MATHEMATICS OF PHYSICAL PROPERTIES OF CRYSTALS 23 



on the direction of its normal ;;. We shall choose a triangular area ds such 

 that an arbitrarily chosen set of mutually perpendicular unit axes Xi , X2 , 

 ;V3 pass through the vertices of the triangle. Let us consider the conditions 

 of equilibrium of the tetrahedral element of volume so formed. The areas 

 normal to .vi , .To , X3 are dsi , dso , dss , respectively, and the forces per unit 

 area acting through these faces are: 



Any body forces (such as gravity) depend on a higher order of smallness 

 (that is on the volume rather than on the area) and hence are negligible. 

 Whence for equilibrium: 



But 



Fds = Fidsi + Fidsi + F^dss 



dsi = Hids, ds2 = n^ds and dsz — nzds 



where I W2 I is the normal to the area ds. Whence we may write: F = fn 

 where/ is the matrix 



//ll/l2/l3\ 

 I /2I /22 /23 I 



Xfdifiifzs/ 



For the body to be in rotational equilibrium the tangential forces must 

 balance, hence /12 = /21 ,/i3 = /ai and/23 = /32 • 



Transformation of A xes 



A change of axes that transforms vectors through F' — aF changes F = 

 fn to aF = afa~ an so that if /' = afa~^ then F' = fn'. 



In order to relate the stress to other quantities through a matrix we wish 

 to convert it into a single column matrix. We put /n = X^ , /22 = X2 , 

 ^33 == Xz , /23 = /32 = Xi , /31 = /i3 = X5 and /12 = /21 = X^ . 



Changing to the X representation we find 



x' = aX' (8.1) 



where a is the matrix eq. (7.2). 



