Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/171

163

SECONDE PARTIE. — SECTION VII.

et différentiant par $\delta ,$ les quantités $x,\,y,\,z$ demeurant constantes,

$\delta \operatorname {F} (\rho ')=\delta \operatorname {F} (r')-x\delta {\frac {\partial \operatorname {F} (r')}{\partial x'}}-y\delta {\frac {\partial \operatorname {F} (r')}{\partial y'}}-z\delta {\frac {\partial \operatorname {F} (r')}{\partial z'}},$

où il faudra mettre pour $x,\,y,\,z$ leurs valeurs

${\begin{aligned}x=&-{\frac {\mathrm {m} 'x'+\mathrm {m} ''x''+\mathrm {m} '''x'''+\ldots }{\mathrm {m} }},\\y=&-{\frac {\mathrm {m} 'y'+\mathrm {m} ''y''+\mathrm {m} '''y'''+\ldots }{\mathrm {m} }},\\z=&-{\frac {\mathrm {m} 'z'+\mathrm {m} ''z''+\mathrm {m} '''z'''+\ldots }{\mathrm {m} }}.\end{aligned}}$

119. Supposons que la force d’attraction $\mathrm {R} '$ soit comme la puissance $\rho '^{\mu }$ de la distance $\rho ',$ on aura

$\operatorname {F} (\rho ')={\frac {\rho '^{\mu +1}}{\mu +1}},$

et la fonction $\operatorname {F} (\rho ')$ sera une fonction homogène de $x',\,y',\,z'$ du degré $\mu +1\,;$ de sorte qu’on aura, par la propriété de ces fonctions,

$x'{\frac {\partial \operatorname {F} (r')}{\partial x'}}+y'{\frac {\partial \operatorname {F} (r')}{\partial y'}}+z'{\frac {\partial \operatorname {F} (r')}{\partial z'}}=(\mu +1)\operatorname {F} (r')\,;$

donc, différentiant par $\delta$ et observant que

${\frac {\partial \operatorname {F} (r')}{\partial x'}}\delta x'+{\frac {\partial \operatorname {F} (r')}{\partial y'}}\delta y'+{\frac {\partial \operatorname {F} (r')}{\partial z'}}\delta z'=\delta \operatorname {F} (r'),$

on aura

$x'\delta {\frac {\partial \operatorname {F} (r')}{\partial x'}}+y'\delta {\frac {\partial \operatorname {F} (r')}{\partial y'}}+z'\delta {\frac {\partial \operatorname {F} (r')}{\partial z'}}=\mu \delta \operatorname {F} (r').$

Donc, si l’on fait, pour abréger,

${\begin{aligned}&(\mathrm {R} ''\,)=x''\,{\frac {\partial \operatorname {F} (r')}{\partial x'}}+y''\,{\frac {\partial \operatorname {F} (r')}{\partial y'}}+z''\,{\frac {\partial \operatorname {F} (r')}{\partial z'}},\\&(\mathrm {R} ''')=x'''{\frac {\partial \operatorname {F} (r')}{\partial x'}}+y'''{\frac {\partial \operatorname {F} (r')}{\partial y'}}+z'''{\frac {\partial \operatorname {F} (r')}{\partial z'}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\end{aligned}}$