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The zeroth law of thermodynamics

To establish the equivalence of our definitions and the conventional thermodynamic ones we shall make contact with the zeroth law of thermodynamics. This law has an analogy with mechanics, where in equilibrium the forces are balanced. In particular if two subsystems (which we label 1 and 2) are in contact and in equilibrium:
T1 = $\displaystyle T_2\rightarrow \hbox{ thermal
P1 = $\displaystyle P_2\rightarrow \hbox{ mechanical equilibrium}$ (7)
$\displaystyle \mu_1$ = $\displaystyle \mu_2 \rightarrow \hbox{ chemical

The zeroth law has a fairly straightforward statistical interpretation and this will allow us to begin to establish the equivalence between the statistical definitions and the conventional thermodynamic ones. Consider two systems that are free to exchange energy but are isolated from the rest of the universe by an ideal insulating surface. The particle numbers N1, N2 and volumes V1, V2 are fixed for each subsystem. The total energy will be constant under our assumptions and we assume further that the two subsystems are sufficiently weakly interacting that

E = E1 + E2, (8)

where E1 and E2 are the energies of the subsystems. Assume that the densities of state g(E), g1(E),g2(E) are coarse grained so that $\Omega=g(E)\delta E$, $\Omega_1=g_1(E1)\delta E$, $\Omega_2=g_2(E_2)\delta E$. We then have

\begin{displaymath}g(E) =\int dE_1 g_2(E-E_1)g_1(E_1),
\end{displaymath} (9)

If the subsystems are sufficiently large, the product g2(E-E1)g1(E1) will be a sharply peaked function of E1. From the definition of the entropy we note that it is a monotonically increasing function of g and that the product g1g2 will be at a maximum when the total entropy

S(E, E1) = S1(E1) + S2(E-E1) (10)

is at a maximum. The most likely value $\langle E_1\rangle$ of E1 is the one for which

\begin{displaymath}\frac{\partial S_1}{\partial E_1}+\frac{\partial S_2}{\partial E_2}
\frac{\partial E_2}{\partial E_1}=0.
\end{displaymath} (11)

Since $\partial E_2/\partial E_1 = -1$, we find using ([*]), that

\end{displaymath} (12)

or T1 = T2. The most probable partition of energy between the two systems is the one for which the two temperatures are the same. Consider next two subsystems that are separated by a movable wall. The two systems are free to exchange energy, but the number of particles is fixed in each subsystem and the total volume V=V1+V2 is constant. We write E=E1+E2. The density of allowed states for the total system is then

\begin{displaymath}g(E,V)=\sum \limits_{volume\; settings}\int

The integrand is sharply peaked for large systems and takes on its maximum value when S1(E1,V1)+S2(E-E1,V-V1)=max. Differentiation using

\begin{displaymath}\left({{\partial S}\over{\partial E}}\right)_{VN}={{1}\over{T...
...eft({{\partial S}\over{\partial V}}\right)_{EN}=

implies that it is overwhelmingly probable that the system will be near a state for which

\begin{displaymath}{{1}\over{T_1}}={{1}\over{T_2}};\;\; {{P_1}\over{T_1}}

or T1=T2,P1=P2. Conventionally, one would say that the pressure in the two compartments must be equal at equilibrium because the forces have to be in balance. The argument now being made is quite different, there are no forces, instead the movable wall is guided to its equilibrium by the invisible hand of the law of large numbers. Similarly, consider two systems 1 and 2 which are free to exchange particles and energy. It is easy to show that the most probable configuration is the one for which $T_1=T_2,\mu _1=\mu _2.$
next up previous
Next: Boltzmann factor Up: Boltzmann statistics Previous: Statistical definition of thermodynamic
Birger Bergersen