Speed of light c = 1 - √(ε₀ μ₀)

WHAT — the claim

The astonishing part: $\varepsilon_0$ came from electrostatics (Coulomb's law) and $\mu_0$ came from magnetism (forces between wires). Light was thought to be a totally separate thing. Maxwell combined them and out popped a speed equal to the measured speed of light — proving light is electromagnetism.

WHY does this combination give a speed?

HOW — derive it from Maxwell's equations (from scratch)

We derive the wave equation for $\vec E$ and read off the speed. Work in vacuum (no charges $\rho=0$, no currents $\vec J=0$). Maxwell's equations:

$\nabla\cdot\vec E = 0,\qquad \nabla\cdot\vec B = 0$ $\nabla\times\vec E = -\frac{\partial \vec B}{\partial t}\quad(\text{Faraday})$ $\nabla\times\vec B = \mu_0\varepsilon_0\frac{\partial \vec E}{\partial t}\quad(\text{Ampère–Maxwell})$

Step 1 — take the curl of Faraday's law. $\nabla\times(\nabla\times\vec E) = -\frac{\partial}{\partial t}(\nabla\times\vec B)$ Why this step? We want a single equation in $\vec E$ alone. Taking the curl lets us substitute the other Maxwell equation for $\nabla\times\vec B$ and eliminate $\vec B$.

Step 2 — use the vector identity $\nabla\times(\nabla\times\vec E) = \nabla(\nabla\cdot\vec E) - \nabla^2\vec E$. Since $\nabla\cdot\vec E = 0$ in vacuum, the first term dies: $-\nabla^2\vec E = -\frac{\partial}{\partial t}(\nabla\times\vec B)$ Why this step? It converts the awkward double curl into the Laplacian $\nabla^2$, which is what appears in a wave equation.

Step 3 — substitute Ampère–Maxwell $\nabla\times\vec B = \mu_0\varepsilon_0\,\partial\vec E/\partial t$: $-\nabla^2\vec E = -\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\vec E}{\partial t}\right) = -\mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}$ Why this step? This is the move that couples the two source-free Maxwell equations and makes the constants $\mu_0\varepsilon_0$ appear together.

Step 4 — clean up: $\boxed{\;\nabla^2\vec E = \mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}\;}$

By taking the curl of Ampère–Maxwell instead, the identical equation appears for $\vec B$ — so $\vec E$ and $\vec B$ travel together at the same speed $c$.

Plug in the numbers

Common mistakes

Recall checkpoint

Flashcards

Connections

• Maxwell's Equations
• Displacement Current
• Faraday's Law of Induction
• Ampère–Maxwell Law
• Electromagnetic Waves
• Refractive Index and n = c/v
• Permittivity and Permeability
• Wave Equation

Concept Map

Hinglish (regional understanding)