Electromagnetic Waves

By the 1860s physicists had four hard-won laws — Gauss's law for electric fields, Gauss's law for magnetism, Faraday's law of induction, and Ampère's law as completed by Maxwell's displacement current. Each described how charges and currents make fields. Nobody expected these four equations to predict light. Then James Clerk Maxwell did something almost casual: he asked what the equations allow in empty space, where there are no charges and no currents at all.

The answer that fell out is the single idea of this page: in vacuum, Maxwell's equations combine into a wave equation. The electric field obeys \nabla^2\mathbf{E} = \mu_0\varepsilon_0\,\partial^2\mathbf{E}/\partial t^2, the magnetic field obeys the very same equation, and the wave speed that pops out is c = 1/\sqrt{\mu_0\varepsilon_0}. Plug in the numbers measured in a lab full of batteries and magnets and you get 3\times 10^8\ \text{m/s} — the speed of light. Maxwell's own words land like a thunderclap: light "is an electromagnetic disturbance." The thing your eyes detect, the thing radios and X-ray machines emit, is a ripple in the electromagnetic field.

The four equations, in vacuum

Everything we need starts from Maxwell's equations with no sources — no charge density (\rho = 0) and no current (\mathbf{J} = 0):

Notice the beautiful symmetry: a changing \mathbf{B} curls up an \mathbf{E} (Faraday), and a changing \mathbf{E} curls up a \mathbf{B} (Ampère–Maxwell). Each field, by changing, keeps the other alive. That mutual feeding is the engine of the wave — and it is Maxwell's displacement-current term on the right of the last equation that makes it possible. Without that term, no light.

The derivation: take the curl of Faraday's law

We want an equation for \mathbf{E} alone. The trick is to take the curl of both sides of Faraday's law, so we can feed the other equations in. Start from

\nabla\times(\nabla\times\mathbf{E}) = \nabla\times\left(-\frac{\partial\mathbf{B}}{\partial t}\right) = -\frac{\partial}{\partial t}\left(\nabla\times\mathbf{B}\right).

(Space and time derivatives commute, so the curl slid inside the time derivative.) Now hit each side with a known identity or equation.

Left side — the "curl of a curl" identity. For any vector field, vector calculus gives

\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}.

But in vacuum \nabla\cdot\mathbf{E} = 0, so the first term vanishes, leaving just -\nabla^2\mathbf{E}.

Right side — substitute Ampère–Maxwell. Replace \nabla\times\mathbf{B} with \mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t:

-\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}\right) = -\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}.

Set the two rewritten sides equal, and the minus signs cancel:

-\nabla^2\mathbf{E} = -\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2} \quad\Longrightarrow\quad \boxed{\;\nabla^2\mathbf{E} = \mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}\;}

That is a wave equation. Doing the identical dance starting from Ampère–Maxwell (take its curl, use \nabla\cdot\mathbf{B} = 0, substitute Faraday) gives the twin result for the magnetic field:

\nabla^2\mathbf{B} = \mu_0\varepsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2}.

Both fields obey the same wave equation with the same speed. Neither field is "the cause" — they travel together as one object.

Reading off the speed — the punchline

The generic wave equation is \nabla^2 f = \dfrac{1}{v^2}\dfrac{\partial^2 f}{\partial t^2}, where v is the speed of the wave. Matching that template to what we derived, the coefficient \mu_0\varepsilon_0 must be 1/v^2:

\frac{1}{v^2} = \mu_0\varepsilon_0 \quad\Longrightarrow\quad v = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \equiv c.

Now put in the two constants, both measured with ordinary lab equipment long before anyone connected them to light:

Multiply and take the reciprocal square root:

\mu_0\varepsilon_0 = (1.2566\times 10^{-6})(8.85\times 10^{-12}) \approx 1.112\times 10^{-17}\ \text{s}^2/\text{m}^2, c = \frac{1}{\sqrt{1.112\times 10^{-17}}} \approx 2.998\times 10^{8}\ \text{m/s}.

That is the measured speed of light, to the digits you keep. Two constants from electricity and magnetism, combined, are the speed of light. This is the moment physics discovered that optics is a branch of electromagnetism.

The plane wave: what the solution looks like

The simplest solution of the wave equation is a plane wave travelling in, say, the +z direction. Take the electric field pointing along x:

\mathbf{E}(z,t) = E_0\cos(kz - \omega t)\,\hat{\mathbf{x}},

where k = 2\pi/\lambda is the wavenumber and \omega = 2\pi f the angular frequency. Feed this into the wave equation and you find the two must satisfy \omega/k = c — the wave marches forward at exactly c. Faraday's law then forces the magnetic field to point along y:

\mathbf{B}(z,t) = \frac{E_0}{c}\cos(kz - \omega t)\,\hat{\mathbf{y}}.

Read off four facts that define electromagnetic radiation:

The graph shows the two fields along the propagation axis. They rise and fall in step; drag the phase slider to send the wave travelling. (The magnetic curve is drawn as cB so it shares the electric field's height — physically |\mathbf{B}| is smaller by the enormous factor c.)

Worked examples

Example 1 — from frequency to wavelength. A radio station broadcasts at f = 100\ \text{MHz} = 1.0\times 10^8\ \text{Hz}. All electromagnetic waves in vacuum obey c = f\lambda, so

\lambda = \frac{c}{f} = \frac{3.0\times 10^8}{1.0\times 10^8} = 3.0\ \text{m}.

FM radio waves are a few metres long — which is why the antenna is that sort of size.

Example 2 — the magnetic field of sunlight. Bright sunlight has an electric-field amplitude of about E_0 = 1000\ \text{V/m}. Because |\mathbf{E}| = c|\mathbf{B}|,

B_0 = \frac{E_0}{c} = \frac{1000}{3.0\times 10^8} \approx 3.3\times 10^{-6}\ \text{T}.

Just a few microtesla — weaker than a fridge magnet. This is why we barely notice the magnetic side of light: divided by c, it is tiny in ordinary units.

Example 3 — wavenumber to frequency. A green laser has wavelength \lambda = 500\ \text{nm} = 5.0\times 10^{-7}\ \text{m}. Its wavenumber and angular frequency are

k = \frac{2\pi}{\lambda} = \frac{2\pi}{5.0\times 10^{-7}} \approx 1.26\times 10^{7}\ \text{rad/m}, \omega = ck = (3.0\times 10^8)(1.26\times 10^7) \approx 3.8\times 10^{15}\ \text{rad/s}.

Equivalently f = c/\lambda = 6.0\times 10^{14}\ \text{Hz} — visible light oscillates hundreds of trillions of times a second.

One equation, the whole spectrum

Nothing in the derivation fixed the frequency. Any \omega and k obeying \omega = ck is a valid solution — so electromagnetic waves come in every frequency, all travelling at the same c. That continuum is the electromagnetic spectrum: from long-wavelength radio and microwaves, through infrared, the thin band of visible light (400–700\ \text{nm}), to ultraviolet, X-rays, and gamma rays. Radio and gamma rays are the same physics — the same \mathbf{E} and \mathbf{B} feeding each other — differing only in how fast they wiggle.

A very natural guess — and wrong for a travelling wave. In a mass on a spring, kinetic and potential energy trade off, so one peaks when the other is zero (90° out of phase). People expect \mathbf{E} and \mathbf{B} to do the same. They do not: in a plane wave in vacuum they carry the identical \cos(kz-\omega t), so they peak together and are zero together — perfectly in phase.

Why? Faraday's law relates the space derivative of E to the time derivative of B. For a cosine both derivatives are the same sine, so no phase offset appears. (The 90°-out-of-phase picture does apply to a standing wave, e.g. inside a cavity — but that is a sum of two travelling waves, a different situation.) For light streaming freely through space: same phase, always.

Nineteenth-century physicists were so uneasy about this that they invented a substance — the "luminiferous ether" — for light to wave in, by analogy with sound needing air. But look back at the derivation: we set \rho = 0 and \mathbf{J} = 0 and the wave survived anyway. The fields themselves are the medium. A wiggle in \mathbf{E} creates a wiggle in \mathbf{B} (via Ampère–Maxwell), which recreates \mathbf{E} (via Faraday), which recreates \mathbf{B}… on and on through utterly empty space, no ether required.

The Michelson–Morley experiment (1887) hunted for the ether and found nothing, and Einstein's relativity later explained why there is none to find. Light needs no material to travel through — that is exactly how starlight crosses the vacuum of space to reach your eye.