Page:PoincareDynamiqueJuillet.djvu/12

x₀, y₀, z₀:

$\Delta ={\frac {\partial (x,\ y,\ z)}{\partial (x_{0},\ y_{0},\ z_{0})}}.$

If ε, x₀, y₀, z₀ remain constant, we give to t an increasement ∂t; to x, y, z the increasements ∂x₀, ∂y₀, ∂z₀ will result; and to Δ the increasement ∂Δ, and there will be:

$\partial x=\xi \partial t,\quad \partial y=\eta \partial t,\quad \partial z=\zeta \partial t,$ $\Delta +\partial \Delta ={\frac {\partial (x+\partial x,\ y+\partial y,\ z+\partial z)}{\partial (x_{0},\ y_{0},\ z_{0})}}$

hence

$1+{\frac {\partial \Delta }{\Delta }}={\frac {\partial (x+\partial x,\ y+\partial y,\ z+\partial z)}{\partial (x,\ y,\ z)}}={\frac {\partial (x+\xi \partial t,\ y+\eta \partial t,\ z+\zeta \partial t)}{\partial (x,\ y,\ z)}}$

We deduce:

(12)	${\frac {1}{\Delta }}{\frac {\partial \Delta }{\partial t}}={\frac {d\xi }{dx}}+{\frac {d\eta }{dy}}+{\frac {d\zeta }{dz}}$

The mass of each electron is invariable, we have:

(13)	${\frac {\partial \rho \Delta }{\partial t}}=0$

where:

${\frac {\partial \rho }{\partial t}}+\sum \rho {\frac {d\xi }{dx}}=0,\quad {\frac {\partial \rho }{\partial t}}={\frac {d\rho }{dt}}+\sum \xi {\frac {d\rho }{dx}},\quad {\frac {d\rho }{dt}}+\sum {\frac {d\rho \xi }{dx}}=0.$

These are the different forms of the continuity equation with respect of variable t. We find similar forms with respect to the variable ε. Either:

$\delta U={\frac {\partial U}{\partial \epsilon }}\delta \epsilon ,\quad \delta V={\frac {\partial V}{\partial \epsilon }}\delta \epsilon ,\quad \delta W={\frac {\partial W}{\partial \epsilon }}\delta \epsilon ;$

it follows:

(11bis)

$\delta U={\frac {dU}{d\epsilon }}\delta \epsilon +\delta U{\frac {dU}{dx}}+\delta V={\frac {dU}{dy}}+\delta W{\frac {dU}{dz}},$

(12bis)

${\frac {1}{\Delta }}{\frac {\partial \Delta }{\partial \epsilon }}=\sum {\frac {dU}{d\epsilon }},\quad {\frac {\partial \rho \Delta }{\partial \epsilon }}=0,$

(13bis)

$\delta \epsilon {\frac {\partial \rho }{\partial \epsilon }}+\sum \rho {\frac {d\rho U}{dx}}=0,\quad {\frac {\partial \rho }{\partial \epsilon }}={\frac {d\rho }{d\epsilon }}+\sum {\frac {\delta U}{\delta \epsilon }}{\frac {d\rho }{dx}},\quad \delta \rho +{\frac {d\rho \delta U}{dx}}=0.$

Note the difference between the definition of $\delta U={\tfrac {dU}{d\epsilon }}\delta \epsilon$ and that of $\delta \rho ={\tfrac {d\rho }{d\epsilon }}\delta \epsilon$, we note that it is this definition of δU that suits to formula (10).

This equation will allow us to transform the first term of (9); we find in fact:

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