SPHEROIDS, AND ON THE OCEAN TIDES UPON A YIELDING NUCLEUS. 19 
But nt is the coefficient of viscosity, and in treating the tides of the purely viscous 
19v 
spheroid we put tan e= —X coefficient of viscosity; therefore adopting the same 
notation here, we have tan y=tan t/z+tan e. 
If the modulus of relaxation t be zero, whilst the coefficient of rigidity n becomes 
infinite, but nt finite, the substance is purely viscous, and we have xfj= 0 and y=e, so 
that the solution reduces to the case already considered. If t be infinite, the sub¬ 
stance is purely elastic, and we have \l/=y, y= - and since co h ^ = therefore 
1 J T 2 ^ 2 COS ip Sill -xjr 
a I' 
cos (vl+ V ). 
19% V CL 
But according to Thomson’s notation* --=-, so that cr= -S cos (vt-V-ri), which is 
& 2 gwa 2 r + 2 v 
the solution of Thomson’s problem of the purely elastic spheroid. 
The present solution embraces, therefore, both the case considered by him, and that 
of the viscous spheroid. 
9. Ocean tides on an elastico-viscous nucleus. 
If r — a-\-u be the equation to the ocean spheroid, we have, as in sec. (6), that the 
height of tide relatively to the nucleus is given by 
a? 2 
U— (T— —S COS — -cr, 
9 5 
and substituting the present value of cr, 
U — cr=~ - S 
52 
\ cos y i 
cos (vt+rj) — —-^cos ( vt + V + 'l J —x) 
2 a ~ sin (y —G) • / , . \ 
= — ^-S- , 7 sm (vt + rj —x)- 
5 2 cos \ / a.s 
If the nucleus had been rigid the rise and fall would have been given by H cos {yt-\-rj), 
2 a 
where 11 = ^ - S ; therefore on the yielding nucleus it is given by 
5 2 
u — cr= — TI 8111 ^ ^ sin (vt-\-7)—x) 
cos ip ' ‘ /v/ 
= — H cos y (tan y—tan xjj) sin (vt- bp—y) 
= — H cos y tan e sin (ftf+p —y). 
* ‘ Nat. Phil.’, § 840. 
D 2 
