1290 
is required to keep constant the volume of a solid body when 
gaining heat. 
1. Let us consider a row of n equidistant points. Be the elongation 
in the direction x (the direction of the row being taken as an axis) 
for the vt point §. Then the force exerted by the vth molecule on 
the (y---1)* will be represented by 
( 
S= f (§&.—&-1) + EE JE 5 3 - 2 5 (1) 
The total potential energy, then, can be represented by 
eas phe a Aen . 
ay (Es, Za + 6 ay (EE) . . . ( ) 
Eq == 5 
where the sum has to be extended over all molecules. 
Now for a stationary state, S, the time-average of S, will be 
equal for all points. Therefore, adding (1) for all points, we get n 
times the time-average of S. Thus 
<9 a me ee = ene 
NIS == A) =S (Gy Gy) os ee 
2 2 
the mean of the first term in (1) being zero, as the mean length is 
invariable, and as the taking of the mean and of the sum may be 
interchanged. 
For the mean value of & we have 
re eo ee 
j= $2685) 
the mean value of the second term being zero. We thus find 
nS eS 
5) 2/ 
for 2a 4 &, where é, represents the kinetie and « the total 
energy. Putting g =n?c,, f=ne,, we find 
= a ee ee 
2¢, 
For the dilatation taken from the absolute zero, we find 
0, — 
pee os le 
being the relation of GRUNEISEN. 
2. We shall now consider the same problem, approximating this 
time the problem for a row of points by that of a continuum. 
Therefore we have to do with a bar in which the elastic qualities 
depart from Hooker’s law, 
