Text established from Boltzmann, L. (1872). "Weitere Studien über das Wärmegleichgewicht unter Gasmolekülen," Sitzungsberichte der kaiserlichen Akademie der Wissenschaften in Wien, Mathematisch-Naturwissenschaftliche Classe, LXVII, 275–370. Original pagination is preserved. Question marks mark lacunae in the manuscript left unresolved by Boltzmann. German technical terms are noted in brackets at first occurrence. Equation (17a), referenced below, appears on p. 113 of the original; the variational proof in the footnote supplies the missing demonstration.
From these equations it can again be proved that
must always decrease [immer abnehmen muss], unless $u_1^2 = u_1 u_3,\; u_2^2 = u_1 u_3,\; u_3^2 = u_2 u_4 \dots$ (that is, unless all expressions multiplied by the coefficients $B$ in equation (35)) vanish. Equations (35) have an inconvenient feature: while they can be written compactly with summation formulae, they resist being written out fully and explicitly. For clarity, we therefore begin with the simplest case, then move stepwise toward the general one.
First let $p = 3$; the molecules can have only three kinetic energies, $\epsilon$, $2\epsilon$, and $3\epsilon$. Then the system of equations (35) reduces to the following three equations:
and the expression for $E$ becomes
Differentiating gives
or, after rearranging,
The sum in parentheses vanishes by equations (36). Thus $dE/dt$ is obtained by multiplying the first equation by $\log u_1$, the second by $\log u_2$, the third by $\log u_3$, then adding. This yields
or
Consider the two factors multiplying $B_{11}^{22}$. When $u_2^2 > u_1 u_3$, the first factor is positive and the logarithm is negative. When $u_2^2 < u_1 u_3$, the first factor is negative and the logarithm is positive. In either case their product is always negative. Since $B_{11}^{22}$ is positive, $dE/dt$ is always negative or zero. Equality holds only when $u_2^2 = u_1 u_3$.
It is also easy to see that $E$ cannot run to minus infinity. None of $u_1,u_2,u_3$ may be negative or imaginary. For positive $u$, the quantity $u\log u$ cannot be smaller than $-1/e$. Hence $E$ cannot be smaller than
where $e$ is the base of natural logarithms. Therefore, since its derivative cannot be positive, $E$ must move steadily toward a minimum where $dE/dt = 0$, namely where $u_2^2 = u_1 u_3$.
The same proof does not carry over unchanged when $p > 3$. Here I treat only the case $p = 4$. In this case equations (35) reduce to:
p. 301The question marks in equation (37) preserve lacunae in Boltzmann's original manuscript (Sitzungsberichte, LXVII, 301). These terms were not resolved in 1872. The coefficient $B_{22}^{44}$ in the third line suggests the term should read $(u_4^2 - u_2^2)$ for detailed balance of 2–2 to 4–4 collisions, while the fourth line likely intends $(u_1u_4 - u_3^2)$. Boltzmann later corrected related points in his 1877 reply to Loschmidt's reversibility objection.
For $E$ one finds
Substituting the derivatives from equations (37), then rearranging terms, gives
The change in the order of summands required here is analogous to our earlier transformation of definite integrals. From the expression above, $dE/dt$ is again necessarily negative, unless the following hold simultaneously:
which can be written as
Likewise, in the general case, $dE/dt$ is necessarily negative, so $E$ must decrease unless
Since $E$ cannot be smaller than
it must approach a minimum for which equations (38) hold. Thus the system approaches the distribution of states determined by equations (38).
We now have to prove that equations (38) determine the distribution of states uniquely. If we add together all the equations (35), we obtain
hence
In a similar way one finds
where $a$ and $b$ are constants. The meaning is straightforward. The quantity $u_1 + \sqrt{2}u_2 + \sqrt{3}u_3 + \dots = a$ is the total number of molecules in unit volume [Gesamtmolekülzahl], while $b$ is their total kinetic energy [kinetische Energie]. Equations (40) and (41) therefore state that these two totals remain constant.
Assume $a$ and $b$ are given. Set the quotient $u_2/u_1$ equal to $y$. Equations (38) then become
Substituting these values into equations (40) and (41) yields
Because all the $u$’s are positive, $\frac{b}{\epsilon} - a$ must be positive, while $\frac{b}{\epsilon} - pa$ must be negative. Thus $b$ lies between $\epsilon a$ and $p\epsilon a$. In equation (42) the coefficient of $y^{p-1}$ is therefore positive, while the constant term is negative. The polynomial is positive for $y=\infty$ and negative for $y=0$, so it has at least one positive root. It has exactly one, because the sequence of coefficients changes sign only once. Negative or imaginary values of $y$ have no meaning here.
Given $y$, all the $u$’s are determined uniquely, and with them all the $w$’s. Hence, regardless of the initial distribution of states, there is one and only one distribution approached as time increases. This distribution depends only on the constants $a$ and $b$, the total number and total kinetic energy of the molecules, that is, the density and temperature of the gas.
p. 304This theorem was first proved only for the case where the distribution of states is initially uniform. It should also hold when this is not so, provided the molecules are distributed in such a way that mixing occurs over time, so the distribution becomes uniform after a long interval. This fails only in special cases, for example when molecules initially move in a straight line and are reflected back along that same line at the walls. Since we have established the result for arbitrary $p$ and $\epsilon$, we can pass to the case where $1/p$ and $\epsilon$ become infinitesimal.
† For very large $p$, the expression (39) becomes large, of order $p$. In that case one must seek a smaller negative value that $E$ can never exceed. The quantity denoted here by $E$ differs by a constant from the quantity earlier so denoted. To obtain the quantity denoted by $E_1$ in equation (17a), page 113[ref], which again differs only by a constant from the other quantities denoted by this letter, we must add to the present $E$ the term $-\frac{3\log\epsilon}{2}(u_1+\sqrt{2}u_2+\dots)$. Therefore
It is then clear that $E_1$ is a real and continuous function of the $u$’s for all real positive values. If we call a negative quantity “smaller” when its numerical value is larger, then $E$ is not smaller than expression (39), hence $E_1$ is not smaller than
Thus $E_1$ must have a minimum when the $u$’s range over all real positive values compatible with equations (40) and (41). One can further show that at this minimum none of the $u$’s can be zero. The minimum therefore cannot lie on the boundary of the region determined by the $u$’s, and it can be found by the usual rules of differential calculus.
If we add to the total differential of $E_1$ those of equations (40) and (41), multiplying the former by an undetermined multiplier $\lambda$ and the latter by $\mu$, we obtain
At the minimum, the coefficient of each differential must vanish. Eliminating $\lambda$ and $\mu$ gives
or $u_2/u_1 = u_3/u_2 = u_4/u_3 = \dots$, which is the same as equations (38). These equations therefore determine the smallest value $E_1$ can take, given the constraints (40) and (41). Since the $u$’s satisfy (40) and (41) throughout the process, this is the smallest value of $E_1$ attained during the process.
To compute it, set again $u_2 = u_1y,\; u_3 = u_1y^2,\dots$ Then equations (38), (40) and (41) yield a unique positive value of $y$, and this corresponds to the actual minimum of $E_1$. The minimum value is therefore
$E_1$ cannot be smaller than this value. It remains finite for infinitesimal $\epsilon$ and infinite $p$. Accounting for equations (43), it reduces to $a\log C - bh$, or, since $a = 2\int C$ and $b = 2/h$, one can write it as $\frac{1}{2}a(\log C - 1)$, which is finite. Hence $E_1$ cannot be minus infinity. On the other hand, it may be plus infinity. It remains to show that in that case thermal equilibrium cannot occur. That proof, together with an explicit discussion of the exceptional case where $\lim_{\tau\to0} \frac{\epsilon}{\tau}[\dots]$ depends on whether $\epsilon/\tau$ or $\tau/\epsilon$ vanishes, will not be pursued further here.
We have first:
For infinitesimal $\epsilon$ we can again set:
and obtain
which is again the Maxwell distribution [Maxwellsche Verteilung]. Likewise, one can verify that the sum here denoted by $E$ reduces, aside from an additive constant, to the integral in equation (17a). This method therefore reproduces the earlier results obtained by transformations of definite integrals, while being simpler to follow. One must accept, as a transitional abstraction [nur als Uebergangsstufe], that a molecule may take only a finite number of kinetic energies.
If one sets the time derivatives in equations (35) equal to zero, one obtains the condition that the distribution of states is stationary. This corresponds to setting $\frac{\partial f(x,t)}{\partial t} = 0$ in equation (16), yielding:
A solution is
which is the Maxwell distribution. From what has been shown above, there are infinitely many other formal solutions. They are not useful, however, because $f(x)$ becomes negative or imaginary for some values of $x$. This makes clear why Maxwell’s attempt to prove a priori that his solution is the only one must fail. It is not the only solution. It is the only one yielding purely positive probabilities, and therefore the only one of physical use.