Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 39A / b / i / Solution

Each Fourier mode has zero initial velocity, so

ϕ (x, t) = \int_{- \infty}^{\infty} a (k) e^{ik x} cos (ω (k) t) d k .

Along

x = V t

this is

ϕ (V t, t) = \frac{1}{2} \int_{- \infty}^{\infty} a (k) [e^{i t (kV + ω (k))} + e^{i t (kV - ω (k))}] d k .

Put

s = c^{2} - V^{2}, k_{0} = \frac{A V}{s} .

The two phases have stationary points at

- k_{0}

and

k_{0}

, respectively. At those points,

k_{0} V - ω (k_{0}) = - A s, ω^{''} (k_{0}) = \frac{s ^{3}}{A c ^{2}} .

Because the initial field is real,

a (- k) = a (k)^{*}

. Applying the stationary phase method to the two conjugate contributions gives the stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray:

ϕ (V t, t) \sim (\frac{2 π A c ^{2}}{t s ^{3}})^{1/2} Re {a (k_{0}) e^{- i (A s t + π /4)}} .

Equivalently, if

a (k_{0}) = ∣ a (k_{0}) ∣ e^{i δ}

,

ϕ (V t, t) \sim ∣ a (k_{0}) ∣ (\frac{2 π A c ^{2}}{t s ^{3}})^{1/2} cos (A s t + \frac{π}{4} - δ) .

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