Electromagnetic Wave-surface. 413 
to the wave equation. As, at the commencement, we found 
the induction and the displacement to be perpendicular to the 
normal, so now we find that the corresponding forces are per- 
pendicular to the ray. There was no difficulty in reaching 
the index equation before, when we had a single normal with 
two values of v the normal velocity, and two rays differently 
inclined to the normal. There should then be no difficulty, 
by parallel reasoning, in arriving at the wave-surface equation 
from analogous equations which express that the ray is per- 
pendicular to the magnetic and electric forces, considering 
two parallel rays travelling with different ray-velocities with 
two differently inclined wave-fronts. 
Now, as we got the index equation from 
BOING ee Beis we a 5's PE ZO)) OtS 
Cs Dec ne 71 2 a ie a iba Aran O10) IC 
we must have two corresponding equations for one ray- 
direction. Let M be a unit vector defining the direction of 
the ray, and w be the ray-velocity, so that 
Cae 12 aie tian Mammen homme L (58) 
Operate on (25) and (26) by VM, giving 
VMVNH=—vVMcH, 
VMVNE= vVMyH. 
Now use the formula of transformation (18), giving 
N(HM)—H(MN)=—vVMcH, 
N(EM)—H(MN) = vVMzeH. 
But HM=0 and EM=(, as proved before. Also v=w(MN), 
or the wave-velocity is the normal component of the ray- 
velocity. Hence 
Mwy MoM ce! aa) 
oe ON Fe Se) ceria Re) 
which are the required analogues of (25) and (26). Or, 
by (69), 
levies ee Pea ee ae 
— HV pill... ab Silay cape be Mare 
are the analogues of (28) and (29). The rest of the work is 
plain. Eliminating H and H successively, we obtain 
O=EK+ Vpe-Vock, 
0O=H+VpcVppyH ; 
Phil. Mag. 8. 5. Vol. 19. No. 121. June 1885. 2E 
