﻿of Elastic Stresses in an Isotropic Body. 211 



as for vectors, and we deduce for second differential co- 

 efficients 



Ql = , ' 1, M ! "'' l 'Bl', +2 '' !i trs)" (4 " 4, S' 

 n$ = ^mt^ 3 ^hz + u ™ farr w~ (**-*«) §^ 2 - 



It is required to establish that in (2) 1? 6 2 , 3 , yfr lt i/r 2 , yjr 3 

 act as stresses. Actually I shall assume that the operators 

 n i? Xl 2 > X2 3 act upon them as upon the elements of stress, 

 and first examine this hypothesis when the three operators 

 separately act on the six equalities in (2). 



Selecting the first equality in (2) and applying the 

 operator Xlj on each of the two sides of the equality, 



then 



fljP = 



and ^6 2 ~d 2 0, 9o V*i 



d: 2 d/y 2 dyd- 



becomes 



2 y*i , o3 8 f. , «,y «?.-<?,) 2 B 2 (^-^s) 

 9« 2 By 2 " ctyd^ d.yd^ 



B 2 



-.&-£)* 



which is null, and the hypothesis is not negated. 



In fact, the hypothesis is not negated in any case. 



The argument is then as follows : — If we had written 

 did, n 2 #> etc. in the equations, and had proceeded to find 

 these 18 quantities, we should have 18 linear equations 

 from which to find them. The solution is therefore unique, 

 and cannot differ from that employed by hypothesis and 

 found to satisfy the 18 tests. 



We shall therefore regard 1? 6 2 i @z-> ^1? ^2? ^3 as acting 

 as elements of a stress, though the} T have not the proper 

 dimensions. 



It is proposed to find the conditions which must exist 

 between these primary arbitrary stresses in order that the 

 elements P, Q, R, S, T, U may be elements of an elastic 

 stress. 



