306 



LIPKA. 



Adding F2 to both members of this equation, we evidently have 



— — (P2F1 — P1F2) = — — (P2F3 — P3F2) 



dpi dps 



or 



dai2 , da-is _ 



dpi dps 

 By similar procedure we finally get the conditions 



(49) 



50:12 , do!2s _ „ da:z , dou _ 



dpi dps 



dpo dpi 



dan , dai2 _ das 4 , 5a 13 _ ^ . 



dasi , da^i _ 

 dps dpi 

 da]3 dan 



dpi dp2 

 dan . da2i _ „ 



dpi dpi 

 da2i da23 



dps dp2 



= 0; 



dpi dp2 dpi dps 



Hence the as are given by 



= 0. 



(50) 



"12= Pi 01— Pi (j>2; «23= PS 4>2— Pi 4>3', 0!Si= Pi 4>3— P3 04*, 

 "41= Pi4>i— P44>U "13= Pz4>l— Pl(i>Z\ Ot2i= pi(t>2— P2<t>i', 



where the 0's are arbitrary junctions of the coordinales xi, X2, xs, x.\ only. 

 Combining (50) and (46), we must have 



(51) 



•^'1 — 01 F2 — </>2 Fs — 03 Fi — 04 



Pi P2 PS Pi 



Letting K stand for these equal ratios, (51) may be written 

 (52) Fi^<l)i + piK, (i=l,2,3,4). 



Multiplying by pi and summing with respect to i, we get 



I, PiFi= I, Pi 4>i + A' 2 PC, 



i i i 



and remembering that 



S(a;'i)2=l, Zxix'i^O 



or 



^p^=l, XpiFi=0, 



