Page:PoincareDynamiqueJuillet.djvu/34

that is to say that the equations (10) only differ from equations (5) by the accentuation of the letters, it must be assumed:

l = 1.

Suppose now that we have η = ζ = 0, where ξ = V, ${\frac {d\xi }{dt}}={\frac {dV}{dt}}$; the equations (5) take the form:

(5bis)

$-{\frac {d}{dt}}{\frac {dH}{d\xi }}={\frac {dD}{dV}}{\frac {d\xi }{dt}}=X_{1},\quad -{\frac {d}{dt}}{\frac {dH}{d\eta }}={\frac {D}{V}}{\frac {d\eta }{dt}}=Y_{1}.$

We can also pose:

${\frac {dD}{dV}}=f(V)=f(\xi ),\quad {\frac {D}{V}}=\varphi (V)=\varphi (\xi ).$

If the equations of motion are not altered by the Lorentz transformation, we must have:

$f(\xi ){\frac {d\xi }{dt}}=X_{1},$ $\varphi (\xi ){\frac {d\eta }{dt}}=Y_{1},$ $f(\xi ^{\prime }){\frac {d\xi ^{\prime }}{dt^{\prime }}}=X_{1}^{\prime }=l^{-2}\mu ^{-1}(X_{1}+\epsilon \sum X_{1}\xi )=l^{-2}\mu ^{-1}X_{1}(1+\epsilon \xi )=l^{-2}X_{1},$ $\varphi (\xi ^{\prime }){\frac {d\eta ^{\prime }}{dt^{\prime }}}=Y_{1}^{\prime }=l^{-2}k^{-1}\mu ^{-1}Y_{1},$

and therefore:

(11)	${\begin{cases}f(\xi ){\frac {d\xi }{dt}}=l^{2}f(\xi ^{\prime }){\frac {d\xi ^{\prime }}{dt^{\prime }}},\\\\\varphi (\xi ){\frac {d\eta }{dt}}=l^{2}k\mu \varphi (\xi ^{\prime }){\frac {d\eta ^{\prime }}{dt^{\prime }}}.\end{cases}}$

But we have:

${\frac {d\xi ^{\prime }}{dt^{\prime }}}={\frac {d\xi }{dt}}{\frac {1}{k^{3}\mu ^{3}}},\quad {\frac {d\eta ^{\prime }}{dt^{\prime }}}={\frac {d\eta }{dt}}{\frac {1}{k^{2}\mu ^{2}}},$

where:

$f(\xi ^{\prime })=f\left({\frac {\xi +\epsilon }{1+\xi \epsilon }}\right)=f(\xi ){\frac {k^{3}\mu ^{3}}{l^{2}}},$ $\varphi (\xi ^{\prime })=\varphi \left({\frac {\xi +\epsilon }{1+\xi \epsilon }}\right)=\varphi (\xi ){\frac {k\mu }{l^{2}}};$

whence, by eliminating l², we find the functional equation:

$k^{2}\mu ^{2}{\frac {\varphi \left({\frac {\xi +\epsilon }{1+\xi \epsilon }}\right)}{\varphi (\xi )}}={\frac {f\left({\frac {\xi +\epsilon }{1+\xi \epsilon }}\right)}{f(\xi )}},$