Page:Elementary Principles in Statistical Mechanics (1902).djvu/113

In like manner, let us denote by $V_{q}$ the extension-in-configuration below a certain limit of potential energy which we may call $\epsilon _{q}$. That is, let

$V_{q}=\int \ldots \int \Delta _{\dot {q}}^{\frac {1}{2}}dq_{1}\ldots dq_{n},$

(271)

the integration being extended (with constant values of the external coördinates) over all configurations for which the potential energy is less than $\epsilon _{q}$. $V_{q}$ will be a function of $\epsilon _{q}$ with the external coördinates, an increasing function of $\epsilon _{q}$, which does not become infinite (in such cases as we shall consider^[1]) for any finite value of $\epsilon _{q}$. It vanishes for the least possible value of $\epsilon _{q}$, or for $\epsilon _{q}=-\infty$, if $\epsilon _{q}$ can be diminished without limit. It is not always a continuous function of $\epsilon _{q}$. In fact, if there is a finite extension-in-configuration of constant potential energy, the corresponding value of $V_{q}$ will not include that extension-in-configuration, but if $\epsilon _{q}$ be increased infinitesimally, the corresponding value of $V_{q}$ will be increased by that finite extension-in-configuration.

Let us also set

$\phi _{q}=\log {\frac {dV_{q}}{d\epsilon _{q}}}.$

(272)

The extension-in-configuration between any two limits of potential energy $\epsilon _{q}'$ and $\epsilon _{q}''$ may be represented by the integral

$\int _{\epsilon _{q}'}^{\epsilon _{q}''}e^{\phi _{q}}\,d\epsilon _{q}$

(273)

whenever there is no discontinuity in the value of $V_{q}$ as function of $\epsilon _{q}$ between or at those limits, that is, whenever there is no finite extension-in-configuration of constant potential energy between or at the limits. And in general, with the restriction mentioned, we may substitute $e^{\phi _{q}}\,d\epsilon _{q}$ for $\Delta _{\dot {q}}dq_{1}\ldots dq_{n},$ in an $n$-fold integral, reducing it to a simple integral, when the limits are expressed by the potential energy, and the other factor under the integral sign is a function of

1. If $V_{q}$ were infinite for finite values of $\epsilon _{q}$, $V$ would evidently be infinite for finite values of $\epsilon$.