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An electromagnetic waveguide confines an electromagnetic wave across its axis while allowing it to travel along the axis. Conducting walls or changes of refractive index select transverse normal modes: reproducible cross-sectional field patterns with their own longitudinal wavenumbers. A hollow metal pipe and an optical fiber are different realizations of this principle. Their usefulness is to deliver electromagnetic energy along a chosen route instead of letting it spread freely.

For a uniform vacuum-filled conducting guide, write a normal mode's axial and time dependence as $e^{i(\beta z-\omega t)}$. Here $\omega >0$ is angular frequency, $\beta$ is the longitudinal wavenumber, and $k_{\perp }^2$ is the transverse eigenvalue selected by the wall boundary conditions. The electromagnetic wave equation separates intoThe geometry consumes part of the available squared wavenumber: when $\omega /c<k_{\perp }$, this branch is an evanescent wave rather than a propagating normal mode. The waveguide cutoff frequency is therefore a property of a specified branch and geometry. A transverse electromagnetic mode in a suitable two-conductor geometry can have $k_{\perp }=0$ and no positive geometric cutoff. An optical fiber needs its own dielectric boundary conditions, rather than the metal-wall formulas below.

The rectangular electromagnetic waveguide makes these ideas explicit with a single sine profile. The scalar-potential conducting waveguide modes give a broader construction. At an open aperture, a confined field can radiate into the electromagnetic aperture angular spectrum; it need not remain one guided normal mode.

In vacuum without sources, Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-\dot{B}$, and $\nabla \times B=\dot{E}/c^2$. For $E^{\prime }=cB$, $B^{\prime }=-E/c$, both divergences remain zero, andThis proves the quarter-turn electromagnetic duality transformation.

A perfect conductor has zero electric field in its bulk. Continuity of tangential electric field therefore gives $n\times E=0$ at its surface. Faraday's law makes a nonzero-frequency magnetic field vanish in that bulk, and continuity of its normal component gives $n\cdot B=0$ for the wave field. An independently trapped static magnetic flux is not excluded by perfect conductivity alone; the stated condition applies to the oscillating field under consideration.

For amplitudes with dependence $e^{i(kz-\omega t)}$, let ${\gamma }^2={\omega }^2/c^2-k^2$ and suppose $({\Delta }_{\perp }+{\gamma }^2)\psi =0$. With $k\ne 0$ and $\omega \ne 0$, define the transverse-magnetic family of scalar-potential conducting waveguide modes byThe electric divergence is ${\Delta }_{\perp }\psi +ik{e}_z=0$; the magnetic divergence is zero because the transverse field is a rotated gradient. Direct differentiation gives $\nabla \times e=i\omega b$ and $\nabla \times b=-i\omega e/c^2$, where longitudinal differentiation means multiplication by $ik$. Thus all vacuum equations hold, with $\boxed{{\gamma }^2={\omega }^2/c^2-k^2}$.

On a wall parallel to $z$, $\psi =0$ makes both $e_z$ and the boundary-tangential derivative of $\psi$ vanish. It also makes $n\cdot b$ proportional to that tangential derivative, hence zero. This proves the Dirichlet boundary condition supplies both conductor conditions.

For a transverse-electric family, use duality and rescale the potential, or check directly thatsatisfies the same equations. If $t=\hat{z}\times n$ is the cross-sectional wall tangent, then $t\cdot e=n\cdot {\nabla }_{\perp }\psi$ and $n\cdot b=-(k/\omega )n\cdot {\nabla }_{\perp }\psi$. Consequently $\boxed{{\partial }_n\psi =0}$, a Neumann boundary condition, supplies both wall conditions. Duality exchanges the field constructions, but does not preserve electric-conductor boundary conditions without this new boundary test.