Page:Poincaré - Sur la dynamique de l’électron.djvu/13

ou, en intégrant par parties,

(14)

$\int dt\ d\tau \ \psi \delta \rho =\int dt\ d\tau \sum \rho {\frac {d\psi }{dx}}\delta U.$

Proposons-nous maintenant de déterminer

$\delta (\rho \xi )={\frac {d(\rho \xi )}{d\epsilon }}\delta \epsilon .$

Observons que $\rho \Delta$ ne peut dépendre que de $x_{0},$ $y_{0},$ $z_{0};$ en effet, si l’on considère un élément d’électron dont la position initiale est un parallélipipède rectangle dont les arêtes sont $dx_{0},$ $dy_{0},$ $dz_{0}$ la charge de cet élément est

$\rho \Delta dx_{0\ }dy_{0\ }dz_{0\ }$

et, cette charge devant demeurer constante, on a :

(15)

${\frac {\partial \rho \Delta }{\partial t}}={\frac {\partial \rho \Delta }{\partial \epsilon }}=0.$

On en déduit :

(16)

${\frac {\partial ^{2}\rho \Delta U}{\partial t\ \partial \epsilon }}={\frac {\partial }{\partial \epsilon }}\left(\rho \Delta {\frac {\partial U}{\partial t}}\right)={\frac {\partial }{\partial t}}\left(\rho \Delta {\frac {\partial U}{\partial \epsilon }}\right).$

Or on sait que pour une fonction $A$ quelconque on a, par l’équation de continuité,

${\frac {1}{\Delta }}{\frac {\partial A\Delta }{\partial t}}={\frac {dA}{dt}}+\sum {\frac {dA\xi }{dx}}$

et de même

${\frac {1}{\Delta }}{\frac {\partial A\Delta }{\partial \epsilon }}={\frac {dA}{d\epsilon }}+\sum {\frac {dA{\frac {\partial U}{\partial \epsilon }}}{dx}}.$

On a donc :

(17)

${\frac {1}{\Delta }}{\frac {\partial }{\partial \epsilon }}\left(\rho \Delta {\frac {\partial U}{\partial t}}\right)={\frac {d\rho {\frac {\partial U}{\partial t}}}{\partial \epsilon }}+{\frac {d\left(\rho {\frac {\partial U}{\partial t}}{\frac {\partial U}{\partial \epsilon }}\right)}{dx}}+{\frac {d\left(\rho {\frac {\partial U}{\partial t}}{\frac {\partial V}{\partial \epsilon }}\right)}{dy}}+{\frac {d\left(\rho {\frac {\partial U}{\partial t}}{\frac {\partial W}{\partial \epsilon }}\right)}{dz}},$

(17bis)

${\frac {1}{\Delta }}{\frac {\partial }{\partial t}}\left(\rho \Delta {\frac {\partial U}{\partial \epsilon }}\right)={\frac {d\rho {\frac {\partial U}{\partial \epsilon }}}{dt}}+{\frac {d\left(\rho {\frac {\partial U}{\partial t}}{\frac {\partial U}{\partial \epsilon }}\right)}{dx}}+{\frac {d\left(\rho {\frac {\partial V}{\partial t}}{\frac {\partial U}{\partial \epsilon }}\right)}{dy}}+{\frac {d\left(\rho {\frac {\partial W}{\partial t}}{\frac {\partial U}{\partial \epsilon }}\right)}{dz}}.$

Les 2^ds membres de (17) et (17^bis) doivent être égaux et, si l’on se souvient que

${\frac {\partial U}{\partial t}}=\xi ,\quad {\frac {\partial U}{\partial \epsilon }}\delta \epsilon =\delta U,\quad {\frac {d\rho \xi }{d\epsilon }}\delta \epsilon =\delta \rho \xi ,$

il vient :

(18)

$\delta \rho \xi +{\frac {d(\rho \xi \delta U)}{dx}}+{\frac {d(\rho \xi \delta V)}{dy}}+{\frac {d(\rho \xi \delta W)}{dz}}={\frac {d(\rho \delta U)}{dt}}+{\frac {d(\rho \xi \delta U)}{dx}}+{\frac {d(\rho \eta \delta U)}{dy}}+{\frac {d(\rho \zeta \delta U)}{dz}}.$

Transformons maintenant le 2^d terme de (9) ; il vient :

$\int dt\ d\tau \sum F\delta \rho \xi$ $=\int dt\ d\tau \left[\sum F{\frac {d(\rho \delta U)}{dt}}+\sum F{\frac {d(\rho \eta \delta U)}{dy}}+\sum F{\frac {d(\rho \zeta \delta U)}{dz}}-\sum F{\frac {d(\rho \xi \delta V)}{dy}}-\sum F{\frac {d(\rho \xi \delta W)}{dz}}\right].$