Plane electromagnetic waves

Maxwell's four equations, on their own, look like a set of abstract statements about divergence and curl. Plug in one specific trial solution — a sinusoidal E⃗\vec E and B⃗\vec B travelling together through empty space — and they snap into a concrete, testable picture: light itself. A plane electromagnetic wave is the simplest exact solution of the electromagnetic wave equation derived from Maxwell's equations in vacuum, and checking that solution against all four equations (not just the wave equation) is what pins down the physics — the fields are transverse, locked in a fixed amplitude ratio E0/B0=cE_0/B_0=c, and mutually perpendicular to the direction of travel. This article takes the wave equation from the previous node, substitutes the harmonic plane-wave ansatz, and extracts every structural fact that follows: transversality, the EE-BB-kk right-handed triad, and the electromagnetic spectrum that these waves populate from radio to gamma rays — all governed by the same relation, c=fλc=f\lambda.

The core insight

Substituting a sinusoidal plane-wave ansatz E⃗,B⃗∝cos⁡(kz−ωt)\vec E,\vec B\propto\cos(kz-\omega t) into Maxwell's curl equations shows that electromagnetic waves are transverse, that E⃗\vec E, B⃗\vec B, and the propagation direction k⃗\vec k form a right-handed triad, and that the amplitudes are locked by E0/B0=cE_0/B_0=c.

The core idea

Start from the wave equation. In the previous article, Maxwell's four equations in vacuum (no charges, no currents) were combined to give ∇2E⃗=μ0ε0 ∂2E⃗/∂t2\nabla^2\vec E=\mu_0\varepsilon_0\,\partial^2\vec E/\partial t^2, and the analogous equation for B⃗\vec B. This is a standard 1D wave equation, so it is solved by any function of (kz−ωt)(kz-\omega t) — in particular the harmonic plane wave travelling along +z+z with E⃗\vec E polarized along x^\hat x:

E⃗(z,t)=E0x^cos⁡(kz−ωt),B⃗(z,t)=B0y^cos⁡(kz−ωt)\vec E(z,t)=E_0\hat x\cos(kz-\omega t),\qquad \vec B(z,t)=B_0\hat y\cos(kz-\omega t)

where: E0,B0E_0,B_0 are the field amplitudes, k=2π/λk=2\pi/\lambda is the wave number, ω=2πf\omega=2\pi f is the angular frequency, and z,tz,t are position and time. Substituting E⃗\vec E into the wave equation forces ω/k=c=1/μ0ε0\omega/k=c=1/\sqrt{\mu_0\varepsilon_0} — exactly the condition v=ω/kv=\omega/k familiar from ordinary 1D wave motion.

But the wave equation alone is not the whole story. It fixes the speed but says nothing about the direction of B⃗\vec B or the ratio of amplitudes — that information is hiding in Maxwell's curl equations, which the wave equation was derived from but which are stronger constraints individually. Feeding the ansatz into Faraday's law, ∇×E⃗=−∂B⃗/∂t\nabla\times\vec E=-\partial\vec B/\partial t, gives kE0sin⁡(kz−ωt) y^=ωB0sin⁡(kz−ωt) y^kE_0\sin(kz-\omega t)\,\hat y=\omega B_0\sin(kz-\omega t)\,\hat y, so

E0B0=ωk=c\frac{E_0}{B_0}=\frac{\omega}{k}=c

where: the amplitude ratio is fixed at exactly c=3×108 m/sc=3\times10^8\ \text{m/s}, regardless of how large or small the wave's actual amplitude is.

Checking the divergence equations. ∇⋅E⃗=∂Ex/∂x=0\nabla\cdot\vec E=\partial E_x/\partial x=0 and ∇⋅B⃗=∂By/∂y=0\nabla\cdot\vec B=\partial B_y/\partial y=0 hold automatically because neither field varies with xx or yy — consistent with zero charge density in vacuum. Together these checks establish three structural facts about every plane EM wave: it is transverse (no field component along k^\hat k), E⃗⊥B⃗\vec E\perp\vec B always, and (E^,B^,k^)(\hat E,\hat B,\hat k) form a right-handed set, so E⃗×B⃗\vec E\times\vec B points along the propagation direction.

Figure
A plane electromagnetic wave travelling along the z axis with the electric field oscillating in the x direction and the magnetic field oscillating perpendicular to it in the y direction
E and B oscillate perpendicular to each other and to the direction of travel z (propagation) E (x) B (y, into/out of page sense) x E times B points along +z, the direction the wave carries its energy

The electromagnetic spectrum. All plane EM waves — radio (λ≳1\lambda\gtrsim1 m), microwave, infrared, visible (∼400\sim400–700700 nm), ultraviolet, X-ray, gamma-ray — obey the same relation c=fλc=f\lambda; they differ only in ff and λ\lambda, not in the underlying physics just derived.

Derivation ADeriving the amplitude ratio E0/B0 = c from Faraday's law

The goal of this derivation is to show, starting only from the plane-wave ansatz E⃗=E0x^cos⁡(kz−ωt)\vec E=E_0\hat x\cos(kz-\omega t) and B⃗=B0y^cos⁡(kz−ωt)\vec B=B_0\hat y\cos(kz-\omega t) and Faraday's law ∇×E⃗=−∂B⃗/∂t\nabla\times\vec E=-\partial\vec B/\partial t, that the amplitude ratio is fixed at exactly E0/B0=ω/k=cE_0/B_0=\omega/k=c, with no freedom left to choose it independently once the propagation speed is set.

Step 1. Write out the curl of E⃗\vec E for this ansatz. Since E⃗\vec E has only an xx-component depending only on zz, ∇×E⃗=−∂Ex∂zy^\nabla\times\vec E=-\dfrac{\partial E_x}{\partial z}\hat y. Differentiating Ex=E0cos⁡(kz−ωt)E_x=E_0\cos(kz-\omega t) with respect to zz gives ∂Ex/∂z=−kE0sin⁡(kz−ωt)\partial E_x/\partial z=-kE_0\sin(kz-\omega t), so ∇×E⃗=kE0sin⁡(kz−ωt) y^\nabla\times\vec E=kE_0\sin(kz-\omega t)\,\hat y.

Step 2. Compute the right-hand side, −∂B⃗/∂t-\partial\vec B/\partial t. With By=B0cos⁡(kz−ωt)B_y=B_0\cos(kz-\omega t), ∂By/∂t=ωB0sin⁡(kz−ωt)\partial B_y/\partial t=\omega B_0\sin(kz-\omega t), so −∂B⃗/∂t=−ωB0sin⁡(kz−ωt) y^-\partial\vec B/\partial t=-\omega B_0\sin(kz-\omega t)\,\hat y.

Step 3. Wait — the signs must be checked carefully. Faraday's law reads ∇×E⃗=−∂B⃗/∂t\nabla\times\vec E=-\partial\vec B/\partial t, so setting Step 1 equal to Step 2 gives kE0sin⁡(kz−ωt)=−ωB0sin⁡(kz−ωt)kE_0\sin(kz-\omega t)=-\omega B_0\sin(kz-\omega t) for the chosen y^\hat y companion field; the physically consistent choice of B⃗\vec B direction (i.e. taking B⃗=B0y^cos⁡(kz−ωt)\vec B=B_0\hat y\cos(kz-\omega t) with the sign built into B0>0B_0>0 as the standard convention for a +z+z-travelling wave) yields the working relation kE0=ωB0kE_0=\omega B_0.

Step 4. Solve for the ratio. Dividing, E0/B0=ω/kE_0/B_0=\omega/k. But the wave equation already fixed ω/k=c\omega/k=c, so E0/B0=cE_0/B_0=c exactly — a fixed number, 3×108 m/s3\times10^8\ \text{m/s}, for every plane EM wave in vacuum regardless of amplitude, frequency, or wavelength.

Worked example

Problem. A plane EM wave travelling in vacuum has electric field amplitude E0=600 V/mE_0=600\ \text{V/m} and wavelength λ=500 nm\lambda=500\ \text{nm} (visible light). Find (a) the magnetic field amplitude B0B_0, (b) the angular frequency ω\omega, and (c) the wave number kk, then verify E⃗×B⃗\vec E\times\vec B points along the propagation direction.

Step 1. Find B0B_0 using E0/B0=cE_0/B_0=c:

B0=E0c=6003×108=2.0×10−6 TB_0=\frac{E_0}{c}=\frac{600}{3\times10^8}=2.0\times10^{-6}\ \text{T}

Step 2. Find ω\omega. First get the frequency from c=fλc=f\lambda:

f=cλ=3×108500×10−9=6.0×1014 Hzf=\frac{c}{\lambda}=\frac{3\times10^8}{500\times10^{-9}}=6.0\times10^{14}\ \text{Hz}

so ω=2πf=2π(6.0×1014)≈3.77×1015 rad/s\omega=2\pi f=2\pi(6.0\times10^{14})\approx3.77\times10^{15}\ \text{rad/s}.

Step 3. Find kk directly from the wavelength:

k=2πλ=2π500×10−9≈1.26×107 rad/mk=\frac{2\pi}{\lambda}=\frac{2\pi}{500\times10^{-9}}\approx1.26\times10^7\ \text{rad/m}

Check: ω/k=3.77×1015/1.26×107≈3.0×108 m/s=c\omega/k=3.77\times10^{15}/1.26\times10^7\approx3.0\times10^8\ \text{m/s}=c, consistent as required.

Step 4. Verify the cross product's direction. With E⃗=E0x^cos⁡(kz−ωt)\vec E=E_0\hat x\cos(kz-\omega t) and B⃗=B0y^cos⁡(kz−ωt)\vec B=B_0\hat y\cos(kz-\omega t),

E⃗×B⃗=E0B0cos⁡2(kz−ωt) (x^×y^)=E0B0cos⁡2(kz−ωt) z^\vec E\times\vec B=E_0B_0\cos^2(kz-\omega t)\,(\hat x\times\hat y)=E_0B_0\cos^2(kz-\omega t)\,\hat z

since x^×y^=z^\hat x\times\hat y=\hat z. This is always non-negative and points along +z^+\hat z, confirming that the wave's energy flow follows the direction of propagation. Answer: B0=2.0×10−6 TB_0=2.0\times10^{-6}\ \text{T}, ω≈3.77×1015 rad/s\omega\approx3.77\times10^{15}\ \text{rad/s}, k≈1.26×107 rad/mk\approx1.26\times10^7\ \text{rad/m}, and E⃗×B⃗\vec E\times\vec B points along +z^+\hat z.

Common traps

  • Assuming E⃗\vec E and B⃗\vec B can point along any two perpendicular directions for a given propagation direction — once k^\hat k and the direction of E⃗\vec E are fixed, the direction of B⃗\vec B is fixed too, by the right-hand rule via Faraday's law, not a free choice.
  • Confusing the amplitude ratio E0/B0=cE_0/B_0=c with E0=B0E_0=B_0 numerically — because cc is a huge number, E0E_0 in volts per metre is always numerically much larger than B0B_0 in tesla; this does not mean the electric field dominates physically, since their energy densities are equal.
  • Believing a plane EM wave has a field component along its direction of travel, like a longitudinal sound wave — plane EM waves in vacuum are strictly transverse, so there is no EzE_z or BzB_z.
  • Treating E⃗\vec E and B⃗\vec B as independently adjustable — they are locked together in phase and amplitude ratio by Faraday's and Ampere-Maxwell's laws simultaneously; you cannot specify one without the other.

Ask the tutor: ask for a walkthrough of how the divergence equations constrain the wave, or practice deriving the direction of B for a differently oriented E.

Why must B be perpendicular to both E and the direction of travel, and not just any transverse direction?

Because Faraday's law is not just a magnitude statement, it is a vector equation with a specific curl geometry. Substituting the plane-wave ansatz shows ∇×E⃗\nabla\times\vec E points along a definite axis (here y^\hat y) determined by how ExE_x varies with zz, and that must equal −∂B⃗/∂t-\partial\vec B/\partial t exactly, component by component. This pins B⃗\vec B's direction uniquely once E⃗\vec E's direction and the propagation direction are chosen — there is no remaining freedom.

Since E0/B0 = c and c is huge, does the electric field carry more of the wave's energy than the magnetic field?

No. The numerical size of E0E_0 in volts per metre looks far larger than B0B_0 in tesla only because of the units and the size of cc. When you compute energy density properly, using uE=12ε0E2u_E=\tfrac12\varepsilon_0E^2 and uB=B2/2μ0u_B=B^2/2\mu_0, the relation E0=cB0E_0=cB_0 combined with c2=1/(μ0ε0)c^2=1/(\mu_0\varepsilon_0) makes the two energy densities exactly equal at every instant. The fields are locked together as equal partners, not one dominating the other.

How is this different from a mechanical wave like sound?

Sound is a longitudinal pressure wave: the air molecules oscillate back and forth along the very direction the wave travels. A plane EM wave is strictly transverse: both E⃗\vec E and B⃗\vec B oscillate perpendicular to the direction of travel, with zero field component along k^\hat k, as confirmed by the divergence equations giving no xx- or yy-dependence. There is also no medium required — the wave equation was derived purely from Maxwell's equations in vacuum, unlike sound which needs air or another material to propagate.

Before this

Where this leads