 Solution

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.