(a) If we’re only interested in the work done on free charges and currents, Eq. 8.6 becomes
\frac{d W}{d t}=\int_{ \nu }\left( E \cdot J _{f}\right) d \tau . \text { But } J _{f}= \nabla \times H -\frac{\partial D }{\partial t}(\text { Eq. } 7.56), \text { so } E \cdot J _{f}= E \cdot( \nabla \times H )- E \cdot \frac{\partial D }{\partial t}
\begin{matrix} \text { (i) } \nabla \cdot D =\rho_{f}, & \text { (iii) } \nabla \times E =-\frac{\partial B }{\partial t} \text {, } \\ \text { (ii) } \nabla \cdot B =0, & \text { (iv) } \nabla \times H = J _{f}+\frac{\partial D }{\partial t} \text {. } \end{matrix} (7.56)
From product rule #6, \nabla \cdot( E \times H )= H \cdot( \nabla \times E )- E \cdot( \nabla \times H ), \text { while } \nabla \times E =-\frac{\partial B }{\partial t} , so
E \cdot( \nabla \times H )=- H \cdot \frac{\partial B }{\partial t}- \nabla \cdot( E \times H ) . \text { Therefore } E \cdot J _{f}=- H \cdot \frac{\partial B }{\partial t}- E \cdot \frac{\partial D }{\partial t}- \nabla \cdot( E \times H ) , and hence
\frac{d W}{d t}=-\int_{ \nu }\left( E \cdot \frac{\partial D }{\partial t}+ H \cdot \frac{\partial B }{\partial t}\right) d \tau-\oint_{ S }( E \times H ) \cdot d a.
This is Poynting’s theorem for the fields in matter. Evidently the Poynting vector, representing the power per unit area transported by the fields, is S = E×H, and the rate of change of the electromagnetic energy density is \frac{\partial u_{ em }}{\partial t}= E \cdot \frac{\partial D }{\partial t}+ H \cdot \frac{\partial B }{\partial t} .
For linear media, D =\epsilon E \text { and } H =\frac{1}{\mu} B , \text { with } \epsilon \text { and } \mu constant (in time); then
\frac{\partial u_{ em }}{\partial t}=\epsilon E \cdot \frac{\partial E }{\partial t}+\frac{1}{\mu} B \cdot \frac{\partial B }{\partial t}=\frac{1}{2} \epsilon \frac{\partial}{\partial t}( E \cdot E )+\frac{1}{2 \mu} \frac{\partial}{\partial t}( B \cdot B )=\frac{1}{2} \frac{\partial}{\partial t}( E \cdot D + B \cdot H ) .
so u_{ em }=\frac{1}{2}( E \cdot D + B \cdot H ) . \quad \text { qed }
(b) If we’re only interested in the force on free charges and currents, Eq. 8.13 becomes f =\rho_{f} E + J _{f} \times B.
F =\int_{ \nu }( E + v \times B ) \rho d \tau=\int_{ \nu }(\rho E + J \times B ) d \tau (8.13)
\text { But } \rho_{f}= \nabla \cdot D , \text { and } J _{f}= \nabla \times H -\frac{\partial D }{\partial t}, \text { so } f = E ( \nabla \cdot D )+( \nabla \times H ) \times B -\left(\frac{\partial D }{\partial t}\right) \times B . Now
\frac{\partial}{\partial t}( D \times B )=\frac{\partial D }{\partial t} \times B + D \times\left(\frac{\partial B }{\partial t}\right), \text { and } \frac{\partial B }{\partial t}=- \nabla \times E , \text { so } \frac{\partial D }{\partial t} \times B =\frac{\partial}{\partial t}( D \times B )+ D \times( \nabla \times E ) , and
\text { hence } f = E ( \nabla \cdot D )- D \times( \nabla \times E )- B \times( \nabla \times H )-\frac{\partial}{\partial t}( D \times B ) . As before, we can with impunity add the term H ( \nabla \cdot B ) , so
f =\{[ E ( \nabla \cdot D )- D \times( \nabla \times E )]+[ H ( \nabla \cdot B )- B \times( \nabla \times H )]\}-\frac{\partial}{\partial t}( D \times B )
The term in curly brackets can be written as the divergence of a stress tensor (as in Eq. 8.19), and the last term is (minus) the rate of change of the momentum density, g = D × B.
f =\nabla \cdot \overleftrightarrow{ T }-\epsilon_{0} \mu_{0} \frac{\partial S }{\partial t} (8.19)