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

88 6. To reduce a given bivector $$\mathfrak{r}$$ to the above form, we may set where $$a$$ and $$b$$ are scalars, which we may regard as known. The value of $$q$$ may be determined by the equation 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  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 It is eepecially to be noticed that from this equation we cannot conclude that  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 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 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 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 This shows that the directional ellipses of the two bivectors are similar and the angular direction from the real to the imaginary