PART II. POLAR MAGNETIC PHENOMENA AND TERRE1.LA EXPERIMENTS. CHAP. VI. 705 



ve obtain 



dy dx r* dr 



(9) 7,7 y -,,- 



From (8) and (9) it follows, that 



r 3 fdrV _ ( AM 1 , \* 



4(1 -f P) (<* r) U J ' U + ** ' r ") 



The equation of the kinetic energy gives 



According to (5) and (6) we have 



dy ^S 

 f /;-~2 

 nd from (7) we obtain 



/hence 



- 3r)- 



( 1 _L A 



Consequently 



/(2c 2 3r) 2 r\ /\ 2 2 



2/t 



From (10) and (u) we obtain by elimination of : 



By multiplication with 4 ( 1 -|- 2 ) a (c- r)r- this equation assumes the form 



( 1 + 2 ) r 4 ( 2(i + yr) (4c 2 3*-) (A^ + a (1 + 2 ) r) 2 =- . 

 If r is not constant, this equation must be identically satisfied. Consequently 



But, when c = it follows from (6), that r = 0. 



Consequently r must be constant in all cases. Then it follows from (4) and (5) that x, y and z 

 ust be constant, and it is seen from the original system of differential equations that x, y and z 

 ust be 0. 



