Page:Elementary Principles in Statistical Mechanics (1902).djvu/161

But since

$\Sigma (c_{1}P_{1})=P$

and

$\Sigma c_{1}=1,$

$\Sigma (c_{1}Q_{1})=\Sigma (c_{1}P_{1}\log P_{1})-P\log P.$

(450)

This proves (448), and shows that the sign $=$ will hold only when

$P_{1}=P,\qquad P_{2}=P,\qquad \mathrm {etc.}$

for all phases, i. e., only when the distribution in phase of the original ensembles are all identical.

Theorem IX. A uniform distribution of a given number of systems within given limits of phase gives a less average index of probability of phase than any other distribution.

Let $\eta$ be the constant index of the uniform distribution, and $\eta +\Delta \eta$ the index of some other distribution. Since the number of systems within the given limits is the same in the two distributions we have

$\int \ldots \int e^{\eta +\Delta \eta }\,dp_{1}\ldots dq_{n}=\int \ldots \int e^{\eta }\,dp_{1}\ldots dq_{n},$

(451)

where the integrations, like those which follow, are to be taken within the given limits. The proposition to be proved may be written

$\int \ldots \int (\eta +\Delta \eta )e^{\eta +\Delta \eta }\,dp_{1}\ldots dq_{n}>\int \ldots \int \eta e^{\eta }\,dp_{1}\ldots dq_{n},$

(452)

or, since $\eta$ is constant,

$\int \ldots \int (\eta +\Delta \eta )e^{\Delta \eta }\,dp_{1}\ldots dq_{n}>\int \ldots \int \eta \,dp_{1}\ldots dq_{n}.$

(453)

In (451) also we may cancel the constant factor $e^{\eta }$, and multiply by the constant factor $(\eta +1)$. This gives

$\int \ldots \int (\eta +1)e^{\Delta \eta }\,dp_{1}\ldots dq_{n}=\int \ldots \int (\eta +1)\,dp_{1}\ldots dq_{n}.$

The subtraction of this equation will not alter the inequality to be proved, which may therefore be written

$\int \ldots \int (\Delta \eta -1)e^{\Delta \eta }\,dp_{1}\ldots dq_{n}>\int \ldots \int -\,dp_{1}\ldots dq_{n}$