1293 
T= eyey + 0,6, + ee, — 4" 4 es” + €,°) 
1, =eest + 4 (E40 04 — 020 — 02%) 
e,...e, being the “components” of the strain (changes in length 
and angle). 
The energy of a volume is found by multiplying the expression 
(17) with the element dr, and integrating. 
From (17) the normal stress in the direction of « can be deduced, 
using the formula 
ae 
é, 
We can only observe the time-average of this force and, taking 
the mean value, the linear parts issuing from the terms with A and 
B will fall out. We obtain therefore for the average value of the 
tension in the direction of z 
S=38Cl,? + DI, (e, He) + Di, + Etec, — Hes’). 
This force is again equal for all points, we can therefore integrate 
over the volume 1, and interchange the integration and the taking 
of the average. Taking into account the isotropy, it is easily seen 
that ¢,e, — de,” is equal to +/,, so that we get for S 
2 4 q 3 8 
S= UIC HED DAEB Ald... (08) 
Now determining the mean value of the potential energy, the 
terms with C, D, and E will be found to fall out, and we get two 
parts, relating respectively to the longitudinal and transverse waves, 
as appears from the meaning of the invariants Z, and /,. These 
parts are 
EADE ir ED Pe ve) 
and 
By gee EY Tees nk steet eo ND 
In the stationary state the potential energy is distributed in a 
given way over these waves. 
For the thermal pressure we now find 
3C+4D— NEUES 
El — 5 Er Pee eer mere KE 
eae B 
1) This is the usual formula, which is, however, not correct if the second powers 
of the deformations are. taken into account, as has been done in the above by 
introducing C, D, and Z. In the third contribution we shall show that even in 
case of Hooxe’s-law being true, a coefficient of expansion will be found, if the 
correct formula for S is used. As far as numerical values are known, they seem 
to indicate that the influence of the terms neglected here could sometimes be 
sensible. 
CS 
