Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/219

Rh If we differentiate with respect to $$\alpha_{1}, \beta_{1}, \gamma_{1},$$ and write $$\alpha_{2}, \beta_{2}, \gamma_{2}$$ for $$d\alpha_{1}, d\beta_{1}, d\gamma_{1},$$ we obtain If we differentiate with respect to $$\alpha_{2}, \beta_{2}, \gamma_{2},$$ and write $$\alpha_{2}, \beta_{2}, \gamma_{2}$$ for $$d\alpha_{2}, d\beta_{2}, d\gamma_{2},$$ we obta  The equation derived by differentiating with respect to $$\alpha_{2}, \beta_{2}, \gamma_{2}$$ and writing $$\alpha_{1}, \beta_{1}, \gamma_{1},$$ for $$d\alpha_{2}, d\beta_{2}, d\gamma_{2},$$ is identical with (19). We should also observe that equations (18) and (20) by addition give equation (16), which therefore will not need to be considered in addition to the last three equations.

14. The geometrical signification of our equations may now be simplified by a suitable choice of the position of the origin of coordinates, which is as yet wholly arbitrary.

We shall hereafter suppose that the origin is placed in a plane of maximum or minimum displacement, if such there are. In the case of circular polarization, in which the displacements are everywhere equal, its position is immaterial. The lines $$\rho_{1}$$ and $$\rho_{2},$$ of which $$\alpha_{1}, \beta_{1}, \gamma_{1}$$ and $$\alpha_{2}, \beta_{2}, \gamma_{2}$$ are respectively the components, will now be the semi-axes of the displacement-ellipse, and therefore at right angles. (See § 9.) The case of circular polarization will not constitute any exception. Hence, and by § 9,  where we are to read $$+$$ or $$-$$ in the last member according as the system of displacements has the character of a right-handed or a left-handed screw. 15. Equation (19) is now reduced to the form