 Solution

ID: past-exam-of-the-collisionless-plasma-physics-course-of-the-university-of-oxford/2025/2/h/solution

The barrier wavenumber is

κ = ω_{p e 0} x_{T}^{2} - x^{2} / (c L)

. The substitution

x = x_{T} sin u

gives

η = \frac{ω _{p e 0}}{c L} \int_{- x_{T}}^{x_{T}} x_{T}^{2} - x^{2} d x = \frac{ω _{p e 0} x _{T}^{2}}{c L} \int_{- π /2}^{π /2} cos^{2} u d u = \frac{π ω _{p e 0} L}{2 c} (1 - \frac{ω ^{2}}{ω _{p e 0}^{2}}) .

T ≃ exp [- \frac{π ω _{p e 0} L}{c} (1 - \frac{ω ^{2}}{ω _{p e 0}^{2}})] .

At fixed

ω / ω_{p e 0}

, increasing the column width or peak plasma frequency suppresses transmission exponentially. Raising

ω

towards

ω_{p e 0}

narrows the evanescent region and increases transmission. However, when

η

becomes order one or smaller the two wave turning points interact; the opaque-barrier WKB approximation cannot be extrapolated to claim

T \to 1

ω = ω_{p e 0}

. Smooth matching to an exterior is also implicit in the leading result; at

∣ x ∣ = L

the stated profile joins continuously to vacuum.

The shaded area in the lower panel is

η

after restoring the factor

ω_{p e 0} L / c

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