Page:PoincareDynamiqueJuillet.djvu/11

δρ = δρξ = 0 and the second integral is zero. Because δJ must vanish, we should have:

(8)	$f+{\frac {dF}{dt}}+{\frac {d\psi }{dx}}=0$

It remains in the general case:

(9)	$\delta J=\int dt\ d\tau \left(\psi \delta \rho -\sum F\delta \rho \xi \right).$

It remains to determine the forces acting on the electrons. To do this we must suppose that a supplementary force -Xdτ, -Ydτ, -Zdτ applies to each element of an electron, and write that this force is in equilibrium with the forces of electromagnetic origin. Let U, V, W be components of the displacement of the element dτ of the electron, where the displacement is counted from an arbitrary initial position. Let δU, δV, δW be the variations of this displacement; the virtual work corresponding to the supplementary force is:

$-\int \sum X\delta U\ d\tau ,$

so that the equilibrium condition about which we have spoken can be written:

(10)	$\delta J=-\int \sum X\delta U\ d\tau \ dt.$

It's about the transformation of δJ. To begin the search for the continuity equation, we express how the charge of an electron is preserved by the variation.

Let x₀, y₀, z₀ be the initial position of an electron. Its current position is:

$x=x_{0}+U,\quad y=y_{0}+V,\quad z=z_{0}+W.$

We also introduce an auxiliary variable ε, which produces changes in our various functions, so that for any function A we have:

$\delta A=\delta \epsilon {\frac {dA}{d\epsilon }}$

It is indeed convenient to switch from the notation of variation calculus to that of ordinary calculus, or vice versa.

Our functions should be regarded: 1° as dependent on five variables x, y, z, t, ε, so that we can remain at the same place when ε and t vary alone: we then indicate their derivatives by the ordinary d; 2° as dependent on five variables x₀, y₀, z₀, t, ε so that we may always follow a single electron when t and ε vary alone, then we denote their derivatives by ∂. We will have then:

(11)	$\xi ={\frac {\partial U}{\partial t}}={\frac {dU}{dt}}+\xi {\frac {dU}{dx}}+\eta {\frac {dU}{dy}}+\zeta {\frac {dU}{dz}}={\frac {\partial x}{\partial t}}.$

Denote now by Δ the functional determinant of x, y, z with respect to