Energy Conservation

The equation of motion of point charges is

$$m_i \ddot{{\bf r}}_i = \int d{\bf r} \left\{ e_i \delta({\bf r}- {\bf r}_i (t)) {\bf E} + e_i \delta({\bf r}- {\bf r}_i (t)) \dot{{\bf r}}_i \times{\bf B}\right\}$$

$${\rm If \ we \ apply} (\sum_i {\bf v}_i\cdot ) \ {\rm from \ the \ left}, \ {\rm Here} \ {\bf v}_i = \dot{{\bf r}_i}$$

$$\sum_i m_i {\bf v}_i \cdot \dot{{\bf v}}_i = \sum_i \int d{\bf r} \left\{ e_i \delta({\bf r}- {\bf r}_i (t)) {... ...r}_i (t)) \underbrace{ {\bf v}_i\cdot [ {\bf v}_i\times{\bf B}] }_{=0} \right\}$$

$$= \sum_i \int d{\bf r} e_i \delta({\bf r}- {\bf r}_i (t)) {\bf v}_i\cdot {\bf E}$$ (8)

From

$${\bf J} = \sum_i e_i \dot{{\bf r}}_i (t) \delta ({\bf r} - {\bf r}_i (t))$$ (9)

$$= {\rm rot}{\bf H} - \frac{\partial {\bf D}} {\partial t}$$ (10)

Then

$$\sum_i \frac{d}{dt}\left( \frac{1}{2}m_i {\bf v}_i^2 \right) = \int d{\bf r} \left( {\rm rot}{\bf H} - \frac{\partial {\bf D}}{\partial t}\right) \cdot {\bf E}$$ (11)

$$\frac{1}{2}\frac{\partial }{\partial t} ({\bf E}\cdot{\bf D}+{\bf B}\cdot{\bf H}) = {\bf E}\cdot\frac{\partial {\bf D}}{\partial t} + {\bf H}\cdot\frac{\partial {\bf B}}{\partial t}$$

$$= {\bf E}\cdot\frac{\partial {\bf D}}{\partial t} - {\bf H}\cdot{\rm rot}{\bf E}$$

$$\frac{d}{dt} \left( \sum_i \frac{1}{2} m_i {\bf v}_i^2 \right) = \int d{\bf r}\left[-\frac{1}{2}\frac{\partial }{\partial t} ({\bf E}\cdot{\bf D}+{\bf B}\cdot{\bf H}) \right.$$

$$\left. \underbrace{ - {\bf H}\cdot{\rm rot}{\bf E} +{\bf E}\cdot{\rm rot}{\bf H} }_{ = - {\rm div} ({\bf E}\times{\bf H}) } \right]$$

$$\frac{d}{dt} \left[ \underbrace{\sum_i \frac{1}{2}m_i {\bf v}_i^2... ...B}\cdot{\bf H}) }_{{\rm total \ energy \ of \ electromagnetic \ field}} \right] = - \int dS \underbrace{[{\bf E}\times{\bf H}]}_{{\rm Poynting \ Vector}} \cdot {\bf n}$$ (12)

The Poynting vector means the energy flux going out from the system.