ROTATIONS IN HYPERSPACE. 677 



same for all positions which r can take by the rotation. Then {M\ • r) • 

 (Ml • r) is constant and equal to the square of the projection of r on 

 the fixed plane Mi. Likewise (Mo • r) • (M2 • r) is equal to the square 

 of the projection of r on the plane il/2. It follows at once then that 

 ds is a constant since the expression in the bracket in (41) is constant. 

 The curvature of the path curve is 



(42) 



^ dr d-r f dr\dt fdt 



^ = ds=d?=[''-d~s)d-s = ''-^''-Als 



= [miKMi-(Mi-r) + mo^J/o- (J/o- r)] (|J 



The vectors Mi • (Mi • r) and Mo • (Mo • r) are the projections of r on the 

 fixed planes Mi and Mo respectively and therefore constant in length. 

 Hence: The path curves are curves of constant scalar curvature. 

 The unit vector in the direction of the curvature C is 



M-(M-r) M-(M-r) 



V[M-(M-r)y VC-C 



The ^'ecto^s r and c are unit vectors and perpendicular to each other 

 Hence t X c will be the unit osculating plane to the path cur^e. The 

 first torsion of the path curve is the rate of change of this plane with 

 respect to the arc. 



rf , , dr , dc dc 

 ds as ds as 



dr 

 ds 



values above we have 



Since -^ = cV[M ■ (31 ■ r)]- the product -^xc = 0. Substituting the 

 ds ds 



T =(M-r)x{M'[M-(M-r)]]('^^^' ^ 



^dsJ VC-C 

 = (miMi + m^M^MmiHIriMi-lMi-r)] 



+ .M/.[3/.(3/.r)](|J^l^ 



But since Mi-(Mi-r) is the projection of r on Mi, Mi-[Mi-(3Ii-r)] 

 is a vector in Mi perpendicular to this projection and consequently is 

 equal to J/i • r. Similarly for the second term. Then 



T = mim2(mo^ - mi'-) (Mi-r)x(M2'r) 



