Partition function related to number of microstatesby Troy124 Tags: function, microstates, number, partition 

#1
Feb2013, 03:04 PM

P: 3

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 Shannonentropy ##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. 



#2
Feb2013, 03:30 PM

HW Helper
P: 1,391

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
Feb2013, 04:44 PM

P: 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. 


Register to reply 
Related Discussions  
calculate number of microstates of n harmonic oscillators  Advanced Physics Homework  1  
Are the number of microstates of a gas just equivalent to pressure  Classical Physics  9  
Introductory Statistical Mechanics  counting number of microstates  Advanced Physics Homework  3  
Related to partition theory, ways of representing a number  Linear & Abstract Algebra  8  
Definition of the number of microstates accessible to a system  Classical Physics  2 