02 · Intermediate → advanced

Monochromatic plane waves

Fix AoN's harmonic convention, read propagation from constant-phase planes, derive k=ω/c and the transverse Maxwell amplitude constraints, recover the signed E–B–k triad and vacuum wave impedance, then express the most general transverse complex amplitude as the bridge to polarization.

155–200 minIntermediate → advanced

In this lesson

Interpret k and propagation direction from AoN's exp(+jωt−jk·r) convention.
Derive the vacuum dispersion relation k=ω/c from the monochromatic plane-wave ansatz.
Derive k·E₀=k·B₀=0 and k×E₀=ωB₀ directly from Maxwell.
Determine signed E, B and propagation directions, including reverse propagation.
Derive H₀=(1/η₀)k̂×E₀ and check the vacuum wave-impedance units.
Write the general transverse complex amplitude and identify relative phase as the input to polarization.
Distinguish an ideal infinite plane wave from a finite beam or source near field.
Solve wavelength, wavenumber and E/B/H amplitude problems with independent dimensional checks.

A wave equation says which space-time patterns can propagate; Maxwell's first-order equations say which of those patterns are electromagnetic. This chapter builds one monochromatic plane wave carefully: phase convention first, then constant-phase planes, the dispersion relation, transverse amplitudes and the signed E–B–k triad. Polarization is introduced only as the freedom left inside the transverse plane; its full classification belongs to the next chapter.

0102.1 — Fix the harmonic convention before interpreting k

AoN uses the physical real field as the real part of a complex representation with time factor exp(+jωt). A monochromatic plane-wave phasor therefore carries the spatial factor exp(−jk·r). The sign pair matters: changing the time convention changes the matching spatial sign and several intermediate curl formulas, while the physical real field is unchanged when the convention is transformed consistently.

For a real cosine wave, the phase is ωt−k·r+φ₀. A surface on which this phase is constant is a plane perpendicular to k. Increasing time requires k·r to increase to remain on the same phase surface, so the wave travels in the +k direction under this convention. The vector k therefore contains both propagation direction and spatial phase rate.

E(r,t)=Re{E₀ exp[j(ωt−k·r)]}

Ẽ(r)=E₀ exp(−jk·r), E(r,t)=Re{Ẽ(r)exp(jωt)}

ωt−k·r+φ₀=constant

k=k k̂, k=2π/λ

0202.2 — Insert the plane-wave ansatz into the wave equation

For a constant complex amplitude E₀, each spatial derivative acting on exp(−jk·r) supplies a factor −jk. Consequently the Laplacian supplies −k². Two time derivatives of exp(+jωt) supply −ω². Substitution into the vacuum wave equation leaves an algebraic compatibility condition.

A nonzero plane-wave amplitude therefore requires k²=ω²/c². In vacuum the phase speed is ω/k=c, the wavelength is λ=2π/k, and f=ω/(2π). This is the nondispersive vacuum relation. It is not a license to use k=ω/c inside an arbitrary material; material constitutive response can change both magnitude and complex character of k.

∇²Ẽ=−k²Ẽ

∂²E/∂t²=−ω²E

(−k²+ω²/c²)E=0

k=ω/c, λ=2π/k=c/f, f=ω/(2π)

0302.3 — Maxwell removes the longitudinal degrees of freedom

The vector wave equation treats Cartesian components independently, so by itself it allows a mathematical solution with E₀ parallel to k. Maxwell does not. Applying ∇·E=0 to the plane-wave phasor gives −jk·E₀=0; hence k·E₀=0. The same argument gives k·B₀=0.

Faraday's law supplies the relation between field amplitudes. With the AoN exp(+jωt) convention, ∇×E contributes −jk×E and −∂B/∂t contributes −jωB. Their equality gives k×E₀=ωB₀. Thus E, B and k are mutually orthogonal in a vacuum plane wave, and B is not an independent arbitrary amplitude.

∇·Ẽ=−j k·Ẽ=0 ⇒ k·E₀=0

∇·B̃=−j k·B̃=0 ⇒ k·B₀=0

−j k×E₀=−jωB₀ ⇒ k×E₀=ωB₀

B₀=(1/c) k̂×E₀, |B₀|=|E₀|/c

0402.4 — Signed geometry: reversing propagation reverses the magnetic direction

Take propagation along +x and choose E along +y. The right-handed cross product x̂×ŷ=ẑ gives B along +z, out of the page. The existing AoN figure below has already been checked against this signed geometry: E points +y, k and S point +x, and B is +z.

Now reverse only the propagation direction while keeping the same electric-field direction. With k̂=−x̂ and E along +y, the relation B=(1/c)k̂×E gives B along −z. This is an important diagnostic: k and E cannot be reversed or retained independently while B is left untouched. The Maxwell cross product fixes the handedness.

The reverse statement is also useful: if E and B are known at an instant for a plane wave, E×B points in the propagation direction. Full power-density derivation is postponed to the later energy-flow chapter.

E : +y ↑

k, S : +x →

B = k×E/ω

S = E×B/μ₀

E ⊥ B ⊥ k

Traveling wave: E, B and propagation form a right-handed triad

For the convention cos(kx−ωt), choose k∥+x, E∥+y and B∥+z (out of page). Maxwell–Faraday gives B=(1/ω)k×E, so E×B∥+x and S=E×B/μ₀ travels with the wave. The arrows depict one instant when E>0, not a longitudinal electric field or a spatial wave profile. In vacuum |B₀|=|E₀|/c.

+x̂×+ŷ=+ẑ

k̂=+x̂, E₀=E₀ŷ ⇒ B₀=(E₀/c)ẑ

k̂=−x̂, E₀=E₀ŷ ⇒ B₀=−(E₀/c)ẑ

E×B ∥ k

0502.5 — Introduce H and the vacuum wave impedance without hiding the model

In vacuum B=μ₀H. Combining this with B=(1/c)k̂×E and c=1/√(μ₀ε₀) gives H=(1/η₀)k̂×E, where η₀=√(μ₀/ε₀) is the vacuum wave impedance. Its unit is ohms because E/H has units (V/m)/(A/m)=V/A.

Numerically η₀ is approximately 376.73 Ω. Unlike c, this decimal is not an independently exact defining constant in the modern SI because μ₀ is no longer assigned an exact numerical value. In an ideal homogeneous isotropic lossless medium one can derive a related wave impedance only after specifying ε and μ; lossy or dispersive media require a more careful complex-frequency-domain treatment.

B=μ₀H

η₀=√(μ₀/ε₀)≈376.73 Ω

H₀=(1/η₀) k̂×E₀

|E₀|/|H₀|=η₀, [E/H]=Ω

0602.6 — General transverse amplitude and the handoff to polarization

Choose propagation along +z. Maxwell requires only that E₀ have no z component, so the most general monochromatic electric amplitude can be written E₀=Eₓx̂+Eᵧŷ, where Eₓ and Eᵧ may be complex. Their magnitudes and relative phase determine how the real electric vector moves in the transverse xy plane as time advances.

If Eₓ and Eᵧ have the same phase, the real field remains on one fixed transverse line. A nonzero relative phase can make the field tip trace a circle or an ellipse. We do not classify handedness here because that classification depends on the declared time convention, viewing direction and phase sign. Chapter 03 will derive linear, circular and elliptical polarization from these complex components without relying on memorized pictures.

Finally, a mathematical plane wave has infinite transverse extent and perfectly planar phase fronts. Real antennas and optical/THz beams are finite. Plane waves remain powerful local/far-field or spectral building blocks, but they should not be confused with an exact finite beam or a source near field.

k=k ẑ ⇒ E₀=E_x x̂+Eᵧ ŷ

E_x=|E_x|exp(jφ_x), Eᵧ=|Eᵧ|exp(jφᵧ)

Δφ=φᵧ−φ_x

B₀=(1/c) ẑ×E₀=(1/c)(E_x ŷ−Eᵧ x̂)

0702.7 — Numerical audit at 1 THz

Consider a vacuum wave at f=1.00 THz propagating along +x with electric peak amplitude 120 V/m along +y. Its wavelength is λ=c/f≈2.9979×10⁻⁴ m=299.79 μm, and k=2π/λ≈2.0958×10⁴ rad/m.

The magnetic peak amplitude is E₀/c≈4.003×10⁻⁷ T=0.400 μT and points +z. The H-field peak amplitude is E₀/η₀≈0.319 A/m. These are peak amplitudes because the chosen E₀ is a peak amplitude; mixing peak E with RMS H would be inconsistent.

Independent checks: kλ=2π, E/B has dimensions m/s and returns c, while E/H has dimensions ohms and returns η₀. These checks are often faster at catching a missing c or inverted cross product than redoing the algebra.

f=1.00 THz, E₀=120 V/m

λ=c/f≈299.79 μm

k=2π/λ≈2.0958×10⁴ rad/m

B₀=E₀/c≈4.003×10^-7 T, H₀=E₀/η₀≈0.319 A/m

Practice — plane-wave reasoning and signed checks

Under the AoN convention E=Re{E₀exp[j(ωt−kz)]}, show from constant phase that positive k corresponds to propagation toward +z.
Starting from ∇·E=0, derive the amplitude constraint for a plane wave and explain why a nonzero Ez is impossible when k=k ẑ in source-free vacuum.
A wave propagates along +z and E points +x. Determine the signed directions of B and E×B.
Keep E along +x but reverse propagation from +z to −z. What happens to B? Verify using a cross product rather than a drawing.
For a 300 GHz vacuum wave, calculate λ and k.
A +x vacuum plane wave has peak E=75 V/m along +y. Calculate peak B and H and give their signed directions.
For +z propagation write E₀=3x̂+4e^{jπ/2}ŷ V/m. Compute B₀ symbolically from ẑ×E₀ and state what feature of E₀ chapter 03 must interpret.
Explain why an exact monochromatic plane wave is not the same object as a finite Gaussian beam or an antenna near field, even though plane waves are useful building blocks.

Worked solutions

1. Constant phase means ωt−kz+φ₀=C, so z=(ω/k)t+constant. For k>0 and ω>0 the phase plane moves toward increasing z; its speed is ω/k=c in vacuum.

2. For Ẽ=E₀exp(−jkz), ∇·Ẽ=−jkE₀,z exp(−jkz)=0. With k≠0, E₀,z=0. Equivalently k·E₀=0: the electric amplitude is transverse.

3. ẑ×x̂=ŷ, so B points +y. Then x̂×ŷ=ẑ, so E×B points +z, matching propagation.

4. For k̂=−ẑ, B=(1/c)(−ẑ×x̂)=−ŷ E/c. B reverses sign; E×B then points −z as required.

5. λ=c/f=299792458/(300×10⁹)≈9.993×10⁻⁴ m=0.9993 mm. k=2π/λ≈6.288×10³ rad/m.

6. B=75/c≈2.502×10⁻⁷ T along +z because x̂×ŷ=ẑ. H=75/376.73≈0.199 A/m, also +z in the corresponding H direction.

7. B₀=(1/c)[3(ẑ×x̂)+4e^{jπ/2}(ẑ×ŷ)]=(1/c)[3ŷ−4e^{jπ/2}x̂] T. The complex relative phase between x and y components controls the polarization trajectory and handedness, treated in chapter 03.

8. A plane wave has exactly planar phase fronts and infinite transverse extent. A finite beam has transverse structure and diffraction; an antenna near field can contain reactive/longitudinal components. Plane waves can approximate far-field patches or serve as Fourier/angular-spectrum components without making the full field an exact single plane wave.

Chapter 03 will keep the same harmonic convention and derive polarization from the two complex transverse components: linear, circular and elliptical states, handedness, polarization ellipse and basis changes.