Page:PoincareDynamiqueJuillet.djvu/7

It is easy to see that:

${\frac {d}{dt^{\prime }}}={\frac {k}{l}}\left({\frac {d}{dt}}-\epsilon {\frac {d}{dx}}\right),\quad {\frac {d}{dx^{\prime }}}={\frac {k}{l}}\left({\frac {d}{dx}}-\epsilon {\frac {d}{dt}}\right),\quad {\frac {d}{dy^{\prime }}}={\frac {1}{l}}{\frac {d}{dy}},\quad {\frac {d}{dz^{\prime }}}={\frac {1}{l}}{\frac {d}{dz}}$

and we conclude:

(9)

$\left\{{\begin{aligned}f^{\prime }={\frac {1}{l^{2}}}f,&g^{\prime }={\frac {k}{l^{2}}}(g+\epsilon \gamma ),&h^{\prime }={\frac {k}{l^{2}}}(h-\epsilon \beta ),\\\alpha ^{\prime }={\frac {1}{l^{2}}}\alpha ,&\beta ^{\prime }={\frac {k}{l^{2}}}(\beta -\epsilon h),&\gamma ^{\prime }={\frac {k}{l^{2}}}(\gamma +\epsilon g).\end{aligned}}\right.$

These formulas are identical to those of Lorentz.

Our transformation does not alter the equations (I). Indeed, the continuity condition, and the equations (6) and (8), already provided us with some of the equations (I) (except the accentuation of letters).

Equations (6) close to the continuity condition give:

(10)	${\frac {d\psi ^{\prime }}{dt^{\prime }}}+\sum {\frac {dF^{\prime }}{dx^{\prime }}}=0.$

It remains to establish that:

${\frac {df^{\prime }}{dt^{\prime }}}+\rho ^{\prime }\xi ^{\prime }={\frac {d\gamma ^{\prime }}{dy^{\prime }}}-{\frac {d\beta ^{\prime }}{dz^{\prime }}},\quad {\frac {dz^{\prime }}{dt^{\prime }}}={\frac {dg^{\prime }}{dz^{\prime }}}-{\frac {dh^{\prime }}{dy^{\prime }}},\quad \sum {\frac {df^{\prime }}{dx^{\prime }}}=\rho ^{\prime }$

and it is easy to see that these are necessary consequences of equations (6), (8) and (10).

We must now compare the force before and after transformation.

Let X, Y, Z be the force before, and X', Y', Z' the force after transformation, both related to unit volume. In order for X' to satisfy the same equations as before the transformation, we must have:

${\begin{aligned}X^{\prime }=\rho ^{\prime }f^{\prime }+\rho ^{\prime }(\eta ^{\prime }\gamma ^{\prime }-\zeta ^{\prime }\beta ^{\prime }),\\Y^{\prime }=\rho ^{\prime }g^{\prime }+\rho ^{\prime }(\zeta ^{\prime }\alpha ^{\prime }-\xi ^{\prime }\gamma ^{\prime }),\\X^{\prime }=\rho ^{\prime }h^{\prime }+\rho ^{\prime }(\xi ^{\prime }\beta ^{\prime }-\eta ^{\prime }\alpha ^{\prime }),\end{aligned}}$

or, replacing all quantities by their values (4), (4^bis) and (9) and taking into account equations (2):

(11)	$\left\{{\begin{aligned}X^{\prime }&={\frac {k}{l^{5}}}(X+\epsilon \sum X\xi ),\\Y^{\prime }&={\frac {1}{l^{5}}}Y,\\Z^{\prime }&={\frac {1}{l^{5}}}Z.\end{aligned}}\right.$

If we represent the components of the force X₁, Y₁, Z₁, not per unit volume, but per unit of electric charge of the electron, and X'₁, Y'₁, Z'₁ are the same quantities after the transformation, we would have:

$X_{1}=f+\eta \gamma -\zeta \beta ,\ X_{1}^{\prime }=f^{\prime }+\eta ^{\prime }\gamma ^{\prime }-\zeta ^{\prime }\beta ^{\prime },\ X=\rho X_{1,}\ X^{\prime }=\rho ^{\prime }X_{1}^{\prime }$