IliTCiTOOCK — A Stiidij of the Vect.oi- Product V(\)a()^. 33 



If we now compare (20) with (15j, we have, axiomatically, 



V\W{)^ = - VI ViW<i>fi, (22) 



whence by transposing and factoring 



F/3[ni0-V^ + 7r«/3] = 0, (23) 



that is, the vector in brackets is parallel to j3. Suppose 



m0-i0/3 + 7r(*/3 = «/3, (24) 



where « is a scalar. I shall now show that this scalar is one of Joly's 

 invariants. The function tt will then be fully determined. 



4. Determination of the scalar a. 



To determine the value of a, return to (14), and write 6'X in place of p, 

 giving 



-// VOXdji + 0' Vti\<t>i3 = - VldirOX. (25) 



The first term on the left, by (17j, is equal to ■m<^'B'~'^ FA/3 ; the second term, 

 by (21), is m FA0 i(/)j3 ; whence (25) becomes 



m.^'6>"i FAjS + m VXe-^i.fd = - FjSvr&A. (26j 



To see the meaning of the left side we may write 



'nid~^(j> = 5, a linear vector function ; (27) 



and therefore 



vKj)'!)''^ = S,', the conjugate function. (28) 



Equation (26) will now read 



r FA^ + FA$j3 = - VfintiX ■ (29) 



but by Hamilton's relation referred to in the introduction, 



r FXj3 + VX^ji = x" FAj3 - r?X/3, (30) 



where x" is the first invariant of S, namely, 



„ SXfx^v + Sfxv^X + SvX^/j. r■'^\ 



Comparing (29) and (30), 



.;•" FA^ - F?A/3 = - V\iirQX, (32) 



which may be written 



F/3 [x"X - ^\ - TT^A] = 0, (33) 



that is, the vector in brackets (which does not involve /3) must be parallel to 

 any vector /3 (which is impossible), or else must vanish idL'iitically, i.e., 



x"X - ?A - ttOX = : (34) 



