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In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

The phase has wavevector $\mathbf{k}=k{\mathbf{e}}_z$. The electric field has fixed direction $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, so it has linear polarization at $45^{\circ }$ to the $x$ and $y$ axes, perpendicular to the wavevector. Substitution into the electromagnetic wave equation givesFor the usual positive-frequency convention, $\omega =c|k|$ and propagation is in the direction $\mathrm{sgn}(k){\mathbf{e}}_z$. If $k>0$, this reduces to $\omega =ck$. The following wave formulas assume the nonconstant case $\omega \ne 0$.

In a source-free vacuum, Maxwell equations areTaking a curl of the last two equations and using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ gives the electromagnetic wave equationsSubstitution of the complex plane wave amplitudes into Maxwell equations, then taking real parts, givesFor a nonzero propagating wave these imply the dispersion relation $\boxed{{\omega }^2=c^2|\mathbf{k}{|}^2}$. Conversely choose any nonzero transverse complex $\mathbf{e}$ and set $\mathbf{b}=\mathbf{k}\times \mathbf{e}/\omega$ to obtain such a wave. The static uniform-field sector at $\omega =|\mathbf{k}|=0$ is separate from propagation.

At a perfect conductor, the oscillatory boundary conditions are $\mathbf{n}\times \mathbf{E}=0$ and $\mathbf{n}\cdot \mathbf{B}=0$. Surface charge and surface current allow nonzero normal electric and tangential magnetic fields. The magnetic condition here concerns the time-dependent wave field; a perfect conductor can retain a pre-existing static magnetic field.

Let $R=I-2\mathbf{n}{\mathbf{n}}^T$ be the orthogonal reflection across the boundary plane, so $\det R=-1$. The proposed reflected amplitudes are ${\mathbf{k}}^{\prime }=R\mathbf{k}$, ${\mathbf{e}}^{\prime }=-R\mathbf{e}$, and ${\mathbf{b}}^{\prime }=R\mathbf{b}$. For an orthogonal reflection, $(R\mathbf{a})\times (R\mathbf{d})=-R(\mathbf{a}\times \mathbf{d})$. HenceAlso $|{\mathbf{k}}^{\prime }|=|\mathbf{k}|$, so the reflected wave obeys the same dispersion relation. At $\mathbf{n}\cdot \mathbf{x}=0$, the incident and reflected phases coincide because ${\mathbf{k}}^{\prime }-\mathbf{k}$ is parallel to $\mathbf{n}$. Their summed amplitudes satisfyThe first has zero tangential part and the second zero normal part, proving both boundary conditions. Finally ${\mathbf{k}}^{\prime }\cdot \mathbf{n}=-\mathbf{k}\cdot \mathbf{n}>0$, as required for reflection into the vacuum.

The outgoing Green function of the electromagnetic wave equation has support on the past light cone. Convolution with that retarded Green function yields the retarded electromagnetic potentialwhere the source is evaluated at the retarded timeThis is the causal solution with no incoming radiation.

In empty space the fields are divergence-free. Taking the curl of Faraday's law and usinggivesThe same calculation, starting from the Ampère-Maxwell equation, gives the corresponding equation for $B$. Thus every component satisfies the electromagnetic wave equation with speedIts agreement with the measured speed of light identifies light as an electromagnetic wave.