54G 
MR. G. H. DARWIN ON PROBLEMS CONNECTED 
body. Hence the effective force due to inertia on a unit of volume of the interior of 
the earth at a point r, 6, ft is wr sin 6 y-, and it acts in a small circle of latitude from 
west to east. The sum of the moments of these forces about the axis of Z is of course 
equal to J3, and therefore this bodily force would equilibrate the surface forces found 
in (6), if the earth were rigid. 
The components of the bodily force parallel to the axes are in rectangular co¬ 
ordinates. 
13 13 
wy-^, —ivx— 0 
(?) 
The problem is therefore reduced to that of finding the state of flow in the interior 
of a viscous sphere, which is subject to a bodily force of which the components are (7) 
and to the surface stresses of which the components are (6). 
Let a, ft, y be the component velocities of flow at the point x, y, z, and v the 
coefficient of viscosity. Then neglecting inertia because the motion is very slow, the 
equations of motion are 
dp 0 43 
■■—~ -f- v V ~ oiw-yyy — 0 
dx C J 
d'p . 13 _ 
— -r--\-vVB—iv—x=0 
dy 0 
-y+vV 2 y =0 
dor dij dz 
( 8 ) 
We have to find a solution of these equations, subject to the condition above stated, 
as to surface stress. 
Let a', ft', y, p be functions which satisfy the equations (8) throughout the sphere. 
Then if we put a=a'-j-a / , ft=/3'-\- l 8 / , y=y'+y / , p=p'-{-p / , we see that to complete 
the solution we have to find a /5 ft /} y t , p t , as determined by the equations 
+ W*a =0, < l y ‘ &C. = 0, 
dx dy 
dp 
dz 
l+f+f=° 
• (9) 
which they are to satisfy throughout the sphere. They must also satisfy certain 
equations to be found by subtracting from the given surface stresses (6), components 
of surface stress to be calculated from a', ft'. y , p'r 
We have first to find a', ft', y, p'. 
Conceive the symbols in equations (8) to be accented, and differentiate the first 
* TLis statement of method is taken from Thomson and Tait’s ‘ Nat. Phil.,’ § 783. 
