Page:Scientific Memoirs, Vol. 1 (1837).djvu/465

(I)

${\begin{aligned}&{\frac {d\epsilon }{dy}}=-q{\frac {dF}{dy}}+q{\frac {dG}{dy}}+q{\frac {dG_{1}}{dy}}+q{\frac {dG_{2}}{dy}}\ldots \ldots +q{\frac {dG_{\nu }}{dy}}+\mathrm {etc.} \\&{\frac {d\epsilon }{dz}}=-q{\frac {dF}{dz}}+q{\frac {dG}{dz}}+q{\frac {dG_{1}}{dz}}+q{\frac {dG_{2}}{dz}}\ldots \ldots +q{\frac {dG_{\nu }}{dz}}+\mathrm {etc.} \end{aligned}}$

in which $G_{1}$, $G_{2}$, … $G_{\nu }$, &c. denote the quantities analogous to $G$ which correspond with the different molecules 1, 2 … $\nu$, &c.

Let us likewise put

${\begin{aligned}&\Phi =\iiint {\frac {gq'\,dx'\,dy'\,dz'}{\{x'-xi)^{2}+(y'-\eta )^{2}+(z'-\zeta )^{2}\}^{\frac {1}{2}}}}\\&\Gamma _{\nu }=\iiint {\frac {\gamma {\bar {\omega }}_{\nu }\;d\xi _{\nu }\;d\eta _{\nu }\;d\zeta _{\nu }}{\{(\xi _{\nu }-\xi )^{2}+(\eta _{\nu }-\eta )^{2}+(\zeta _{\nu }-\zeta )^{2}\}^{\frac {1}{2}}}}\end{aligned}}$

where $\gamma$ denotes the force of repulsion existing among the molecules of matter at the distance assumed as unity.

The equations for the equilibrium of a molecule, if we take into consideration the motion of its centre of gravity only, will be

(II)

${\begin{aligned}&\iint \epsilon \,d\eta \,d\zeta =\iiint {\bar {\omega }}{\frac {d\Phi }{d\xi }}\,d\xi \,d\eta \,d\zeta -\Sigma \iiint {\bar {\omega }}{\frac {d\Gamma _{\nu }}{d\xi }}\,d\xi \,d\eta \,d\zeta \\&\iint \epsilon \,d\xi \,d\zeta =\iiint {\bar {\omega }}{\frac {d\Phi }{d\eta }}\,d\xi \,d\eta \,d\zeta -\Sigma \iiint {\bar {\omega }}{\frac {d\Gamma _{\nu }}{d\eta }}\,d\xi \,d\eta \,d\zeta \\&\iint \epsilon \,d\xi \,d\eta =\iiint {\bar {\omega }}{\frac {d\Phi }{d\zeta }}\,d\xi \,d\eta \,d\zeta -\Sigma \iiint {\bar {\omega }}{\frac {d\Gamma _{\nu }}{d\zeta }}\,d\xi \,d\eta \,d\zeta \end{aligned}}$

The sum $\Sigma$ is to be extended to all the numbers $\nu$, that is to say, to all the molecules except that one the equilibrium of which we are considering; the double integral is to be extended to the whole surface of this molecule, and the triple integrals to its whole volume.

4. Let us begin by considering the equilibrium of the æther. The elasticity possessed by the æther at any point of space can be only the result of the reciprocal action of the contiguous parts: hence we are led, by considerations analogous to those employed by Laplace in reference to the repulsion of caloric, in the 12th book of the Mécanique Céleste, to conclude that, in a fluid considered as a continuous mass, the elasticity is proportional to the square of the density. If then $k$ represents a constant coefficient, we shall have $\epsilon ={\frac {1}{2}}kq^{2}$, and by substituting this value in the equations (I) we shall derive the following:

$k{\frac {dq}{dx}}=-{\frac {dF}{dx}}+{\frac {dG}{dx}}+{\frac {dG_{1}}{dx}}+{\frac {dG_{2}}{dx}}\ldots \ldots +{\frac {dG_{\nu }}{dx}}+\mathrm {etc.}$