Page:Scientific Papers of Josiah Willard Gibbs.djvu/362

Let us now examine the special condition of equilibrium which relates to a line at which three different masses meet, when one or more of these masses is solid. If we apply the method of pages 316, 317 to a system containing such a line, it is evident that we shall obtain in the expression corresponding to (660), beside the integral relating to the surfaces, a term of the form

$\int \textstyle \sum \displaystyle (\sigma \delta T)Dl$

to be interpreted as the similar term in (611), except so far as the definition of $\sigma$ has been modified in its extension to solid masses. In order that this term shall be incapable of a negative value it is necessary that at every point of the line

$\textstyle \sum \displaystyle (\sigma \delta T)\geqq 0$

(671)

for any possible displacement of the line. Those displacements are to be regarded as possible which are not prevented by the solidity of the masses, when the interior of every solid mass is regarded as incapable of motion. At the surfaces between solid and fluid masses, the processes of solidification and dissolution will be possible in some cases, and impossible in others.

The simplest case is when two masses are fluid and the third is solid and insoluble. Let us denote the solid by $S$, the fluids by $A$ and $B$, and the angles filled by these fluids by $\alpha$ and $\beta$ respectively. If the surface of the solid is continuous at the line where it meets the two fluids, the condition of equilibrium reduces to

$\sigma _{AB}\cos \alpha =\sigma _{BS}-\sigma _{AS}.$

(672)

If the line where these masses meet is at an edge of the solid, the condition of equilibrium is that

	$\sigma _{AB}\cos \alpha \leqq \sigma _{BS}-\sigma _{AS};$	$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$ (673)
and	$\sigma _{AB}\cos \beta \leqq \sigma _{AS}-\sigma _{BS};$

which reduces to the preceding when $\alpha +\beta =\pi$. Since the displacement of the line can take place by a purely mechanical process, this

satisfying this condition cannot form a closed figure, the crystal will be bounded by two or three kinds of surfaces determined by the same condition. The kinds of surface thus determined will probably generally be those for which $\sigma$ has the least values. But the relative development of the different kinds of sides, even if unmodified by gravity or the contact of other bodies, will not be such as to make $\textstyle \sum \displaystyle (\sigma s)$ a minimum. The growth of the crystal will finally be confined to sides of a single kind.
It does not appear that any part of the operation of removing a layer of molecules presents any especial difficulty so marked as that of commencing a new layer; yet the values of $\mu _{1}''$ which will just allow the different stages of the process to go on must be slightly different, and therefore, for the continued dissolving of the crystal the value of $\mu _{1}''$ must be less (by a finite quantity) than that given by equation (665). It seems probable that this would be especially true of those sides for which $\sigma$ has the least values. The effect of dissolving a crystal (even when it is done as slowly as possible) is therefore to produce a form which probably differs from that of theoretical equilibrium in a direction opposite to that of a growing crystal.