Negatively Electrified Corpuscles by Hot Bodies. 255 



Now i£ the electric force is that in a light-wave the mean 

 energy E per unit volume in the wave is equal to the mean 

 value of X 2 /-±7rV 2 ; hence we see that the mean rate at which " 

 the particle is emitting energy (this is the rate of emission of 

 the energy of the scattered light) is 



If there are N corpuscles per unit volume the energy in the 

 scattered light coming from each unit of volume per second 

 is equal to 



'6 m* * 



The scattered light will be polarized in the same way as 

 light reflected from small particles. This scattering of the light 

 will cause the medium to absorb light. We can find the co- 

 efficient of absorption as follows : — Suppose the axis of z is the 

 direction of propagation of the light. Let AB and CD be 

 two planes at right angles to z separated by a distance Bz, 

 CD being in front. Then if A is the area of either of these 

 planes, the rate at which energy is being scattered by the 

 particles between the planes is equal to 



-47T Ne* „ A ~ 

 ^ 2 VE . ABz. 



Now when things are in a steady state this energy must be 

 supplied by the excess of the energy flowing into the region 

 between AB and CD through AB^ over that flowing out 

 through CD: the average rate at which energy flows across 

 AB is AEV; the rate at which it flows out across CD is 

 A(E + SE)V: hence we have 



4-7T Np 4 



-A$EV=^q^VEA^; 



or <ZE ^_47rNe 4 F 



dz 3 rt* ' 



tllUS _ in Ne4 



E = Ce 3 *■*• 



A "V 4 



and thus the coefficient of absorption is — ——z. 



Thus the region round incandescent metals or carbon will, 

 in virtue of the corpuscles coming from these substances, 



