114 BELL SYSTEM TECHNICAL JOURNAL 



For the third and tifth equations, since we must have ei3 = cis ; €23 = f2;> 

 in order to satisfy the symmetry relation, the equations can only be satis- 

 fied if 



e.3 = eo3 = 0. (126) 



Similarly solving the lirst three equations simultaneously, we find 



fl2=0;6u= 622. (127) 



Hence the remaining constants are 



en = 622 ; 633 • (128) 



Similarly for third and fourth rank tensors, for a crystal having Z a trigonal 

 axis, the remaining terms are 



hn , hu = —lh\ , hn = 0; hu , //15 , /'le = — /'22 



/?21 = — /'22, //22 , /'23 = 0, //24 = /'l5 ] hb = — hi , hi = " /?I1 (129) 

 //31 ; ^32 = //31 ; //33 ; /'34 = 0; /;35 = 0; //36 = 



cn ; ^12 ; ^13 ; cu ; fis = ~<^25 ; ^le = 



c\2 ; C21. — c\\ ; C23 = c\i ; C24 = — '"14 ; C25 ; ^26 = 



Cn ; C20 = C\3 ; f33 ; ("34 = 0; czh — ^\ C36 = 



(130) 

 Cu ; ^24 — ~Cu ;czi — ^; cu ; f45 — 0; C46 — c\^ 



C\i = —^25; ^25 ; <'35 = 0; f45 = 0; Css = C44 ; C56 = Cu 



C16 = 0; ^26 — 0; r36 = 0; C46 — C21, ; ("56 "^ Cu ; fee = 2 vn~Ci2)- 



If the Z axis is a trigonal axis and the X a binary axis, as it is in quartz, 

 the resulting constants are obtained by combining the conditions (116), 

 (118), (120) with conditions (128), (129), (130) respectively. The resulting 

 second, third and fourth rank tensors have the following terms 



611 ; 612 = 0; 613 = 



612 = 0; 622 = 6U ; 623 = (131) 



613 = 0; €23 = 0; €33 



flu ; fin = — //ii ; //13 = 0; //i4 ; //I5 = 0; //i6 = 



//21 = 0; //22 = 0; //23 = 0; //24 = 0; h, = -hu ; //26 = -hn (132) 



//3l = 0; //32 = 0; //33 = 0; //34 = 0; hy, - 0; //36 - 



