Page:Zur Elektrodynamik bewegter Systeme II.djvu/10

 Now, we multiply I' by M, II' by E, and sum up; then it follows:

$$-\Gamma(\Sigma)-(\Lambda\cdot\mathsf{E})=\left(\mathsf{E}\cdot\frac{\overline{d\mathfrak{E}}}{dt}\right)+\left(\mathsf{M}\cdot\frac{\overline{d\mathfrak{M}}}{dt}\right)$$,

or by (13):

We neglect the change that will be suffered by $$\epsilon$$ and $$\mu$$ due to the deformation, thus we put $$\tfrac{d\epsilon}{dt}=\tfrac{d\mu}{dt}=0$$; then by III it becomes:

$$\left(\mathsf{E}\cdot\frac{d'\mathfrak{E}}{dt}\right)+\left(\mathsf{M}\cdot\frac{d'\mathfrak{M}}{dt}\right)=\frac{d}{dt}\left\{ \frac{1}{2}(\epsilon\mathsf{E}^{2}+u\mathsf{M}^{2})\right\} +2\left(\Sigma\cdot\frac{d'u}{dt}\right)+\left(\frac{d'\Sigma}{dt}\cdot u\right)$$,

or by V:

Furthermore it is by (16):

Eventually, it is given from (17) by arranging with respect to the components of $$u$$:

We include (21), (22), (23) in (20), and denote by $$\tau$$ a material element of volume, so that

$$\frac{dw}{dt}+w\Gamma(u)=\frac{1}{\tau}\frac{d}{dt}(w\cdot\tau)$$

Then it follows:

where

{{MathForm2|(25)|$$A=-\left(u\cdot\frac{\overline{d\Sigma}}{dt}\right)+\frac{1}{2}\left(\epsilon\mathsf{E}^{2}+\mu\mathsf{M}^{2}\right)\Gamma(u)-S_{i,k}\left\{ \left(\epsilon\mathsf{E}_{i}\mathsf{E}_{k}+\mu\mathsf{M}_{i}\mathsf{M}_{k}\right)\frac{\partial u_{i}}{\partial k}\right\} \qquad\left.{i\atop k}\right\} =x,y,z.$$.}}

In (24), the left-hand side is the decrease of electromagnetic energy, the first member of the right-hand side the radiation, the second member the chemical-thermal energy spent, $$A$$ is thus the work spent (always calculated for the unit of time and of the material volume).