1002 
we find 
x (t,—t,) - leat (Td Te, 6a Tea Li)» 
so that we get z 
[ar [0 (—L) — arr (T,,dp + T,,69 + T,,4q + Tod) = 9. 
Now 
T= DV —g Yen Tu 
and therefore in our case 
cae — Vi=9 Vas Tib ’ 
from which follows 
4, 1 Pst =) 
ET en ER Faes Elis Sen and, sT „== 
P q q 8 
By substituting this and replacing 
[6 (—L) dr by 
(d (ÒL\ IL wp LU ae a (aL) PL), 
sf Ps oe Pp dr 0q' oq haa Fe Os! nde 5e a 
we get, as the coefticients of dp, dq, and ds must be separately zero, 
* (55) — 5p OL r z d (5) OL r° 5 5) 
ee a poe — x —(s - $ 
vr, dr 0 ' Ay ae ~ 32 mee PAU 
Ee, q ’ q q (7) 
In this # must be looked upon as a known function of p, q, and s, 
given by (9). 
The tensor 2,,:/—g possesses the same symmetry properties as gs. 
Of the equations (1) only the first does not pass into an identity, 
but into 
Pet egia(Er+ete+csj=o. .. © 
fe q 5 
if we put Eu A Sne S. 
Then this equation with the three equations (7) form a system 
of four differential equations for the determination of p, q, and s, 
and say P, if, in connection with the nature of the substance, 
we know two more relations between P, Q, and S. If eg. Q=P 
and S=const., we have the case of an incompressible fluid; 
Q= P, S=f(P) represents the case of a compressible liquid or gas. 
