lr + i v V v 



y 



\3x 



3y 



3x 2 



(33) 



z=0 



The second and third terms on the right-hand side have the factor y_ or 



2 

 Fn , while the first term does not, so that one may consider the former 



two terms are of higher order. This is not the case because the first and 



second terms can be transformed into an integral over the plane z = 0. To 



show this, let us express the double body potential in the form 



s+s 



r 9<y Q) i 



'=) 



Q \PQ/J 



d S, 



(34) 



where S is the mirror image surface of S in reference to the plane z = 



* 



and PQ is the distance between P and Q. If the point P is inside S + S , 



2 2 

 then <J> n (P) = 0. Next we consider the Fourier transform of 8 <f> n /3x on the 



* 

 plane z = 0. Since cp = inside S + S , we have, after integrating by 



parts twice with respect to x 



aV 



3x 



■II 



3\ 



3x / z=0 



ik(x cos + y sin 6) 



dxdy 



.2 2 

 = - k cos 



>o)_ 



ik(x cos G + y sin 6) , , 

 e J dxdy 



z=0 



I ... Q . ] ik(x cos 8 + y sin 0) , ,.„,., 

 l-r lk cos 0»<J> I e ' dy (35) 



The value of <J) on the plane z = 0, on the other hand, becomes 



19 



