4.3: Thermal Equilibrium

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Two Systems in Thermal Contact

Consider two systems in thermal contact, as depicted in Figure $\PageIndex{1}$. The two subsystems #1 and #2 are free to exchange energy, but their respective volumes and particle numbers remain fixed. We assume the contact is made over a surface, and that the energy associated with that surface is negligible when compared with the bulk energies $E\ns_1$ and $E\ns_2$. Let the total energy be $E=E\ns_1+E\ns_2$. Then the density of states $D(E)$ for the combined system is

$$D(E)=\int\!\!dE\ns_1\,D\ns_1(E\ns_1)\,D\ns_2(E-E\ns_1)\ .$$

The probability density for system #1 to have energy $E\ns_1$ is then

$$P\ns_1(E\ns_1)={D\ns_1(E\ns_1)\,D\ns_2(E-E\ns_1)\over D(E)}\ .$$

Note that $P\ns_1(E\ns_1)$ is normalized: $\int\!dE\ns_1 \,P\ns_1(E\ns_1)=1$. We now ask: what is the most probable value of $E\ns_1$? We find out by differentiating $P\ns_1(E\ns_1)$ with respect to $E\ns_1$ and setting the result to zero. This requires

$$\begin{split} 0&={1\over P\ns_1(E\ns_1)}\,{dP\ns_1(E\ns_1)\over dE\ns_1} = {\pz\over \pz E\ns_1}\,\ln P\ns_1(E\ns_1)\\ &={\pz\over \pz E\ns_1}\,\ln D\ns_1(E\ns_1) + {\pz\over \pz E\ns_1}\,\ln D\ns_2(E-E\ns_1)\ . \label{maxprob} \end{split}$$

We conclude that the maximally likely partition of energy between systems #1 and #2 is realized when

$${\pz S\ns_1\over \pz E\ns_1}={\pz S\ns_2\over \pz E\ns_2}\ .$$

This guarantees that

$$S(E,E\ns_1)=S\ns_1(E\ns_1) + S\ns_2(E-E\ns_1)$$

is a maximum with respect to the energy $E\ns_1$, at fixed total energy $E$.

Figure $\PageIndex{1}$: Two systems in thermal contact.

The temperature $T$ is defined as

$${1\over T}=\pabc{S}{E}{V,N}\ , \label{Teqn}$$

a result familiar from thermodynamics. The difference is now we have a more rigorous definition of the entropy. When the total entropy $S$ is maximized, we have that $T\ns_1=T\ns_2$. Once again, two systems in thermal contact and can exchange energy will in equilibrium have equal temperatures.

According to Equations \ref{phinrel} and \ref{phiurel}, the entropies of nonrelativistic and ultrarelativistic ideal gases in $d$ space dimensions are given by

$$\begin{align} S\ns{_{\ssr{NR}}}&=\half Nd\,\kB\ln\!\bigg({E\over N}\bigg) + N\kB\ln\!\bigg({V\over N}\bigg) + { const.}\\ S\ns_{\ssr{UR}}&=Nd\,\kB\ln\!\bigg({E\over N}\bigg) + N\kB\ln\!\bigg({V\over N}\bigg) + { const.}\ .\end{align}$$

Invoking Equation $\ref{Teqn}$, we then have

$$E\ns_{\ssr{NR}}=\half N d\,\kT \qquad,\qquad E\ns_{\ssr{UR}}=N d\,\kT\ .$$

We saw that the probability distribution $P\ns_1(E\ns_1)$ is maximized when $T\ns_1=T\ns_2$, but how sharp is the peak in the distribution? Let us write $E\ns_1=E^*_1+\RDelta E\ns_1$, where $E^*_1$ is the solution to Equation $\ref{maxprob}$. We then have

$$\ln P\ns_1(E^*_1+\RDelta E\ns_1)=\ln P\ns_1(E^*_1) + {1\over 2\kB}\, {\pz^2\!S\ns_1\over\pz E_1^2}\bigg|\nd_{E^*_1}(\RDelta E\ns_1)^2 + {1\over 2\kB}\,{\pz^2\!S\ns_2\over\pz E_2^2}\bigg|\nd_{E^*_2} (\RDelta E\ns_1)^2+ \ldots\ ,$$

where $E^*_2=E-E^*_1$. We must now evaluate

$${\pz^2 \!S\over \pz E^2}={\pz\over\pz E}\bigg({1\over T}\bigg)=-{1\over T^2}\pabc{T}{E}{V,N} =-{1\over T^2 \,C\ns_V}\ ,$$

where $C\ns_V=\big(\pz E/\pz T\big)\nd_{V,N}$ is the heat capacity. Thus,

$$P\ns_1 = P^*_1\,e^{-(\RDelta E\ns_1)^2/2k\ns_\RB T^2 {\bar C}\ns_V}\ ,$$

where

$${\bar C}\ns_V={C\ns_{V,1}\,C\ns_{V,2}\over C\ns_{V,1}+C\ns_{V,2}}\ .$$

The distribution is therefore a Gaussian, and the fluctuations in $\RDelta E\ns_1$ can now be computed:

$$\big\langle (\RDelta E\ns_1)^2\big\rangle = \kB T^2\,{\bar C}\ns_V \qquad\Longrightarrow\qquad (\RDelta E_1)\ns_{\ssr{RMS}}=\kT\sqrt{\bar C \ns_V/\kB}\ .$$

The individual heat capacities $C\ns_{V,1}$ and $C\ns_{V,2}$ scale with the volumes $V\ns_1$ and $V\ns_2$, respectively. If $V\ns_2\gg V\ns_1$, then $C\ns_{V,2}\gg C\ns_{V,1}$, in which case ${\bar C}\ns_V\approx C\ns_{V,1}$. Therefore the RMS fluctuations in $\RDelta E\ns_1$ are proportional to the square root of the system size, whereas $E\ns_1$ itself is extensive. Thus, the ratio $(\RDelta E_1)\ns_{\ssr{RMS}}/E\ns_1\propto V^{-1/2}$ scales as the inverse square root of the volume. The distribution $P\ns_1(E\ns_1)$ is thus extremely sharp.

Thermal, mechanical and chemical equilibrium

We have $dS\big|\nd_{V,N}={1\over T}\,dE$ , but in general $S=S(E,V,N)$. Equivalently, we may write $E=E(S,V,N)$. The full differential of $E(S,V,N)$ is then $dE=T\,dS - p\,dV + \mu\,dN$, with $T=\pabc{E}{S}{\,V,N}$ and $p=-\pabc{E}{V}{\,S,N}$ and $\mu=\pabc{E}{N}{\,S,V}$. As we shall discuss in more detail, $p$ is the pressure and $\mu$ is the chemical potential. We may thus write the total differential $dS$ as

$$dS={1\over T}\,dE+ {p\over T}\,dV - {\mu\over T}\,dN\quad.$$

Employing the same reasoning as in the previous section, we conclude that entropy maximization for two systems in contact requires the following:

• If two systems can exchange energy, then $T\ns_1=T\ns_2$. This is thermal equilibrium.
• If two systems can exchange volume, then $p\ns_1/T\ns_1=p\ns_2/T\ns_2$. This is mechanical equilibrium.
• If two systems can exchange particle number, then $\mu\ns_1/T\ns_1=\mu\ns_2/T\ns_2$. This is chemical equilibrium.

Gibbs-Duhem Relation

The energy $E(S,V,N)$ is an extensive function of extensive variables, it is homogeneous of degree one in its arguments. Therefore $E(\lambda S,\lambda V,\lambda N)=\lambda E$, and taking the derivative with respect to $\lambda$ yields

$$\begin{split} E&=S\pabc{E}{S}{V,N}+ V\pabc{E}{V}{S,N}+N\pabc{E}{N}{S,V}\\ &=TS-pV+\mu N\quad. \end{split}$$

Taking the differential of each side, using the Leibniz rule on the RHS, and plugging in $dE=T\,dS-p\,dV+\mu\,dN$, we arrive at the Gibbs-Duhem relation⁵,

$$S\,dT-V dp + N\,d\mu=0\quad.$$

This, in turn, says that any one of the intensive quantities $(T,p,\mu)$ can be written as a function of the other two, in the case of a single component system.