Electromagnetic wave equation

Maxwell's four equations, written down in the previous article, look at first like a static catalogue of how charges and currents produce fields. Combine two of them the right way, however, and something remarkable falls out: a genuine wave equation, with no mechanical medium in sight, predicting that electric and magnetic fields can sustain each other and travel through empty space. Plug in the two constants that set the strength of electrostatic and magnetic forces, μ0\mu_0 and ε0\varepsilon_0, and the wave speed that pops out is c≈3×108 m/sc\approx3\times10^8\ \text{m/s} — the measured speed of light. This article walks through that derivation, restricted strictly to vacuum with no charges or currents present, and shows why it convinced Maxwell that light itself is an electromagnetic phenomenon.

The core insight

Taking the curl of Faraday's law and substituting the Ampère–Maxwell law converts two coupled first-order Maxwell equations into a single second-order wave equation for E⃗\vec E (and identically for B⃗\vec B), with wave speed c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0}.

The core idea

Start from the source-free Maxwell equations. In vacuum, with no charge density (ρ=0\rho=0) and no current (J⃗=0\vec J=0), Maxwell's equations reduce to a fully symmetric set:

∇⋅E⃗=0,∇⋅B⃗=0,∇×E⃗=−∂B⃗∂t,∇×B⃗=μ0ε0∂E⃗∂t\nabla\cdot\vec E=0,\quad \nabla\cdot\vec B=0,\quad \nabla\times\vec E=-\frac{\partial\vec B}{\partial t},\quad \nabla\times\vec B=\mu_0\varepsilon_0\frac{\partial\vec E}{\partial t}

where: E⃗,B⃗\vec E,\vec B are the electric and magnetic fields, μ0\mu_0 is the permeability of free space, and ε0\varepsilon_0 is the permittivity of free space. Notice the coupling — a changing B⃗\vec B curls into an E⃗\vec E field (Faraday), and a changing E⃗\vec E curls into a B⃗\vec B field (Ampère–Maxwell). Neither field can change without generating the other; this mutual feeding is the seed of a self-sustaining wave.

Decouple the two curl equations. Taking the curl of one equation and substituting the other eliminates one field, leaving a single second-order equation purely in E⃗\vec E (or purely in B⃗\vec B). The tool needed is the vector identity for a curl of a curl,

∇×(∇×A⃗)=∇(∇⋅A⃗)−∇2A⃗\nabla\times(\nabla\times\vec A)=\nabla(\nabla\cdot\vec A)-\nabla^2\vec A

where: ∇2\nabla^2 is the vector Laplacian, applied component-wise. Because ∇⋅E⃗=0\nabla\cdot\vec E=0 and ∇⋅B⃗=0\nabla\cdot\vec B=0 in vacuum, the first term on the right vanishes for both fields, so the curl-of-curl collapses cleanly to −∇2E⃗-\nabla^2\vec E or −∇2B⃗-\nabla^2\vec B. This is the algebraic step that makes the whole derivation work — without ρ=0\rho=0, the ∇(∇⋅E⃗)\nabla(\nabla\cdot\vec E) term would survive and spoil the clean wave form.

Read off the wave speed by comparison. Once the derivation below produces ∇2E⃗=μ0ε0 ∂2E⃗/∂t2\nabla^2\vec E=\mu_0\varepsilon_0\,\partial^2\vec E/\partial t^2, compare it to the general 3D wave equation from the mechanical-waves chapter, ∇2ψ=1v2∂2ψ∂t2\nabla^2\psi=\dfrac{1}{v^2}\dfrac{\partial^2\psi}{\partial t^2}. Matching coefficients gives v2=1/(μ0ε0)v^2=1/(\mu_0\varepsilon_0), so

c=1μ0ε0c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}

where: cc is the wave speed. Numerically, with μ0=4π×10−7 T ⋅ m/A\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A} and ε0=8.85×10−12 C2/N ⋅ m2\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2, this evaluates to c≈3×108 m/sc\approx3\times10^8\ \text{m/s} — obtained purely from electric and magnetic constants, with no optical measurement fed in. Every Cartesian component of E⃗\vec E (or B⃗\vec B) independently obeys this scalar wave equation, so all the familiar mechanical-wave solutions carry over unchanged: travelling-wave forms f(z−ct)+g(z+ct)f(z-ct)+g(z+ct) and harmonic plane waves E0cos⁡(kz−ωt+ϕ)E_0\cos(kz-\omega t+\phi) with ω=ck\omega=ck.

Recognise the parallel with the mechanical wave equation. This is exactly the same 1D wave equation encountered for a stretched string, ∂2y/∂t2=v2 ∂2y/∂x2\partial^2y/\partial t^2=v^2\,\partial^2y/\partial x^2, except that E⃗\vec E (or B⃗\vec B) now plays the role of the string's displacement yy, and the statement is three-dimensional rather than restricted to a single axis. What this equation alone does not fix is the direction of E⃗\vec E relative to B⃗\vec B, or either field relative to the propagation direction — that extra structure emerges only once a specific plane-wave solution is substituted back into the full set of Maxwell equations, the topic of the following article.

Figure
Sketch of an electric field and a magnetic field oscillating together as they travel along a propagation axis
E and B satisfy the same wave equation and travel together at speed c propagation axis z, speed c E field B field (perpendicular in reality) Each field independently obeys the wave equation derived from the source-free Maxwell equations.
Derivation ADeriving the electromagnetic wave equation for E

The goal is to start purely from the four source-free Maxwell equations in vacuum, with no assumed wave form and no optical input whatsoever, and to arrive at a genuine second-order wave equation for E⃗\vec E alone, thereby also reading off the wave speed as a specific combination of the two electromagnetic constants μ0\mu_0 and ε0\varepsilon_0.

Step 1. Take the curl of both sides of Faraday's law, ∇×E⃗=−∂B⃗/∂t\nabla\times\vec E=-\partial\vec B/\partial t. This gives

∇×(∇×E⃗)=−∂∂t(∇×B⃗)\nabla\times(\nabla\times\vec E)=-\frac{\partial}{\partial t}(\nabla\times\vec B)

The right side still contains ∇×B⃗\nabla\times\vec B, which is exactly what the Ampère–Maxwell law tells us how to replace.

Step 2. Expand the left side using the vector identity ∇×(∇×E⃗)=∇(∇⋅E⃗)−∇2E⃗\nabla\times(\nabla\times\vec E)=\nabla(\nabla\cdot\vec E)-\nabla^2\vec E. Since the vacuum condition gives ∇⋅E⃗=0\nabla\cdot\vec E=0, the first term vanishes entirely, leaving

−∇2E⃗=−∂∂t(∇×B⃗)-\nabla^2\vec E=-\frac{\partial}{\partial t}(\nabla\times\vec B)

Step 3. Substitute the Ampère–Maxwell law, ∇×B⃗=μ0ε0 ∂E⃗/∂t\nabla\times\vec B=\mu_0\varepsilon_0\,\partial\vec E/\partial t, into the right-hand side:

−∇2E⃗=−μ0ε0∂2E⃗∂t2-\nabla^2\vec E=-\mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}

Cancelling the matching minus signs on both sides gives the electromagnetic wave equation,

∇2E⃗=μ0ε0∂2E⃗∂t2\nabla^2\vec E=\mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}

Step 4. Identify the wave speed by comparing to the standard 3D wave equation ∇2ψ=(1/v2) ∂2ψ/∂t2\nabla^2\psi=(1/v^2)\,\partial^2\psi/\partial t^2. Matching coefficients gives 1/v2=μ0ε01/v^2=\mu_0\varepsilon_0, so v=c=1/μ0ε0≈3×108 m/sv=c=1/\sqrt{\mu_0\varepsilon_0}\approx3\times10^8\ \text{m/s}. The identical sequence of steps, starting from the curl of the Ampère–Maxwell law and substituting Faraday's law, produces ∇2B⃗=μ0ε0 ∂2B⃗/∂t2\nabla^2\vec B=\mu_0\varepsilon_0\,\partial^2\vec B/\partial t^2 — the same equation, same speed, for B⃗\vec B.

Worked example

Problem. Using c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0}, verify that c≈3×108 m/sc\approx3\times10^8\ \text{m/s}, and find the percentage change in cc if ε0\varepsilon_0 were hypothetically 1% larger while μ0\mu_0 stayed fixed.

Step 1. Substitute the constants directly: c=1(4π×10−7)(8.85×10−12)=11.112×10−17=13.335×10−9≈3.00×108 m/sc=\dfrac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}}=\dfrac{1}{\sqrt{1.112\times10^{-17}}}=\dfrac{1}{3.335\times10^{-9}}\approx3.00\times10^8\ \text{m/s}, matching the standard value of the speed of light exactly as the derivation predicts.

Step 2. For the sensitivity, note c∝(ε0)−1/2c\propto(\varepsilon_0)^{-1/2}, so a small fractional change obeys δcc=−12δε0ε0\dfrac{\delta c}{c}=-\dfrac{1}{2}\dfrac{\delta\varepsilon_0}{\varepsilon_0}. With δε0/ε0=+1%=0.01\delta\varepsilon_0/\varepsilon_0=+1\%=0.01, this gives δc/c=−0.005\delta c/c=-0.005.

Step 3. Interpret the result: cc would decrease by 0.5% if ε0\varepsilon_0 alone rose by 1%. Both constants enter through an inverse square root, so μ0\mu_0 and ε0\varepsilon_0 share equal responsibility for fixing the propagation speed of every electromagnetic wave in vacuum — neither one alone "chooses" the speed of light; it is their product that does.

Competitive stretch

If instead both μ0\mu_0 and ε0\varepsilon_0 were each 1% larger simultaneously, the fractional changes in the square-root denominator add: δc/c=−12(δμ0/μ0+δε0/ε0)=−12(0.01+0.01)=−0.01\delta c/c=-\tfrac12(\delta\mu_0/\mu_0+\delta\varepsilon_0/\varepsilon_0)=-\tfrac12(0.01+0.01)=-0.01, a full 1% drop in cc. This kind of first-order propagation-of-error reasoning, treating cc as a power-law combination of two independent constants, is a standard JEE-level follow-up to the basic derivation.

Common traps

  • Writing ∇×(∇×E⃗)=−∇2E⃗\nabla\times(\nabla\times\vec E)=-\nabla^2\vec E directly without first checking ∇⋅E⃗=0\nabla\cdot\vec E=0 — the identity always has the extra ∇(∇⋅E⃗)\nabla(\nabla\cdot\vec E) term, which only vanishes because this derivation is restricted to a source-free vacuum.
  • Curling the same Maxwell equation twice instead of curling one and substituting the other — the derivation only closes when Faraday's law is curled and the Ampère–Maxwell law is substituted in (or vice versa for B⃗\vec B).
  • Treating c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0} as an empirical fit tuned to match the known speed of light — it is a prediction derived purely from electric and magnetic constants that happened to match the already-measured light speed, which is precisely why it convinced Maxwell that light is electromagnetic.
  • Applying this exact derivation inside matter — it holds strictly for ρ=0,J⃗=0\rho=0,\vec J=0 vacuum; a dielectric or conducting medium needs D⃗,H⃗\vec D,\vec H and different constants.

Ask the tutor: if the curl-of-curl vector identity or any single algebraic step feels shaky, ask the AI tutor to walk through it component by component.

Why does the derivation need rho equals zero and J equals zero?

Those are exactly the conditions that make ∇⋅E⃗=0\nabla\cdot\vec E=0 and ∇⋅B⃗=0\nabla\cdot\vec B=0 true, which is what makes the ∇(∇⋅E⃗)\nabla(\nabla\cdot\vec E) term in the curl-of-curl identity vanish. Without that term dropping out, you would be left with a messier equation involving charge and current distributions, not a clean wave equation. That is also why this derivation is confined to vacuum, or regions far from any charges and currents — inside a medium with real sources the simplification breaks down.

Does this derivation tell us the direction of E relative to B?

No. The wave equation only says each Cartesian component of E⃗\vec E, and separately of B⃗\vec B, obeys a scalar wave equation — it says nothing yet about how E⃗\vec E and B⃗\vec B are oriented relative to each other or to the direction the wave travels. That extra structure only appears when you substitute an actual plane-wave solution back into the full set of Maxwell equations, which shows E⃗\vec E, B⃗\vec B and the propagation direction are mutually perpendicular. That step belongs to the next article on plane electromagnetic waves.

Is c really just a coincidence with the measured speed of light?

No — that is precisely the historic significance. Maxwell computed 1/μ0ε01/\sqrt{\mu_0\varepsilon_0} from purely electrical and magnetic laboratory measurements, constants that had nothing to do with optics, and the number matched the independently measured speed of light. A coincidence of that precision, from two totally different branches of physics, is exactly the evidence that convinced Maxwell light itself is an electromagnetic wave rather than some separate phenomenon.

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Where this leads