of Edinburgh, Session 1882-83. 
153 
We represent q by a single letter r, so that 
q-^q} = r , 
giving 
= . 
Thus we get 
Z = S^(rar“^ — a ) . 
Having generally 
(5) rar“i - a = - r"^a) = 2rV. (Vr a), 
we get 
Z = 2S. [(B.rY(Yr.a)]. 
Let ^ be the unit vector of Vr, and let us introduce the factor 
^ 2 = -1 
under the sign S. Then we get 
Z= -2S.K.Ci^.r.V(Vra)]. 
Of course, we see at a glance that Z will vanish when we assume 
TVr = 0, 
in which case we get r = ± 1 , and consequently ^q=g ) : this will 
be the first solution spoken of. But we will continue to examine 
the expression of Z independently of the vanishing of Yr. 
Decomposing into its scalar and its vector, we notice that the 
term in Z depending on will vanish, because the factor of 
will have a vector under the sign S. We have therefore 
Z= -2(TYr)S[CVCi8.r.VCa], 
or replacing r by q~^p^ and grouping the factors, we get 
(6) Z = - 2(TVgij)S[C(Vf/3 . 2-")(i^Vfa)]. 
Thus far we could advance without making any particular 
hypothesis about the versors p and q. How we see that if both 
had the same axis, that common axis would be identical with 
and we could at once deduce some further transformation. 
Let us examine how far the generality of the solution of Z - 0 
will be restricted if we introduce the hypothesis that p and q be 
co-axial, namely, UYp = UVg' 'I 
Considering this incidental question from a purely analytical 
point of view, we observe that the six elements comprised in a and 
/3' are by the two first conditions (1) reduced to four independent 
elements, a and being given. By (3) we introduce six elements 
