x0 , y0 , z0 :
Δ
=
∂
(
x
,
y
,
z
)
∂
(
x
0
,
y
0
,
z
0
)
.
{\displaystyle \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:
∂
x
=
ξ
∂
t
,
∂
y
=
η
∂
t
,
∂
z
=
ζ
∂
t
,
{\displaystyle \partial x=\xi \partial t,\quad \partial y=\eta \partial t,\quad \partial z=\zeta \partial t,}
Δ
+
∂
Δ
=
∂
(
x
+
∂
x
,
y
+
∂
y
,
z
+
∂
z
)
∂
(
x
0
,
y
0
,
z
0
)
{\displaystyle \Delta +\partial \Delta ={\frac {\partial (x+\partial x,\ y+\partial y,\ z+\partial z)}{\partial (x_{0},\ y_{0},\ z_{0})}}}
hence
1
+
∂
Δ
Δ
=
∂
(
x
+
∂
x
,
y
+
∂
y
,
z
+
∂
z
)
∂
(
x
,
y
,
z
)
=
∂
(
x
+
ξ
∂
t
,
y
+
η
∂
t
,
z
+
ζ
∂
t
)
∂
(
x
,
y
,
z
)
{\displaystyle 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)
1
Δ
∂
Δ
∂
t
=
d
ξ
d
x
+
d
η
d
y
+
d
ζ
d
z
{\displaystyle {\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)
∂
ρ
Δ
∂
t
=
0
{\displaystyle {\frac {\partial \rho \Delta }{\partial t}}=0}
where:
∂
ρ
∂
t
+
∑
ρ
d
ξ
d
x
=
0
,
∂
ρ
∂
t
=
d
ρ
d
t
+
∑
ξ
d
ρ
d
x
,
d
ρ
d
t
+
∑
d
ρ
ξ
d
x
=
0.
{\displaystyle {\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:
δ
U
=
∂
U
∂
ϵ
δ
ϵ
,
δ
V
=
∂
V
∂
ϵ
δ
ϵ
,
δ
W
=
∂
W
∂
ϵ
δ
ϵ
;
{\displaystyle \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 )
δ
U
=
d
U
d
ϵ
δ
ϵ
+
δ
U
d
U
d
x
+
δ
V
=
d
U
d
y
+
δ
W
d
U
d
z
,
{\displaystyle \delta U={\frac {dU}{d\epsilon }}\delta \epsilon +\delta U{\frac {dU}{dx}}+\delta V={\frac {dU}{dy}}+\delta W{\frac {dU}{dz}},}
(12bis )
1
Δ
∂
Δ
∂
ϵ
=
∑
d
U
d
ϵ
,
∂
ρ
Δ
∂
ϵ
=
0
,
{\displaystyle {\frac {1}{\Delta }}{\frac {\partial \Delta }{\partial \epsilon }}=\sum {\frac {dU}{d\epsilon }},\quad {\frac {\partial \rho \Delta }{\partial \epsilon }}=0,}
(13bis )
δ
ϵ
∂
ρ
∂
ϵ
+
∑
ρ
d
ρ
U
d
x
=
0
,
∂
ρ
∂
ϵ
=
d
ρ
d
ϵ
+
∑
δ
U
δ
ϵ
d
ρ
d
x
,
δ
ρ
+
d
ρ
δ
U
d
x
=
0.
{\displaystyle \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
δ
U
=
d
U
d
ϵ
δ
ϵ
{\displaystyle \delta U={\tfrac {dU}{d\epsilon }}\delta \epsilon }
and that of
δ
ρ
=
d
ρ
d
ϵ
δ
ϵ
{\displaystyle \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:
∫
d
t
d
τ
ψ
δ
ρ
=
−
∫
d
t
d
τ
ψ
∑
d
ρ
δ
U
d
x
,
{\displaystyle \int dt\ d\tau \ \psi \delta \rho =-\int dt\ d\tau \ \psi \sum {\frac {d\rho \delta U}{dx}},}