554 Prof. J. W. Nicholson on the High-frequency 



The ambiguity of sign is necessary for the inclusion of all 

 possible cases in which (r 1} r 2 , r 3 ) may be positive. Taking- 

 first the negative sign in the ambiguity, we can deduce at 

 once 



ar 1 +br 2 = cr 3 , 



or nfaj + O+^Ol+n) = r 3 ( r i—r 2 ), 



or n 0*2 + 0=0, 



which can only be true for positive values if ? , i = 0, or the 

 outer electron is in the nucleus. Then, of course, we have 

 only two electrons, and a solution obviously exists with 

 r 2 =r s . We are compelled, therefore, to take the positive 

 sign in the ambiguity, and 



ari _ br 2 _ cr 3 

 ¥+7 " ^a 2 " a?+F 2 ' 

 Each ratio is equal to (ar 1 —br 2 )l(a 2 -\-b 2 ), so that 

 ai\ — br 2 — cr z 

 or i\(r 2 + r 3 ) — r 2 (i\ + r 3 ) = r 3 (r L — r 2 ), 



which is an identity. The two equations involved in (5) are 

 therefore equivalent only to one, namely, 



r 2(a 2 + & 2 ) = c r ?> (c — a), 

 or in terms only of radii 



r 2 z +r i 2 (r 2 -r 3 ) + r 2 (r l ±r 2 ) + 5r 1 r 2 r 3 =0, . . (6) 



and the whole set of equations are equivalent to this and 



bc(b 2 + c 2 ) _ c(b 2 + c 2 ) 



since b = a + c. This becomes, in terms of radii, 

 r 1 (a 2 -ca) = ( C -?' 1 )(^ 2 + c 2 ) 

 or ai\ (a — c) = — r 2 (b 2 + c 2 ), 



and finally 



r 2 3 + r{ 2 {r 2 - r.) + r 3 2 (r 1 + rj + 5fyy, = 0, 



which is not distinct from (6). 



Thus all the equations are only equivalent to this single 

 equation, which must always hold between the instantaneous 

 radii of the orbits when the three electrons are in a straight 

 line. If r 2 and r 3 are nearly equal, and not zero, we find 



6r 1 +2r 2 =0 



which cannot be satisfied by any positive value of r L . 



