412 Mr. Oliver Heaviside on the 
should have arrived at 
h'p=B.—( Bolts Bo)eo3 3 a iy ae ee 
h’ being a constant, corresponding to (59); of this no sepa- 
rate proof is needed, as it amounts to exchanging mw and c and 
turning E into H, to make (389) become (40). And from 
(62), multiplying it by 4—’62, we arrive at 
pp "B,=0; oro =0, 2°. re 
corresponding to (61). The ray is thus perpendicular both 
to the electric and to the magnetic force. The first half of the 
demonstration is now completed, but before giving the second 
half we may notice some other properties. 
Thus, to determine the values of the scalar constants h and 
h’. Multiply (59) by o, and use (50) and (51); then 
h= — (B,c7'181) (eno) = —m(B,u~"B;), 
the second form following from (54). Insert in (59), then 
eG Se yee 64. 
Py one mee, 
gives p explicitly in terms of wo and ,, the latter of which is 
known in terms of the former by (49). Multiply this by 
—18,, using (50); then | 
1 
pe'B = — eats 8 Sea (65) 
Similarly we shall find 
ies h! = =n Boe: Bs) 5. «, ins 7 
giving 
co Bo 
= —— —~——; .... (67 
P= aca n(Boc"B2) se) 
and, corresponding to (65), we shall have 
eke. 
n 
Now to resume the argument, stopped at equation (63). 
Up to equation (59) the work is plain and straightforward, 
according to rule in fact, being merely the elimination of the 
differentials, and the getting of an equation between p and oc. 
What to do next is not at all obvious. From (59), or from 
(64), the same with hf eliminated, we may obtain all sorts of 
scalar products containing p and £,, and if we could put B, 
explicitly in terms of p, (60) or (65) would be forms of the 
wave-surface equation. From the purely mathematical point 
of view no direct way presents itself; but (61) and (63), 
considered physically as well as mathematically, guide us at 
once to the second half of the transformation from the index 
