282-284] TRANSFORMATION FORMULA. 511 



Hence the arithmetic mean of the normal pressures on any three 

 mutually perpendicular planes through the point P is the same. 

 We shall denote this mean pressure by p. 



Again, resolving parallel to y, we obtain the third of the 

 following symmetrical system of equations: 



>. + P*K*k, } (3). 



p X y=p l l l m l +p,l. 2 m. 2 +p 3 l 3 m 3 

 These shew that 



Pyz = Pzy > PZX = PXZ > pxy = Pyx > 



as in Art. 281. 



If in the same figure we suppose PA, PB, PC to be drawn 

 parallel to x, y, z respectively, whilst ABC is any plane drawn near 

 P, whose direction -cosines are I, m, n, we find in the same way 

 that the components (ph x , phy, Phz) of the stress exerted across this 

 plane are 



Phx = lp xx + mpxy + np xz , "I 



Phz = Ipzx + mpzy + np zz \ 



284. Now PI, p 2 , p s differ from p by quantities depending 

 on the motion of distortion, which must therefore be functions of 

 a', b', c', only. The simplest hypothesis we can frame on this 

 point is that these functions are linear. We write therefore 



= -p + X (a' + V + c') + 



where X, //, are constants depending on the nature of the fluid, 

 and on its physical state, this being the most general assumption 

 consistent with the above suppositions, and with symmetry. Sub- 

 stituting these values of PI, p^, p 3 in (1) and (3) of Art. 283, and 

 making use of the results of Art. 282, we find 



Pxx = p + X (a + b + c) + 2pa,\ 



Pzz = ~p + X (a -f b + c) + 2/iC 



x y = fyh ............... (3). 



