Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/250

XI.

Reprenons maintenant les équations (D) de l’Article II et faisant, pour abréger,

${\begin{array}{lll}\lambda =ydx-xdy,&\lambda '=y'dx'-x'dy',&\lambda ''=y''dx''-x''dy'',\\\mu =zdx-xdz,&\mu '=z'dx'-x'dz',&\mu ''=z''dx''-x''dz'',\\\nu =zdy\,-ydz,&\nu '=z'dy'\,-y'dz',&\nu ''=z''dy''\,-y''dz'',\end{array}}$

on aura, après avoir multiplié par $dt,$

(O)

$\left\{{\begin{aligned}\mathrm {{\frac {\lambda }{C}}+{\frac {\lambda '}{B}}+{\frac {\lambda ''}{A}}} =&adt,\\\mathrm {{\frac {\mu }{C}}+{\frac {\mu '}{B}}+{\frac {\mu ''}{A}}} =&bdt,\\\mathrm {{\frac {\nu }{C}}+{\frac {\nu '}{B}}+{\frac {\nu ''}{A}}} =&cdt.\end{aligned}}\right.$

Or je trouve, comme plus haut,

${\begin{aligned}\lambda ^{2}+\mu ^{2}+\nu ^{2}=&\left(x^{2}+y^{2}+z^{2}\right)\left(dx^{2}+dy^{2}+dz^{2}\right)-(xdx+ydy+zdz)^{2}\\=&r^{2}u^{2}dt^{2}-(rdr)^{2},\end{aligned}}$

et par analogie

${\begin{aligned}\lambda '\,^{2}+\mu '^{2}\,+\nu '^{2}\,=&r'^{2}\,u'^{2}\,dt^{2}-(r'\,dr'\,)^{2},\\\lambda ''^{2}+\mu ''^{2}+\nu ''^{2}=&r''^{2}u''^{2}dt^{2}-(r''dr'')^{2}\,;\end{aligned}}$

je trouve de même

${\begin{aligned}\lambda \lambda '+\mu \mu '+\nu \nu '=&(xx'+yy'+zz')(dxdx'+dydy'+dzdz')\\&-(x'dx+y'dy+z'dz)(xdx'+ydy'+zdz')\\=&p''v''dt^{2}-d\mathrm {V} d\mathrm {V} '=p''v''dt^{2}-\left({\frac {dp''}{2}}\right)^{2}+\left({\frac {d\rho }{2}}\right)^{2},\end{aligned}}$

et par analogie

${\begin{alignedat}{5}\lambda \lambda ''+&\mu \mu ''&+&\nu \nu ''=&p'v'dt^{2}&-&\left({\frac {dp'}{2}}\right)^{2}&+&\left({\frac {d\rho }{2}}\right)^{2},\\\lambda '\lambda ''+&\mu '\mu ''&+&\nu '\nu ''=&pvdt^{2}&-&\left({\frac {dp}{2}}\right)^{2}&+&\left({\frac {d\rho }{2}}\right)^{2}.\end{alignedat}}$