g n^{2} sin θ cos θ i g S - n^{2} 0 n^{2} sin θ cos θ 0 P - n^{2} sin^{2} θ δ E = 0.

R = S - g, L = S + g, R L = S^{2} - g^{2} .

Here

R

L

the cold-plasma left-handed wave, with their polarizations fixed explicitly in d. In the electron-only response,

R = 1 - \frac{ω _{p e}^{2}}{ω ( ω - Ω _{e} )}, L = 1 - \frac{ω _{p e}^{2}}{ω ( ω + Ω _{e} )} .

N = n^{2}

, expansion of the determinant yields the dispersion relation

A N^{2} - BN + C = 0, A = S sin^{2} θ + P cos^{2} θ, B = R L sin^{2} θ + PS (1 + cos^{2} θ), C = PR L .

Equivalently,

P (N - R) (N - L) cos^{2} θ + (SN - R L) (N - P) sin^{2} θ = 0.

tan^{2} θ = - \frac{P ( n ^{2} - R ) ( n ^{2} - L )}{( S n ^{2} - R L ) ( n ^{2} - P )} .

The undivided determinant equation remains valid when this rational form has simultaneous zero factors.

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