MR. G. F. C. SEARLE ON PROBLEMS IN ELECTRIC CONVECTION. 
683 
Ej — [x (Ha% — H z u. 2 ) 
<W_ 
dx 
Hj -f K (E 2 w 3 — E 3 w 2 ) = 
<m 
dx 
( 21 ). 
together with four other equations of the same type. The quantities 'P and 11 are, 
as we shall see, sufficient to determine the state of the field at every point, and must 
be found in order to get a solution of any problem. 
If we solve this set of six equations for the components of E and H, and remember 
that K/xv 2 = 1, where v is the velocity of an electromagnetic disturbance through the 
medium, we obtain 
Ei (1 — 
r 
Vj“ 
- 1 - 
tt 1 2 \d'P ( u l u i d'¥ } cZ'P , / dfl 
v 2 / dx 
+ 
dy 
+ 
v“ dz 
+ /* 
u 
= -(1 - 
ap \ dSl ! wp/, 2 cZfl UyU% dfl „ { d}V 
v 2 ) dx v 2 dy v 2 dz ' Ui 
d_i 1 
Z dy 
dz 
• ( 22 ), 
_ “ 3 
together with four similar equations. 
By differentiating these equations and using div E = 47rp/K and div H = 0, we 
find 
V 3 ^ — 
+ 
¥ = - 
(24), 
v 3 n — 
v~ 
d , d t d \8 
“1 Tx +U Uy +U >Tz) n=0 
(25). 
These equations become much simpler when the motion takes place parallel to the 
axis of x. We then have u x — u, u 2 = u s = 0, and thus obtain 
L =~§.< 26 >. 
E ^( 1 -f)=-?-^f.( 27 >> 
+ .< 28 >- 
= - f.< 28 >- 
4 s 2 
