WITH THE TIDES OF A VISCOUS SPHEROID. 
561 
Now 24-p5f=4-23, and (2-547)- 1 —(60-4)~ 1 =-376. 
Therefore the loss of energy=[(4’23) 3 — 1 — *376 X 8‘84] X earth’s present k.e. 
= 13*57 X earth’s present k.e. of rotation. 
Hence the whole heat, generated in the earth from first to last, gives a supply of 
heat, at the present rate of loss, for 13'6 X 262 million years, or 3,560 million years. 
This amount of heat is certainly prodigious, and I found it hard to believe that it 
should not largely affect the underground temperature. But Sir W. Thomson pointed 
out to me that the distribution of its generation would probably be such as not 
materially to affect the temperature gradient at the earth’s surface ; this remarkable 
prevision on his part has been confirmed by the results of the following problem, which 
I thought might be taken to roughly represent the state of the case. 
Conceive an infinite slab of rock of thickness 2a (or 8,000 miles) being part of 
an infinite mass of rock ; suppose that in a unit of volume, distant x from the 
medial plane, there is generated, per unit time, a quantity of heat equal to 
f)[320a' 1 —560a 3 x 3 +259x 4 ] ; suppose that initially the slab and the whole mass of rock 
have a uniform temperature V; let the heat begin to be generated according to the 
above law, and suppose that the two faces of the slab are for ever maintained at the 
constant temperature V ; then it is required to find the distribution of temperature 
within the slab after any time. 
This problem roughly represents the true problem to be considered, because if we 
replace x by the radius vector r, we have the average distribution of internal heat- 
generation due to friction; also the maintenance of the faces of the slab at a constant 
temperature represents the rapid cooling of the earth’s surface, as explained by Sii 
W. Thomson in his investigation. 
Let d be temperature, y thermal capacity, k conductivity ; then the equation of heat- 
flow is 
3 2 0 a 4 — 56 0a' 2 x 3 + 2 5 9a; 4 ]. 
Let 320 ^=2L, 560 y=12M, 259 ^—30N, and let the thermometric conductivity 
k=-. Then 
7 
~ +Lake 3 - Make 4 + Nx 6 - K]. 
Let the constant R— (L — M+N)a 6 , and put 
xp =Ij -f Lake 3 —Make 4 +Nx G —It 
= d — La 4 (a 3 — x 2 ) + Ma 3 (a 4 —x 4 ) — N (a 6 —x°). 
Then when x=±a, i/>=d. 
4 c 2 
