Electromagnetic waves — derivation from Maxwell's equations

The starting point: Maxwell's equations in vacuum

• WHAT they say: divergence laws (1),(2) say field lines have no sources in vacuum (no isolated start/end points). Curl laws (3),(4) are the coupling — a time-varying field of one type curls up a field of the other type.
• WHY the $\mu_0\varepsilon_0\,\partial_t\vec E$ term matters: that displacement current term, added by Maxwell, is the missing ingredient. Without it (4) reads $\nabla\times\vec B=0$ in vacuum and no wave exists. Maxwell's correction is literally what makes light possible.

The key vector identity (the engine of the derivation)

Deriving the wave equation for $\vec E$

Step 1. Take the curl of Faraday's law (3). $\nabla\times(\nabla\times\vec E) = \nabla\times\left(-\frac{\partial\vec B}{\partial t}\right)$ Why this step? Curling the $\vec E$ equation brings a $\nabla\times\vec B$ onto the right side — and we have a formula for $\nabla\times\vec B$, namely (4). This is how we close the loop.

Step 2. Left side: apply the curl-of-curl identity, then use $\nabla\cdot\vec E=0$ from (1). $\nabla(\underbrace{\nabla\cdot\vec E}_{=0}) - \nabla^2\vec E = -\nabla^2\vec E$ Why this step? Equation (1) kills the first term — this is the moment the "no charges" assumption pays off, leaving a clean Laplacian.

Step 3. Right side: swap order of space-curl and time-derivative (fields are smooth), then insert (4). $\nabla\times\left(-\frac{\partial\vec B}{\partial t}\right) = -\frac{\partial}{\partial t}(\nabla\times\vec B) = -\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\vec E}{\partial t}\right) = -\mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}$ Why this step? We replace $\vec B$-stuff entirely with $\vec E$-stuff using Ampère–Maxwell — now the equation only involves $\vec E$.

Step 4. Equate the two sides; the minus signs cancel. $\boxed{\ \nabla^2\vec E = \mu_0\varepsilon_0\frac{\partial^2\vec E}{\partial t^2}\ }$

By the identical procedure (take curl of (4), use $\nabla\cdot\vec B=0$) you get $\nabla^2\vec B = \mu_0\varepsilon_0\frac{\partial^2\vec B}{\partial t^2}$

Structure of a plane wave (geometry of $\vec E$, $\vec B$, $\vec k$)

Try a plane wave travelling in $+x$: $\vec E=\vec E_0\,e^{i(kx-\omega t)}$, $\vec B=\vec B_0\,e^{i(kx-\omega t)}$.

• Transverse, from (1): $\nabla\cdot\vec E=0 \Rightarrow ik\,E_{0x}=0 \Rightarrow E_{0x}=0$. So $\vec E$ has no component along the direction of travel. Same for $\vec B$. → EM waves are transverse.
• $\vec E\perp\vec B$, from (3): $\nabla\times\vec E=-\partial_t\vec B$. With $\vec E=E_0\hat y$, the curl points along $\hat z$, forcing $\vec B\parallel\hat z$. So $\vec E,\ \vec B,\ \vec k$ form a right-handed orthogonal triad ($\hat E\times\hat B=\hat k$).
• Amplitude ratio, from (3): $kE_0=\omega B_0 \Rightarrow \dfrac{E_0}{B_0}=\dfrac{\omega}{k}=c$.

Worked examples

Common mistakes (Steel-manned)

Flashcards

Connections

• Maxwell's Equations — the four laws this derivation starts from
• Displacement Current — the term that enables the wave
• Faraday's Law of Induction — equation (3)
• Ampère–Maxwell Law — equation (4)
• Wave Equation — the universal $\nabla^2 f=\frac1{v^2}\partial_t^2 f$ template
• Poynting Vector and EM Energy — $\vec S=\frac{1}{\mu_0}\vec E\times\vec B$ (where the energy goes)
• Electromagnetic Spectrum — same equations, different $\lambda$
• Polarization — consequence of transverse $\vec E$

Concept Map

Hinglish (regional understanding)