

272 Prof. L. Eatanson on the Molecular Theory 



same ; so that E (2) and H (2) are at right angles to each other- 

 The direction of H (2) being at right angles to r, we are- 

 justified in calling H (2) a transverse vector ; with regard to 

 the electric vector, however, equation (10) shows that E n 

 cannot vanish unless s (rl) = 0, a condition fulfilled only in 

 the equatorial plane containing the vibrator and perpendicular 

 to its axis. 



At any point in the equatorial plane the direction of E (2) 

 is parallel to that of +1 and its magnitude is B. For points 

 situated in the (positive or negative) direction of the axis,. 

 H (2) vanishes and the absolute value of E (2) is | A — B | ; the 

 direction of E (2) in this case is concurrent with that of ±1. 

 At any other point 



(E (2 V = B 2 + ( s (rl/) 2 A(A-2B) . . . (13) 



(H^) 2 -(l-( s (rl)) 2 )C 2 . ..... (14) 



so that, in general, the electric and magnetic energy per unit 

 volume have different values. 



The following particular relations may be deduced at once^ 

 from the preceding formulae : — 





1 II x. 



1 1! ih 



111* 



Ef= 



-B + r*A 



rjyA 



r^A 



E» = 



r v r^ A 



-B + rjA 



Xyt z k 



Ef = 



r^r x A 



r^TyA 



-B + r*A 



H»= 







-r,C 



+ r,C 



Hf= 



+ r~C 







-r.C 



H? } = 



-r y C 



+ r x C 







Consider a spherical surface S, of radius r, with centre at 

 \'(.i i , i/ , Zq). Calculate the amount of energy, emitted from 

 the (perfectly isolated) vibrator, which crosses an element d$ 

 of the surface in unit time. Using as before r to represent a 

 unit vector drawn outwards in the direction of the normal 

 to dS, we have for the flow of electromagnetic energy across 

 <ZS, in accordance with Pointing's theorem : 



^ [-BC s (r T (i v (it))) -t s (rl)AC s (r T (r v (rl)))]. (15). 



The value of the second term in square brackets is zero, so 

 that (15) reduces to 



-^BCsin 2 (rl), (16> 



47T 



