Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/215

Rh By such reductions it appears that $$l\text{S}_{}$$ is a linear function of the nine products of $$\text{L, M, N}$$ with Now if we set  we have by (4) and (2)  Therefore $$l\text{S}_{}$$ is a linear function of the nine products of $$\text{L, M, N}$$ with $$\text{L}\Theta, \text{M}\Theta, \text{N}\Theta.$$ That is, $$l\text{S}_{''}$$ is the product of and a quadratic function of $$\text{L, M }$$ and $$\text{N.}$$ We may therefore write  where $$\Phi$$ is a quadratic function of $$\text{L, M }$$ and $$\text{N,}$$ dependent, however, on the nature of the medium and the period of oscillation.

9. It will be useful to consider more closely the geometrical significance of the quantity $$\Theta.$$ For this purpose it will be convenient to have a definite understanding with respect to the relative position of the coordinate axes. We shall suppose that the axes of X, Y, and Z are related in the same way as lines drawn to the right, forward and upward, so that a rotation from X to Y appears clockwise to one looking in the direction of Z.

Now if from any same point, as the origin of coordinates, we lay off lines representing in direction and magnitude the displacements in all the different wave-planes, we obtain an ellipse, which we may call the displacement-ellipse. Of this, one radius vector ($$\rho_{1}$$) will have the components $$\alpha_{1}, \beta_{1}, \gamma_{1}$$ and another ($$\rho_{2}$$) the components $$\alpha_{2}, \beta_{2}, \gamma_{2}.$$ These will belong to conjugate diameters, each being parallel to the tangent at the extremity of the other. The area of the ellipse will therefore be equal to the parallelogram of which $$\rho_{1}$$ and $$\rho_{2}$$ are two sides, multiplied by $$\pi.$$ Now it is evident that $$\beta_{1}\gamma_{2} - \gamma_{1}\beta_{2}, \, \gamma_{1}\alpha_{2} - \alpha_{1}\gamma_{2}, \, \alpha_{1}\beta_{2} - \beta_{1}\alpha_{2}$$ are numerically equal to the projections of this parallelogram on the planes of the coordinate axes, and are each positive or negative according as a revolution from $$\rho_{1}$$ to $$\rho_{2}$$ appears clockwise or counter-clockwise to one looking in the direction of the proper coordinate axis. Hence, $$\Theta$$ will be numerically equal to the parallelogram, that is, to the area of the displacement-ellipse divided by $$\pi,$$ and will be positive or negative