Derivation of the Canonical Ensemble

[Dario56]

One of the common derivations of the canonical ensemble goes as follows:

Assume there is a system of interest in the contact with heat reservoir which together form an isolated system. Heat can be exchanged between the system and reservoir until thermal equilibrium is established and both are at temperature $T$.

Suppose system of interest is in microstate $i$ and that it has energy $E_{i}$, energy of the reservoir is than defined as $E_R = E - E_i$. In the last equation $E$ is the energy of the total isolated system which must be constant.

We want to determine the number of microstates of the total isolated system in which the system of interest is in microstate $i$ or the probability that isolated system will be in such a state. Number of such microstates is denoted as $\Omega_i(E)$ and is equal to the number of microstates of the reservoir $\Omega_R(E - E_i)$ since the system of interest is in particular microstate $i$.

Total number of microstates of the isolated system is defined as: $$\Omega (E) = \sum_{i=1} \Omega_R(E - E_i)$$

that is, it is equal to sum of the number of the reservoir microstates over every microstate of the system of interest $i$.

Not going further into the derivation, number of microstates and probability mentioned previously are given by: $$\Omega_R(E - E_i) = \Omega_R(E)e^{- \beta E_i}$$ $$ P_i = \frac {e^{-\beta E_i}} {\sum_{j=1}^N e^{-\beta E_j}} $$

where the denominator of the previous equation is the canonical partition function $Z$and $N$ is the number of the microstates in which the system can be.Now, we come to my question.

According to the notation given, microstate of the system $i$ is determined by the system's energy $E_i$.

However, we know that the task of statistical thermodynamics is to connect macrostate of the system with microstates in which it can be found. Energy of the system is its macrostate and therefore such a macrostate is compatible with many different microstates in which system may be.

Because of this, energy of the system $E_i$ can't depend only on particular microstate $i$ since the same microstate can be compatible with more than one energy of the system.

In another words, for any $i$ there may be more than one term in the partition function. However, equation form of the partition function doesn't allow more than one term for each microstate.

 

[Lord Jestocost]

One should keep in mind, that the sum in $\Omega (E_{total}) = \sum_{n=1} \Omega_R(E_{total}- E_n)$ is over all the states of the system of interest, $S$, rather than over the energy levels of $S$. In the latter case, one would have to include a factor of $\Omega_S(E_n)$ in the sum to take into account the degeneracy of states with energy $E_n$.

See: "Statistical Physics" by David Tong

Statistical Physics - DAMTP​

 

[vanhees71]

That's why the usual derivation goes like
$$\Omega(E)=\Omega_1(E_1) \Omega_2(E_2)=\Omega_1(E_1) \Omega_2(E-E_1).$$
Then define the entropy
$$S=\ln \Omega(E)=k \ln \Omega_1(E_1) + k \ln \Omega_2(E-E_1).$$
The most probable state is given by the maximum of the entropy, i.e., for
$$1/T_1=\frac{\partial \Omega_1(E_1)}{\partial E_1} = 1/T_2=\left [\frac{\partial \Omega_2(E_2)}{\partial E_2} \right]_{E_2=E-E_1}.$$
That's the "zeroth Law" introducing temperature as a thermodynamical variable of state.

Now we consider the probability for the system of interest to have one of its possible energies $E_1$. This probability is proportional to the number of microstates of system 2 with energy $E-E_1$ times the degeneracy factor of $E_1$, which however is just an overall factor, all of which we lump into a common factor $C$,
$$P(E_1)=C \Omega_2(E-E_1).$$
Expanding the logarithm of this gives
$$\ln P(E_1)=\ln C +\ln \Omega_2(E) -E_1/T=C'-E_1/T$$
and thus
$$P(E_1)=\frac{1}{Z} \exp(-E_1/T), \quad Z=\sum_{E_1} \exp(-E_1/T).$$

 

[Dario56]

One should keep in mind, that the sum in $\Omega (E_{total}) = \sum_{n=1} \Omega_R(E_{total}- E_n)$ is over all the states of the system of interest, $S$, rather than over the energy levels of $S$. In the latter case, one would have to include a factor of $\Omega_S(E_n)$ in the sum to take into account the degeneracy of states with energy $E_n$.

See: "Statistical Physics" by David Tong

Statistical Physics - DAMTP​

Yes, I think this question is clear now. What confused me is that I thought that microstate $i$ can be compatible with more than one energy state of system of interest $E_i$. This really has no sense since microstate is an instantaneous configuration of particles over available energy states.

If energy of the system of interest changes, so do the available energy states of the particles which means that for every energy of the system of interest there is a unique set of microstates in which system may be.