252 Prof. J. J. Thomson on the 
equation becomes 
4a 
Using this frequency equation, and supposing that p=1, 
2. e. that there is only one corpuscle at the centre of the 
hexagon, we get instead of (1), 
048 /3e 58, 
The roots of this equation in q? are both positive, so that q 
is real and the equilibrium is stable. 
Let us now investigate the conditions for stability for 
displacements at right angles to the plane of the orbit. © 
For the motion at right angles to the plane of the rine, 
the frequency equation when k=3 is i 
9 3 J 
te S,2- —- +L,—-L,— my Ni —N,— mg’) =(Mzi—2megq)?. 


ve? pe? 34 e? 2 
og ae oe 

For this to represent the displacement of a stable system q? 
must be positive, so that if p=1 
Vee ene aac? 
bs ONG see 
must be positive ; we have, however, 

2 Q 2 
me? + = -- sl? a5 a 3 
so that for 
yer. em 34)e2 
CO a> Wasa A 
to be positive | 
maw” must be greater than ae ee fe 2.e, “46 a 
Let us now consider the stability of the corpuscle at he 
centre of the ring: if it is displaced through a distance z at 
right angles to the ring, the equation of motion of the 
corpuscles is 
Co. Re es 
Ne re e+ oe 
Thus if the motion is stable 
9 
ve- Ge? 
yori aes eae 
L° Ng 
