102 



with regard to the averages in a canonic ensemble become the 

 right ones. 



du du 



Instead of Ku =■ we can also write in ii =z or u - = 0. 



(It at 



If now we multiply tiie Einstkin-Hopf erination by u and work out 

 the case average, we get 



du 



u — = ti u^ 4- u F 

 dt ' ^ 



Further we shall demonstrate that for sufficiently long periods 

 the second member is identically zero. 



For according to (5) we have got for the first term -— — (3 or — ;r- *» e 



shall now determine the second average For this purpose we multiply 

 (2) by F and work out the average, and arrive at 



uF = u^ e-Pt F -\- e-^t f \e^< F {t) dt 



f 







In this formula the first average in the second member is zero. 

 To determine the second average we must consider that F for the 

 integral refers to a definite time t. And so only those parts of the 

 integral where the argument differs only slightly from ^yield distributions 

 to the average. In the exponent we can again take the argument 

 equivalent to t, so that we can write for the second term 



t 



('F{t)F{t-n)dri or f] 



{t)F{t-ri)dn or \F{%)F{%-ri)dri 



t Yl — 00 



Now this integral is just half of the integral of (4«), as it covers 

 half the region of integration, whilst the integrand is of course 

 symmetrically with regard to §. 



And so in the end w^e find for the second member identically zero 

 as the two averages first neutralize each other. If now we take (5) 

 into consideration for shorter periods the second member becomes 



« »g-2/3t g-2/3t 



This result is also obtained by differentiating (4) with respect to 

 the time. 



du . • I /. 



The case average of u — for finite times not large with reference 



dt 



to - is consequently not strictly zero. Now we can however determine 



