01 · Advanced

Boundary conditions in hollow waveguides

Start from source-free Maxwell equations in a z-invariant PEC guide, derive the wall conditions, separate transverse and longitudinal field content, prove why a simply connected hollow single-conductor guide cannot support a nontrivial TEM mode, and formulate the transverse eigenvalue problem that later generates TE/TM modes and cutoff.

210–270 minAdvanced
In this lesson
  • State the geometry, material and phasor assumptions for a uniform hollow PEC waveguide.
  • Derive the PEC boundary conditions on E, B, surface charge and surface current with an explicit normal convention.
  • Separate fields into transverse and longitudinal components relative to the guide axis.
  • Introduce modal z-dependence without confusing propagation constant with the transverse eigenvalue.
  • Derive the scalar transverse Helmholtz equations for the longitudinal field components.
  • Explain how PEC side walls impose Dirichlet data on E_z and Neumann data on H_z.
  • Prove that a simply connected hollow guide with one conductor boundary cannot support a nontrivial TEM mode.
  • Explain why multi-conductor guides such as coaxial lines evade the no-TEM proof.
  • Distinguish a boundary-value eigenproblem from a memorized rectangular-waveguide formula.
  • Connect the transverse eigenvalue k_c to β²=k²−k_c² as a preview of cutoff without yet specializing to a rectangular guide.

A hollow metallic waveguide is not a two-conductor transmission line with different artwork. Its allowed fields are eigenfunctions of Maxwell equations on a bounded transverse domain. The correct starting point is therefore the wall boundary condition, the guide topology and the separation between transverse and longitudinal field components. Only after that structure is established do the familiar TE/TM labels, cutoff frequencies and rectangular-guide formulas become meaningful.

0101.1 — Declare the guide, material and phasor assumptions

Consider a uniform guide whose cross-section A does not vary with z. The interior is homogeneous, isotropic and source-free with permittivity ε and permeability μ. The side wall is an ideal perfect electric conductor (PEC). We use AoN's exp(+jωt) phasor convention and seek modal fields proportional to exp(−jβz).

Inside the dielectric region, Maxwell's equations are ∇×E=−jωμH, ∇×H=jωεE, ∇·E=0 and ∇·H=0. Define k=ω√(με). Loss, anisotropy, dielectric loading and finite wall conductivity are deliberately excluded from this first eigenproblem.

A mode is not chosen by drawing a field pattern. It is a nonzero field pair that satisfies Maxwell, the wall conditions and the assumed z dependence simultaneously.

∇×E=−jωμH, ∇×H=jωεE
∇·E=0, ∇·H=0
k=ω√(με)
E,H ∝ exp(−jβz) with exp(+jωt)

0201.2 — PEC boundary conditions come from Maxwell interface laws

Define the unit normal n from the PEC metal INTO the guide region. In an ideal PEC the time-varying fields inside the metal vanish. Applying Maxwell interface conditions at the metal/dielectric surface gives zero tangential electric field and zero normal magnetic flux density on the guide side.

Surface charge and surface current support the remaining discontinuities. With the stated normal convention, n·D=ρ_s and n×H=J_s on the guide side. These are surface sources on the conductor; they are not volume sources in the dielectric eigenproblem.

Because E is normal to the PEC surface while H is tangential there, the time-average Poynting vector has no normal component into an ideal wall. A PEC redirects power; it does not dissipate it.

Figure 1.1 — The modal problem lives on the transverse cross-section

One question: where do the PEC boundary conditions act, and how is the normal used in the surface laws defined?

dielectric domain APEC wall ∂An : PEC → guideguide axis: z

The transverse domain A is bounded by the perfect conductor. Here n is defined from the metal into the guided region. Wall conditions act on ∂A; propagation along the guide is carried separately by exp(−jβz).

n×E=0 on PEC
n·B=0 on PEC
n·D=ρs
n×H=Js
⟨S⟩·n=(1/2)Re(E×H*)·n=0 [ideal PEC]

0301.3 — Split the fields into transverse and longitudinal parts

Relative to the invariant guide axis z, write E=E_t+ẑE_z and H=H_t+ẑH_z. The transverse gradient is ∇_t=x̂∂/∂x+ŷ∂/∂y, while ∂/∂z becomes −jβ for one modal phasor.

This decomposition does not yet say TE, TM or TEM. It only separates the unknowns according to geometry. Maxwell equations then couple E_t,H_t to the longitudinal scalars E_z,H_z.

The modal propagation constant β belongs to axial evolution. The transverse eigenvalue introduced next belongs to the cross-sectional boundary-value problem; confusing them hides the origin of cutoff.

E=Et+ẑEz, H=Ht+ẑHz
∇=∇t−jβẑ
∇t=x̂∂x+ŷ∂y

0401.4 — Longitudinal fields satisfy a transverse Helmholtz eigenproblem

Each Cartesian field component satisfies the homogeneous vector Helmholtz equation in the uniform filling. After inserting exp(−jβz), the z derivatives contribute −β². The longitudinal components therefore obey scalar transverse Helmholtz equations.

Define k_c²=k²−β². The quantity k_c is fixed by the transverse geometry and boundary conditions for each eigenfunction. Conversely, once k_c is known, the axial propagation constant follows from β²=k²−k_c².

This is the structural origin of cutoff. Chapter 03 will analyze what happens when k crosses k_c; here we only establish the eigenvalue relation.

(∇²+k2)E=0, (∇²+k2)H=0
(∇t2+kc2)Ez=0
(∇t2+kc2)Hz=0
kc2≡k2−β2
β2=k2−kc2

0501.5 — PEC walls give Dirichlet data for TM candidates

A side wall is parallel to z, so the z direction is tangential to that wall. The PEC condition n×E=0 therefore forces the longitudinal electric component E_z to vanish at the boundary.

For a TM candidate, H_z=0 while E_z is the generating longitudinal scalar. Its eigenproblem is therefore the transverse Helmholtz equation with homogeneous Dirichlet boundary data E_z=0 on ∂A.

A scalar function that satisfies this boundary condition is only a CANDIDATE eigenfunction. It must also satisfy the differential equation with one k_c and generate nonzero transverse fields through Maxwell equations.

TM candidate: Hz=0, Ez≠0
(∇t2+kc2)Ez=0 in A
Ez=0 on ∂A [Dirichlet]

0601.6 — PEC walls give Neumann data for TE candidates

For a TE candidate E_z=0 and H_z is the generating scalar. Maxwell reconstruction gives E_t=(jωμ/k_c²)ẑ×∇_tH_z for the exp(+jωt−jβz) convention.

At a PEC side wall, the tangential component of E_t must vanish. The component of ẑ×∇_tH_z along the wall tangent is proportional to the normal derivative of H_z, so the boundary condition becomes ∂H_z/∂n=0.

Thus the TE scalar eigenproblem is a homogeneous Neumann problem. Its constant k_c=0 solution must be treated carefully: in a simply connected hollow guide it does not generate a nontrivial TE/TEM propagating field.

TE candidate: Ez=0, Hz≠0
Et=(jωμ/kc2) ẑ×∇t Hz
(∇t2+kc2)Hz=0 in A
∂Hz/∂n=0 on ∂A [Neumann]

0701.7 — Why a hollow single-conductor guide has no nontrivial TEM mode

A TEM mode would have E_z=H_z=0. Maxwell then requires β²=k² and the transverse electric field is irrotational and divergence-free in the homogeneous cross-section, so E_t=−∇_tΦ with ∇_t²Φ=0.

PEC tangential E=0 means Φ is constant along each connected conductor boundary. In a simply connected hollow guide there is only one conductor boundary, so Φ has one constant boundary value. The uniqueness theorem for Laplace's equation then gives Φ constant everywhere in A.

Therefore E_t=0, and Maxwell then gives H_t=0: only the trivial field remains. A coaxial or other multi-conductor line escapes this proof because different conductor boundaries can be held at different constants, producing a nontrivial harmonic potential and a genuine TEM field.

TEM: Ez=Hz=0
β2=k2
Et=−∇tΦ, ∇t2Φ=0
Φ=constant on the single PEC boundary ⇒ Φ=constant in A
therefore Et=Ht=0 [trivial]

0801.8 — A boundary-compatible trial function is not yet a full mode

Take a rectangular test domain 0<x<a, 0<y<b only as a boundary audit. The scalar sin(πx/a)sin(πy/b) vanishes on every wall, so it satisfies the TM Dirichlet condition and has transverse eigenvalue k_c²=(π/a)²+(π/b)².

The scalar cos(πx/a), independent of y, has zero normal derivative on all four walls and satisfies a TE-type Neumann condition with k_c=π/a. These examples preview the eigenfunctions that chapter 04 will derive systematically.

But boundary compatibility alone is not a complete electromagnetic solution: the scalar must satisfy its Helmholtz equation, the chosen TE/TM definition, and Maxwell reconstruction. This discipline prevents drawing a plausible field pattern and calling it a mode.

FTM=sin(πx/a)sin(πy/b)
kc,TM2=(π/a)2+(π/b)2
FTE=cos(πx/a)
kc,TE=π/a

Practice — PEC boundaries, transverse eigenproblems and the TEM exclusion

  1. Starting from the general interface laws, recover n×E=0 and n·B=0 at an ideal PEC wall using n from metal into the guide.
  2. Explain why a PEC can support nonzero surface charge and current even though the fields inside the ideal conductor vanish.
  3. Show that the time-average Poynting flux normal to an ideal PEC wall is zero.
  4. Insert exp(−jβz) into the scalar Helmholtz equation and derive (∇_t²+k²−β²)ψ=0.
  5. Why does E_z obey Dirichlet data on the side wall of a uniform PEC guide?
  6. Use E_t=(jωμ/k_c²)ẑ×∇_tH_z to show that tangential E=0 implies ∂H_z/∂n=0 for TE candidates.
  7. Prove the no-TEM result for a simply connected hollow guide using the Laplace uniqueness theorem.
  8. Why can a coaxial line support TEM although a hollow rectangular metal pipe cannot?
  9. Verify that sin(πx/a)sin(πy/b) satisfies Dirichlet conditions on a rectangle and compute its k_c.
  10. Verify that cos(πx/a) satisfies the Neumann wall condition on the same rectangle and state why that alone is not enough to declare a complete TE mode.
Worked solutions

1. Use n×(E_guide−E_PEC)=0 and n·(B_guide−B_PEC)=0 with E_PEC=B_PEC=0. The chosen normal direction does not change the zero-valued conditions, yielding n×E=0 and n·B=0 on the guide side.

2. The fields vanish in the PEC bulk, but discontinuities of normal D and tangential H are supported by surface sources. With n from metal to guide, n·D=ρ_s and n×H=J_s.

3. At the wall E has no tangential component while H has no normal magnetic-flux component; E is normal and H tangential. Their cross product is tangent to the wall, so its normal component vanishes. Hence (1/2)Re(E×H*)·n=0.

4. Write ∇²=∇_t²+∂²/∂z². Acting on ψ_t(x,y)e^(−jβz), the z derivative contributes −β². Thus [∇_t²+(k²−β²)]ψ_t=0 and k_c²=k²−β².

5. The side wall contains the z direction, so E_z is tangential to the PEC. Since every tangential electric component must vanish, E_z=0 on ∂A.

6. Let s be the wall tangent in the transverse plane. The tangential wall field is E_t·s=(jωμ/k_c²)(ẑ×∇_tH_z)·s, which is proportional up to orientation sign to ∂H_z/∂n. Setting E_t·s=0 gives ∂H_z/∂n=0.

7. TEM gives E_t=−∇_tΦ with ∇_t²Φ=0. Tangential E=0 makes Φ constant on the connected PEC boundary. Laplace uniqueness then makes Φ equal that same constant throughout the simply connected cross-section, so E_t=0 and then H_t=0.

8. A coax has two disconnected conductor boundaries. Laplace's equation can take different constant potentials on the two conductors, producing a nonzero transverse electric field between them and hence a TEM mode.

9. The sine product vanishes at x=0,a and y=0,b. Applying ∇_t² gives −[(π/a)²+(π/b)²] times the same function, so k_c²=(π/a)²+(π/b)².

10. The normal derivative of cos(πx/a) vanishes at x=0,a because sin(πx/a)=0, and its y derivative is identically zero on y=0,b. It is therefore Neumann-compatible with k_c=π/a, but Maxwell reconstruction and the TE definition must still be satisfied to form a complete mode.

Chapter 02 will take the longitudinal eigenfunctions as generating functions and reconstruct complete TE/TM field families, derive their modal wave impedances and orthogonality structure, while keeping the rectangular-guide specialization for chapter 04.