Heating and on Double Refraction resulting therefrom. 171 

 with two similar equations ; or with use of (3) and (4) 



/tfc N d (du dv dw\ „ d6 rt ,„ 



< x+ ^£ + ^ + *) + ^-*S= ' . (6) 

 if 



r =(3\ + 2^)* (7) 



One of the simplest cases that can be considered is that of 

 a plate, bounded by infinite planes parallel to xy, and so 

 heated that 6 is a function of z only. If, further, 6 be sym- 

 metrical with respect to the middle surface, the plate will 

 remain unbent ; and if the mean value of be zero, the 

 various plane sections will remain unextended. Assuming, 

 therefore, that u } v vanish while w is variable, we get from 

 (3) and (4) 



R=(\+2/,)g- 7 0=O, .... (8) 



P=Q=xg- 7 0, (9) 



y = T=U=0 (10) 



In (8) R is assumed to vanish, since no force is supposed 

 to act upon the faces. From (8), (9) 



r=^=-ai < u > 



If the plate be examined in the polariscope by light tra- 

 versing it in the direction of y, the double refraction, de- 

 pending upon the difference between R and P, of which the 

 former is zero, is represented simply by (11). Dark bars 

 will be seen at places where = 0. If the direction of the 

 light be across the plate, i. e. parallel to z, there is no ten- 

 dency to double refraction, since everywhere P=Q. 



In the above example where every layer parallel to xy 

 remains unextended, the local alteration of temperature pro- 

 duces its fall effect. But in general the circumstances are 

 such that the plate is able to relieve itself to a considerable 

 extent. A uniform elevation of temperature, for instance, 

 would entail no stress. And again, a uniform temperature 

 gradient, such as would finally establish itself if the two 

 surfaces of the plate were kept at fixed temperatures, is com- 

 pensated by bending and entails no stress. In such cases 

 before calculating the stress by (11) we must throw out the 

 mean value of 9 so as to make \Pdz = 0, and also such a 

 term proportional to the distance from the middle surface as 



N2 



