Page:Elementary Principles in Statistical Mechanics (1902).djvu/159

It appears from the definitions of $\eta _{1}$ and $\eta _{2}$ that (436) may also be written

${\begin{aligned}\int \ldots \int \eta _{1}e^{\eta }\,dp_{1}\ldots dq_{n}+&\int \ldots \int \eta _{2}e^{\eta }\,dp_{1}\ldots dq_{n}\leq \\&\qquad \int \ldots \int \eta e^{\eta }\,dp_{1}\ldots dq_{n},\end{aligned}}$

(440)

or

$\int \ldots \int (\eta -\eta _{1}-\eta _{2})e^{\eta }\,dp_{1}\ldots dq_{n}\geq 0,$

where the integrations cover all phases. Adding the equation

$\int \ldots \int e^{\eta _{1}+\eta _{2}}\,dp_{1}\ldots dq_{n}=1,$

(441)

which we get by multiplying (438) and (439), and subtracting (437), we have for the proposition to be proved

$\mathop {\int \ldots \int } _{\rm {phases}}^{\rm {all}}[(\eta -\eta _{1}-\eta _{2})e^{\eta }+e^{\eta _{1}+\eta _{2}}-e^{\eta }]\,dp_{1}\ldots dq_{n}\geq 0.$

(442)

Let

$u=\eta -\eta _{1}-\eta _{2}.$

(443)

The main proposition to be proved may be written

$\int \ldots \int (ue^{u}+1-e^{u})e^{\eta _{1}+\eta _{2}}\,dp_{1}\ldots dq_{n}\geq 0.$

(444)

This is evidently true since the quantity in the parenthesis is incapable of a negative value.^[1] Moreover the sign $=$ can hold only when the quantity in the parenthesis vanishes for all phases, i. e., when $u=0$ for all phases. This makes $\eta =\eta _{1}+\eta _{2}$ for all phases, which is the analytical condition which expresses that the distributions in phase of the two parts of the system are independent. Theorem VIII. If two or more ensembles of systems which are identical in nature, but may be distributed differently in phase, are united to form a single ensemble, so that the probability-coefficient of the resulting ensemble is a linear function

1. See Theorem I, where this is proved of a similar expression.