1329 - 
If we now represent the number of molecules simultaneously 
exerting forces on the suspended particle, by n’, we find: 
i 0 St, u Oe 
) keen Ds —_ . ETET 
he Hieron. M de 
a it ELON: je 
PU Oe 
Me Onis 
For q we find on an average the value zero. As middle value 
' 
Vt 
of the square of q we find: 
so that 
! 
: tas 
pu nu nm 
For 7’ we shall now introduce the hypothesis that it is equal to 
the number of molecules contained in a spherical layer which sur- 
rounds the Brownian particle and has the thickness of one molecular 
diameter o. [f we think that in a liquid the molecules almost touch 
each other, this means that only the inner layer of liquid molecules 
will exert forces on the particle. Hence if we write for the number 
of molecules per cub. cm., we put: n’ =427a’ on. If a—=1000, 
q 
peur 
hence of the order 10~° em, then —= +100, if a=1000 o this 
ratio becomes + 1000, at least if the density of the suspended 
particle and of the medium is almost equal. 
: ats NE 
Now that we have thus found an expression for =, we shall 
p u 
CEL sat ale 
also try to find one for ——. For this purpose we point out that q 
pu 
consists of terms which all contain a factor w’,v’ or aw’. After col- 
lision of the molecule the quantities w 
/ 
,v’, and w’ have all assumed 
quite new values. If we denote by +t the mean time between two 
collisions for a molecule of the medium, then 72’ molecuies will on 
collisions in the time rt. We shall 
now assume that-in the time rt the quantity g on an average reverses 
its sign once: or at least assumes an entirely new value, independ- 
ent of the value of the beginning of the time 1. If we choose the 
/ 
an average have undergone 7 
before introduced integration interval & now equal to rt, we may 
put (? = q? 7° by approximation. 
