A Treatise on the Theory of Screws

480 the theory of screws. [432- If, however, the vector be left, then Y must be displaced to a distance Yo, defined by H log (XX'OO') = H log (FFoP'-P) ■ we therefore have H log (YY'PP) = // log (FF0 P'P); but from an obvious property of the logarithms, H log (YY.PP) = -H log (YY0PPy, whence, finally, H log (YY'PP) = - Æ log ( YY0PP). We hence have the important result, that the intervene through which a point on one of the common conjugate polars is displaced by one of two heteronymous vectors of equal amplitude, merely differs in sign from the displacement which the same point would receive from the other vector. 433. The Conception of Force in non-Euclidian Space. In ordinary spiicc we aro Quite familiar with the perfect identity which subsists between the composition of small rotations and the composition of forces. We shall now learn that what we so well know in ordinary space is but the survival, in an attenuated form, of a much more complete theory in non-Euclidian space. We have in non-Euclidian space force-motors and force-vectors, just as we have displacement-motors and displacement-vectors. We shall base the Dynamical theory on an elementary principle in the theory of Energy. Suppose that a force of intensity f act on a particle which is displaced in a direction directly opposed to the force through a distance 8, then the quantity of work done is denoted by -/S. 434. Neutrality of Heteronymous Vectors. We are now able to demonstrate a very important theorem which lies at the foundation of all the applications of Dynamics in non-Euclidian space. The virtual moment of a force-vector and a displacement-vector will always vanish when the vectors are homonymous and at right angles. The analogies of ordinary geometry would have suggested this result, and it is easily shown to be true. If, however, the two vectors be not homonymous, the result is extremely remarkable. The two vectors must then have their virtual moment zero under all circumstances. The proof of this singular proposition is very simple. Let the two vectors be what they may, we can always find one pair of conjugate polars which belong to them both. Let the two forces be X, X on the two conjugate polars, and let the displacements be p, - p, then the work done is X/Z — = 0.