103, 104] NEWTONIAN POTENTIAL FUNCTION. 197 



In this case from the solution of any problem for the surface q 3 

 we may by interchanging the functions V and //, obtain the 

 solution of a new problem, as in the case of uniplanar problems. 

 But this is not the full extent of the analogy. We have for the 

 length of the arc of any curve on the surface q s , by 20, 



ds ,_dq* dql 

 ~ V + hf ' 



If we can find any two functions u(q lt q 2 ), v(q ly q 2 ) such that 

 du? + dv* = Mds*, 



where M is a function of the position of the point, and does not 

 involve the differentials dq l} dq 2 , we have 



(8) 



Now each member of the last equation may be factored into 

 complex factors linear in dq 1} dq 2) 



/7 -j\/j -T\ JIT /^i .dq 2 \fdq l .dq 2 \ 

 (9) (du + idv) (du idv) = M I -=*- + 1 ~- 1 1 -^ i -+- 1 . 



Each of the four factors in this equation is linear in dq 

 and dq 2 , the first, for instance, being 



/du ^ v \ f j f^ u ^ v \ j 



Now if a product of two linear forms is identically equal to a 

 product of two others, each factor on one side of the equation 

 must be a multiple of one of those on the other, so that in this 

 case we must have either 



, , . , fdq l . dq 2 \ 

 du + idv = ( -* + 1 -j] <f>, 



or du + idv = 



where < and ty are independent of the differentials dq 1} dq 2 . 

 If we put 



that is 



+ i dq 2 



