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

That is, $\Phi -a{\text{I}}$ is a planar dyadic, which may be expressed by the equation

$\left\vert \Phi -a{\text{I}}\right\vert =0.$

(See No. 140.) Let

$\Phi =\lambda i+\mu j+\nu k;$

the equation becomes

$\left\vert [\lambda -ai]i+[\mu -aj]j+[\nu -ak]k\right\vert =0,$

or,

$[\lambda -ai]\times [\mu -aj].[\nu -ak]=0$

or,

$a^{3}-(i.\lambda +j.\mu +k.\nu )a^{2}+(i.\mu \times \nu +j.\nu \times \lambda +k.\lambda \times \mu )a-\lambda \times \mu .\nu =0.$

This may be written

$a^{3}-\Phi _{\text{S}}a^{2}+\{\Phi ^{-1}\}_{\text{S}}\left\vert \Phi \right\vert a-\left\vert \Phi \right\vert =0.$[1]

Now if the dyadic $\Phi$ is given in any form, the scalars

$\Phi _{\text{S}},$⁠$\{\Phi ^{-1}\}_{\text{S}},$⁠$\left\vert \Phi \right\vert$

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

$\left\vert \Phi -a{\text{I}}\right\vert =0.$

By substitution of such a value, $\Phi -a{\text{I}}$ becomes a planar dyadic, the planes of which may be easily determined.^[2] Let $\alpha$ be a vector normal to the plane of the consequents. Then

${\begin{aligned}\{\Phi -a{\text{I}}\}.\alpha &=0,\\\Phi .\alpha &=a\alpha .\end{aligned}}$

If $\Phi$ is a tonic, we may obtain three equations of this kind, say

$\Phi .\alpha =a\alpha ,$⁠$\Phi .\beta =b\beta ,$⁠$\Phi .\gamma =c\gamma ,$

in which $\alpha ,\beta ,\gamma$ are not complanar. Hence (by No. 108),

$\Phi =a\alpha \alpha '+b\beta \beta '+c\gamma \gamma ',$

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 $\alpha .\alpha '=1.$^[3] 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)

$[\Phi .\alpha ].[\Phi ^{2}.\beta ]\times [\Phi .\beta ]=\left\vert \Phi \right\vert \,\alpha .[\Phi .\beta ]\times \beta ,$

that is,

$a\alpha .[\Phi ^{2}.\beta ]\times [\Phi .\beta ]=\left\vert \Phi \right\vert \,\alpha .[\Phi .\beta ]\times \beta .$

1. [See note on p. 90.]
2. In particular cases, $\Phi -a{\text{I}}$ may reduce to a linear dyadic, or to zero. These, however, will present no difficulties to the student.
3. For the case in which the two planes are perpendicular to each other, see No. 157.