Structure of Monochromatic Plane Waves

Derivation

Structure of Monochromatic Plane Waves

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Statement

In a source-free, linear, homogeneous, isotropic, non-dispersive medium (vacuum unless stated), every monochromatic plane-wave solution of Maxwell's equations, E(r,t) = E₀ e^{i(k·r − ωt)} and B(r,t) = B₀ e^{i(k·r − ωt)}, is strictly transverse: k·E = k·B = 0. The triad (E, B, k) is mutually orthogonal and right-handed, the two fields oscillate in phase, their amplitudes obey |E| = c|B|, and the complex vector E₀ encodes the polarization state — linear, circular, or elliptical.

Why it matters

This is the structural skeleton of light. The wave equation alone tells you a disturbance propagates at speed c; it says nothing about how the fields point. Maxwell's equations, applied to the plane-wave ansatz, fix the geometry completely — and that geometry (transversality, orthogonality, the E–B ratio) underlies polarizers, antennas, the Poynting flux, radiation pressure, and the photon's two helicity states.

It also explains a negative fact of enormous consequence: electromagnetic waves have no longitudinal component in free space. Sound has a longitudinal mode; light does not. That single constraint is why Polaroid sunglasses work, why radio dipoles have a preferred axis, and why a spin-1 photon has only two, not three, independent states.

Assumptions

Source-free region (ρ = 0, J = 0).If free charge or current is present, ∇·E ≠ 0 and ∇×B gains a μ₀J term; the field acquires a longitudinal component and transversality fails locally (e.g. inside a plasma or conductor). A single sharp frequency ω and single wavevector k (monochromatic plane wave).Drop it and E is a Fourier superposition; each component still obeys the relations, but the total field need not be transverse to any one direction, and "the" polarization becomes time-dependent. Linear, homogeneous, isotropic, non-dispersive, lossless medium with real ε, μ.In an anisotropic crystal D = εE is a tensor relation, so D⊥k but E need not be; the wave splits into ordinary and extraordinary rays with different speeds (birefringence), and E acquires a small longitudinal part. Infinite plane wavefronts (idealized geometry).A real beam of finite transverse extent must, by ∇·E = 0, carry a longitudinal field component of order (λ/w) near the axis; strict transversality is the plane-wave limit w → ∞.

Derivation

E(r,t) = E₀ e^{i(k·r − ωt)}, B(r,t) = B₀ e^{i(k·r − ωt)}

Ansatz. Each Cartesian component solves the wave equation derived previously, so this is the general single-mode solution; E₀, B₀ are constant complex vectors. A

∂_t → −iω, ∇ → ik

For the plane-wave phase, each spatial derivative brings down a factor ik and the time derivative a factor −iω. This replaces every differential operator by algebra. A

∇·E = 0 ⟹ ik·E₀ e^i(⋯) = 0 ⟹ k·E = 0

Gauss's law in vacuum (ρ = 0) with the substitution of Step 2. The exponential never vanishes, so the bracket must. A

∇·B = 0 ⟹ k·B = 0

No-monopoles law, identical algebra. Steps 3–4 establish transversality: both fields lie in the plane perpendicular to k. A

∇×E = −∂_tB ⟹ ik×E = iωB

Faraday's law with Step 2 on both sides (∇× → ik×, ∂_t → −iω, and the leading minus cancels). A

B = (k×E) / ω = (k̂×E) / v_p, v_p = ω/k

Divide Step 5 by iω. This fixes B from E: it is perpendicular to both k and E, and since k, ω are real, B carries the same complex phase as E — the fields are in phase. B

|B| = k|E| / ω = |E| / v_p

Take magnitudes of Step 6 using k ⊥ E, so |k×E| = k|E|. In vacuum v_p = c, giving |E| = c|B|. B

∇×B = μ₀ε₀ ∂_tE ⟹ ik×B = −iωμ₀ε₀E

Ampère–Maxwell law in vacuum (J = 0). This is not independent information — it must be consistent with Step 6 for a non-trivial solution to exist. B

k×(k×E) = −ω²μ₀ε₀E ⟹ −k²E = −ω²μ₀ε₀E

Substitute Step 6 (ωB = k×E) into Step 8 and expand with the BAC–CAB identity k×(k×E) = (k·E)k − k²E, dropping k·E = 0 from Step 3. C

k² = ω²μ₀ε₀ ⟹ v_p = ω/k = 1/√(μ₀ε₀) = c

The consistency condition is exactly the dispersion relation; it confirms the phase speed and closes the system. All four equations are satisfied by the transverse triad. C

E₀ = E_x x̂ + E_y ŷ, E_x = a_xe^iφ_x, E_y = a_ye^iφ_y

Choose k = k ẑ. Transversality (Step 3) confines E₀ to the x–y plane, so two complex numbers exhaust the freedom: two real amplitudes and two phases. B

E(z,t) = Re[E₀e^i(kz−ωt)] = a_xcos(ψ+φ_x)x̂ + a_ycos(ψ+φ_y)ŷ

Take the physical (real) part, with ψ = kz−ωt. The relative phase δ = φ_y−φ_x and amplitude ratio a_y/a_x alone determine the tip's trajectory. B

δ = 0, π → linear; a_x=a_y, δ=±π/2 → circular; else → elliptical

Eliminating ψ between the two components of Step 12 gives a conic (an ellipse) traced by E in the transverse plane; the special phase/amplitude values degenerate it to a line or a circle. B

Result

k·E = k·B = 0, B = (k̂×E)/c, |E| = c|B|, E₀ = a_xx̂ + a_ye^iδŷ

Reading. A vacuum plane wave is transverse: E and B both lie in the wavefront, perpendicular to the propagation direction k̂. They are mutually perpendicular, oscillate in step (same phase, peaking and vanishing together), and (E, B, k̂) forms a right-handed triad, so E×B points along k̂ — the direction of energy flow. Their magnitudes are locked at |E| = c|B|. All remaining freedom is the polarization, carried by the single complex phase δ and amplitude ratio.

Units check. [E] = V m⁻¹ = kg·m·s⁻³·A⁻¹; [c|B|] = (m s⁻¹)(T) = (m s⁻¹)(kg·s⁻²·A⁻¹) = kg·m·s⁻³·A⁻¹. The two sides match, confirming |E| = c|B| is dimensionally consistent. Likewise [k×E/ω] = (m⁻¹·V·m⁻¹)/(s⁻¹) = V·s·m⁻² = T = [B]. ✓

Limiting cases

Static limit (ω → 0, k → 0). From B = k×E/ω the ratio survives, but no propagating wave exists — recovers decoupled electrostatics/magnetostatics.
Circular polarization (a_x = a_y, δ = ±π/2). |E| is constant while its direction rotates at ω; carries definite spin angular momentum ±ℏ per photon (helicity states).
Linear polarization (δ = 0 or π). E oscillates along a fixed line; the ellipse collapses to a segment; equal superposition of the two circular states.
One amplitude zero (a_y = 0). Pure x̂-polarized wave; B along ŷ; the elementary textbook plane wave.
Medium with index n (v_p = c/n). Structure is unchanged but |E| = v_p|B| = (c/n)|B|, so B is larger for given E.

Breaks when

Inside a conductor or plasma. A finite conductivity σ (or bound-charge response below the plasma frequency) makes k complex and the fields no longer in phase; E leads B, |E| ≠ c|B|, and the wave is evanescent or attenuated. A longitudinal plasma oscillation (Langmuir mode) can even propagate — genuinely non-transverse.
Anisotropic (birefringent) media. With a dielectric tensor, only D⊥k is guaranteed; E tilts out of the wavefront (the extraordinary ray), the Poynting vector no longer parallels k, and the two polarizations travel at different speeds.
Bounded waveguides. Metallic boundary conditions force TE/TM modes with a genuine longitudinal E or B component and a cutoff frequency; the free-space transverse (TEM) mode cannot exist in a simply-connected hollow guide.
Near-field / finite beams. Within a wavelength of a source, or on the axis of a tightly focused Gaussian beam, ∇·E = 0 demands a longitudinal component of order (λ/w); strict transversality holds only in the far field / plane-wave limit.

Failure modes

Writing |E| = |B| in SI. Confuses Gaussian units (where E and B share dimensions) with SI. In SI |E| = c|B|, a factor of ~3×10⁸.
Thinking E and B are 90° out of phase. False in a lossless medium — they peak together. The quarter-cycle lag is a near-field / conductor phenomenon, not a plane-wave one. (Also different from spatial perpendicularity.)
Treating polarization as a property of B. By convention polarization refers to the E vector's trajectory; B is fixed by E via B = k̂×E/c.
Getting the handedness of the triad wrong. The order is (E, B, k̂) right-handed, i.e. E×B ∥ k̂. Swapping gives energy flowing backwards.
Assuming any two of "orthogonal / in phase / |E|=c|B|" imply the third. Each comes from a different Maxwell equation; all four equations are needed for the full structure.
Forgetting to drop k·E in the BAC–CAB step. Retaining it leaves a spurious longitudinal term and a wrong dispersion relation.

Discussion

The deep content of this derivation is that Maxwell's equations are not four independent constraints on the plane wave but a tightly interlocking set. Gauss's laws kill the longitudinal parts; Faraday's law then slaves B to E (fixing direction, phase, and magnitude in one stroke); and the Ampère–Maxwell law, rather than adding new geometry, merely demands self-consistency — and that consistency is precisely the dispersion relation ω = ck. The counting is elegant: E starts with three complex components (6 real numbers), transversality removes one (leaving 4), an overall phase and amplitude are conventional, and what remains — two numbers — is exactly the two-dimensional space of polarization states, the classical shadow of the photon's two helicities.

The right-handed triad (E, B, k̂) is not decoration: the Poynting vector S = E×B/μ₀ points along k̂, so energy flows in the propagation direction, and its time average S̄ = |E₀|²/(2μ₀c) uses the |E| = c|B| relation directly. Because E and B are in phase, this flux never reverses — a lossless wave carries a steady, unidirectional stream of energy and momentum, which is why light exerts radiation pressure p = S̄/c.

Polarization is the physically richest part of the result. The complex amplitude E₀ = a_xx̂ + a_ye^iδŷ maps onto the surface of the Poincaré sphere: the equator is linear polarization at all angles, the poles are the two circular (helicity) states, and every other point is an ellipse. Superposition of the two circular states with a controllable relative phase is the basis of every wave plate, optical isolator, and circular dichroism measurement.

Viewed relativistically, transversality and the phase-lock are frame-independent statements about the field tensor F^μν: for a plane wave both Lorentz invariants vanish, E·B = 0 and E² − c²B² = 0. Because these are invariants, no observer can boost into a frame where E and B are non-perpendicular or where |E| ≠ c|B|; the null character of the wave is absolute. This is the classical seed of the photon's masslessness and its lightlike four-momentum k^μk_μ = 0.

Common misconceptions. (i) "E and B are perpendicular because one creates the other 90° later." No — spatial perpendicularity (a geometric fact from ∇×) is distinct from temporal phase (they are in phase). (ii) "The wave needs a medium to be transverse." Transversality here is a vacuum property, following from ∇·E = 0, not from any elastic restoring force. (iii) "Unpolarized light has E=0 on average so it isn't transverse." Unpolarized light is a rapidly varying, incoherent mixture of transverse states; each instantaneous field is still transverse.

Worked examples

Find B from a given E. A vacuum wave has E = 300 x̂ cos(kz − ωt) V m⁻¹, λ = 500 nm. Find B(z,t).

Propagation is +ẑ, E along x̂; use B = k̂×E/c. A

k̂×x̂ = ẑ×x̂ = ŷ ⟹ B = (300/c) ŷ cos(kz−ωt)

Symbolic direction and amplitude fixed before numbers. A

B₀ = 300 / (2.998×10⁸) = 1.00×10⁻⁶ T

Insert c. A

B = 1.00 ŷ cos(kz − ωt) μT, k = 2π/λ = 1.26×10⁷ m⁻¹

Reading. A 300 V m⁻¹ optical field pairs with a mere 1 μT magnetic field — the factor c makes B feel tiny, yet they carry equal energy density (ε₀E²/2 = B²/2μ₀). B lies along ŷ, in phase with E.

Units check. V m⁻¹ ÷ (m s⁻¹) = V s m⁻² = T. ✓

Identify the polarization state. E₀ = E₀(x̂ + iŷ), wave along +ẑ with phase (kz − ωt). Classify it and give the sense of rotation seen by a receiver.

Compare with a_xx̂ + a_ye^iδŷ: here a_x = a_y = E₀ and e^iδ = i, so δ = +π/2. B

E(0,t) = Re[E₀(x̂+iŷ)e^−iωt] = E₀[cosωt x̂ + sinωt ŷ]

Take the real part at z = 0; equal amplitudes and 90° phase ⟹ constant |E| = E₀, so circular. B

at ωt=0: E∥x̂; at ωt=π/2: E∥ŷ ⟹ x̂ → ŷ rotation

As t increases the tip turns from +x toward +y. A receiver looks back along −ẑ (wave coming at it). C

Circular polarization; left-circular (positive helicity) — E×B ∥ +ẑ, spin +ℏ per photon

Reading. Equal components with a +90° phase give a field of constant magnitude rotating at ω. Facing the incoming wave, the receiver sees counter-clockwise rotation — the optics convention for left-circular / positive helicity. The wave carries angular momentum +ℏ per photon.

Units check. Purely geometric (phases and unit vectors); |E| = E₀ keeps units of V m⁻¹ throughout. ✓

Problems

A vacuum plane wave has B = 2.0×10⁻⁷ T amplitude. Find the amplitude of E and the time-averaged intensity S̄.
Solution
E₀ = cB₀ = (3.00×10⁸)(2.0×10⁻⁷) = 60 V m⁻¹. Intensity S̄ = E₀B₀/(2μ₀) = (60)(2.0×10⁻⁷)/(2·4π×10⁻⁷) = 1.2×10⁻⁵/(2.513×10⁻⁶) ≈ 4.8 W m⁻². (Equivalently S̄ = cε₀E₀²/2 = (3×10⁸)(8.85×10⁻¹²)(3600)/2 ≈ 4.8 W m⁻².)
Show explicitly that E = E₀(x̂ + ẑ) e^i(kz−ωt) cannot be a valid vacuum plane wave.
Solution
With k = kẑ, Gauss's law requires k·E = 0. Here k·E₀ = k(ẑ·(x̂+ẑ))E₀ = kE₀ ≠ 0. The ẑ (longitudinal) component violates ∇·E = 0, so it is not an allowed free-space wave. Only the x̂ part survives; the ẑ part would require a charge density ρ = ε₀∇·E ≠ 0.
A wave travels along +ẑ with E₀ = 3x̂ + 4e^iπŷ (V m⁻¹). Classify its polarization and give the orientation of the line if linear.
Solution
Relative phase δ = π (from e^iπ = −1), so E = 3cosψ x̂ − 4cosψ ŷ = (3x̂ − 4ŷ)cosψ. Both components share the factor cosψ ⟹ linear polarization along the fixed direction (3, −4). Angle from x̂: θ = arctan(−4/3) = −53.1° (i.e. 53.1° below the x-axis). Amplitude √(3²+4²) = 5 V m⁻¹.
Light of intensity 1.0×10³ W m⁻² (roughly solar) hits a perfectly absorbing black surface. Using the plane-wave structure, find the peak E-field and the radiation pressure.
Solution
S̄ = cε₀E₀²/2 ⟹ E₀ = √(2S̄/(cε₀)) = √(2·10³/((3×10⁸)(8.85×10⁻¹²))) = √(2000/2.655×10⁻³) = √(7.53×10⁵) ≈ 868 V m⁻¹. Peak B₀ = E₀/c ≈ 2.9 μT. For full absorption, pressure p = S̄/c = 10³/(3×10⁸) ≈ 3.3×10⁻⁶ Pa. (Doubles for a perfect reflector.)
Two vacuum plane waves of equal amplitude E₀, both along +ẑ, are superposed: one x̂-polarized, one ŷ-polarized with a variable phase lag δ. Sketch the polarization as δ runs 0 → π/2 → π and give the state at each value.
Solution
Total E₀ = E₀(x̂ + e^iδŷ), equal amplitudes. δ = 0: E ∥ (x̂+ŷ), linear at +45°. δ = π/2: a_x=a_y with quarter-wave lag ⟹ circular (one handedness). δ = π: E ∥ (x̂−ŷ), linear at −45°. For intermediate δ the state is elliptical, its major axis rotating and its ellipticity peaking (circular) at δ = π/2 — precisely how a quarter-wave plate (δ = π/2) converts 45° linear light into circular light.