Rigidity, of the Earth. 489 



The shear strain in the plane of r and the symmetrical 

 axis is 



and since the shear stress is zero at the surface, it follows 

 that the shear strain is also zero at the surface. Hence 



i|5+r|^=0whenr=a (22) 



r ^0 dr 



This is one boundary condition. 

 The radial tension at any point is 



(*-«n)8 + 2n|^ 



u Or 



and this must be zero when r = a. But since n is infinitely 

 small compared with k, and k8 is finite, this condition can 

 be written 



*8 + 2n!^ = wbenr=a (23) 



This is the second boundary condition. It remains to 

 express the two equations (22) and (23) in more convenient 

 forms. 



Substituting for e from (6) and for rj from (9) in (22), 

 we get (f(r) being dropped because it is zero) 



— 2e cos 6 sin 6 — 1 cos 6 sin 6 r-r < -= --,- (r 3 e) > = 



dr [_r 2 dr x y J 





 when r = a. 



That is, 



6e+r t{?i^} = whcnr=a - - • ( 24) 



This is a very convenient form for the first boundary 

 condition. 



Now, since (23) is true for all values of 0, we may 

 differentiate all through with respect to 6. Hence 



a|| + 2w^-=0 whenr=a. . . . (25) 



But since h is zero and n finite, the equation of equilibrium 

 (12) may be written 



k fe +2n h (r ^-f^ = (26) 



This last equation is true at every point of the sphere, and 



