EQUATIONS OF PEOPAGATION OF ELECTRIC WAVES. 
II 
In these expressions is any solution of the equation 
_ 3 ^(t>0 
0/2 ~ 0r2 • 
( 19 ). 
It is worth while to observe that the components of the cnrl of the vector repre¬ 
sented hy (L8) are 
r~ dr^ 
0 . 
7’'’ 
0r2 
y 
r- dr^ 
( 20 ). 
12. We may use the results just obtained to describe two types of sources of electro¬ 
magnetic disturbances. We sliall take (pQ to be a function of c/ — r, say 
(f)Q = (^(c/ — r) = (l> .(21). 
In the case of sources of the first type, the magnetic force is axial, and the lines 
of electric force are circles about the axis ; when the axis is the axis of .r, the vector 
potential has the form 
( 22 ) 
the magnetic displacement has the form 
- - 3(</ + f■/) + 
XV 
- S'/+ 3 7/■ + 73/>. 
+ 3 7 , 4 
. . (23) 
and the electric displacement lias the form 
— ( ^ + 7, 
V 
(24). 
In these formulm the dots denote partial differentiation with respect to tlie time. 
The axis of the source is the axis of x, corresponding to = .t in § II ; when the 
axis of the source is in the direction (/', m', n), the result will be obtained by adding 
the expressions for the component vectors due to sources in the directions of the 
axes, and given hy putting l'(f), m'cf) and n'(f) in place of </> (with cyclical inter¬ 
changes of the letters x, y, z). If the source is at the point [x, y', z), instead of the 
origin, we have to write x — x, y — y', z — z in place of :r, y, z, and take r to be the 
distance of ix, y, z) from (:r', y\ z). 
13. In the case of sources of the second type, the electric force is axial, and the 
lines of magnetic force are circles about the axis ; when the axis is the axis of x, the 
vector potential has the form 
c 2 
