140 Mr. G. F. C. Searle on the 



Comparing the series for TT with those for F and E and 

 using /xKr 2 =l, we obtain 



W= 1 ^(2F-E-^). 



The momentum carried off by the pulse is, by symmetry, 

 parallel to the direction of motion and hence, by (36), is 

 given by 



p_ /jLiniQ 2 rr sin 2 6 cos 6 d<p dfy 

 Sirb JJ (1 — ncas0) a 



fiunQ 2 f*C smdyco^^jr— sin 3 $ cos 3 \^ 7 7 

 = ~M)j '. (1-n sin * cost) 2 ^ *' 



where, as before, </> goes from to 7r and ^ from to 27r. 

 Expanding the denominator and integrating term by term, 

 we find 





(2 Q 2 /o 7 / 1 V 1 0.1-1/1.3Y , , 4 15/1.3.5V , 1 



Thus, as was proved in § 18, the momentum carried off in 

 the pulse is proportional to u 3 for small values of ujv. 

 Comparing the series for P with those for F and E and using 

 jjlKv 2 =1, we obtain 



Since the momentum of the field, when the disk is in steady 

 motion, is parallel to the direction of u, we can employ the 

 formulae of § 15. Thus, since 



U =^7rQ 2 /4&K, 

 we obtain 



U-iHP + U = U 7r- 1 {(3-7i 2 )F-E} = U (l + 7nV61+...), 

 T=W---JwP=Uo7r- 1 {(l + n 2 )F-E}===n 2 U (3/4 + 7n732 + .'..). 



Comparing these results with those of § 26, we see that the 

 magnetic energy of a disk moving in its own plane is, for 

 small speeds, greater than the magnetic energy of the disk 

 moving parallel to its axis in the proportion of 3 to 2. 



§ 28. To show how the increase of speed affects the electric 

 and magnetic energies when the disk moves (1) along its axis, 

 (2) in its plane, the following table is given. {Since the 

 " transverse mass, 5 ' m t , is equal to 2T/ir, it will be convenient 

 to tabulate mt/m , where w is the value of m t for infinitesimal 



