Page:Scientific Papers of Josiah Willard Gibbs.djvu/358

respectively. The value of the fraction is therefore equal to that of the differential coefficient

${\frac {d\textstyle \sum \displaystyle (\sigma s)}{dv}}$

as determined by the displacement of a particular side while the other sides are fixed. The condition of equilibrium for the form of a crystal (when the influence of gravity may be neglected) is that the value of this differential coefficient must be independent of the particular side which is supposed to be displaced. For a constant volume of the crystal, $\textstyle \sum \displaystyle (\sigma s)$ has therefore a minimum value when the condition of equilibrium is satisfied, as may easily be proved more directly.

When there are no foreign substances at the surfaces of the crystal, and the surrounding fluid is indefinitely extended, the quantity $\textstyle \sum \displaystyle (\sigma s)$ represents the work required to form the surfaces of the crystal, and the coefficient of $s\,\,\delta N$ in (664) with its sign reversed represents the work gained in forming a mass of volume unity like the crystal but regarded as without surfaces. We may denote the work required to form the crystal by

$W_{S}-W_{V},$

$W_{S}$ denoting the work required to form the surfaces {i.e., $\textstyle \sum \displaystyle (\sigma s)$}, and $W_{V}$ the work gained in forming the mass as distinguished from the surfaces. Equation (664) may then be written

$-\delta W_{V}+\textstyle \sum \displaystyle (\sigma \delta s)=0.$

(667)

Now (664) would evidently continue to hold true if the crystal were diminished in size, remaining similar to itself in form and in nature, if the values of $\sigma$ in all the sides were supposed to diminish in the same ratio as the linear dimensions of the crystal. The variation of $W_{S}$ would then be determined by the relation

$dW_{S}=d\textstyle \sum \displaystyle (\sigma \delta s)={\tfrac {3}{2}}\textstyle \sum \displaystyle (\sigma \delta s),$

and that of $W_{V}$ by (667). Hence,

$dW_{S}={\tfrac {3}{2}}dW_{V},$

and, since $W_{S}$ and $W_{V}$ vanish together,

$W_{S}={\tfrac {3}{2}}W_{V},$

$W_{S}-W_{V}={\tfrac {1}{3}}W_{S}={\tfrac {1}{2}}W_{V},$

(668)

—the same relation which we have before seen to subsist with respect to a spherical mass of fluid as well as in other cases. (See pages 257, 261, 298.) The equilibrium of the crystal is unstable with respect to variations in size when the surrounding fluid is indefinitely extended, but it may be made stable by limiting the quantity of the fluid.