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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.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.