D O C . 4 2 7 Q U A N T U M T H E O R Y O F I D E A L G A S 4 2 5
(19)
If function ψ is given, then the right-hand side of the equation can be calculated for
any value of χ. By inversion, one hence also obtains χ as a function of the right-
hand side. Thus the problem is, in fact, reduced to the question of function ψ.
§7. Relation of These Results to the Classical Theory as Well as to the Quan-
tum Theory of the Ideal Gas Given by Me
We shall examine the case in which constant h comes out of the distribution law.
To abbreviate, we set
From (1) and (18) one recognizes that h comes out of the expression for dn when
and only when is independent of u. In this case, let us call this function .
Then, with a suitable choice of the function φ, an equation of the form
(20)
must be valid. If one takes the logarithm of this equation and differentiates it twice
(for u and v), one easily realizes that lgψ must be a linear function. φ then also eas-
ily results. It emerges that ψ must, in fact, be the exponential function (Maxwell
velocity distribution). —
For the classical theory, the corresponding statement is
(21)
for the statistical theory developed by me, the statement is
(22)
Thus, Planck’s function takes the place of the exponential function with a negative
exponent.3)
I have shown in a recently published paper that, unlike (21), statement
(22) sufficiently satisfies Nernst’s
theorem.[10]
Two goals have been achieved by the present analysis. First, a general condition
(equation (18)) has been found that must satisfy any theory of the ideal gas. Second,
from the above it is evident that the equation of state derived by me is not disturbed
by adiabatic compression or by conservative force fields.
3)
This easily follows from (18), (20), and (21) of the above-cited treatise.
ψ(x χ)x1 + /2 xd
x= 0
x=∞
³
Nh3
2π( 2mκT)3 /2V
-------------------------------------. =
u
h3N
mκT)3 ( /2V
-------------------------- -, = v
L
κT
------. =
1
u
--ψ - ψ(v)
ψ( v φ(u)) + uψ(v) =
[p. 25]
ψ(v) e–v, =
ψ(v)
1
ev 1 –
-------------. =