= Solution
Because $\varepsilon_{ij}$ and $\mu_{ij}$ are symmetric and time-independent, the <linear anisotropic dielectric> relations give
$$
E_i\partial_tD_i
=\varepsilon_{ij}E_i\partial_tE_j
=\frac12\partial_t(\varepsilon_{ij}E_iE_j),
$$
and similarly
$$
H_i\partial_tB_i
=\frac12\partial_t(\mu_{ij}H_iH_j).
$$
Integrate the identity from part (b) over $V$ and apply the <divergence theorem> to obtain
$$
\boxed{
\frac{\partial}{\partial t}\int_V
\frac12(\varepsilon_{ij}E_iE_j+\mu_{ij}H_iH_j)\,dV
+\int_S(\mathbf E\times\mathbf H)\cdot d\mathbf S
=-\int_V\mathbf E\cdot\mathbf J\,dV}.
$$
When $\mathbf J=0$, this is conservation of electromagnetic energy. The <Poynting theorem in a linear anisotropic medium> identifies
$$
\boxed{u=\frac12(\varepsilon_{ij}E_iE_j+\mu_{ij}H_iH_j)}
$$
as the energy density and
$$
\boxed{\mathbf N=\mathbf E\times\mathbf H}
$$
as the energy flux, or <Poynting vector>.
Back to article page
