434 Mr. R. Hargreaves on 



Thus the seolian surfaces are similar quadrics, viz. 

 Up = (e + A,)A P , 

 and the external potential is 



^=^f^ | [l-2(P^ + 2P^)/A p (, + X)]. 

 But 



J A (6 + X)f 3(6 + X)Ja (6 + 7C)f 



which with ^(P^ 2 + 2P / ( //r)=:(e-f X)A, gives 



^ e ~ "6Ja (6 + X)t ~ 3(e + X)i ~ 3{2(1V + 2P'^~) f* 

 _ ^'PT 



Thus the external potential is the same as for a conductor 

 with equal charge, and the action is that of a point charge at 

 the centre under the law by which kju v * replaces 1/r. The use 

 of the infinite upper limit requires e to be positive. 



§ 26. We now prove a theorem akin to Maelaurin's theorem 

 in isotropic potentials. Let u* x , u a2 be two peolians derived 

 from u a by the parameters X l5 and X x +X 2 respectively. We 

 shall show that a 2 may be derived from a { by the parameter X 2 , 

 a x and X 2 taking the places of a and X in the scheme (81) ; and 

 shall then establish a relation between the discriminants used 

 in getting at a 2 according as we set out from a or from a x . 

 For a 2 . . . as derived from the base a by the parameter X x + X 2 , 

 the first equation of (81) is 



a 2 ( A+pXA, + pX 2 A a ) + yliV' + r%A a + r'X 2 A a ) 



+ /3 2 '{i3' + q'\ 1 A a + q%A a ) =A a . 



Multiply each member by A„j and use both relations (84) 

 with X x for the X there ; then removing the common factor 

 A(X X ), there remains 



« 2 (A ai +pX 2 A ai ) + r /(0; : + r / X 2 A ai ) +/3 2 '(B' mi + g'\ 2 A ai ) =A ai , (97) 



which reproduces the form of (81) with X 2 for parameter and 

 a x ... for base. Now use the first of (84), viz. A a A(\) = A* 

 or its equivalent the first of (85) A u J(a, X) =A a in the three 

 forms required to get <% 2 and « T from the base a, and to get a 2 

 from the base a x% They are A KL J(a, X a ) =A a — A* 2 J (a, X Y + X 2 ), 

 and A^J^x, X 2 ) = A a] ; and their product gives 



J(a,X 1 +X 2 )=J(a,X 1 ) x J(«i,\ 2 ), . . (98) 



