Electromagnetic Wave Equation in Vacuum

Derivation

Electromagnetic Wave Equation in Vacuum

D-164 Home PU-204 Threads waves · light · fields Depends on Assembly of Maxwell's Equations

Statement

In a source-free vacuum (no free charge, ρ = 0; no free current, J = 0), Maxwell's four equations decouple into two identical second-order wave equations, one for each field: ∇²E = μ₀ε₀ ∂²E/∂t² and ∇²B = μ₀ε₀ ∂²B/∂t². Both describe disturbances propagating at a single speed c = 1/√(μ₀ε₀), fixed entirely by the electric and magnetic constants of the vacuum.

Why it matters

This is the moment electromagnetism becomes optics. Maxwell's equations were assembled from static and quasi-static laboratory experiments — Coulomb, Ampere, Faraday — none of which mention light. Decoupling them reveals a self-sustaining travelling disturbance whose speed is a combination of two measured electrical constants, and that number matched the measured speed of light. Light is an electromagnetic wave; the visible, radio, X-ray, and gamma bands are one phenomenon.

The derivation also fixes the structure of all subsequent radiation physics: the wave equation's speed is frame-independent by construction (it contains no reference to any medium's rest frame), which is the seed of special relativity, and its linearity guarantees superposition, the basis of interference, diffraction, and Fourier optics.

Assumptions

Vacuum, source-free region.If ρ ≠ 0 or J ≠ 0 the divergence of E and the curl of B pick up source terms; the equations no longer decouple cleanly and you obtain inhomogeneous wave equations with the sources as drivers (the basis of radiation from antennas).

Linear, non-dispersive, isotropic medium — here literally empty space, so μ₀, ε₀ constant.In matter with frequency-dependent ε(ω), μ(ω) the single speed splits into phase and group velocities and the wave equation must be written frequency by frequency.

Fields are twice continuously differentiable (C²) in space and time.The vector identity and the interchange of ∇× with ∂/∂t both require mixed partials to commute; at a sharp boundary or shock front the classical PDE fails pointwise and must be replaced by matching conditions or a distributional formulation.

Cartesian components, so ∇² acts componentwise on the vector.The identity ∇×(∇×A) = ∇(∇·A) − ∇²A defines the vector Laplacian; in curvilinear coordinates ∇² of a vector is not the componentwise scalar Laplacian and extra curvature terms appear.

Derivation

∇·E = 0 ∇·B = 0 ∇×E = −∂B/∂t ∇×B = μ₀ε₀ ∂E/∂t

Maxwell's equations with ρ = 0, J = 0 (from maxwell-equations-assembly). These are the starting data. A

∇×(∇×E) = ∇×(−∂B/∂t)

Take the curl of Faraday's law. Applying the same operator to both sides preserves equality. B

∇(∇·E) − ∇²E = −∂/∂t(∇×B)

Left: vector identity ∇×(∇×E) = ∇(∇·E) − ∇²E. Right: ∇× and ∂/∂t commute because they act on independent variables and the field is C² (Clairaut). C

−∇²E = −∂/∂t(∇×B)

The gradient term vanishes: ∇·E = 0 in vacuum, so ∇(∇·E) = 0. A

−∇²E = −∂/∂t(μ₀ε₀ ∂E/∂t)

Substitute the Ampere–Maxwell law ∇×B = μ₀ε₀ ∂E/∂t into the right-hand side. This is where the displacement-current term ε₀∂E/∂t is indispensable — without it the RHS is zero and there is no wave. B

∇²E = μ₀ε₀ ∂²E/∂t²

Constants μ₀, ε₀ pull outside the time derivative; multiply both sides by −1. This is the wave equation for E. B

∇×(∇×B) = μ₀ε₀ ∂/∂t(∇×E) ⇒ −∇²B = μ₀ε₀ ∂/∂t(−∂B/∂t)

Repeat steps 2–5 starting from the curl of Ampere's law, using ∇·B = 0 to kill the gradient term and Faraday's law ∇×E = −∂B/∂t on the right. B

∇²B = μ₀ε₀ ∂²B/∂t²

Identical structure to step 6. Comparing with the canonical wave equation ∇²ψ = (1/v²)∂²ψ/∂t² identifies 1/v² = μ₀ε₀, hence v = c = 1/√(μ₀ε₀). A

Result

∇²E = μ₀ε₀ ∂²E/∂t² ∇²B = μ₀ε₀ ∂²B/∂t² c = 1/√(μ₀ε₀)

Reading. Each Cartesian component of E and of B obeys the same three-dimensional wave equation. A disturbance in the fields propagates without a medium at the fixed speed c, determined solely by the two vacuum constants. The two fields are not independent: they are locked together through the original first-order equations (which force E ⊥ B ⊥ propagation direction and E = cB in magnitude), but each separately satisfies the second-order wave equation.

Units check. [μ₀] = H·m⁻¹ = kg·m·A⁻²·s⁻²; [ε₀] = F·m⁻¹ = A²·s⁴·kg⁻¹·m⁻³. Product [μ₀ε₀] = s²·m⁻², so [1/√(μ₀ε₀)] = m·s⁻¹, a speed. The PDE checks too: [∇²E] = (V/m)/m², and [μ₀ε₀ ∂²E/∂t²] = (s²/m²)(V/m)/s² = (V/m)/m². Consistent.

Limiting cases

Static limit (∂/∂t → 0): the RHS vanishes and both equations collapse to Laplace's equation ∇²E = 0 — recovering electrostatics/magnetostatics, no propagation.
Displacement current off (ε₀∂E/∂t dropped, pre-Maxwell): the RHS of the E-equation is zero and there is no wave at all — the term Maxwell added is exactly what makes light possible.
Plane-wave / 1-D limit (fields depend on z, t only): reduces to ∂²E/∂z² = (1/c²)∂²E/∂t², with general d'Alembert solution f(z − ct) + g(z + ct).
In a linear dielectric (μ₀ε₀ → με): same form with speed v = 1/√(με) = c/n, defining the refractive index n = √(με/μ₀ε₀).

Breaks when

Sources are present (ρ ≠ 0 or J ≠ 0). The gradient term ∇(∇·E) = ∇(ρ/ε₀) no longer vanishes and the Ampere term carries μ₀J; you get inhomogeneous wave equations driven by the sources, best handled through the potentials with retarded solutions, not this clean homogeneous form.
Dispersive or nonlinear media. If ε depends on frequency, a single time-domain wave equation with one speed is wrong; each Fourier component moves at its own phase velocity. If ε depends on field strength (intense laser fields), the equation becomes nonlinear and superposition fails — harmonic generation, self-focusing, solitons.
Boundaries and interfaces. At a conductor surface or dielectric interface the fields are not C² across the boundary; the bulk wave equation holds only on each side and must be stitched together with boundary conditions on the tangential/normal components.
Curved spacetime / non-inertial strong-gravity settings. The flat-space ∇² and constant c must be promoted to a curved d'Alembertian; light no longer travels in straight lines at globally constant coordinate speed.

Failure modes

Dropping ∇(∇·E) too early. Students set it to zero before invoking ∇·E = 0, or forget the vector-identity term entirely and write ∇×(∇×E) = −∇²E as if it were automatic. It is only true in source-free vacuum.
Sign error in the double curl. Faraday carries a minus sign and step 6 flips sign again; miscounting gives ∇²E = −μ₀ε₀∂²E/∂t², a Helmholtz-type equation with exponential (non-propagating) solutions.
Forgetting the displacement current. Using the magnetostatic ∇×B = μ₀J = 0 in vacuum kills the RHS and "proves" light cannot exist.
Treating ∇² of a vector as the componentwise Laplacian in curvilinear coordinates. In spherical/cylindrical coordinates this omits the curvature terms in the vector Laplacian and gives the wrong radial equation.
Believing the two second-order equations are the whole story. They admit solutions (e.g. E and B parallel, or longitudinal) that the original first-order Maxwell equations forbid; the wave equations are necessary but not sufficient — the first-order constraints still apply.

Discussion

The decisive physical content is that c emerges from two constants measured in benchtop circuit and magnetostatics experiments, with no optics involved. Weber and Kohlrausch had measured the ratio of electrostatic to electromagnetic units (numerically 1/√(μ₀ε₀)) and found it close to Fizeau's measured speed of light. Maxwell's derivation turned that numerical coincidence into an identity: light is an electromagnetic disturbance. This is arguably the first great unification in physics.

Notice what the wave equation does not contain: any velocity of a source or observer, and any material medium. The speed c is written purely in terms of vacuum constants. If the equations are to hold in every inertial frame — and there is no reason singled out for one — then c must be the same in all of them. Einstein took exactly this seriously in 1905, and special relativity is the consequence of insisting Maxwell's equations be frame-independent.

The two second-order equations are equivalent to the first-order Maxwell system only together with the constraints. Solving ∇²E = μ₀ε₀∂²E/∂t² alone would permit longitudinal waves; it is ∇·E = 0 that forces transversality (k·E₀ = 0 for a plane wave), and Faraday's law that ties B = (k×E)/ω, fixing the mutual orientation and the amplitude ratio E₀ = cB₀.

At a deeper level the wave equations are the Euler–Lagrange equations of the free electromagnetic field, whose Lagrangian density is −(1/4μ₀)F_μνF^μν. In covariant form the two vector wave equations plus the Lorenz-gauge condition collapse to the single equation ∂_μ∂^μA^ν = 0, i.e. □A^ν = 0, manifestly Lorentz-invariant. The masslessness of the photon is precisely the absence of an m²A^ν term; a nonzero photon mass would add such a term (the Proca equation) and give a frequency-dependent speed, tightly bounded by experiment.

Common misconceptions. The wave equation does not need a medium — there is no ether; the fields themselves are the "substance" that oscillates. And E and B are not 90° out of phase in a travelling plane wave in vacuum (a common confusion with the standing-wave or reactive near-field case): they oscillate in phase, peaking together, with energy flowing steadily in the propagation direction.

Worked examples

Example 1 — Predict the speed of light from the vacuum constants.

c = 1/√(μ₀ε₀)

Result of the derivation; symbolic form first. A

μ₀ = 1.25664×10⁻⁶ H/m, ε₀ = 8.8542×10⁻¹² F/m

Insert measured constants (SI). Units H/m and F/m. A

μ₀ε₀ = 1.25664×10⁻⁶ × 8.8542×10⁻¹² = 1.11265×10⁻¹⁷ s²/m²

Multiply; the H·F product reduces to s². A

c = 1/√(1.11265×10⁻¹⁷) = 1/(3.3356×10⁻⁹ s/m)

Take the square root of the product. A

c = 2.998×10⁸ m/s

Reading. Two purely electrical constants predict the speed of light to four significant figures — the match that identified light as electromagnetic.

Example 2 — Verify a plane wave is a solution and extract its frequency.

E_x(z,t) = E₀ cos(kz − ωt)

Trial plane wave for green laser light, propagating in +z. B

∂²E_x/∂z² = −k²E_x, ∂²E_x/∂t² = −ω²E_x

Differentiate twice in space and in time. B

−k²E_x = μ₀ε₀(−ω²)E_x ⇒ ω/k = 1/√(μ₀ε₀) = c

Substitute into the wave equation; it holds for all z,t iff the dispersion relation ω = ck is satisfied. So it is a solution, and the phase speed is c. B

λ = 532 nm ⇒ f = c/λ = (2.998×10⁸)/(532×10⁻⁹)

Use c = fλ with the given wavelength; metres cancel to give s⁻¹. A

f = 5.64×10¹⁴ Hz (period T ≈ 1.77 fs)

Reading. Any waveform of the form f(z − ct) solves the equation; a monochromatic component obeys ω = ck, and 532 nm green light oscillates at 564 THz.

Problems

Starting from Ampere's law ∇×B = μ₀ε₀∂E/∂t, derive the wave equation for B explicitly, stating where each vacuum Maxwell equation is used.
Solution
Take the curl: ∇×(∇×B) = μ₀ε₀∂/∂t(∇×E). Left side via the vector identity is ∇(∇·B) − ∇²B; use ∇·B = 0 to drop the first term, giving −∇²B. Right side: substitute Faraday ∇×E = −∂B/∂t, giving μ₀ε₀∂/∂t(−∂B/∂t) = −μ₀ε₀∂²B/∂t². Equate and cancel the minus signs: ∇²B = μ₀ε₀∂²B/∂t². Equations used: Ampere (start), ∇·B = 0 (kill gradient), Faraday (RHS).
For the plane wave E = E₀ cos(kz − ωt) x̂, use Faraday's law to find B, and confirm E₀ = cB₀.
Solution
Faraday: ∇×E = −∂B/∂t. With E along x̂ depending on z, ∇×E = ∂E_x/∂z ŷ = −kE₀ sin(kz − ωt) ŷ. So ∂B/∂t = kE₀ sin(kz − ωt) ŷ. Integrate in time: B = (kE₀/ω) cos(kz − ωt) ŷ. Thus B₀ = kE₀/ω = E₀/c (since ω/k = c), i.e. E₀ = cB₀. Note B is along ŷ, perpendicular to both E (x̂) and propagation (ẑ), and in phase with E.
A parallel-plate capacitor's plane-wave analogue: take an X-ray of wavelength 0.10 nm. Compute its frequency and angular frequency.
Solution
f = c/λ = (2.998×10⁸ m/s)/(0.10×10⁻⁹ m) = 3.0×10¹⁸ Hz. Angular frequency ω = 2πf = 2π(3.0×10¹⁸) = 1.9×10¹⁹ rad/s. Wavenumber k = ω/c = 2π/λ = 6.3×10¹⁰ rad/m.
Show that in a linear dielectric with permittivity ε = ε_rε₀ and permeability μ ≈ μ₀, the wave speed is c/n with n = √ε_r. For water, ε_r ≈ 1.77 at optical frequencies; find n and v.
Solution
Repeating the derivation in a linear medium replaces μ₀ε₀ by με, so v = 1/√(με). With μ ≈ μ₀: v = 1/√(μ₀ε_rε₀) = c/√ε_r, hence n = √ε_r. For water at optical frequencies n = √1.77 = 1.33, and v = (2.998×10⁸)/1.33 = 2.25×10⁸ m/s. (The value ε_r ≈ 80 for static water gives the wrong optical index because water is strongly dispersive — this is why the non-dispersive assumption matters.)
Explain, using the derivation, why the magnetostatic Ampere law ∇×B = μ₀J (without displacement current) cannot support electromagnetic waves in vacuum, and state what conserved quantity the displacement current restores.
Solution
In vacuum J = 0, so the pre-Maxwell law gives ∇×B = 0. At step 5 the right-hand side ∂/∂t(∇×B) would be zero, yielding ∇²E = 0 (Laplace) — no time dependence, no propagation. The displacement current ε₀∂E/∂t supplies the nonzero RHS that couples spatial curvature of E to its second time derivative, producing the wave. Physically Maxwell added it to make the equations consistent with charge conservation: taking the divergence of the full Ampere–Maxwell law and using ∇·E = ρ/ε₀ gives the continuity equation ∇·J + ∂ρ/∂t = 0; without displacement current one would require ∇·J = 0 always, which fails for a charging capacitor.