﻿Quaternionic Form of Relativity. 807 



We know already that the electromagnetic bivector P is a 

 (scalarless) ^-quaternion. Hence, by (IV.), i£ we multiply 

 it, on the left side, by any physical quaternion cov. q, the 

 resulting product will again be transformed like q. Now, 

 the current-quaternion C being precisely such a quaternion, 

 consider the product 



P = CF, (24) 



which, by the above, will again be transformed by Q[ ]Q. 

 Develop it, by (10) and (19) ; then 



=p\ iJVE-f E-f -pM--pE > 



or, remembering that the full product AB is VAB— (AB), 



P = P, + *P W , (25) 



where P e , P TO are the quaternions 



P^p^(pE) + E+iVpMJ. . . (25.) 



P TO =p|^(pM)4-M-iv P E~j. . . (25m) 



The vector of P e is the well-known ponderomotive force, per 

 unit volume, and the scalar of P e is i\c times the activity of 

 this force, while P m is the magnetic analogue of P e . Notice 

 that the whole P, (25), though having with ^the transformer 

 Q[ ]Q in common, has not the structure of the standard q, 

 inasmuch as it is a full biquaternion *. (And how each of 

 its constituents, P e , P m , which have the structure of q, are 

 transformed, we do not as yet know, — though we shall know 

 in a moment.) 



Similarly, the complementary electromagnetic bivector Gr 

 being a (scalarless) i£-quaternion, multiply it on the right 

 side by C. Then the product G-C will, by (IV.), again be 

 transformed by Q[ ]Q, i. e. again like q. Develop it ; then, 

 by (10) and (19a), 



GC:=p|^+iMp-E+^Ep \, 



and this is precisely, with the same meanings of P e and P,„ 

 as above, equal to 



aG=-P e -JriP m (26) 



This again is a full biquaternion. 



* In Ilamiltoris, of course, and not in Clifford's meaniDg of the word. 



