404 Mr. Oliver Heaviside on the 
But when referred to any rectangular axes, we have 
—_ Pufkr2— Me 
CIE Te ST es 
was / eal ke: ssita) 
Hi fe2feg ibe Mikes 
by solution of (12). The accents belong to the inverse coefii- 
cients. The rest may be written down symmetrically, by 
cyclical changes of the figures. In the index surface the 
operators are inverse to those in the wave-surface. 
Conjugate Property.—The following property will occur 
frequently. A and B being any vectors 
Ag Be Bea os ou <r 
or the scalar product of A and wB equals that of B and pA. 
It only requires writing out the full scalar products to see its 
truth, which results from the identities #4,=f, &c. Similarly, 
ApcB=pAcB=cpAB, &e., 
AB=AppB=pAp7'B, &e., 
where in the first line c is another self-conjugate operator. 
D is expressed in terms of E similarly to (12) by coefficients 
Ci15 Ciz, &C.3 or, as in (13), by the principal capacities &, ¢2, ¢3. 
Theorem.—The following important theorem will be re- 
quired. A and B being any vectors, 
fy pops VAB = eV uALB. Mpa Geren) 0 CLT 
For completeness a proof is now inserted, adapted from that 
given by Tait. Since VAB is perpendicular to A and B, by 
definition of a vector product, therefore 
AVAB=0, and BVAB=0, 
by definition of a scalar product. Therefore 
ApuwVAB=0, and Buu-'VAB=0, 
by introducing wu-'=1. Hence 
pAw-'VAB=0, and pBu-!VAB=0 
by the conjugate property ; that is, ~-!VAB is perpendicular 
to vA and to wB. Or 
hwo VAB=VypApB, 
where hisascalar. Or 
AVAB=pVypAuB 
by operating by w. To find h, multiply by any third vector 0 
(not to be in the same plane as A and B), giving 
ACVAB=CyuVuApb; 
