556 
ME, G. H. DARWIN ON PROBLEMS CONNECTED 
Now if P, Q, If, S, T, U be the stresses across three mutually rectangular planes at 
x, y, z, estimated in the usual way, then the work done per unit time on a unit of 
volume situated at x, y, z is 
p l+Q|+ R l+ s (S + l) +T (S + £)+ u (l + ^) !i 
But P= — S = i/^ + y j, and Q, R, T, U have symmetrical forms. There¬ 
fore, substituting in the expression for the work (which will be called — !, and 
remembering that 
we have 
dt 
cU.dJ3_ , c h_ 0 
daTdy ^dz ’ 
a. (*ijl 
v ( 
+ {d/j j j + lA+rfb +U+*j + \dy + dxj 
Now from (22) 
2 
K 2 
= 12cch/ 3 = fr 4 sin 4 0[l — cos 4(<£ —cat)] 
and from (23) 
1 
K 2 
d/3 ,r/ 7 \S /riy .d«V (d« d/3\* 
dl + dy) 
= 9z 2 (ar+1/ 2 ) + [8 (a 3 —P 3 —y 2 ) — 5s 2 ] 3 
= 9r 4 sin 3 0 cos 3 ^+(8a 3 — 5r 2 — 3r 3 sin 3 Of . 
(24) 
(25) 
Adding (24) and (25) and rearranging the terms 
1 tfE 
K' 2 v dt 
= — -fr 4 sin 4 6 cos 4(<£ — 6M) + (8« 3 — 5r 3 ) 2 — fr 3 sin 3 #[32a 3 —r 2 (26 + sin 3 0)_|. 
The first of these terms is periodic, going through its cycle of changes in six lunar 
hours, and therefore the average rate of work, or the average rate of heat generation, 
is given by 
<4 y=sin 2ej [(8a 2 — 5r 3 ) 3 —fr 3 sin 3 0{S2a z — r 3 (26 + sin 2 #)}] . . (26) 
It will now be well to show that this formula leads to the same results as those 
given in the paper on “Precession.” 
In order to find the whole heat generated per unit time throughout the sphere, we 
must find the integral j j ~~ r 3 sin 6drd0d<j>, from r=a to 0, 0=n to 0, r/>= 2 tt to 0. 
* Thomson and Tait, ‘Nat, Phil.,’ § 670. 
