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):

$=\left({\mathsf {E}}\cdot {\frac {d'{\mathfrak {E}}}{dt}}\right)+\left({\mathsf {M}}\cdot {\frac {d'{\mathfrak {M}}}{dt}}\right)+({\mathsf {E}}\cdot {\mathfrak {E}}_{def})+({\mathsf {M}}\cdot {\mathfrak {M}}_{def})$.

(20)

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:

$={\frac {dw}{dt}}-\left(u\cdot {\frac {d'\Sigma }{dt}}\right)$.

(21)

Furthermore it is by (16):

$({\mathsf {E}}\cdot {\mathfrak {E}}_{def})+({\mathsf {M}}\cdot {\mathfrak {M}}_{def})=\left(({\mathsf {E}}\cdot {\mathfrak {E}})+({\mathsf {M}}\cdot {\mathfrak {M}})\right)\Gamma (u)+({\mathsf {E}}\cdot {\mathfrak {E}}_{\delta })+{\mathsf {M}}\cdot {\mathfrak {M}}_{\delta })$ $=w\Gamma (u)+{\frac {1}{2}}(\epsilon {\mathsf {E}}^{2}+\mu {\mathsf {M}}^{2})\Gamma (u)+\epsilon ({\mathsf {E}}\cdot {\mathsf {E}}_{\delta })$ $+\mu ({\mathsf {M}}\cdot {\mathsf {M}}_{\delta })-({\mathsf {E}}\cdot [u{\mathsf {M}}]_{\delta })+({\mathsf {M}}\cdot [u{\mathsf {E}}]_{\delta })$.

(22)

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

$-({\mathsf {E}}\cdot [u{\mathsf {M}}]_{\delta })+[{\mathsf {M}}\cdot [u{\mathsf {E}}]_{\delta })=-(u\cdot \Sigma _{def})=-\left(u\cdot {\frac {\overline {d\Sigma }}{dt}}\right)+\left(u\cdot {\frac {d'\Sigma }{dt}}\right)$.

(23)

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:

$-{\frac {1}{\tau }}{\frac {d(w\tau )}{dt}}=\Gamma (\Sigma )+(\Lambda \cdot {\mathsf {E}})+{\mathsf {A}}$,

(24)

where

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

(25)

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).