PIEZOELECTRIC CR YSTA LS IN TENSOR FORM 91 



Finally multiplying through the last of equation (32) by 9 we can write 

 them as 



Si = snTi + 512^2 + suTz + SuTi + Si^T^ + suT^ + oci dQ 



Si = SieTi + -^267^2 + ■^36^3 + SisT4 + 5667^6 + •^662^6 + OC^ dO 



dQ = Q d(T = 6[aiTi + q:27'2 + otsTs + 0474 + ai,T^ + a^Te] + pCpdQ 



since ©t^ is the total heat capacity of the unit volume at constant stress, 



which is equal to pCp, where p is the density and Cp the heat capacity at 

 constant stress per gram of the material. 



To get the adiabatic elastic constants which correspond to no heat loss 

 from the element, or dQ = 0, dQ can be eliminated from (35) giving 



^1 = s'nTi + 5127^2 + SnTs + 3^X4 + s[f,Tf, + s'^Tf, + (ai/pCp) dQ 

 (36) 



Se = s'uTi + sIbT^ + SuTs + s'teTi + sl^T^ + s^Te + (as/pCp) dQ 



where 



,-, = s% - «-i^. (37) 



pLp 



For example for quartz, the expansion coeffxients are 

 ai = 14.3 X 10"V°C; 02 = 14.3 X 10"V°C; a, = 7.8 X 10"V°C; 



The density and specific heat at constant pressure are 



p — 2.65 grams/cm ; Cp= 7.37 X 10^ergs/cm^ 

 Hence the only constants that differ for adiabatic and isothermal values are 



•^11 = 522 ; .^12 ; -^13 ; -^33 • 

 Taking these values as 



sn = 127.9 X 10~'* cmVdyne; Su = -15.35 X 10"''; 



su = 11.0 X 10"'*; 533 = 95.6 X 10"''. 



We find that the corresponding isothermal values are 



sfi = 128.2 X 10"'*; 5?2 = -15.04 X 10"'*; 



5?3 = 10.83 X 10"'*; s% = 95.7 X 10"'* cmVdyne 



^See "Quartz Crystal Applications" Bell System Technical Journal, Vol. XXII> 

 No. 2, July 1943, W. P. Mason. 



