Partition function related to number of microstates

[Troy124]

Hi,

I have a question about the partition function.

It is defined as $Z = \sum_{i} e^{-\beta \epsilon_{i}}$ where $\epsilon_i$ denotes the amount of energy transferred from the large system to the small system. By using the formula for the Shannon-entropy $S = - k \sum_i P_i \log P_i$ (with $k$ a random constant or $k_B$ in this case), I end up with the following: $$ S = - k \sum_i P_i \log P_i = (k \sum_i P_i \beta \epsilon_i) + (k \sum_i P_i \log Z) = \frac{U}{T} + k \log Z $$

This simplifies to $Z = e^{-\beta F}$ by using the Helmholtz free energy defined as $F = U - T S$. But Boltzmann's formula for entropy states $S = k \log \Omega$, where $\Omega$ denotes the number of possible microstate for a given macrostate. So we will get $$ \Omega = e^{S/k} = e^{\beta (U - F)} = Z e^{\beta U} $$

So the partition function is related to the number of microstates, but multiplied by a factor $e^{\beta U}$. And this bring me to my question: why is it multiplied by that factor? Maybe the answer is quite simple, but I can't seem to think of anything.

 

[Mute]

Hi,

I have a question about the partition function.

It is defined as $Z = \sum_{i} e^{-\beta \epsilon_{i}}$ where $\epsilon_i$ denotes the amount of energy transferred from the large system to the small system. By using the formula for the Shannon-entropy $S = - k \sum_i P_i \log P_i$ (with $k$ a random constant or $k_B$ in this case), I end up with the following: $$ S = - k \sum_i P_i \log P_i = (k \sum_i P_i \beta \epsilon_i) + (k \sum_i P_i \log Z) = \frac{U}{T} + k \log Z $$

This simplifies to $Z = e^{-\beta F}$ by using the Helmholtz free energy defined as $F = U - T S$. But Boltzmann's formula for entropy states $S = k \log \Omega$, where $\Omega$ denotes the number of possible microstate for a given macrostate. So we will get $$ \Omega = e^{S/k} = e^{\beta (U - F)} = Z e^{\beta U} $$

So the partition function is related to the number of microstates, but multiplied by a factor $e^{\beta U}$. And this bring me to my question: why is it multiplied by that factor? Maybe the answer is quite simple, but I can't seem to think of anything.

Boltzmann's formula $S = k_B \ln \Omega$ is applicable only to the case of a microcanonical ensemble - a system in which every microstate is equally likely. Note that setting $P_i = 1/\Omega$ in $S = -k_B \sum_{i=1}^\Omega P_i \ln P_i$ gives Boltzmann's formula.

The partition function $Z = \sum_i \exp(-\beta \epsilon_i)$ corresponds to a canonical ensemble. The microstates in a canonical ensemble are not equally likely, so Boltzmann's formula $S = k_B \ln \Omega$ does not apply. (However, the more general formula, $S = -k_B \sum_{i=1}^\Omega P_i \ln P_i$, does still apply).

You can thus not equate $\Omega$ to $Ze^{\beta U}$, as the two formulas you used for entropy are not simultaneously true.

 

[Troy124]

Hi,

Thanks for your reply. I finally figured out that I mixed up the entropy of the environment with the entropy of the system, because my idea was that the total system, so environment + system, could be described by the microcanonical ensemble and I could use Boltzmann's formula, but then you will end up with something different:

The system including its environment can be described as a microcanonical ensemble. The number of possible configurations for this ensemble are $\Omega_{total} = \sum_i w_i$ where $w_i$ denotes the number of possible configurations given an $\epsilon_i$.

We know $$w_i = \Omega (E-\epsilon_i) \Omega (\epsilon_i)$$ (with $\Omega (\epsilon_i) = 1$, $\Omega (E-\epsilon_i)$ the number of microstates of the system when its energy equals $E-\epsilon_i$ and $\Omega (E)$ the number of microstates of the environment, when it is not thermally connected to another system) and thus $$ \Omega_{total} = e^{S_{total}/k} = e^{S/k} e^{S_{env}/k} = e^{\beta (U - F)} \Omega_{env} = \sum_i \Omega (E - \epsilon_i) = \Omega (E) \sum_i e^{-\beta \epsilon_i} = \Omega (E) e^{-\beta F}$$

This simplifies to $$ \Omega_{env} = \Omega (E) e^{-\beta U}$$

Do you know if this is correct, because I have never seen this result before. It does seem okay to me though.