WITH THE TIDES OF A VISCOUS SPHEROID. 
555 
axis of the tidal spheroid, and that of z is the earth’s axis of rotation, and if W be the 
effective disturbing potential estimated per unit volume, we have 
W = wr~[ S — g- ) — wt sin 2 e.xy 
( 20 ) 
It was also shown in the paper on “Tides” that the solution of Sir W. Thomson’s 
problem of the state of internal strain of an elastic sphere, devoid of gravitation, as 
distorted by a bodily force, of which the potential is expressible as a solid harmonic 
function of the second degree, is identical in form with the solution of the parallel 
problem for a viscous spheroid. 
That solution is as follows :— 
19a 
dx\ ■ 
with symmetrical expressions for /3 and y. 
Since -y-f—: ) = w -y -— ;W, the solution may he written 
dx\r>) r 5 dx r 7 J 
a — 
38a 
(8a 2 -5f-)'p+4, t W 
/3 = &C., y=&C. 
Then substituting for W from (20) we have 
a — vw sin 2 e [(8 of — 5r 2 ) y+ 4x 2 y] 
ooa 
/3= sin 2e [(8a 3 — 5>' 3 )a:-{- 4.x// 3 ] - 
sin 2e ixyz 
( 21 ) 
WT 
Putting K=-— sin 2e, we have 
d “=-Kxy, |=iK[8a=-(#+15/+54 3 )]. §=-5K^ 
dx 
tf/3 
dx 
=iK[8a 2 -(15x 2 +y 3 +52 2 )], xy, ~=-5Kxz 
dy 
dy 
dy 
— 2 K zx, 
dz 
dy 
dz 
= 2K xy 
j 
And 
( 22 ) 
cl (3 d f V _T 7 I ^^ O T7" I cJ (3 TTpo / 0 O 0\ r on 
=- 3Kzx ' *+ f fe=- 8l W*,+/fa= K P(«-*-r)-5U • 
(23) 
* See Thomson and Tait’s ‘Hat. Phil.,’ § 834, or “Tides,” Section 3. 
