Forced Vibrations of Electromagnetic Systems. 395 



We know therefore by the preceding, the complete solution 

 due to sudden starting of the single vortex line. That is, we 

 know the individual waves in detail produced by e 1} e 3 , &c. 

 The resultant travelling disturbance is therefore confined 

 between two spherical surfaces of radii vt—a and vt + a, after 

 the centre has been reached, or of radii a—vt and a + vt 

 before the centre is reached. But it cannot occupy the whole 

 of either of the regions mentioned. 



The actual shape of the boundaries, however, may be easily 

 found. It is sufficient to consider a plane section through the 

 axis of the sphere. Let A and B be the points on this plane 

 cut by the vortex line. Describe circles of radius vt with A 

 and B as centres. If vt<a, the circles do not intersect ; the 

 disturbance is therefore wholly within them. But when 

 vt is > a, the intersecting part contains no H, and only the E 

 of the steady field due to the vortex line, which we know by 

 §24. 



That within the part common to both circles there is no H 

 we may prove thus. The vortex line in question may be 

 imagined to be a line of latitude on any spherical surface 

 passing through A and B, and centred upon the axis. Let 

 a 1 be the radius of any sphere of this kind. Then, at a time 

 making vt > a, the disturbance must lie between the surfaces 

 of spheres of radii vt— % and vt + a x , whose centre is that of 

 the sphere %. Now this excludes a portion of the space 

 between the vt—a and vt-i-a circles, referring to the plane 

 section ; and by varying the radius a, we can find the whole 

 space excluded. Thus, find the locus of intersections of 

 circles of radius 



vt-(a 2 + z 2 f, 



with centre at distance z from the origin, upon the axis. 

 The equation of the circle is 



or 



x 2 +y 2 -2x;2=vH 2 + a 2 -2vt(a? + z 2 )i. . . (172) 

 Differentiate with respect to z, giving 



z(vH 2 -x 2 f=ace, (173) 



and eliminate z between (173) and (172). After reductions, 

 the result is 



x 2 + {y±a) 2 = vH 2 , (174) 



indicating two circles, both of radius vt, whose centres are at 

 A and B. Within the common space, therefore, the steady 

 electric field has been established. 



