Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/75

Rh 127. Def.—We shall write $$\Phi^{-1}$$ for the reciprocal of any (complete) dyadic $$\Phi,$$ also $$\Phi^2$$ for $$\Phi. \Phi$$ etc., and $$\Phi^{-2}$$ for $$\Phi^{-1}. \Phi^{-1}$$ etc. It is evident that $$\Phi^{-n}$$ is the reciprocal of $$\Phi^n$$,

128. In the reduction of equations, if we have we may cancel the $$\Phi$$ (which is equivalent to multiplying by $$\Phi^{-1}$$) if $$\Phi$$ is a complete dyadic, but not otherwise The case is the same with such equations as  To cancel an incomplete dyadic in such cases would be analogous to cancelling a zero factor in algebra.

129. Def.—If in any dyadic we transpose the factors in each term, the dyadic thus formed is said to be conjugate to the first. Thus are conjugate to each other. A dyadic of which the value is not altered by such transposition is said to be self-conjugate. The conjugate of any dyadic $$\Phi$$ may be written $$\Phi_{\text{C}}$$ It is evident that $$\Phi_{\text{C}}. \rho$$ and $$\Phi. \rho$$ are conjugate functions of $$\rho.$$ (See No. 106.) Since $$\{ \Phi_{\text{C}} \}^2 = \{ \Phi^2 \}_{\text{C}},$$ we may write $$\Phi^2_{\text{C}},$$ etc., without ambiguity.

130. The reciprocal of the product of any number of dyadics is equal to the product of their reciprocals taken in inverse order. Thus {{MathForm2||$$\{ \Phi. \Psi. \Omega \}_{\text{C}} = \Omega_{\text{C}}. \Psi_{\text{C}}. \Phi_{\text{C}} \}^{-1} = \Omega^{-1}. \Psi^{-1}. \Phi^{-1}.$$}} The conjugate of the product of any number of dyadics is equal to the product of their conjugates taken in inverse order. Thus Hence, since  and we may write $$\Phi_{\text{C}}^{-1}$$ without ambiguity.

131. It is sometimes convenient to be able to express by a dyadic taken in direct multiplication the same operation which would be effected by a given vector ($$\alpha$$) in skew multiplication. The dyadic $$I \times \alpha$$ will answer this purpose. For, by No. 117,

The same is true of the dyadic $$\alpha \times I,$$ which is indeed identical with $$I \times \alpha,$$ as appears from the equation $$I. \{ \alpha \times I \} = \Phi. \{ I \times \alpha \}. I.$$