Page:Scientific Memoirs, Vol. 1 (1837).djvu/477

is done by substituting the volume $v$ for $\iiint d\,\xi \,d\eta \,d\zeta$, the term which stands under the sign $\Sigma$ in the first of the equations (II) will be represented by ${\bar {\omega }}{\frac {d\,\Gamma _{\nu }}{d\mathrm {x} }}$.

If we now write in their places all the expressions just found for the integrals which constitute the first of the equations (II), we shall have

$kqv{\frac {d\,\mathrm {q} }{d\,\mathrm {x} }}={\bar {\omega }}v{\frac {d\Phi }{d\mathrm {x} }}-{\bar {\omega }}v\Sigma {\frac {d\Gamma _{\nu }}{d\mathrm {x} }}.$

By similar substitutions the second and the third equation will give respectively

${\begin{aligned}k\mathrm {q} v{\frac {d\mathrm {q} }{d\mathrm {y} }}&={\bar {\omega }}v{\frac {d\Phi }{d\mathrm {y} }}-{\bar {\omega }}v\Sigma {\frac {d\Gamma _{\nu }}{d\mathrm {y} }}\\k\mathrm {q} v{\frac {d\mathrm {q} }{d\mathrm {z} }}&={\bar {\omega }}v{\frac {d\Phi }{d\mathrm {z} }}-{\bar {\omega }}v\Sigma {\frac {d\Gamma _{\nu }}{d\mathrm {z} }}.\end{aligned}}$

These three equations must hold good for the particular values $\mathrm {x}$, $\mathrm {y}$, $\mathrm {z}$; $\mathrm {x} _{1}$, $\mathrm {y} _{1}$, $\mathrm {z} _{1}$;… … …$\mathrm {x} _{\nu }$, $\mathrm {y} _{\nu }$, $\mathrm {z} _{\nu }$, &c., which answer to the centre of the molecules in their state of equilibrium; and as each molecule furnishes three similar equations, the whole collectively will be sufficient to enable us to determine the unknown quantities.

If from the formulæ marked (III)″, (IV)′, (V)′ we derive, by means of the changes already indicated, the expressions for ${\frac {d\mathrm {q} }{d\mathrm {x} }}$, ${\frac {d\Phi }{d\mathrm {x} }}$, ${\frac {d\Gamma _{\nu }}{d\mathrm {x} }}$, we find

${\begin{aligned}{\frac {d\mathrm {q} }{d\mathrm {x} }}&=-{\frac {1}{k}}\Sigma v_{\nu }(g{\bar {\omega }}_{\nu }+f\mathrm {q} _{\nu }){\frac {(1+\alpha r_{\nu })e^{-\alpha r_{\nu }}}{r_{\nu }^{2}}}\centerdot {\frac {\mathrm {x} _{\nu }-\mathrm {x} }{r_{\nu }}}\\{\frac {d\Phi }{d\mathrm {x} }}&={\frac {g}{f}}\Sigma v_{\nu }(g{\bar {\omega }}_{\nu }+f\mathrm {q} _{\nu })\left\{{\frac {(1+\alpha r_{\nu }e^{-\alpha r_{\nu }}}{r_{\nu }^{2}}}-{\frac {1}{r_{\nu }^{2}}}\right\}{\frac {\mathrm {x} _{\nu }-\mathrm {x} }{r_{\nu }}}\\{\frac {d\Gamma _{\nu }}{d\mathrm {x} }}&=-v^{\nu }(\gamma {\bar {\omega }}_{\nu }+g\mathrm {q} _{\nu }){\frac {1}{r_{\nu }^{2}}}\centerdot {\frac {\mathrm {x} _{\nu }-\mathrm {x} }{r_{\nu }}}\end{aligned}}$

and shall obtain ${\frac {d\mathrm {q} }{d\mathrm {y} }}$, ${\frac {d\Phi }{d\mathrm {y} }}$, ${\frac {d\Gamma _{\nu }}{d\mathrm {y} }}$; ${\frac {d\mathrm {q} }{d\mathrm {z} }}$, changing in these formulæ $\mathrm {x}$ into $\mathrm {y}$ and into $\mathrm {z}$.

If we introduce these expressions into the foregoing equations, recollecting that, according to the hypothesis of Franklin and Æpinus, we must make $f=g$, and take $\gamma$ a little less than $g$, the result will be