Mr Havelock, On the continuous spectrum. 177 



where A, m, jju are constants ; 



6 = absolute temperature of the gas, 



**=&- &y + fa- v*y +(&-&*, 



(&> Vi> £1) ail d (£>, V-2, §2) being the velocities of the two molecules. 



If we consider Maxwell's law of distribution of velocities still 

 to hold, the number of such encounters with relative velocity r 

 in a short time is proportional to 



Ntpre-t tfi«-Hh»+$i'+&W+fifl d^d Vl d^d^d V2 d^ . . .(11), 



where N x density of the gas, 



£ oc 1/0. 



Now any physical analysis of the radiation is an average taken 

 over a short, but sensible, time ; so we can find the energy of each 

 pulse and then sum up for all the pulses. 



We have seen that for a pulse of type (10) the energy is 



A 



A 2 C u 



E = -^-r^ m ~^ e~Wdu (12). 



tip Jo 



Then multiplying by (11) and summing for all the molecules of 

 the gas, we find for the average intensity of the radiation the 

 expression 



+00 



E = CN^e- 1 j du ffffl fr 2 ™-^' W er* W+iiH-iiH-tf +*•+&» 



_«, d^ 1 dr} 1 d^ 1 d^dr)»d^ 2 ...(l'S). 



If in the last six integrals we change to new variables given by 



$, + !.= ff; vi + v>=V; Z + ^W)- ,(14) ' 



we can perform the integrations with respect to U, V, W. Then 

 changing to variables given by 



u = -»|r sin <fi sin s ; v = yjr cos </> sin s ; w = \jr cos s... (15), 



we can integrate with respect to (f> and s. 



And we are left with 



E=CN*p\ du yn+ie-^'-p-d-f 



Jo Jo 



= CN 2 /3%~ m du\ x 2m+1 e~^ ~& dx (16). 



