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Next: Temperature Up: Statistical thermodynamics Previous: Introduction

Thermal interaction between macrosystems

Let us begin our investigation of statistical thermodynamics by examining a purely thermal interaction between two macroscopic systems, $A$ and $A'$, from a microscopic point of view. Suppose that the energies of these two systems are $E$ and $E'$, respectively. The external parameters are held fixed, so that $A$ and $A'$ cannot do work on one another. However, we assume that the systems are free to exchange heat energy (i.e., they are in thermal contact). It is convenient to divide the energy scale into small subdivisions of width $\delta E$. The number of microstates of $A$ consistent with a macrostate in which the energy lies in the range $E$ to $E+\delta E$ is denoted ${\mit\Omega}(E)$. Likewise, the number of microstates of $A'$ consistent with a macrostate in which the energy lies between $E'$ and $E'+\delta E$ is denoted ${\mit\Omega}'(E')$.

The combined system $A^{(0)} = A + A'$ is assumed to be isolated (i.e., it neither does work on nor exchanges heat with its surroundings). It follows from the first law of thermodynamics that the total energy $E^{(0)}$ is constant. When speaking of thermal contact between two distinct systems, we usually assume that the mutual interaction is sufficiently weak for the energies to be additive. Thus,

E + E' \simeq E^{(0)} = {\rm constant}.
\end{displaymath} (157)

Of course, in the limit of zero interaction the energies are strictly additive. However, a small residual interaction is always required to enable the two systems to exchange heat energy and, thereby, eventually reach thermal equilibrium (see Sect. 3.4). In fact, if the interaction between $A$ and $A'$ is too strong for the energies to be additive then it makes little sense to consider each system in isolation, since the presence of one system clearly strongly perturbs the other, and vice versa. In this case, the smallest system which can realistically be examined in isolation is $A^{(0)}$.

According to Eq. (157), if the energy of $A$ lies in the range $E$ to $E+\delta E$ then the energy of $A'$ must lie between $E^{(0)}- E -\delta E$ and $E^{(0)}- E$. Thus, the number of microstates accessible to each system is given by ${\mit\Omega}(E)$ and ${\mit\Omega}'(E^{(0)}-E)$, respectively. Since every possible state of $A$ can be combined with every possible state of $A'$ to form a distinct microstate, the total number of distinct states accessible to $A^{(0)}$ when the energy of $A$ lies in the range $E$ to $E+\delta E$ is

{\mit\Omega}^{(0)} = {\mit\Omega}(E)\, {\mit\Omega}'(E^{(0)} - E).
\end{displaymath} (158)

Consider an ensemble of pairs of thermally interacting systems, $A$ and $A'$, which are left undisturbed for many relaxation times so that they can attain thermal equilibrium. The principle of equal a priori probabilities is applicable to this situation (see Sect. 3). According to this principle, the probability of occurrence of a given macrostate is proportional to the number of accessible microstates, since all microstates are equally likely. Thus, the probability that the system $A$ has an energy lying in the range $E$ to $E+\delta E$ can be written

P(E) = C\, {\mit\Omega}(E) \,{\mit\Omega}'(E^{(0)} - E),
\end{displaymath} (159)

where $C$ is a constant which is independent of $E$.

We know, from Sect. 3.8, that the typical variation of the number of accessible states with energy is of the form

{\mit\Omega} \propto E^f,
\end{displaymath} (160)

where $f$ is the number of degrees of freedom. For a macroscopic system $f$ is an exceedingly large number. It follows that the probability $P(E)$ in Eq. (159) is the product of an extremely rapidly increasing function of $E$ and an extremely rapidly decreasing function of $E$. Hence, we would expect the probability to exhibit a very pronounced maximum at some particular value of the energy.

Let us Taylor expand the logarithm of $P(E)$ in the vicinity of its maximum value, which is assumed to occur at $E= \tilde{E}$. We expand the relatively slowly varying logarithm, rather than the function itself, because the latter varies so rapidly with the energy that the radius of convergence of its Taylor expansion is too small for this expansion to be of any practical use. The expansion of $\ln {\mit\Omega}(E)$ yields

\ln {\mit\Omega}(E) = \ln{\mit\Omega}(\tilde{E}) + \beta(\tilde{E})\,\eta -
\end{displaymath} (161)

$\displaystyle \eta$ $\textstyle =$ $\displaystyle E - \tilde{E},$ (162)
$\displaystyle \beta$ $\textstyle =$ $\displaystyle \frac{\partial \ln {\mit\Omega}}{\partial E},$ (163)
$\displaystyle \lambda$ $\textstyle =$ $\displaystyle -\frac{\partial ^2 \ln {\mit\Omega}}{\partial E^2}
= - \frac{\partial\beta}
{\partial E}.$ (164)

Now, since $E' = E^{(0)}-E$, we have
E' - \tilde{E}' = - (E-\tilde{E}) = -\eta.
\end{displaymath} (165)

It follows that
\ln{\mit\Omega}'(E') = \ln{\mit\Omega}'(\tilde{E}')+\beta'(\...
\end{displaymath} (166)

where $\beta'$ and $\lambda'$ are defined in an analogous manner to the parameters $\beta$ and $\lambda$. Equations (161) and (166) can be combined to give
\ln\,[{\mit\Omega}(E)\,{\mit\Omega}'(E')] = \ln\,[{\mit\Omeg...
\end{displaymath} (167)

At the maximum of $\ln\,[{\mit\Omega}(E) \,{\mit\Omega}'(E')]$ the linear term in the Taylor expansion must vanish, so
\beta (\tilde{E}) = \beta'(\tilde{E}'),
\end{displaymath} (168)

which enables us to determine $\tilde{E}$. It follows that
\ln P(E) = \ln P(\tilde{E}) - \frac{1}{2}\,\lambda_0 \,\eta^2,
\end{displaymath} (169)

P(E)=P(\tilde{E}) \exp\left[-\frac{1}{2}\,\lambda_0\, (E-\tilde{E})^2\right],
\end{displaymath} (170)

\lambda_0 = \lambda (\tilde{E})+\lambda'(\tilde{E}').
\end{displaymath} (171)

Now, the parameter $\lambda_0$ must be positive, otherwise the probability $P(E)$ does not exhibit a pronounced maximum value: i.e., the combined system $A^{(0)}$ does not possess a well-defined equilibrium state as, physically, we know it must. It is clear that $\lambda(\tilde{E})$ must also be positive, since we could always choose for $A'$ a system with a negligible contribution to $\lambda_0$, in which case the constraint $\lambda_0>0$ would effectively correspond to $\lambda(\tilde{E})>0$. [A similar argument can be used to show that $\lambda'(\tilde{E}')$ must be positive.] The same conclusion also follows from the estimate ${\mit\Omega}\propto E^f$, which implies that
\lambda(\tilde{E}) \sim \frac{f}{\tilde{E}^2}>0.
\end{displaymath} (172)

According to Eq. (170), the probability distribution function $P(E)$ is a Gaussian. This is hardly surprising, since the central limit theorem ensures that the probability distribution for any macroscopic variable, such as $E$, is Gaussian in nature (see Sect. 2.10). It follows that the mean value of $E$ corresponds to the situation of maximum probability (i.e., the peak of the Gaussian curve), so that

\bar{E} = \tilde{E}.
\end{displaymath} (173)

The standard deviation of the distribution is
{\mit\Delta}^\ast E = \lambda_0^{-1/2}\sim \frac{\bar{E}}{\sqrt{f}},
\end{displaymath} (174)

where use has been made of Eq. (160) (assuming that system $A$ makes the dominant contribution to $\lambda_0$). It follows that the fractional width of the probability distribution function is given by
\frac{{\mit\Delta}^\ast E}{\bar{E}}\sim \frac{1}{\sqrt{f}}.
\end{displaymath} (175)

Hence, if $A$ contains 1 mole of particles then $f\sim N_A\simeq 10^{24}$ and ${\mit\Delta}^\ast E /\bar{E}\sim 10^{-12}$. Clearly, the probability distribution for $E$ has an exceedingly sharp maximum. Experimental measurements of this energy will almost always yield the mean value, and the underlying statistical nature of the distribution may not be apparent.

next up previous
Next: Temperature Up: Statistical thermodynamics Previous: Introduction
Richard Fitzpatrick 2006-02-02