438 Mr. 0. Heaviside on Electromagnetic Waves, and the 

 When H is circular, the operator E/H is given by 



H ±Trk + cp J lr -yG»' ' ' ' • ^ ; 



where ?/ is undetermined. When E is circular, the operator 

 E/H is given by 



E __ s J lr —yG }r cqij.\ 



H Ank + cpJ^-yOt*, [ °^ } 



The use of these operators greatly facilitates and syste- 

 matizes investigation. The meaning is that (313) or (314) 

 is the characteristic equation connecting E and H. 



51. Longitudinal Impressed E.M.F. in a thin Conducting 

 Tube. — Let an infinitely long thin conducting tube of radius 

 a have conductance K per unit of its surface to longitudinal 

 current, and be bounded by a dielectric on both sides. Strictly 

 speaking the tube should be infinitely thin, in order to obtain 

 instantaneous magnetic penetration, and yet be of finite con- 

 ductance without possessing infinite conductivity, because 

 that would produce opacity. In this tube let impressed 

 electric force, of intensity e per unit length, act longitudinally, 

 e being any function of t and z. We have to connect e with 

 E and H internally and externally. 



The magnetic force being circular, (313) is the resistance 

 operator required. Within the tube take y = if the axis is 

 to be included ; else find y by some internal boundary con- 

 dition. Outside the tube take y — i when the medium is 

 homogeneous and boundless, because that is the only way to 

 prevent waves from coming from infinity ; else find y by 

 some outer boundary condition. There is no difficulty in 

 forming the y to suit any number of coaxial cylinders pos- 

 sessing different electrical constants, by the continuity of E 

 and H at each boundary, which equalizes the E/H's of its 

 two sides, and so expresses the y of one side in terms of that 

 on the other ; but this is useless for our purpose. For the 

 present take y = inside, and leave it unstated outside. 



At r = a, E« has the same value on both sides of the tube, 

 on account of its thinness. In the substance of the tube 

 e 4- E a is the force of the flux. On the other hand H is dis- 

 continuous at the tube, thus 



47rK(. + E)=H (out) -H U n)-(|(out)-|(in))E . (315) 



use (313), and the conjugate property (307), and 

 ce obtain 



L 47tK.? 7rsaJ ()a (J 0(7 — ybr 0n )-i 



