206 SEE— TEMPERATURE, SECULAR COOLING [April 20, 



y-axis, and may be calculated when a, a', e and e' are known, by 

 substituting the value of y in the original equation of the ellipse (9). 

 Now to show that no change of volume occurs by this propaga- 

 tion of heat from the center outwards, we may observe : 



1. That although the major axis of the heat ellipsoid enlarges 

 with the time, the minor axis diminishes, so that except for the slight 

 loss at the surface the volume of the heat ellipsoid remains un- 

 changed, and no heat escapes into outer space; hence there is a 

 change of distribution but not in the total quantity of heat. 



2. As this transfer takes place under the confinement of the 

 crust and subjacent layers resting upon the nucleus, the change is 

 necessarily made in such a way as not to do work upon bodies 

 external to the thermodynamic system; nor is work done upon the 

 thermodynamic system itself by external bodies. In either case the 

 constancy of the amount of energy stored up would be altered, 

 which is contrary to the hypothesis. 



Any spherical shell ^irr-dr, outside of the sphere of no change 

 of temperature, already defined, would have another corresponding 

 shell inside, in which the changes of volume would be equal but of 

 opposite sign to that of the first. Therefore in general we should 

 have 



47r 2 r\dx.(T,^-'d%. = 0, (12) 



i=— t 



where d®i is the change of temperature experienced by any shell, 

 of density ai, and specific heat U- Hence denoting the variation of 

 the elements of volume by 8, we should have, since the change of 

 volume is directly proportional to the change of temperature, 



'f's(rfF;) = o. (13) 



1=— t 



The conclusions thus briefly indicated may be stated otherwise 

 as follows : 



1. The inner shells of the nucleus, having lost heat, might shrink 

 under the pressure to which they are subjected. 



2. The outer shells of the nucleus, having gained heat, would 

 have to expand, or the internal tension and latent heat would have 

 to increase. This would be relieved either by bursting the crust, or 



