108 Arthur Holmes — Radio-activity 



loss of heat from the earth is supplied from radio-thermal energy, 

 and that the distribution of the radio-elements falls off exponentially 

 in depth. Their final equation (No. 85, p. 95) expresses 0, the 

 temperature at any depth x, and at any time t, in terms of the 

 following data : — 



S, the initial temperature at, or just below, the surface. 



m, the initial temperature gradient. 



k, the thermal conductivity of average rock. 



h z = k/cp, the diffusivity of average rock. 



c, the specific heat of average rock. 



p, the density of average rock. 



A, the heat production of unit volume of average rock per second. 



a, the decrease of heat generation per unit distance in depth. 



For the details of the equation and the treatment leading up to it 

 the reader is referred to the original work. Mr. Zobel has very kindly 

 differentiated this equation for me, and much of the mathematical 

 treatment which follows below is due to his help, for which I wish 

 here to express my gratitude. Differentiating equation 85, the 

 temperature gradient for any time t, and at any depth x, is obtained. 

 The temperature gradient d0/dx is actually measured at the present 

 time at the surface, so that x — 0. The equation then becomes 



S A r a ' hH 2 r 00 -^ ~\ 



deldx = m + j—7= + -y M — e • ^= / * 7 ^7 



ahVt 

 For large values of ahyjt, the equation may be expressed more 

 simply as 



d^dx = m + ~= + -. [l - — 1=] (10) 



The unknowns in this equation are m, S, and a, and obviously two 

 of them must be assumed to be known. It is highly improbable that 

 the initial temperature of the earth was uniform throughout. The 

 work of Barus shows that the temperature of fusion increases with 

 pressure and therefore with depth. Although this law may not hold 

 to any great depth, yet the temperature must also increase with 

 depth as a result of compression. Originally, then, the temperature 

 e at any depth x would be = mx + S. 

 It is assumed here that: 



m = 0-00005° C. per cm. ; S = 1,000° C. 

 The known factors in equation 10 are : — 

 deldx = 0-00038° C. per cm. 



t = 1,600 million years = 5-05 X 10 16 seconds. 

 A = 63 - 9 X 10- 14 calories per second per c.c. of rock. 

 h = 0-005° C. per cm. 

 P = 2-8. 

 C = 0-25. 



h- = k/cp = 0-0071. 

 h = 0-084. 

 Equation 10 may now be solved for a, which is found to be 4 X 10" . 



1 See Part I, Geol. Mag., Dec. VI, Vol. II, p. 70, 1915. 



