E=-^; F = 





(in and out) T 



tt I Olr 



H= — -T-cpe. 



Sdoa 



i 



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



force in the second to be the negative of the electric force 

 of field due to e in the first tube when the second is non- 

 existent. That is, we virtually abolish the conductance of 

 the second tube and make it perfectly transparent. 



55. Perfectly Reflecting Barrier. Its effects. Vanishing 

 of Conduction Current. — To produce nodal surfaces of E out- 

 side the tube containing the vibrating impressed force, we 

 require an external barrier, which shall prevent the passage 

 of energy or its absorption, by wholly reflecting all distur- 

 bances which reach it. Thus, let there be a perfect conductor 

 a,tr=x. This makes E=0 there. This requires that the y 

 in (317), (318) shall have the value J ^/Gr x, whereas without 

 any bound to the dielectric it would be i. We can now choose 

 m and n so as to make J 0;c = 0. This reduces those equations to 



(328) 



This solution is now the same inside and outside the tube 

 containing the impressed force, and there is no current in 

 the tube, that is, no longitudinal current. 



To understand this case, take away the impressed force and 

 the tube. Then (328) represents a conservative system in 

 stationary vibration. Now, by the preceding, we may intro- 

 duce the tube at a nodal surface of E without disturbing 

 matters, provided there be no impressed force in the tube. 

 But if we introduce the tube anywhere else, where E is not 

 zero, we require, by the preceding, an impressed force which 

 is at every moment the negative of the undisturbed force of 

 the field, in order that no change shall occur. Now this is 

 precisely what the solution (328) represents, e in the tube 

 being cancelled by the force of the field, so that there is no 

 conduction-current. The remarkable thing is that it is the 

 impressed force in the tube itself that sets up the vibrating 

 field, and gradually ceases to work, so that in the end it and 

 the tube may be removed without altering the field. That a 

 perfect conductor as reflector is required is a detail of no 

 moment in its theoretical aspect. 



Shifting the tube, with a finite impressed force in it, 

 towards a nodal suface of E, sends up the amplitude of the 

 vibrations to any extent. 



56. K = and K = co . — If the tube have no conductance, 

 e produces no effect. This is because the two surfaces of 



