Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/79

Rh Now $$\tfrac{1}{2} \{ \Phi + \Phi_{\text{C}} \}$$ is self-conjugate, and (See No. 131.)

138. To illustrate the use of dyadics as operators, let us suppose that a body receives such a displacement that $$\rho$$ and \rho ' being the position-vectors of the same point of the body in its initial and subsequent positions. The same relation will hold of the vectors which unite any two points of the body in their initial and subsequent positions. For if $$\rho_{1}, \rho_{2}$$ are the original position-vectors of the points, and $$\rho_{1}', \rho_{2}'$$ their final position-vectors, we have whence  In the most general case, the body is said to receive a homogeneous strain. In special cases, the displacement reduces to a rotation, lines in the body initially straight and parallel will be straight and parallel after the displacement, and surfaces initially plane and parallel will be plane and parallel after the displacement.

139. The vectors (\sigma, \sigma ') which represent any plane surface in the body in its initial and final positions will be linear functions of each other. (This will appear, if we consider the four sides of a tetrahedron in the body.) To find the relation of the dyadics which express $$\sigma '$$ as a function of $$\sigma,$$ and $$\rho '$$ as a function of $$\rho,$$ let Then, if we write $$\lambda ', \mu ', \nu '$$ for the reciprocals of $$\lambda, \mu, \nu$$ the vectors $$\lambda ', \mu ', \nu '$$ become by the strain $$\alpha, \beta, \gamma$$ Therefore the surfaces $$\mu ' \times \ nu', \nu ' \times \lambda ', \lambda ' \times \mu '$$ become $$\beta \times \gamma, \gamma \times \alpha, \alpha \times \beta.$$ But $$\mu ' \times \nu ', \nu ' \times \lambda ', \lambda ' \times \mu '$$ are the reciprocals of $$\mu \times \nu, \nu \times \lambda, \lambda \times \mu.$$ The relation sought is therefore  140. The volume $$\lambda '. \mu ' \times \nu '$$ becomes by the strain $$\alpha. \beta \times \gamma.$$ The unit of volume becomes therefore $$(\alpha . \beta \times \gamma)(\lambda . \mu \times \nu).$$

Def.—It follows that the scalar product of the three antecedents multiplied by the scalar product of the three consequents of a dyadic expressed as a trinomial is independent of the particular form in which the dyadic is thus expressed. This quantity is the determinant of the coefficients of the nine terms of the form