Thermodynamics without partition function

[cryptist]

Is there a way to derive entropy or free energy without using partition function?

 

[DrClaude]

Yes.

If you want a more complete answer, you'll have to post a more detailed question.

 

[Jorriss]

Thermodynamics came before statistical mechanics.

 

[cryptist]

As you all know, N=∑n_i, U=∑ε_i, F=Nμ-kT∑Z and S=(U-F)/T

Here, I do not want to use partition function Z. How do I write F and S then?

 

[DrClaude]

The Helmoholtz free energy is defined as
$$
F \equiv U - TS
$$
Entropy you can get from the heat capacity:
$$
C_V \equiv T \left( \frac{\partial S}{\partial T} \right)_V
$$

 

[cryptist]

Let me specify the problem. I am using grand canonical ensemble and I am in Fermi-Dirac statistics.

Considering these, entropy is written as S=k[lnZ+β(E-μN)]. How do I write S, without using Z?

 

[cryptist]

The Helmoholtz free energy is defined as
$$
F \equiv U - TS
$$
Entropy you can get from the heat capacity:
$$
C_V \equiv T \left( \frac{\partial S}{\partial T} \right)_V
$$

Heat capacity is a derived quantity. It is really complicated to extract S from Cv.

 

[cryptist]

Thermodynamics came before statistical mechanics.

Ok. Then, how do I derive S without using partition function (a statistical mechanics tool)? Btw, by deriving S, I mean I'll calculate the entropy of a system over momentum states, I am not talking about S=klnΩ which apparently does not include partition function.

 

[cryptist]

Anyway, I found the answer by myself. Thread can be closed.

 

[Useful nucleus]

Anyway, I found the answer by myself. Thread can be closed.

It would be nice to share with us the answer you found