Page:Scientific Papers of Josiah Willard Gibbs.djvu/355

Rh the solid with isotropic stresses. It seems probable, however, that if the fluid in contact with the solid is not renewed, the system will generally find a state of equilibrium in which the outermost portion of the solid will be in a state of isotropic stress. If at first the solid should dissolve, this would supersaturate the fluid, perhaps until a state is reached satisfying the condition of equilibrium with the stressed solid, and then, if not before, a deposition of solid matter in a state of isotropic stress would be likely to commence and go on until the fluid is reduced to a state of equilibrium with this new solid matter.

The action of gravity will not affect the nature of the condition of equilibrium for any single point at which the fluid meets the solid, but it will cause the values of $$p$$ and $$\mu_{1}$$ in (661) to vary according to the laws expressed by (612) and (617). If we suppose that the outer part of the solid is in a state of isotropic stress, which is the most important case, since it is the only one in which the equilibrium is in every sense stable, we have seen that the condition (661) is at least sensibly equivalent to this:—that the potential for the substance of the solid which would belong to the solid mass at the temperature $$t$$ and the pressure $$p + (c_{1} + c_{2})\sigma$$ must be equal to $$\mu_{1}$$. Or, if we denote by $$(p')$$ the pressure belonging to solid with the temperature $$t$$ and the potential equal to $$\mu_{1}$$, the condition may be expressed in the form Now if we write $$\gamma $$ for the total density of the fluid, we have by (612)  By (98) and by (617) whence Accordingly we have  and  $$z$$ being measured from the horizontal plane for which $$(p') = p''$$. Substituting this value in (662), we obtain