Visual walkthrough — Speed of light c = 1 - √(ε₀ μ₀)

We will lean on Maxwell's Equations, and in particular Faraday's Law of Induction, the Ampère–Maxwell Law, and its crucial Displacement Current term. The result is the Wave Equation for Electromagnetic Waves.

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Step 0 — The vocabulary, drawn before we use it

Before any calculus, let us earn every symbol.

The little arrow on top of $\vec{E}$ and $\vec{B}$ (called a vector) just means "this quantity has a direction, not only a size." Look at the figure: at every dot in space there is one $E$-arrow (blue) and one $B$-arrow (yellow). That is all a field is — a rule that pins an arrow to every point.

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Step 1 — The two starting laws, as pictures

Look at the figure: the two laws are mirror images. Each field's swirl is fed by the other field's wobble. This is the "self-building dominoes" idea, now written exactly. We also keep two more facts from Maxwell's Equations: in empty space $\nabla \cdot \vec{E}=0$ and $\nabla \cdot \vec{B}=0$ (the divergence, "do arrows spread out from a point?", is zero because there is no charge to spread from).

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Step 2 — Take the curl of Faraday's law

The figure shows the plan as a loop: $E\to$ (its swirl) $\to B$'s wobble $\to$ (its swirl) $\to E$'s wobble. Following that loop is what will close the equation onto $\vec{E}$ alone.

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Step 3 — Untangle the double curl

Now use $\nabla \cdot \vec{E}=0$ (Step 1, no charges): the spreading part vanishes.

$\nabla \times (\nabla \times \vec{E})=-{\nabla }^2\vec{E}$

The figure contrasts the two pieces: on the left, arrows fanning out from a point (divergence — zero here, crossed out in red); on the right, arrows curving like a wave crest (the ${\nabla }^2$ piece we keep).

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Step 4 — Substitute Ampère–Maxwell and eliminate $\vec{B}$

Cancel the two minus signs:

$\boxed{{\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}}$

The figure marks each surviving symbol: blue ${\nabla }^2\vec{E}$ = "how curved in space," yellow ${\partial }^2/\partial t^2$ = "how accelerating in time," and the red ${\mu }_0{\epsilon }_0$ = "the exchange rate between space-curvature and time-acceleration."

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Step 5 — Read off the speed by matching the wave equation

Set our boxed equation beside it, term against term. The space piece ${\nabla }^2$ matches ${\partial }^2/\partial x^2$; the time piece matches; so the multiplier must match:

$\frac{1}{v^2}={\mu }_0{\epsilon }_0⟹v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\equiv c$

WHY the reciprocal and the square root: the constant sits on the time side of the wave equation as $1/v^2$. To free $v$ we flip it (reciprocal) and take a root. The figure overlays a sine wave $f(x-vt)$ sliding right and labels the slope $v$: bigger ${\mu }_0{\epsilon }_0$ (stiffer space) $\Rightarrow$ smaller $v$ $\Rightarrow$ slower crest.

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Step 6 — The edge cases (never leave a gap)

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The one-picture summary

The final figure is the whole derivation in one loop: Faraday turns $B$'s wobble into $E$'s swirl; curl + identity reshape it into ${\nabla }^2\vec{E}$; Ampère–Maxwell turns $E$'s wobble into $B$'s swirl; substituting closes the loop onto $\vec{E}$ alone, leaving ${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0{\partial }^2\vec{E}/\partial t^2$; and matching the Wave Equation extracts $c=1/\sqrt{{\mu }_0{\epsilon }_0}$.

Recall Feynman retelling — the whole walkthrough in plain words

Two facts of nature: a changing magnetic push makes the electric field curl around it, and a changing electric push makes the magnetic field curl around it. I take the first fact and curl it once more, which lets me swap in the second fact and throw the magnetic field out of the maths completely. What's left is an equation that says the electric field's shape in space equals the field's acceleration in time, multiplied by two little numbers describing how stiff empty space is to electricity and to magnetism. But that exact form is the equation of any travelling wave, where the multiplier is one-over-speed-squared. So I flip it and take a square root — and the speed pops out: one over the square root of ${\epsilon }_0$ times ${\mu }_0$, which is 300 million metres a second. Kill the displacement-current term and the loop never closes, so there'd be no light at all. Put the numbers of glass in instead of empty space, and the same machine spits out the slower speed $c/n$. Nobody chose this speed; it's just how springy the vacuum is.

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Recall checkpoint