16 
MR. J. H. JEANS ON THE INSTABILITY OF THE PEAR-SHAPED 
The value of S 3 can be written down from symmetry ; that of <5 4 , which is of the 
same general type, will not be required in the present investigation. 
13. This completes the solution of the particular potential problem which we have 
had in hand. It might naturally be feared that some mistake might have been made 
either in principle or in detail, and it must be remembered that even one mistake 
might invalidate the answer to the whole problem. I have, therefore, both here and 
elsewhere, taken the utmost care to check the accuracy of my work in every way. 
The following will, I think, show that no error need be feared in the solution which 
has just been obtained. 
The value of V { , if obtained accurately, ought to satisfy V 2 V t - = — 4 tt/ 3, and the term 
— 47 rp must come entirely from the terms independent of e in V { . Thus the terms in 
V f which are multiplied by e, e* and e 3 ought separately to be spherical harmonics. 
It was verified in the previous paper that the terms in e and- e 2 were in actual 
fact of this form. The terms in e 3 will be harmonic if the quantity on the right of 
equation (57) is harmonic, and the conditions for this are expressed by the equations 
10C n + C 12 + C 1?> — 0,. (68) 
6C22 + 3 Cio + C2.3 = 0, • • • u .( 69 ) 
6 C 33 +3C 13 + C 23 = 0,.(70) 
3Hx+ti 2 +tr3 = 0 .(71) 
I have inserted the values just obtained for C n , C 12 , ... tq, ... in these equations, and 
have verified that they are all satisfied. (The necessary transformations of the various 
integrals are tedious, but involve no special difficulties.) 
It follows that the solution we have obtained gives accurately the potential of some 
solid of uniform density p. By the method explained in the earlier sections of the 
previous paper, it is easy to work the problem backwards and to verify that the 
equation of the boundary of the solid in question is obtained by putting X = 0 in the 
equation f + <p = 0. Thus we verify that our solution gives accurately the potential 
of the solid of boundary 
/+eP + e 2 Q + e 3 R = 0..(72) 
Conditions that Pear-shaped Figure shall he one of Equilibrium for a Rotating 
Liquid. 
14. In order that the figure determined by equation (72) shall be one of equi¬ 
librium for a rotating liquid, the potential at the boundary plus (x~ +y 2 ) must, as in 
equation (98) of the former paper, be identical with 
— irpahcO {f + cP + e 2 Q + e 3 Pt} +a constant. 
