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



in value. I did not, however, anticipate that the magnetic 

 energy in the travelling shell would turn out to be constant, 

 viz. £Ui during the whole journey, from t = a/v to « = » , so 

 that it is the electric energy in the shell which gradually 

 decreases to ^Ui- Integrate the square of H according to 

 (142) to verify. 



23. The most convenient way of reckoning the work done, 

 and also the most appropriate in this class of problems, is by 

 the integral of the scalar product of the curl of the impressed 

 force and the magnetic force. Thus, in our problem 

 2Ui = j iW2,eT = J dttlL curl e/4?r 



-SPsrS** < 154 > 



where rfS is an element of the surface r = a. So we have to 

 calculate the time-integral of the magnetic force at the place 

 of vorticity of e, the limits being and 2a/v. This can be 

 easily done without solving the full problem, not only in the 

 case of wi=l, but m— any integer. The result is, if U m 

 be the electric energy of the steady field due to f m , 



and, therefore, by surface-integration according to (154), 



^U m is the magnetic energy in the wth travelling shell. I 

 have entered into detail in the case of m=l, because of its 

 relative importance, and to avoid repetition. In every case 

 the magnetic field of the primary wave outward is cancelled 

 by that of the reflexion of the primary wave inward, pro- 

 ducing a travelling shell of depth 2a, within which is the 

 final steady field. There are, however, some differences in 

 other respects, according as m is even or odd. 



Thus, in the case m = 2, we have, by (110) to (113), 



i( u.-w.)w=i{«.<~' (i-j+£) ; 

 -^ i+ J + jw)}*( i+ £ + £)- < 157 > 



Making this operate upon /„, zero before and constant after 

 t = 0, we obtain, by (132), (140), and Taylor's theorem, 



