Page:PoincareDynamiqueJuillet.djvu/13

 or, by partial integration,

Let us propose now to determine

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

Note that &rho;Δ does not depend on x0, y0, z0; indeed, if we consider an electron whose initial position is a rectangular parallelepiped whose edges are dx0, dy0, dz0, the charge of this element is

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

and this charge should remain constant, then:

We deduce:

Now we know that for any function A, we have by the continuity equation,

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

and also

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

We thus have:

The right-hand sides of (17) and (17bis) must be equal, and if one remembers that

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

we get:

Transforming now the second term of (9); we get: