Mechanical interaction between macrosystems

Let us now examine a purely mechanical interaction between macrostates, where one or more of the external parameters is modified, but there is no exchange of heat energy. Consider, for the sake of simplicity, a situation where only one external parameter $x$ of the system is free to vary. In general, the number of microstates accessible to the system when the overall energy lies between $E$ and $E+\delta E$ depends on the particular value of $x$, so we can write ${\mit\Omega} \equiv {\mit\Omega}(E, x)$.

When $x$ is changed by the amount $dx$, the energy $E_r(x)$ of a given microstate $r$ changes by $(\partial E_r/\partial x)\, dx$. The number of states $\sigma(E,x)$ whose energy is changed from a value less than $E$ to a value greater than $E$ when the parameter changes from $x$ to $x+dx$ is given by the number of microstates per unit energy range multiplied by the average shift in energy of the microstates. Hence,

$$\sigma (E,x) = \frac{{\mit\Omega}(E,x)}{\delta E}\, \overline{\frac{\partial E_r}{\partial x}}\, \,dx,$$ (190)

where the mean value of $\partial E_r/\partial x$ is taken over all accessible microstates (i.e., all states where the energy lies between $E$ and $E+\delta E$ and the external parameter takes the value $x$). The above equation can also be written

$$\sigma (E,x) = - \frac{{\mit\Omega}(E,x)}{\delta E}\, \bar{X}\,\,dx,$$ (191)

where

$$\bar{X}(E,x) = -\overline{\frac{\partial E_r}{\partial x}}$$ (192)

is the mean generalized force conjugate to the external parameter $x$ (see Sect. 4.4).

Consider the total number of microstates between $E$ and $E+\delta E$. When the external parameter changes from $x$ to $x+dx$, the number of states in this energy range changes by $(\partial {\mit\Omega}/\partial x)\,dx$. This change is due to the difference between the number of states which enter the range because their energy is changed from a value less than $E$ to one greater than $E$ and the number which leave because their energy is changed from a value less than $E+\delta E$ to one greater than $E+\delta E$. In symbols,

$$\frac{\partial {\mit\Omega}(E,x)}{\partial x}\,dx = \sigma(E... ...\delta E)\simeq -\frac{\partial \sigma}{\partial E}\,\delta E,$$ (193)

which yields

$$\frac{\partial {\mit\Omega}}{\partial x} = \frac{\partial({\mit\Omega} \bar{X})}{\partial E},$$ (194)

where use has been made of Eq. (191). Dividing both sides by ${\mit\Omega}$ gives

$$\frac{\partial \ln {\mit\Omega}}{\partial x} = \frac{\partia... ...}}{\partial E}\, \bar{X} +\frac{\partial \bar{X}}{\partial E}.$$ (195)

However, according to the usual estimate ${\mit\Omega}\propto E^f$, the first term on the right-hand side is of order $(f/\bar{E})\,\bar{X}$, whereas the second term is only of order $\bar{X}/\bar{E}$. Clearly, for a macroscopic system with many degrees of freedom, the second term is utterly negligible, so we have

$$\frac{\partial\ln {\mit\Omega}}{\partial x} = \frac{\partial \ln {\mit\Omega}}{\partial E} \,\bar{X} = \beta \,\bar{X}.$$ (196)

When there are several external parameters $x_1, \cdots, x_n$, so that ${\mit\Omega}\equiv {\mit\Omega}(E, x_1,\cdots,$ $x_n)$, the above derivation is valid for each parameter taken in isolation. Thus,

$$\frac{\partial \ln{\mit\Omega}}{\partial x_\alpha} = \beta\, \bar{X}_\alpha,$$ (197)

where $\bar{X}_\alpha$ is the mean generalized force conjugate to the parameter $x_\alpha$.