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Each Fourier mode has zero initial velocity, soAlong $x=Vt$ this isPutThe two phases have stationary points at $-k_0$ and $k_0$, respectively. At those points,

Because the initial field is real, $a(-k)=a(k{)}^{\ast }$. Applying the stationary phase method to the two conjugate contributions gives the stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray:Equivalently, if $a(k_0)=|a(k_0)|e^{i\delta }$,