Page:Scientific Papers of Josiah Willard Gibbs.djvu/326

290 of discontinuity meet. We may limit ourselves to the case in which there are three such surfaces, this being the only one of frequent occurrence, and may treat them as meeting in a straight line. It will be convenient to commence by considering the equilibrium of a system in which such a line is replaced by a filament of a different phase.

Let us suppose that three homogeneous fluid masses, $$A, B$$, and $$C$$ are separated by cylindrical (or plane) surfaces, $$\text{A-B, B-C, C-A}$$, which at first meet in a straight line, each of the surface-tensions $$\sigma_{AB}, \sigma_{BC}, \sigma_{CA}$$ being less than the sum of the other two. Let us suppose that the system is then modified by the introduction of a fourth fluid mass $$D$$, which is placed between $$A, B$$, and $$C$$, and is separated from them by cylindrical surfaces $$\text{D-A, D-B, D-C}$$ meeting $$\text{A-B, B-C}$$, and $$\text{C-A}$$ in straight lines. The general form of the surfaces is shown by figure 14, in which the full lines represent a section perpendicular to all the surfaces. The system thus modified is to be in equilibrium, as well as the original system, the position of the surfaces $$\text{A-B, B-C, C-A}$$ being unchanged. That the last condition is consistent with equilibrium will appear from the following mechanical considerations. Let $$v_{D}$$ denote the volume of the mass $$D$$ per unit of length or the area of the curvilinear triangle $$abc$$. Equilibrium is evidently possible for any values of the surface tensions (if only $$\sigma_{AB}, \sigma_{BC}, \sigma_{CA}$$ satisfy the condition mentioned above, and the tensions of the three surfaces meeting at each of the edges of $$D$$ satisfy a similar condition) with any value (not too large) of $$v_{D}$$, if the edges of $$D$$ are constrained to remain in the original surfaces $$\text{A-B, B-C}$$, and $$\text{C-A}$$, or these surfaces extended, if necessary, without change of curvature. (In certain cases one of the surfaces $$\text{D-A, D-B, D-C}$$ may disappear and $$D$$ will be bounded by only two cylindrical surfaces.) We may therefore regard the system as maintained in equilibrium by forces applied to the edges of $$D$$ and acting at right angles to $$\text{A-B, B-C, C-A}$$. The same forces would keep the system in equilibrium if $$D$$ were rigid. They must therefore have a zero resultant, since the nature of the mass $$D$$ is immaterial when it is rigid, and no forces external to the system would be required to keep a corresponding part of the original system in