Page:Scientific Memoirs, Vol. 1 (1837).djvu/476

If this expression for $q^{2}$ be put in the integral ${\frac {1}{2}}k\int q^{2}\,d\eta \,d\zeta$, and the limits extended to the whole surface of the molecule, it is easy to see that it is reduced to $k\mathrm {q} {\frac {d\mathrm {q} }{d\mathrm {x} }}\iint \xi \,d\eta \,d\zeta$. But $\iint \xi \,d\eta \,d\zeta$, expresses the volume $v$ of the molecule, which is equal to ${\frac {4\pi }{3}}\delta ^{3}$; the term on the right in the first of the equations (II) will therefore be simply represented by $kv\mathrm {q} {\frac {d\mathrm {q} }{d\mathrm {x} }}$. It is proper to remark, that in the value of $q{\frac {d\mathrm {q} }{d\mathrm {x} }}$, we are not to include the term which, in the expression for $q$ marked (III)″ is due to the molecule whose equilibrium we are considering, because this term undergoes a change of sign at the two opposite sides of the surface of the molecule, and vanishes within the limits between which the integral is extended.

The inspection of the triple integral which gives the value $\Phi$ is sufficient to show that this integral must be given by the same function that represents $F$, in which $f$, $x$, $y$, $z$ may be replaced by $g$, $\xi$, $\eta$, $\xi$. If, because of the smallness of the dimensions of the molecule, we consider in the differential ${\frac {d\Phi }{d\xi }}$, the coordinates $\xi$, $\eta$, $\zeta$, which answer to any point of the surface as being constant, and substitute for them $\mathrm {x}$, $\mathrm {y}$, $\mathrm {z}$ which answer to the centre, then, it being observed that $\iiint d\xi \,d\eta \,d\zeta$, represents the volume $v$ of the molecule, the first integral of the second member of the first of the equations (II) may be represented ${\bar {\omega }}{\frac {d\Phi }{d\mathrm {x} }}$.

The value of $\Phi$ being deduced from the expression for $F$, such as it is given by the equation (IV)′, will contain, as we have already observed, a surplus of action, due to the æther which is supposed to occupy the place of the molecules also. It will therefore be necessary to make a compensation here also, by adding to the contrary action of the molecules an equal quantity; that is to say, by changing in the triple integral represented by $\Gamma _{\nu }$ the mass $\gamma {\bar {\omega }}$ into the mass $\gamma {\bar {\omega }}_{\nu }+g\mathrm {q} _{\nu }$. If we conceive this change made, the expression for $\Gamma _{\nu }$ will be of the same form as that for $G$ marked (V)′, except that $x$, $y$, $z$ and $g{\bar {\omega }}+f\mathrm {q}$ will be replaced by $\xi$, $\eta$, $\zeta$, and $\gamma {\bar {\omega }}_{\nu }+g\mathrm {q} _{\nu }$, and $\mathrm {x}$, $\mathrm {y}$, $\mathrm {z}$ by $\mathrm {x} _{\nu }$, $\mathrm {y} _{\nu }$, $\mathrm {z} _{\nu }$. Let us then, by approximation, introduce into the differential ${\frac {d\Gamma }{d\xi }}$ instead of the coordinates ($\xi$, $\eta$, $\zeta$) of the surface, the coordinates ($\mathrm {x}$, $\mathrm {y}$, $\mathrm {z}$) of the centre considered as constant; if we perform the integration, which