Derivation of temperature with presumption of microcanonical ensemble?

Question

From many textbooks (i.e. Pathria and Kittel), the link between temperature, $T$ and Statistical/Boltzmann entropy, $S$, is established by allowing two systems to exchange energy through the diathermal wall and then we calculate the number of microstates for the two systems. Supposed that system 1 and system 2 have microstates of $\Omega_1(E_1)$ and $\Omega_2(E_2)$ respectively, then in equilibrium,

$$ \begin{equation} \frac{\partial\hspace{0.1cm} ln(\Omega_1)}{ \partial E_1} = \frac{\partial\hspace{0.1cm} ln(\Omega_2)}{ \partial E_2} \end{equation} $$

at this point, most textbooks have not introduced the calculation of states by integrating over the phase space volume yet, so the Boltzmann's entropy, $S= k \hspace{0.1cm} ln \Omega$ is simply introduced here and they conclude that the definition of temperature must be: $$ \begin{equation} \frac{1}{kT} = \frac{\partial S}{ \partial E} \end{equation} $$ But from the Wiki page of microcanonical sensemble and Robertson's Statistical Thermophysics, it seems like Boltzmann's entropy is one of the identities of a microcanonical ensemble.

So it's tempting for me to make the following statement:

To find the relationship between temperature, $T$ and Statistical/Boltzmann entropy, $S$, one has to invoke the microcanonical ensemble.

But it seems like this statement contradicts the fact that microcanonical ensemble is made up of identical systems with the same fixed energy (pg 110, Robertson's Statistical Thermophysics) and those systems are isolated. But the definition of temperature requires the exchange of energy! Did my reasoning go wrong somewhere?

Answer 1

Your reasoning is correct but incomplete. A microcanonical ensemble indeed describes an isolated system at constant energy $E$. However, this truth does not prevent $\ln(\Omega)$ to depend on $E$. One has to think of different isolated systems, each with its own energy and correspondingly its number of available microstates.

At this point, the usual way of obtaining the relation with temperature is the analysis of the equilibrium condition for two subsystems of a total system after removing the constraint that the two subsystems cannot exchange energy. By looking for the condition of maximizing the number of microstates after removing the constraints, one arrives at the equilibrium condition $$ \frac{\partial{\ln \Omega_1}}{\partial{E_1}}=\frac{\partial{\ln \Omega_2}}{\partial{E_2}}. $$ This is all one needs to say that $\frac{\partial{\ln \Omega}}{\partial{E}}$ is a possible temperature for the system (zeroth principle of thermodynamics requires the equality of temperatures as a probe of thermal equilibrium).

Notice that this argument is basically a thermodynamic argument when thermodynamics is encoded into the dependence of entropy on the extensive variables describing the system. As such, it does not depend on the statistical ensemble, although the most natural ensemble to discuss it is microcanonical. The only delicate point from the side of statistical mechanics is identifying $\ln(\Omega)$ with the thermodynamic entropy.

Answer 2

Your are right that the microcanonical ensemble is working with a fixed energy ($E$ as a parameter). But after obataining the entropy with Boltzmann relation $$ S = k \ln \Omega(E). $$

Then your take the derivetive of entropy w.r.t the parameter $E\to E+dE$. In order to change energy your have to thermally conduct the system to a reservoir with "temperature" slightly higher than the system, and allow exchange of energy between the system and reservoir.

In short, what you calculate in the microcanonical ensemble is a general relation between entropy S (relate to the number of micorstates) and the energy $E$, $S(E)$. Then, you calculate the effect of energy change, in macrostate regime, to the entropy using this general relation. This (energy variation) is where the tempreature comes in to play the role.