6i, 62] The Double-Star Problem 57 



be neglected. Or we may, if we please, regard the problem as one in which 

 the primary is constrained to remain ellipsoidal, in which case the forces 

 of constraint must be just sufficient to neutralise the omitted terms in 

 formula (112). 



The coefficients in formula (112) are precisely those which have been 

 already evaluated in equations (110) and (111). For we have 



dv dv dv dv 



^ = -^f etc., so that -^ = ^ = 0, 

 ox ox oy oz 



and similarly d*V/da? = 3 2 F/d#' 2 . 



62. Let f, f ' denote for the moment the distances of the centres of the 

 primary and secondary from the axis of rotation. Following our previous 

 procedure, the primary must be in equilibrium under a statical field of force 

 of potential 



or 0,2 (a? + f) m *% x 4. a constant .......... ........ (113) 



and the condition for equilibrium, as in 51, is that 



(114) 



over the boundary. The term in x on the right of this equation is removed if 



(115). 



The similar equation for the secondary is 



(116). 



Since %+%' = R, it is clear that equations (115) and (116) suffice to 

 determine f, f ' and co 2 . The ratio f/f ' is obtained at once by division of 

 corresponding sides of the two equations. 



These equations, as has been seen, refer to masses which are constrained 

 to remain ellipsoidal by the supposed application of small external forces. 

 Had the bodies been rotating freely in space, the ratio f/f would have been 

 given directly by the condition that the centre of gravity of the two masses 

 must be on the axis of rotation, namely 



M' M~M+M f 



The two values of f /(' obtained in these two different ways are not found 

 to be identical, their difference giving a measure of the error introduced in 

 supposing the masses to remain ellipsoidal. If we put 



r- r ^ = r 



' ' ' " 



_ rt , (a 2 + X) A ' ^_ a , (a' 2 + X) A' 



