Maxwell's equations

Remarks

Example: with $\rho = 0$ and $J = 0$, for $E_0 \in \R$, $k > 0$ and $\omega = ck$, the plane wave $$E = \big(0, E_0 \cos(kx - \omega t), 0\big), \qquad B = \big(0, 0, \tfrac{E_0}{c} \cos(kx - \omega t)\big)$$ is a solution. Both fields depend only on $x$ and point across the $x$ direction, so both divergences vanish. One has $(\nabla \times E)_z = \partial_x E_y = -k E_0 \sin(kx - \omega t) = -\partial_t B_z$, and $(\nabla \times B)_y = -\partial_x B_z = \tfrac{k E_0}{c} \sin(kx - \omega t) = \tfrac{1}{c^2} \partial_t E_y$.

Non-example: $E = (-y, x, 0)$ and $B = 0$ with $\rho = 0$ and $J = 0$ is not a solution, because $\nabla \times E = (0, 0, 2)$ while $-\partial_t B = 0$.

The first two equations are Gauss's laws, the third is Faraday's law, and the fourth is the Ampère–Maxwell law. Taking the divergence of the fourth equation gives the continuity equation $\partial_t \rho + \nabla \cdot J = 0$ as a condition on the sources. In matter one also uses the fields $D$ and $H$, related to $E$ and $B$ by constitutive relations.