[ ±S6 ] 



LII. Ignoration Problem. Explicit form of Result. 

 By R. Hargreaves *. 



I FIND that the coefficient of each term in the result of 

 the elimination required for the ignoration problem can 

 be expressed by a simple determinant. In respect to one of 

 three types of coefficient this is already known; the present 

 note deals with all types, and I think the use of the values 

 given here adds something to the symmetry and complete- 

 ness of the solution of this important problem. 



§ 1. A kinetic energy T involving n velocities is given in 

 the form 



2T =a 11 i 1 2 + 2a 12 k^ 2 +...,. . . . (1) 



and the variables are divided into two groups, x 1 to x m and 



a m +i to x n . If we write £= -^ for a momentum, the 



QjX 



problem is to express each of the sums %x P ljp and % x m+r jj m+r 



P * m 



in terms of x p and f w + r , that is in terms of velocities of the 

 first group and momenta of the second group. Half the 

 sum of these expressions is the energy T , and half the dif- 

 ference is the kinetic potential L, in the modified or reduced 

 system. We shall find that 



2#y£,= 2T+I, 2 ^n+rfm+r =2K — I,*) 



p r K . (2) 



so that T = T + K, and L = T-K + I; J 



where, if A denotes the determinant (fl« l +i, m+ i...a B ,») and 

 A m+r ,m+s a first minor, the forms of T, K, and I are 



2 AT = 2 b pp Xp + 2%b Pjq XpXq, \ 



P P,q ' 1 



*" r,s I 



&L = 2< Cp fm j rT XpQ m j r . r . i 



p,r 



The position as regards K is well known t; it will appear 

 that b PfQ is the determinant A with a border added, and 

 Cp }m+ r is the same determinant with a row altered. 



§2. To establish this we must solve n equations of which 

 the first is 



a 11 i- 1 + fl 12 ^ 2 + ... +a lM a? n =f l5 ... (4) 



* Communicated by the Author. 



t Cf. Thomson and Tait, vol. i. p. 320 sgq. 



