Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/104

6. To reduce a given bivector ${\mathfrak {r}}$ to the above form, we may set

${\begin{aligned}{\mathfrak {r}}.{\mathfrak {r}}&=(\cos q+\iota \sin q)^{2}[\alpha +\iota \beta ].[\alpha +\iota \beta ]\\&=(\cos 2q+\iota \sin 2q)(\alpha .\alpha -\beta .\beta )\\&=a+\iota b,\end{aligned}}$

where $a$ and $b$ are scalars, which we may regard as known. The value of $q$ may be determined by the equation

$\tan 2q={\frac {b}{a}},$

the quadrant to which $2q$ belongs being determined so as to give $\sin 2q$ and $\cos 2q$ the same signs as $b$ and $a.$ Then $\alpha$ and $\beta$ will be given by the equation

$\alpha +\iota \beta =(\cos q-\iota \sin q){\mathfrak {r}}.$

The solution ia indeterminate when the real and imaginary parts of the given bivector are perpendicular and equal in magnitude. In this case the directional ellipse is a circle, and the bivector may be called circular. The criterion of a circular bivector is

${\mathfrak {r}}.{\mathfrak {r}}=0.$

It is eepecially to be noticed that from this equation we cannot conclude that

${\mathfrak {r}}=0,$

as in the analysis of real vectors. This may also be shown by expressing ${\mathfrak {r}}$ in the form $xi+yj+zk,$ in which $x,y,z$ are biacalars. The equation then becomes

$x^{2}+y^{2}+z^{2}=0,$

which evidently does not require $x,y,$ and $z$ to vanish, as would be the case if only real values are considered.

7. Def.—We call two vectors $\rho$ and $\sigma$ perpendicular when $\rho .\sigma =0.$ Allowing the same analogy, we shall call two bivectors ${\mathfrak {r}}$ and ${\mathfrak {s}}$ perpendicular, when

${\mathfrak {r}}.{\mathfrak {s}}=0.$

In considering the geometrical signification of this equation, we shall first suppose that the real and imaginary components of ${\mathfrak {r}}$ and ${\mathfrak {s}}$ lie in the same plane, and that both ${\mathfrak {r}}$ and ${\mathfrak {s}}$ have not real directions. It is then evidently possible to expreas them in the form

$m[\alpha +\iota \beta ],$$m'[\alpha '+\iota \beta '],$

where $m$ and $n'$ are biscalar, $\alpha$ and $\beta$ are at right angles, and $\alpha '$ parallel with $\beta .$ Then the equation ${\mathfrak {r}}.{\mathfrak {s}}=0$ requires that

$\beta .\beta '=0,$and$\alpha .\beta '+\beta .\alpha '=0.$

This shows that the directional ellipses of the two bivectors are similar and the angular direction from the real to the imaginary