Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/87

Rh That is, $$\Phi - a \text{I}$$ is a planar dyadic, which may be expressed by the equation (See No. 140.) Let  the equation becomes  or,  or,  This may be written  Now if the dyadic $$\Phi$$ is given in any form, the scalars  are easily determined We have therefore a cubic equation in $$a,$$ for which we can find at least one and perhaps three roots. That is, we can find at least one value of $$a,$$ and perhaps three, which will satisfy the equation By substitution of such a value, $$\Phi - a \text{I}$$ becomes a planar dyadic, the planes of which may be easily determined. Let $$\alpha$$ be a vector normal to the plane of the consequents. Then If $$\Phi$$ is a tonic, we may obtain three equations of this kind, say  in which $$\alpha, \beta, \gamma$$ are not complanar. Hence (by No. 108), where $$\alpha ', \beta ', \gamma '$$ are the reciprocals of $$\alpha, \beta, \gamma.$$ In any case, we may suppose $$a$$ to have the same sign as $$\left\vert \Phi \right\vert,$$ since the cubic equation must have such a root. Let $$\alpha$$ (as before) be normal to the plane of the consequents of the planar $$\Phi - a\text{I},$$ and $$\alpha '$$ normal to the plane of the antecedents, the lengths of $$\alpha$$ and $$\alpha '$$ being such that $$\alpha. \alpha ' = 1.$$ Let $$\beta$$ be any vector normal to $$\alpha ',$$ and such that $$\Phi. \beta$$ is not parallel to $$\beta.$$ (The case in which $$\Phi . \beta$$ is always parallel to $$\beta,$$ if $$\beta$$ is perpendicular to $$\alpha ',$$ is evidently that of a tonic, and needs no farther discussion.) $$\{ \Phi - a\text{I} \}. \beta$$ and therefore $$\Phi. \beta$$ will be perpendicular to $$\alpha '.$$ The same will be true of $$\Phi^2. \beta.$$ Now (by No. 140) that is,