(61) 



of Electrons in an Elastic Solid yEtlier, 151 



Hence for a stationary value two of E } F, G are equal, 

 say F and G. Thus either 



E = F = Gr, l-a-QE-dv- 1)717 = 0, 

 or else 



F = G=v/(l-10, E = l-*. 



Expressing in terms of e,f, g we have, either 



e=f=g, (e+^y+il-ky.e-kl^O, . . (59) 

 or else 



f=g=</(l-il)-l, «=-/• (60) 



At this stage take k and I small, and, to get the same 

 notation as before* take the roots of (59) to be — 7t, — A + c, 

 and the solution (60) to be f=g=—h — b, e=—h + a. 

 Finally, express everything in terms of 6, c. Thus 



a=b(b + 2c)/(2b + c) 1 



3A=&+2c 

 M=(b+2c)(-b+c)/(2b+c) \ ' 



3Z=26 + c 3 J 



w /B = ASg-(l- f/ ' + .; f y7^ )EFG 

 \ d(26+c) / 



-^W E '-'- l)(F2 - 1)(G2 - l} ' • • • (62) 



and the stationary values give 



<?+7i=/+A=£+A=0 or c, . . . (63) 

 or else 



f+h=g+h=— 6, e+7i=a=&(&+2c)/(26+c). (61) 



Use (57) ; also reject ??i', m v , ;>*'" from w; also put v for 

 the modified value of if. Thus 



v/B = l(^\/ v -\-Vv) -8/4(2ft + c) . S(YV>/)x(YV??). 



. . . (65) 



Here not ?; = 0, but y= —hp, gives the condition for free 

 aether. Put v + hp = Vo> again reject constants and multiples 

 of Sy>7o &c. from v and call the modified form r ; 



r /B=-HS 2 V>?o- 4/ /+,;.V^,o) - 4{2 l +e) S(VV%) X o(VVW- 



• • • (66) 

 On account of the great similarity between the results of 



* By comparing our present cubic term efg %vith the former 

 ( — e-\rf+g)( )( ) we must expect our present c to correspond to our 

 former — e-\-f+g. 



