Boltzmann: Thermal Equilibrium of Gas Molecules

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From these equations it can again be proved that

$$E = u_1 \log u_1 + \sqrt{2}\,u_2 \log u_2 + \dots + \sqrt{p}\,u_p \log u_p$$

must always decrease unless $u_1^2 = u_1 u_3,\; u_2^2 = u_1 u_3,\; u_3^2 = u_2 u_4 \dots$ (in other words all the expressions multiplied by the coefficients $B$ in equation (35)) vanish. Equations (35) have the inconvenient feature that while they can be written with summation formulae they cannot be written out completely explicitly. It would undoubtedly be an aid to clarity, therefore, if we begin with the simplest case and then proceed gradually to the general case.

First let $p = 3$; the molecules can only have three different kinetic energies, $\epsilon$, $2\epsilon$, and $3\epsilon$. Then the system of equations (35) reduces to the following three equations:

$$ \begin{aligned} \sqrt{1}\,\frac{du_1}{dt} &= B_{11}^{22}\,(u_2^2 - u_1 u_3) \\ \sqrt{2}\,\frac{du_2}{dt} &= 2B_{11}^{22}\,(u_1 u_3 - u_2^2) \\ \sqrt{3}\,\frac{du_3}{dt} &= B_{11}^{22}\,(u_1 u_3 - u_2^2) \end{aligned} \tag{36} $$

and the expression for $E$ reduces to

$$E = u_1 \log u_1 + \sqrt{2}\,u_2 \log u_2 + \sqrt{3}\,u_3 \log u_3$$

The differentiation gives

$$\frac{dE}{dt} = (\log u_1 + 1)\frac{du_1}{dt} + \sqrt{2}(\log u_2 + 1)\frac{du_2}{dt} + \sqrt{3}(\log u_3 + 1)\frac{du_3}{dt}$$
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or, with a different arrangement of the terms,

$$\frac{dE}{dt} = \log u_1 \frac{du_1}{dt} + \sqrt{2}\log u_2 \frac{du_2}{dt} + \sqrt{3}\log u_3 \frac{du_3}{dt} + \left( \frac{du_1}{dt} + \sqrt{2}\frac{du_2}{dt} + \sqrt{3}\frac{du_3}{dt} \right)$$

The sum of the last three terms vanishes according to equations (36) so that one obtains $dE/dt$ by multiplying the first of these equations by $\log u_1$, the second by $\log u_2$, and the third by $\log u_3$, and adding all three together. If one does this he obtains

$$\frac{dE}{dt} = B_{11}^{22}\,(u_2^2 - u_1 u_3)\bigl(\log u_1 + \log u_3 - 2\log u_2\bigr)$$

or

$$\frac{dE}{dt} = B_{11}^{22}\,(u_2^2 - u_1 u_3)\,\log\!\left(\frac{u_1 u_3}{u_2^2}\right)$$

Of the two factors multiplying $B_{11}^{22}$ on the right-hand side of this equation, the first is positive and the second is negative when $u_2^2 > u_1 u_3$, whereas the first is negative and the second is positive when $u_2^2 < u_1 u_3$. Hence their product is always negative, and since $B_{11}^{22}$ must be positive, $dE/dt$ is always negative or zero. The latter is true when $u_2^2 = u_1 u_3$. Now it can easily be shown that $E$ cannot become negatively infinite. Obviously none of the three quantities $u_1, u_2, u_3$ may be negative or imaginary. For positive $u$, however, $u\log u$ cannot have a larger negative value than $-1/e$, hence $E$ cannot have a larger negative value than

$$-\frac{1 + \sqrt{2} + \sqrt{3}}{e}$$

where $e$ is the base of natural logarithms.

Therefore $E$, since its derivative cannot be positive, must continually approach a minimum for which $dE/dt = 0$, and for which $u_2^2 = u_1 u_3$.

The proof cannot be carried out in just the same way when $p > 3$. I consider here only the case $p = 4$. In this case equations (35) reduce to

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$$ \begin{aligned} \sqrt{1}\frac{du_1}{dt} &= B_{11}^{22}(u_2^2 - u_1u_3) + B_{11}^{33}(u_3^2 - u_1u_4) \\ \sqrt{2}\frac{du_2}{dt} &= B_{11}^{22}(u_1u_3 - u_2^2) + (B_{12}^{23}+B_{11}^{24})(u_1u_4 - u_2u_3) \\ \sqrt{3}\frac{du_3}{dt} &= B_{11}^{33}(u_1u_4 - u_3^2) + (B_{12}^{23}+B_{11}^{24})(u_2u_3 - u_1u_4) + 2B_{22}^{44}(u_4^2 - u_2^2) \\ \sqrt{4}\frac{du_4}{dt} &= (B_{12}^{23}+B_{11}^{24})(u_2u_3 - u_1u_4) + B_{11}^{33}(u_1u_4 - u_2^2) \end{aligned} \tag{37} $$

For $E$ one finds

$$E = u_1\log u_1 + \sqrt{2}u_2\log u_2 + \sqrt{3}u_3\log u_3 + \sqrt{4}u_4\log u_4$$ $$\frac{dE}{dt} = \log u_1\frac{du_1}{dt} + \sqrt{2}\log u_2\frac{du_2}{dt} + \sqrt{3}\log u_3\frac{du_3}{dt} + \sqrt{4}\log u_4\frac{du_4}{dt}$$

If one substitutes here for $\frac{du_1}{dt},\frac{du_2}{dt},\frac{du_3}{dt},\frac{du_4}{dt}$ their values from equations (37), he obtains, with a suitable rearrangement of terms,

$$ \begin{aligned} \frac{dE}{dt} = B_{11}^{22}(u_2^2-u_1u_3)\log\frac{u_1u_3}{u_2^2} &+ B_{11}^{33}(u_3^2-u_1u_4)\log\frac{u_1u_4}{u_3^2} \\ &+ (B_{12}^{23}+B_{11}^{24})(u_1u_4 - u_2u_3)\log\frac{u_2u_3}{u_1u_4} \end{aligned} $$
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I remark that the change in the order of the summands, which is necessary here, is analogous to our previous transformation of definite integrals. From the above expression one sees at once that $dE/dt$ is again necessarily negative, unless simultaneously we have

$$u_2^2 = u_1u_3,\qquad u_3^2 = u_2u_4,\qquad u_2u_3 = u_1u_4$$

which can also be written

$$u_3 = \frac{u_2^2}{u_1},\qquad u_4 = \frac{u_2^3}{u_1^2}.$$

Likewise one finds in the general case that $dE/dt$ is necessarily negative so that $E$ must decrease unless

$$u_3 = \frac{u_2^2}{u_1},\quad u_4 = \frac{u_2^3}{u_1^2},\quad\ldots\tag{38}$$

Since $E$ cannot have a larger negative value than

$$-\frac{1+\sqrt{2}+\sqrt{3}+\dots+\sqrt{p}}{e}\tag{39}$$

it must necessarily approach a minimum value for which equations (38) hold. Thus it continually approaches the distribution of states determined by equations (38).

We now have to prove that equations (38) uniquely determine the distribution of states. If we add together all the equations (35), we obtain

$$\frac{d}{dt}\bigl(u_1 + \sqrt{2}u_2 + \sqrt{3}u_3 + \dots + \sqrt{p}u_p\bigr) = 0$$

hence

$$u_1 + \sqrt{2}u_2 + \sqrt{3}u_3 + \dots + \sqrt{p}u_p = a \tag{40}$$
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In a similar way we find that

$$u_1 + 2\sqrt{2}u_2 + 3\sqrt{3}u_3 + \dots + p\sqrt{p}u_p = \frac{b}{\epsilon} \tag{41}$$

where $a$ and $b$ are constants. The meaning of these equations is obvious. In particular,

$$u_1 + \sqrt{2}u_2 + \sqrt{3}u_3 + \dots = a$$

is the total number of molecules in unit volume, while $b$ is their total kinetic energy. Equations (40) and (41) therefore tell us that these two quantities are constant.

Suppose that the two quantities $a$ and $b$ are given. Then we set the quotient $u_2/u_1$ equal to $y$. Equations (38) then reduce to

$$u_3 = y^2 u_1,\quad u_4 = y^3 u_1,\quad\ldots,\quad u_p = y^{p-1}u_1$$

If one substitutes these values into equations (40) and (41), then he finds easily

$$ \begin{aligned} 0 &=\; (p-1)\frac{b}{\epsilon} - pa\; y^{p-1} + \bigl((p-2)\frac{b}{\epsilon} - (p-1)a\bigr)y^{p-2} + \dots \\ &\quad + \bigl(3a-\frac{b}{\epsilon}\bigr)3y^2 + \bigl(2a-\frac{b}{\epsilon}\bigr)2y + \bigl(a-\frac{b}{\epsilon}\bigr) \tag{42} \end{aligned} $$

Since all the $u$'s are necessarily positive, we see immediately that $\frac{b}{\epsilon} - a$ must be positive while $\frac{b}{\epsilon} - pa$ must be negative. Hence $b$ must lie between $\epsilon a$ and $p\epsilon a$. Hence, in equation (42) the coefficient of $y^{p-1}$ is positive, while the term independent of $y$ must be negative. The polynomial is therefore positive for $y = \infty$, and negative for $y = 0$; therefore there is one and only one positive root for $y$, since the series of coefficients changes sign only once. Negative or imaginary values for $y$ are of course meaningless. But from $y$ we can determine uniquely all the $u$'s and also all the $w$'s. Hence, whatever may be the initial distribution of states, there is one and only one distribution which it approaches with increasing time. This distribution depends only on the constants $a$ and $b$, the total number and total kinetic energy of the molecules (density and temperature of the gas).

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This theorem was proved first only for the case that the distribution of states is initially uniform. It must also hold, however, when this is not true, provided only that the molecules are distributed in such a way that they tend to become mixed as time progresses, so that the distribution becomes uniform after a very long time. This will always happen with the exception of certain special cases, for example, when the molecules move initially in a straight line and are reflected back in this straight line at the walls. Since we have established this for arbitrary $p$ and $\epsilon$, we can immediately go to the case where $\frac{1}{p}$ and $\epsilon$ become infinitesimal.

† Footnote: For very large $p$, the expression (39) will be very large, of order $p$. In this case it is necessary to look for a smaller negative value that $E$ can never exceed. The quantity denoted here by $E$ differs by a constant from the one earlier so denoted. If we wish to obtain the quantity denoted by $E_1$ in equation (17a), page 113, then we must add to our present $E$, $-\frac{3\log\epsilon}{2}(u_1+\sqrt{2}u_2+\dots)$. Therefore: $$E_1 = u_1\log\frac{u_1}{\epsilon^{3/2}} + \sqrt{2}u_2\log\frac{u_2}{\epsilon^{3/2}} + \dots$$ It is clear now that $E_1$ is a real and continuous function of the $u$'s for all real positive values. $E_1$ is not smaller than $-\frac{1}{e}(1+\sqrt{2}+\dots+\sqrt{p}) - \frac{3}{2}a\log\epsilon$. Hence, $E_1$ must have a minimum... At the minimum, the factor of each differential must vanish: $\log u_2 - \log u_1 = \log u_3 - \log u_2 = \dots$ which leads to the same equations (38). This minimum value of $E_1$ is $a\log C - bh$, which is a finite quantity.
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We have first:

$$w_k = \sqrt{k}u_k = u_1\sqrt{k}y^{k-1}$$

For infinitesimal $\epsilon$ we can again set

$$\epsilon = dx,\quad k\epsilon = x,\quad y = e^{-h\epsilon},\quad \frac{1}{\sqrt{\epsilon}} = C \tag{43}$$ $$y^k = e^{-hk\epsilon} = e^{-hx}$$

and obtain

$$w_k = C\,\sqrt{x}\,e^{-hx}\,dx$$

which is again the Maxwell distribution. Likewise one can convince himself that the sum which we have here denoted by $E$ reduces, aside from a constant additive term, to the integral in equation (17a).

If one sets the time derivatives in equations (35) equal to zero, he obtains the conditions that the distribution of states does not change with time but is stationary. The condition that the distribution be stationary is obtained by setting

$$\frac{\partial f(x,t)}{\partial t} = 0$$

in equation (16). This gives:

$$ 0 = \int_0^\infty \int_0^{x+x'} \Big[ f(\xi)f(x+x'-\xi) - f(x)f(x') \Big] \sqrt{\frac{xx'}{\xi(x+x'-\xi)}} \; \psi(x,x',\xi) \; d\xi \, dx' $$
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A solution of this equation is

$$f(x) = C\sqrt{x}\,e^{-hx}$$

which is the Maxwell distribution. From what has been said previously it follows that there are infinitely many other solutions, which are not useful however since $f(x)$ comes out negative or imaginary for some values of $x$. Hence, it follows very clearly that Maxwell's attempt to prove a priori that his solution is the only one must fail, since it is the only one that gives purely positive probabilities.

The Kinetic Theory of Gases — Boltzmann