Page:Scientific Papers of Josiah Willard Gibbs.djvu/228

192 discontinuity by the matter on one side and determined by its state of strain shall be equal and opposite to that exerted by the matter on the other side. Since we may also write

where the signs of $$\alpha ', \beta ', \gamma '$$ may be determined by the normal on either side of the surface of discontinuity.

Equation (371) expresses the mechanical condition of equilibrium for a surface where the solid meets a fluid. It involves the separate equations

the fraction $$\frac{Ds}{Ds'}$$ denoting the ratio of the areas of the same element of the surface in the strained and unstrained states of the solid. These equations evidently express that the force exerted by the interior of the solid upon an element of its surface, and determined by the strain of the solid, must be normal to the surface and equal (but acting in the opposite direction) to the pressure exerted by the fluid upon the same element of surface.

If we wish to replace $$\alpha$$ and $$Ds$$ by $$\alpha ', \beta ', \gamma',$$ and the quantities which express the strain of the element, we may make use of the following considerations. The product $$\alpha Ds$$ is the projection of the element $$Ds$$ on the Y-Z plane. Now since the ratio $$\frac{Ds}{Ds'}$$ is independent of the form of the element, we may suppose that it has any convenient form. Let it be bounded by the three surfaces $$x' = \text{const.}$$, $$y' = \text{const.}$$, $$z' = \text{const.}$$, and let the parts of each of these surfaces included by the two others with the surface of the body be denoted by $$L, M,$$, and $$N$$, or by $$L', M',$$, and $$N'$$, according as we have reference to the strained or unstrained state of the body. The areas of $$L', M',$$, and $$N'$$ are evidently $$\alpha ' Ds', \beta ' Ds',$$, and $$\gamma ' Ds'$$; and the sum of the projections of $$L, M,$$, and $$N$$ upon any plane is equal to the projection of $$Ds$$ upon that plane, since $$L, M,$$, and $$N$$ with $$Ds$$ include a solid figure. (In propositions of this kind the sides of surfaces must be distinguished. If the normal to $$Ds$$ falls outward from the small solid figure, the normals to $$L, M,$$, and $$N$$ must fall inward, and vice versa.) Now $$L'$$ is a right-angled triangle of which the perpendicular sides may be called $$dy'$$ and $$dz'$$.