Orbits in the Field of a Doublet. 999 



momentum is given, in general, by the equation 



A 2 = 1i 1 2 +2fL{cos0 -cos 0,); .... (5) 



so that if V be the transverse velocity at any point, and in 

 particular V] the transverse velocity at (r l} &i), 



r 2 Y 2 = r^V i 2 + 2 f i(cos 0-cos 0{). 



Since /r cannot be negative, motion can only take place in 

 the part of the plane in which 



cos 6 > cos x — h Y 2 \2\i. 



It is now evident that there is an important, division of the 

 possible orbits in the field of the doublet. For if the initial 

 conditions are such that 



h x 2 > 2^(1 + COS0!), 



the condition that h 2 the square of the angular momentum 

 is not negative is satisfied at all points of the plane, and the 

 motion is therefore not restricted to any region of the plane, 

 If, however, 



- 1 < cos 6 X - hffi/j, = m<l, 



the expression for h 2 is positive or zero only when cos — in 

 is positive or zero ; and this plainly limits the motion to the 

 sector of the plane (including the line 6 = 0) bounded by 

 the radius vectors = a. and 6 — —a, where ??i = cos a. Thus 

 in this case, if cos X — n\\2\x lies between zero and unity, 

 the motion is restricted to the sector 0=±cc, ;md a is not 

 greater than ir/2. If lt 2 \2jx — cos X lies between zero and 

 unity, the motion is restricted to a sector 0=+a, and in 

 this case « is not less than 7r/2. 



The initial transverse velocity of projection therefore 

 determines the orbits to this extent. If 



r^V^fifi > l + cos^, 



motion is possible all over the plane; if, however, 



f*i 2 Vi 2 /2/Lfc = cos 0i — m 



and 



— 1 < m = cos « < 1, 



the motion is restricted to a sector 0= j-u. Plainly, the 

 larger r^V^ 2 is, the larger is the sector; in the limiting case 

 when r 1 2 V 1 2 /2yLt=l + cos U the sector is the whole plane. 



This discussion exhausts all possible cases, since plainly 

 we cannot have 



r l 2 V 1 2 /2fji < cos 0-1. 



