or, by partial integration,
(14bis )
∫
d
t
d
τ
ψ
δ
ρ
=
∫
d
t
d
τ
∑
ρ
d
ψ
d
x
δ
U
.
{\displaystyle \int dt\ d\tau \ \psi \delta \rho =\int dt\ d\tau \sum \rho {\frac {d\psi }{dx}}\delta U.}
Let us propose now to determine
δ
(
ρ
ξ
)
=
d
(
ρ
ξ
)
d
ϵ
δ
ϵ
{\displaystyle \delta (\rho \xi )={\frac {d(\rho \xi )}{d\epsilon }}\delta \epsilon }
.
Note that ρΔ 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
ρ
Δ
d
x
0
d
y
0
d
z
0
{\displaystyle \rho \Delta dx_{0\ }dy_{0\ }dz_{0\ }}
and this charge should remain constant, then:
(15)
∂
ρ
Δ
∂
t
=
∂
ρ
Δ
∂
ϵ
=
0.
{\displaystyle {\frac {\partial \rho \Delta }{\partial t}}={\frac {\partial \rho \Delta }{\partial \epsilon }}=0.}
We deduce:
(16)
∂
2
ρ
Δ
U
∂
t
∂
ϵ
=
∂
∂
ϵ
(
ρ
Δ
∂
U
∂
t
)
=
∂
∂
t
(
ρ
Δ
∂
U
∂
ϵ
)
.
{\displaystyle {\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).}
Now we know that for any function A, we have by the continuity equation,
1
Δ
∂
A
Δ
∂
t
=
d
A
d
t
+
∑
d
A
ξ
d
x
{\displaystyle {\frac {1}{\Delta }}{\frac {\partial A\Delta }{\partial t}}={\frac {dA}{dt}}+\sum {\frac {dA\xi }{dx}}}
and also
1
Δ
∂
A
Δ
∂
ϵ
=
d
A
d
ϵ
+
∑
d
A
∂
U
∂
ϵ
d
x
{\displaystyle {\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:
(17)
1
Δ
∂
∂
ϵ
(
ρ
Δ
∂
U
∂
t
)
=
d
ρ
∂
U
∂
t
d
ϵ
+
d
(
ρ
∂
U
∂
t
∂
U
∂
ϵ
)
d
x
+
d
(
ρ
∂
U
∂
t
∂
V
∂
ϵ
)
d
y
+
d
(
ρ
∂
U
∂
t
∂
W
∂
ϵ
)
d
z
,
{\displaystyle {\frac {1}{\Delta }}{\frac {\partial }{\partial \epsilon }}\left(\rho \Delta {\frac {\partial U}{\partial t}}\right)={\frac {d\rho {\frac {\partial U}{\partial t}}}{d\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 )
1
Δ
∂
∂
t
(
ρ
Δ
∂
U
∂
ϵ
)
=
d
ρ
∂
U
∂
ϵ
d
t
+
d
(
ρ
∂
U
∂
t
∂
U
∂
ϵ
)
d
x
+
d
(
ρ
∂
V
∂
t
∂
U
∂
ϵ
)
d
y
+
d
(
ρ
∂
W
∂
t
∂
U
∂
ϵ
)
d
z
.
{\displaystyle {\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}}.}
The right-hand sides of (17) and (17bis ) must be equal, and if one remembers that
∂
U
∂
t
=
ξ
,
∂
U
∂
ϵ
δ
ϵ
=
δ
U
,
d
ρ
ξ
d
ϵ
δ
ϵ
=
δ
ρ
ξ
{\displaystyle {\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:
(18)
δ
ρ
ξ
+
d
(
ρ
ξ
δ
U
)
d
x
+
d
(
ρ
ξ
δ
V
)
d
y
+
d
(
ρ
ξ
δ
W
)
d
z
=
d
(
ρ
δ
U
)
d
t
+
d
(
ρ
ξ
δ
U
)
d
x
+
d
(
ρ
η
δ
U
)
d
y
+
d
(
ρ
ζ
δ
U
)
d
z
{\displaystyle \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}}}
Transforming now the second term of (9); we get:
∫
d
t
d
τ
∑
F
δ
ρ
ξ
{\displaystyle \int dt\ d\tau \sum F\delta \rho \xi }
=
∫
d
t
d
τ
[
∑
F
d
(
ρ
δ
U
)
d
t
+
∑
F
d
(
ρ
η
δ
U
)
d
y
+
∑
F
d
(
ρ
ζ
δ
U
)
d
z
−
∑
F
d
(
ρ
ξ
δ
V
)
d
y
−
∑
F
d
(
ρ
ξ
δ
W
)
d
z
]
.
{\displaystyle =\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].}