Partition function related to number of microstates

In summary, the partition function, which is used to calculate the thermodynamic properties of a system, is related to the number of possible microstates by a factor of e^{\beta U}. This factor is necessary because the partition function corresponds to a canonical ensemble, where the microstates are not equally likely. Additionally, the total system, including its environment, can be described as a microcanonical ensemble, with the number of possible configurations being equal to the number of microstates of the environment multiplied by e^{-\beta U}. This result has not been commonly seen before, but it appears to be correct.
  • #1
Troy124
3
0
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.
 
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  • #2
Troy124 said:
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.
 
  • #3
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.
 

1. What is the partition function in thermodynamics?

The partition function in thermodynamics is a mathematical function that helps calculate the number of microstates or possible arrangements of particles in a system at a given energy level. It is denoted by the letter Z and is used to determine the thermodynamic properties of a system, such as entropy and free energy.

2. How is the partition function related to the number of microstates?

The partition function is directly proportional to the number of microstates in a system. This means that as the value of Z increases, the number of possible arrangements of particles also increases. In other words, a higher value of Z indicates a higher level of disorder or randomness in the system.

3. What is Boltzmann's constant and how is it used in the partition function?

Boltzmann's constant (k) is a physical constant used in the calculation of the partition function. It relates the average kinetic energy of particles in a system to the temperature of the system. In the partition function, it is used to convert the energy levels of the system from joules to kelvins, making it easier to work with in calculations.

4. Can the partition function be used for both classical and quantum systems?

Yes, the partition function can be used for both classical and quantum systems. In classical systems, the partition function is calculated using the positions and momenta of particles, while in quantum systems, it is calculated using the energy levels of particles. However, the concept of microstates and the relationship between Z and the number of microstates remains the same for both types of systems.

5. What is the significance of the partition function in statistical mechanics?

The partition function is a fundamental concept in statistical mechanics, as it is used to calculate various thermodynamic properties of a system. It provides a way to connect the microscopic behavior of particles to macroscopic properties of the system. It also allows for the prediction of how a system will behave at different temperatures and energies, making it a crucial tool in understanding and analyzing thermodynamic systems.

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