Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 17B (Electromagnetism)

Take the curl of the vacuum Faraday law and use the vacuum Ampere-Maxwell law:Since the electric divergence vanishes, the double-curl identity makes the left side $-\Delta \mathbf{E}$. Therefore the electromagnetic wave equation isThe same calculation interchanging the fields gives the wave equation for $\mathbf{B}$.

Substitute the monochromatic plane waves into the two divergence equations of Maxwell's equations. They become $ik{E}_{0z}=0$ and $ik{B}_{0z}=0$. For nonzero wavenumber,Thus both fields are transverse to the direction of propagation. The wave equation also gives the vacuum dispersion relation ${\omega }^2=c^2{k}^2$.

For these plane waves, the vacuum Faraday law readsCancel the nonzero factor $i\omega$ to obtainFor positive-frequency propagation in the positive $z$ direction, $k/\omega =1/c$.

Vacuum Maxwell's equations are invariant under spatial rotations. Rotate the preceding coordinate system so that ${\mathbf{e}}_z$ becomes the propagation direction $\hat{\mathbf{k}}$, and rotate the transverse electric direction to a unit polarization vector $\hat{\mathbf{n}}$. Dot and cross products transform covariantly under this rotation, so transversality and the field relation are unchanged. With $\omega =c|\mathbf{k}|$,These complex fields encode real fields through their real parts. The displayed form describes linear electromagnetic polarization; a general transverse complex amplitude can describe elliptical or circular electromagnetic polarization as well.

Use real physical fields in the quadratic energy expressions. The vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ and vacuum Maxwell's equations giveIntegrating over a fixed volume and applying the divergence theorem proves Poynting's theorem:The left side is outward energy flux through the closed surface, not merely local intensity.

For a linearly polarized wave with phase $\chi =\mathbf{k}\cdot \mathbf{r}-\omega t$, the real fields give Poynting vector $\mathbf{S}={\epsilon }_{0}c{E}_0^2{\cos }^2\chi \hat{\mathbf{k}}$. Local intensity oscillates at $2\omega$. In the closed-surface integral, the time-independent part cancels because ${\oint }_A\hat{\mathbf{k}}\cdot d\mathbf{A}=0$. The remaining flux generally oscillates at $2\omega$, has zero time average, and may vanish identically for special volume geometries.

For circular polarization, take $\mathbf{E}=E_0({\mathbf{e}}_1\cos \chi +{\mathbf{e}}_2\sin \chi )$ with orthonormal transverse vectors. Then $E^2=E_0^2$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, so $\mathbf{S}={\epsilon }_{0}c{E}_0^2\hat{\mathbf{k}}$ is spatially and temporally constant. Therefore the closed-surface Poynting flux of a plane wave isThere is still a nonzero steady local energy flow: equal energy enters and leaves every fixed closed volume. Calling the net closed-surface flux a nonzero constant would confuse these two quantities.