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 ,