Page:Scientific Memoirs, Vol. 1 (1837).djvu/474

The expression for $F$ will then be reduced to

(5)′

${\begin{aligned}F&=4\pi f{\frac {1}{r}}\int _{0}^{r}(T_{0}e^{\alpha r^{\prime }}+V_{0}e^{-\alpha r^{\prime }})r'\,dr'\\&{}+4\pi f\int _{r}^{\infty }(T_{0}e^{\alpha r^{\prime }}+V_{0}e^{-\alpha r^{\prime }})dr'\end{aligned}}$

All the quantities $T_{n}$ and $V_{n}$ being null, except $T_{0}$ and $V_{0}$, the values of $Q_{n}$ will also be null, except that of $Q_{0}$: the formula (2)′ will then give $q-q_{0}={\frac {T_{0}e^{\alpha r}+V_{0}e^{-\alpha r}}{r}}$

When $r=\infty$ we must have $q=q_{0}$; we must then also have $T_{0}=0$, and there will remain only $q=q_{0}+{\frac {V_{0}}{r}}e^{-\alpha r}$.

By performing the integrations of the formula (5)′ within the limits indicated, and observing that $T_{0}=0$, we shall obtain

$F=-k{\frac {V_{0}}{r}}\left(e^{-\alpha r}-1\right);$

As, in the differential expression for $F$, we may change $x'$ into $x'-\mathrm {x}$, and x into $x-\mathrm {x}$, without any change taking place in its value, and as a similar change may be made in respect to the other coordinates, it follows that, by taking the point $\mathrm {x}$, $\mathrm {y}$, $\mathrm {z}$, as the origin of the coordinates, we shall be able, in the two preceding formulas, to put

$r={\sqrt {(x-\mathrm {x} ^{2}+(y-\mathrm {y} ^{2}+(z-\mathrm {z} ^{2}}}$

or, generally,

$r_{\nu }={\sqrt {(x-\mathrm {x} _{\nu })^{2}+(y-\mathrm {y} _{\nu })^{2}+(z-\mathrm {z} _{\nu })^{2}.}}$

Now if, by placing the origin of the coordinates in the centre of each molecule respectively, we substitute these expressions of $F$ and $q$, and that previously found for $G$ in the equation (III)′, and take successively for $V_{0}$ as many constants as there are molecules, we shall find that the equation

$\Sigma kV_{0}^{(\nu )}{\frac {e^{-\alpha r_{\nu }}}{r_{\nu }}}=\Sigma kV_{0}^{(\nu )}{\frac {e^{-\alpha r_{\nu }}-1}{r_{\nu }}}+\Sigma {\frac {4\pi (g{\bar {\omega }}_{\nu }+f\mathrm {q} _{\nu })\delta _{\nu }^{3}}{3r_{\nu }}}$

will be satisfied by taking for each molecule

$V_{0}^{(\nu )}={\frac {4\pi }{3k}}(g{\bar {\omega }}_{\nu }+f\mathrm {q} _{\nu })\delta _{\nu }^{3}.$

By substituting for $V_{0}^{(\nu )}$ the value just found, we shall finally have

(IV)′

$F=-{\frac {4\pi }{3}}\Sigma (g{\bar {\omega }}_{\nu }+f\mathrm {q} _{\nu })\delta _{\nu }^{3}{\frac {e^{-\alpha r_{\nu }}-1}{r_{\nu }}}$