Electromagnetic Waves. 155 



To compare the results found here with those previously 

 given by Prof. Lq\q* it will be convenient to resolve first 

 the components (X, Y, Z) perpendicular to and along the 

 axis # = : we obtain the values 



X sin 6 + Y cos 6 = sm0 ° os0 (3/-f 3r/' + r 2 / ") . 



X cos 6 - Y sin 6 = -5^ (3/+ Srf ■+ i*f) + ? (/+ rf% 



K4'4) 



Finally, to obtain the Cartesian components, multiply the 

 first line of (4*4) by cos <f> and sin <\> ; then writing 



x = r sin 6 cos <£, y = r sin # sin 0, 2: = r cos #, 



we obtain the three components of electric force as : 



%&f+bf + *>/'), f (3/+ 3r/' + , •>?"), ] 



-(V+ ? / 2 ) 2 J* • ( 4 * 5 ) 



V ^ + ^ + r2 - n + .3 (/+ rf), 

 r . ? J 



agreeing with Prof. Love's formulae. 



The Cartesian components of the magnetic force are found 

 by multiplying the last formula of (4*3) by ( — sin<£, cos</>, 0) ; 

 thus they are given by 



-Js (/' + '/"), ^(f' + rf), 0. . . (4-6) 



As a further example, we can readily calculate the radiation 

 of energy in the field (4*3) ; for the Poynting vector has 

 the radial component 



±c*v-zfi) =£|(/+<+-y")(r+r/"). 



Integrating over a sphere of radius r we obtain for the 

 whole rate of radiation 



3^3 (/W + r 3 /") (/' + '•/") 



= i(n 2 +^{(/ + nf)W) 2 },. • • (4-7) 



which agrees with known formulae. 



* Phil. Trans. Koy. Soc. A, vol. 197. p. 12 (1901), see formulae (28) ; 

 other equivalent formulas were given by Drude, Pht/sik des Aethers, 

 1894, p. 415 (102). Drude's solution is found by writing- \J=f^et—r)/r, 

 in Hertz's general solution given in (7*1) below. 



