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# 4.3: Thermal Equilibrium

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$$\newcommand\Vsigma ParseError: invalid DekiScript (click for details) Callstack: at (Template:MathJaxArovas), /content/body/div/p[1]/span[26], line 1, column 1 at template() at (Under_Construction/Arovas_Texts/Book:_Thermodynamics_and_Statistical_Mechanics_(Arovas)/04:_Statistical_Ensembles/4.03:_Thermal_Equilibrium), /content/body/p/span, line 1, column 23 $$
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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 $$E$$.

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.