Page:PoincareDynamiqueJuillet.djvu/5

If we now set:

$\square ^{\prime }=\sum {\frac {d^{2}}{dx^{\prime 2}}}-{\frac {d^{2}}{dt^{\prime 2}}},$

it follows:

$\square ^{\prime }=\square l^{-2}.$

Consider a sphere entrained with the electron in a uniform translational motion, and

$(x-\xi t)^{2}+(y-\eta t)^{2}+(z-\zeta t)^{2}=r^{2},\,$

is the equation of that moving sphere whose volume is ${\tfrac {4}{3}}\pi r^{2}$.

The transformation will change it into an ellipsoid, and it is easy to find the equation. It is easily deduced because of equations (3):

(3bis)

$x={\frac {k}{l}}(x^{\prime }-\epsilon t^{\prime }),\ t={\frac {k}{l}}(t^{\prime }-\epsilon x^{\prime }),\ y={\frac {y^{\prime }}{l}},\ z={\frac {z^{\prime }}{l}}.$

The equation of the ellipsoid becomes:

$k^{2}(x^{\prime }-\epsilon t^{\prime }-\xi t^{\prime }+\epsilon \xi x^{\prime })^{2}+(y^{\prime }-\eta kt^{\prime }+\eta k\epsilon x^{\prime })^{2}+(z^{\prime }-\zeta kt^{\prime }+\zeta k\epsilon x^{\prime })^{2}=l^{2}r^{2}.$

This ellipsoid moves in uniform motion; for t' = 0, it reduces to

$k^{2}x^{\prime 2}(1+\xi \epsilon )^{2}+(y^{\prime }+\eta k\epsilon x^{\prime })^{2}+(z^{\prime }+\zeta k\epsilon x^{\prime })^{2}=l^{2}r^{2},$

and has the volume:

${\frac {4}{3}}\pi r^{3}{\frac {l^{3}}{k(1+\xi \epsilon )}}.$

If we want that the charge of an electron is not altered by the transformation, and when we call ρ' the new electrical density, it follows:

(4)	$\rho ^{\prime }={\frac {k}{l^{3}}}(\rho +\epsilon \rho \xi ).$

Those are the new velocities ξ', η', ζ '; we must have:

$\xi ^{\prime }={\frac {dx^{\prime }}{dt^{\prime }}}={\frac {d(x+\epsilon t)}{d(t+\epsilon x)}}={\frac {\xi +\epsilon }{1+\epsilon \xi }},$ $\eta ^{\prime }={\frac {dy^{\prime }}{dt^{\prime }}}={\frac {dy}{kd(t+\epsilon x)}}={\frac {\eta }{k(1+\epsilon \xi )}},\quad \zeta ^{\prime }={\frac {\zeta }{k(1+\epsilon \xi )}},$

where:

4bis	$\rho ^{\prime }\xi ^{\prime }={\frac {k}{l^{3}}}(\rho \xi +\epsilon \rho ),\quad \rho ^{\prime }\eta ^{\prime }={\frac {1}{l^{3}}}\rho \eta ,\quad \rho ^{\prime }\zeta ^{\prime }={\frac {1}{l^{3}}}\rho \zeta$

Here I should mention for the first time a discrepancy with Lorentz.

Lorentz poses (with different notations) (loco citato, page 813, formulas 7 and 8):

$\rho ^{\prime }={\frac {1}{kl^{3}}}\rho ,\quad \xi ^{\prime }=k^{2}(\xi +\epsilon ),\quad \eta ^{\prime }=k\eta ,\quad \zeta ^{\prime }=k\zeta .$