Then 



of the Earth? s Crust in Cooling. 



583 





-z*/4h*t 



Vtt 21iti 



Then performing the integration we find 



Qt * 2 ' V7T £ 



- J|^ \{e+e'*)e-* + e'mM{2qe-* % + rt{l-- Er£ g)} 

 4V/3{V* a Er£g+^(l-Er£gi/2)}]. (6) 



The terms depending on /z, have been omitted. 



When q is zero, this is negative, but when q is large 

 enough, 'dK/'dt is positive. For one value of q.'dK/'dt is 

 zero, and the rock is being neither stretched nor crumpled ; 

 this value of q corresponds to the " level of no strain." 

 Evidently if then we put "d~Kfdt = 0, and regard (6) as an 

 equation to find q, the solution will have 1/c as a factor. 

 Hence in accordance with the approximations already made, 

 q may be put zero inside the square brackets in making an 

 approximation to a solution. 



Then q has to be found from 



0=[e+e'« + e{mh^rt + /3)2q7r-*]Pq/t'n* 

 3/3A 



CsJ {irt) 



[e + e'a + e'mh J (irt) + e'/3/ */2] . 



With the data already given, this gives q = 0*207. In 

 accordance with Fizeau's results, e/e' is taken to be 300° C. 



Hence the depth of the level of no strain is 79 kilometres. 

 This is greater than the depth of the radioactive layer, so 

 that /jl is zero, as has already been assumed. 



In addition to the level of no strain, we require the 

 amount of compression at the surface. This is obtained 

 directly by putting q = in c)K/d£, and then integrating 

 from t = to the present time. The approximation is not 

 good in the early part of the period of integration, but 

 as the compression must in any case vanish with the time, 

 that short interval may safely be neglected. 



Then 



K= - 6 ^(^J\e^e l a ^e l mhy/{t7r)^e'^jx/2}. 



jl 



