= Solution
In charge- and current-free <vacuum>, <Maxwell equations> reduce to
$$
\nabla\cdot\mathbf E=\nabla\cdot\mathbf B=0,\qquad
\nabla\times\mathbf E=-\partial_t\mathbf B,\qquad
\nabla\times\mathbf B=\mu_0\epsilon_0\partial_t\mathbf E.
$$
Taking a <curl> and using $\nabla\times\nabla\times\mathbf E=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$ gives
$$
\nabla^2\mathbf E=\frac1{c^2}\partial_t^2\mathbf E,\qquad
c=(\mu_0\epsilon_0)^{-1/2},
$$
and the same wave equation for $\mathbf B$. For a <plane wave> $\mathbf E=\operatorname{Re}[\mathbf E_0e^{i(\mathbf k\cdot\mathbf r-\omega t)}]$, $\omega=ck$, $\mathbf k\cdot\mathbf E_0=\mathbf k\cdot\mathbf B_0=0$ and $\mathbf B_0=\hat{\mathbf k}\times\mathbf E_0/c$. Thus the fields are transverse, mutually perpendicular and in phase; polarization describes the electric-field trajectory at a fixed point.
The <energy> <density> and <Poynting vector> are
$$
u=\frac{\epsilon_0E^2}{2}+\frac{B^2}{2\mu_0}=\epsilon_0E^2,\qquad
\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}=cu\hat{\mathbf k}.
$$
For a linearly polarized sinusoid with peak electric <amplitude> $E_0$, the mean <intensity> is $\epsilon_0cE_0^2/2$. Electromagnetic momentum <density> is $\mathbf S/c^2$; normal absorption produces <radiation pressure> $\langle S\rangle/c$, while ideal retroreflection doubles it. Linear <superposition> permits interference and standing waves, whose local electric and <magnetic fields> need not have the traveling-plane-wave equality point by point.
Back to article page
