Lord Rayleigh on Bells. 17 



formed by half the hyperboloid and a crown stretching across 

 in the plane of symmetry z = (fig. 2). The deformation of 

 this crown can take place only in the direction perpendicular to 

 its plane, so that 8r = 0, 8(/> = 0. These conditions must apply 

 also to the hyperboloid at the place of attachment to the 

 crown. Hence Sep must vanish with z, or, which is the same, 

 with ^. Accordingly A = in (11) ; and dropping the con- 

 stant multiplier we may take as the solution 



8<j) = sin s% cos s(j>, (12) 



and in correspondence therewith by (2) and (3) 



Br = sr sin s% sin scf) (13) 



a 2 

 bz = — j- {cos s% + s tan ^ sin ,s^} sin scf>. . . (14) 



It is evident from these equations that, whatever may be 

 the value of s, there is no circle of latitude over which both 

 Sep (or 8r) and Sz vanish *. Hence there can be no circu- 

 lar nodal line in the absolute sense. But just as there are 

 meridians (sin scf> = 0) on which the normal motion vanishes, 

 so there may be nodal circles in this more limited sense. The 

 condition to be satisfied is obviously 



§r/8z = dr/dz ; 

 or in the present case 



sin2x + 2stansx(sin 2 x + 67a 2 ) = °- . . (15) 



In this equation the range of % is from to \ir ; and thus 

 there can be solutions only when tan s% is negative. 

 In the case 5 = 2 the equation reduces to 



l + 2sin 2 % + 4& 2 /a 2 = 0, 



which can never be satisfied. 



When 8 ='6, the roots, if any, must lie between ^=30° 

 and x = 60°. A more detailed consideration shows that 

 there is but one root, and that it occurs when % is a little 

 short of 60°. 



* A corresponding proposition may be proved more generally, that is 

 without limitation to the hyperboloid. 



Phil. Mag. S. 5. Vol. 29. No. 176. Jan. 1890. 



