1291 
The force exercised in this bar by the part to the right of a 
section on the part to the left, will be represented by 
0s c, (DEN? 
a DE RENE NT ee 
Ade ie 2 (=) 2) 
The total potential energy then amounts to 
air os i e, 0g Vy : 
Eq — af =) U + = { (5) Kea ° e : . ( ) 
where the integration has to be extended over the length of the 
bar, which has been put equal to unity. 
For this case the mean value S of the force is again equal for 
0 
all points of the bar, and . being zero, we have 
xv 
=z 0&\? 
Integrating this result over the bar, we get 
ake Shey Cay AET COENE 
OE =((=) da == sf (3) da. e e e . (10) 
as also in this case the integrating and the taking of the mean 
value may be interchanged. 
Now, just as in the discrete problem, we determine «,, and 
=) bei find 
5, ) being zero we fin 
„if (EE) OP UE caer ces A 
2 dx 
ik VE 
S= Le == AEP ihe rt SDR 12 
Oe gue (12) 
and from it 
from which the relation of GRÜNeIsEN again follows. 
In calculating ¢ we can use the quantum-theory, but the formula 
(12) is evidently independent of it. 
3. We will show that the same result is obtained by applying 
the method of normal vibrations. The differential equation for the 
motion of the bar is expressed by 
0°§ 0° 0§ 075 
Bi Og On Oa ST att ty hse: hy ere Ree 
where 9 represents the density. Properly speaking the equation (14), 
N 
being non-linear, possesses no normal vibrations as a solution for 
