Visual walkthrough — Electromagnetic waves — derivation from Maxwell's equations

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Step 0 — The three arrows we will use forever

Before any physics, meet the three actors. They are arrows (vectors) — an arrow has a length (how strong) and a direction (which way it points).

• $\vec{E}$ — the electric field arrow. Draw it blue. If a tiny positive charge sat there, it would feel a push along this arrow. Longer arrow = harder push.
• $\vec{B}$ — the magnetic field arrow. Draw it orange. It's the field a compass needle lines up with.
• $\vec{k}$ — the travel arrow (the wavevector). Green. It points the way the wave is going. Its length, written $k$ (no arrow), will later encode "how squished the ripple is". Keep this in mind: bold-arrow $\vec{k}$ = direction-and-length; plain $k$ = just its length.

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Step 1 — The four laws, drawn as pictures

WHAT. We stand in empty space: no charges ($\rho =0$), no currents ($\vec{J}=0$). Maxwell's four rules survive even here:

$\underset{(1)}{\underbrace{\nabla \cdot \vec{E}=0}}\underset{(2)}{\underbrace{\nabla \cdot \vec{B}=0}}\underset{(3)}{\underbrace{\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}}}\underset{(4)}{\underbrace{\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}}}$

Term by term, in plain words:

• $\nabla \cdot \vec{E}$ — the divergence of $\vec{E}$ (the dot product of $\nabla$ with $\vec{E}$). Read it as: "do the blue arrows spread out from this point, like water from a tap?" Equation (1) says $=0$: no taps, no drains. Field lines never start or stop in vacuum.
• $\nabla \cdot \vec{B}=0$ — same story for orange. (Also: no lone magnetic poles ever.)
• $\nabla \times \vec{E}$ — the curl of $\vec{E}$ (the cross product of $\nabla$ with $\vec{E}$): "do the blue arrows swirl around this point, like water round a plughole?" Equation (3) says that swirl equals $-\partial \vec{B}/\partial t$, i.e. a magnetic field that is changing in time makes electric arrows swirl.
• $\partial \vec{B}/\partial t$ — "how fast the orange arrow is changing per second." The minus sign is nature's direction bookkeeping (Lenz).
• Equation (4) is the mirror: a changing electric field makes magnetic arrows swirl. The ${\mu }_0{\epsilon }_0$ out front is the small conversion factor between the two — remember it, it becomes the speed of light.

WHY these four. (1),(2) forbid sources; (3),(4) are the coupling — each changing field births a swirl of the other. That mutual feeding is the whole engine.

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Step 2 — The trick: curl the swirl again

WHAT. Take equation (3) and hit both sides with another curl operator $\nabla \times$: $\nabla \times (\nabla \times \vec{E})=\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)$

WHY. Look at the right side: it now contains $\nabla \times \vec{B}$ — and equation (4) is a ready-made formula for exactly $\nabla \times \vec{B}$. So a second curl lets us substitute (4) into (3) and throw $\vec{B}$ out entirely. Two equations tangled together become one equation about $\vec{E}$ alone. This is the Faraday→Ampère–Maxwell handshake.

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Step 3 — Untangle the left side with one identity

WHAT. There is a fixed algebraic fact, true for any arrow field $\vec{F}$: $\nabla \times (\nabla \times \vec{F})=\nabla (\nabla \cdot \vec{F})-{\nabla }^2\vec{F}$

Term by term:

• $\nabla (\nabla \cdot \vec{F})$ — take the "spreading-out" number, then find its slope. The source-driven piece.
• ${\nabla }^2\vec{F}$ — the Laplacian, defined as ${\nabla }^2\equiv \nabla \cdot \nabla =\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$. For a vector field it acts componentwise: ${\nabla }^2\vec{F}=({\nabla }^2{F}_x,{\nabla }^2{F}_y,{\nabla }^2{F}_z)$. Read it as "how much does $\vec{F}$ at this point differ from the average of its neighbours?" This is the curvature-in-space term — the one that shows up in every wave equation.

Apply it to $\vec{E}$, then use equation (1), $\nabla \cdot \vec{E}=0$: $\nabla \times (\nabla \times \vec{E})=\nabla (\underset{=0}{\underbrace{\nabla \cdot \vec{E}}})-{\nabla }^2\vec{E}=-{\nabla }^2\vec{E}$

WHY. The "no charges" law (1) erases the first term. This is the exact moment the vacuum assumption pays off — we're left with the clean curvature term only.

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Step 4 — Feed in Ampère–Maxwell (the right side)

WHAT. Work on the right side of Step 2. First swap the order of the space-curl and the time-derivative (allowed, the fields are smooth), then plug in equation (4): $\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)=-\frac{\partial }{\partial t}(\nabla \times \vec{B})=-\frac{\partial }{\partial t}\left({\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$

Term by term:

• $\frac{\partial }{\partial t}(\nabla \times \vec{B})$ — the swirl of $\vec{B}$, watched changing in time.
• After (4): ${\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$ — two time-derivatives of $\vec{E}$ (the "acceleration" of the field), scaled by ${\mu }_0{\epsilon }_0$.

WHY. This is where $\vec{B}$ vanishes for good. Everything is now written in $\vec{E}$.

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Step 5 — Set the two sides equal: the wave is born

WHAT. Left side (Step 3) $=$ right side (Step 4). The minus signs on both sides cancel: $\boxed{{\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}}$

Redo the whole thing starting from (4) instead of (3), using $\nabla \cdot \vec{B}=0$, and you get the identical shape for $\vec{B}$: ${\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}$

WHY it's a wave. Compare to the master Wave Equation: ${\nabla }^{2}f=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}$

• ${\nabla }^{2}f$ — how curved the shape is in space.
• $\frac{1}{v^2}{\partial }_t^{2}f$ — how fast the shape accelerates in time. Any quantity whose space-curvature equals its time-acceleration (up to a constant) is a travelling wave. Ours matches — so $\vec{E}$ and $\vec{B}$ travel.

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Step 6 — Plug in the numbers: it's the speed of light

WHAT. Put in the two lab constants: $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}},{\mu }_0=1.2566\times 10^{-6}\frac{\text{T⋅m}}{\text{A}},{\epsilon }_0=8.854\times 10^{-12}\frac{\text{F}}{\text{m}}$ $c=\frac{1}{\sqrt{(1.2566\times 10^{-6})(8.854\times 10^{-12})}}\approx 3.00\times 10^8\text{m/s}$

WHY it's stunning. ${\mu }_0$ was measured with magnets and wires; ${\epsilon }_0$ was measured with static charges. Neither experiment ever involved light. Yet multiply, invert, square-root — out pops the measured speed of light. That single coincidence is what told Maxwell: light is an electromagnetic wave.

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Step 7 — The geometry: E, B, k form a right-handed triad

Setup convention. From here we choose our axes so the wave travels along $+x$. That means the wavevector points that way: $\vec{k}=k\hat{x}$, and from now on the plain symbol $k$ means its length only. With that choice a plane wave is: $\vec{E}={\vec{E}}_0{e}^{i(kx-\omega t)},\vec{B}={\vec{B}}_0{e}^{i(kx-\omega t)}$

• ${\vec{E}}_0$, ${\vec{B}}_0$ — constant vector amplitudes (they carry both a size and a direction). Their magnitudes (lengths) are written $E_0=|{\vec{E}}_0|$ and $B_0=|{\vec{B}}_0|$ — plain, no arrow. Watch this: arrow = vector, plain = its length.
• $k$ — the wavenumber (length of $\vec{k}$): radians of wiggle per metre. Big $k$ = tightly squished wave.
• $\omega$ — the angular frequency: radians of wiggle per second.

Three facts drop out:

1. Transverse, from (1): $\nabla \cdot \vec{E}=0$. For our plane wave $\nabla \cdot \vec{E}=ik{E}_{0x}$, so $ik{E}_{0x}=0\Rightarrow E_{0x}=0$. The field has no piece along the travel direction $\hat{x}$. (See Polarization for what the remaining sideways freedom means.)
2. $\vec{E}\perp \vec{B}$, from (3): pick the surviving $\vec{E}$ to lie along $\hat{y}$, so $\vec{E}=E_0\hat{y}{e}^{i(kx-\omega t)}$. Then Faraday's law needs $\nabla \times \vec{E}$. Computing the cross product with $\nabla =ik\hat{x}$ (only $x$-derivatives survive): $\nabla \times \vec{E}=(ik\hat{x})\times (E_0\hat{y})e^{i(kx-\omega t)}=ik{E}_0(\hat{x}\times \hat{y})e^{i(kx-\omega t)}=ik{E}_0\hat{z}{e}^{i(kx-\omega t)}$ because $\hat{x}\times \hat{y}=\hat{z}$. Faraday says this equals $-{\partial }_t\vec{B}=i\omega \vec{B}$, so $\vec{B}$ must point along $\hat{z}$. Hence $\vec{E}\parallel \hat{y}$, $\vec{B}\parallel \hat{z}$, $\vec{k}\parallel \hat{x}$: three mutually perpendicular arrows with $\hat{E}\times \hat{B}=\hat{k}$ — a right-handed set.
3. Amplitudes, from (3): matching lengths in that same equation, $k{E}_0=\omega B_0\Rightarrow \frac{E_0}{B_0}=\frac{\omega }{k}=c$, hence $E_0=c{B}_0$ (lengths only).

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Step 8 — The degenerate case: no displacement current, no wave

WHAT. Suppose we forget Maxwell's term and use the old law $\nabla \times \vec{B}=0$ in vacuum. Redo Step 4: $\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)=-\frac{\partial }{\partial t}(\underset{=0}{\underbrace{\nabla \times \vec{B}}})=0$ Equating to the left side (Step 3): ${\nabla }^2\vec{E}=0$

WHY it matters. ${\nabla }^2\vec{E}=0$ has no time in it — it's a static, frozen condition, not a wave. There is no travelling ripple, no $c$, no light. So the one term Maxwell inserted is exactly the difference between a dead universe and a lit one. This is the single most important edge case in the whole story.

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Worked examples (each re-checked)

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The one-picture summary

Recall The whole walkthrough compressed into six moves

(1) Meet three arrows and one tool $\nabla$. (2) Write the four vacuum laws; two forbid sources, two couple $\vec{E}$ and $\vec{B}$. (3) Curl Faraday's law twice so the magnetic side can be swapped in via Ampère–Maxwell. (4) The curl-of-curl identity plus $\nabla \cdot \vec{E}=0$ collapses the left side to $-{\nabla }^2\vec{E}$. (5) Ampère–Maxwell turns the right side into $-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}$; equate to get ${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}$ — the wave equation, whose constant reads off as $c=1/\sqrt{{\mu }_0{\epsilon }_0}\approx 3\times 10^8$ m/s. (6) Feeding a plane wave back in fixes the geometry: $\vec{E}$, $\vec{B}$, $\vec{k}$ mutually perpendicular, right-handed, with $E_0=c{B}_0$ — and dropping Maxwell's term (Step 8) destroys the wave entirely.