Klein-Gordon packet velocities ($v_p{v}_g=c^2$)

A useful stipulated dispersive medium obeys the Klein-Gordon equation $u_{tt}-c^2{u}_{xx}+{\omega }_0^{2}u=0$, with $c>0$ and ${\omega }_0>0$. Its positive-angular frequency dispersion relation isHere ${\omega }_0$ is the gap at zero wavenumber, not the central angular frequency ${\omega }_c=\omega (k_0)$ of a wave packet; the parameter in the preceding formula is $A={\omega }_0/c$. For $k>0$, the phase velocity satisfies $v_p>c$, the group velocity satisfies $0<v_g<c$, and $v_p{v}_g=c^2$. In this stipulated medium $c$ is the limiting propagation speed of the model; it need not mean the vacuum speed of light. The same square-root dispersion relation appears in waveguide phase and group velocities, with a waveguide cutoff frequency playing the role of ${\omega }_0$.^[ref]

For the original dimensionless choice $c={\omega }_0=k_0=1$, a carrier wave has phase velocity $v_p=\sqrt{2}$, while a narrow wave packet has leading group velocity $v_g=1/\sqrt{2}$. After time $t=8$, the carrier's zero-wave phase reference has moved to $x=8\sqrt{2}\simeq 11.31$, whereas the leading wave envelope centre is at $x=8/\sqrt{2}\simeq 5.66$. A carrier wave crest therefore moves through the wave envelope; it does not carry an identifiable piece of the envelope with it.

The origin-to-curve secant on the dispersion diagram gives phase velocity, while its local tangent gives group velocity. As $k\to 0^{+}$, $v_p\to \infty$ and $v_g\to 0$; at $k=0$ the phase velocity itself is undefined. As $k\to \infty$, both approach $c$, from opposite sides. On the same positive-angular frequency branch with $k<0$, both signed velocities are negative and their product remains $c^2$. These conventions distinguish direction from phase speed. Reversing both $k$ and $\omega$ represents the same real plane wave; changing just one sign changes its direction.

For an original finite-bandwidth check, take equal-amplitude plane waves at $k_1=0.8$ and $k_2=1.2$, still with $c={\omega }_0=1$. The exact two-tone wave envelope speed iswhile its carrier wave has phase velocity $(\sqrt{1+0.8^2}+\sqrt{1+1.2^2})/2\simeq 1.421337$. Their product is exactly $1$: rationalizing the difference gives $v_{\mathrm{beat}}=2/({\omega }_1+{\omega }_2)$. Nevertheless $v_{\mathrm{beat}}\ne {\omega }^{\prime }(1)$; the derivative describes the limit of vanishing separation, not an arbitrary pair of tones. The two-tone wave envelope repeats indefinitely, so it has no isolated first arrival and cannot measure front velocity.