52 
QUANTUM THEORY 
The same law can be deduced for the distribution of velocities of atoms 
and molecules. The argument is the same if there is any process of dis 
sociation and combination of atoms analogous to the ionisation and capture 
of electrons, provided that radiation is involved. It is not necessary that 
this process should play an important part in distributing velocities ; the 
argument from Law I is that the distribution of velocities however con 
trolled must be such that this process will not under any circumstances 
disturb it. Alternatively we can proceed as follows. In any assemblage 
there will be some free electrons. Let n r , n s be the numbers of electrons 
with kinetic energies y r , y s ; and n/, n/ be the numbers of atoms with 
these kinetic energies. Let a r$ be the probability that 1 marked atom in 
a cu. cm. with energy y r meets 1 marked electron in a cu. cm. with energy 
Xs and that the two energies are interchanged. Then balancing direct and 
reverse processes in equilibrium 
a rs n r 'n s = a sr n'n r 
so that as far as the factor involving temperature is concerned 
= n r ln s . 
Again nothing is discovered as to the weight factor. 
The deduction of Maxwell’s Law from Einstein’s equation indicates 
that radiative processes alone would drive an assemblage to take up the 
Maxwellian distribution of velocities apart from the collisions investigated 
in the usual proofs. It should be stated, however, that the great length 
and difficulty (and perhaps imperfect rigour) of the usual proofs arises 
in connection with the weight factor dudvdw, which is not considered here. 
40. If we prefer not to make the assumption leading to (37-3) the 
factor 
_ C hl a i2 ^23 
123 ~ a a a ’ 
w-13 «-21 M 32 
must be inserted on the right of (37-4). The proof that c 23 is a definite 
natural constant C proceeds as before except that four equations corre 
sponding to four different temperatures must be used to eliminate the 
other unknowns. 
It then follows by taking T infinite that 
ffi.23 = 1 + G If (0), 
so that instead of (37*8) we have 
(1 + Gif (a)} {1 + Gif m = {1 + C/f (0)} {1 + Gif (a + /?)}. 
And the solution is 
U + G If {a)} = ae ka ,