- #1

cryptist

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

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- Thread starter cryptist
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In summary, the conversation discussed ways to derive entropy and free energy without using the partition function in statistical mechanics. The Helmoholtz free energy and the heat capacity were mentioned as tools to calculate these quantities. However, it was noted that extracting the entropy from the heat capacity can be complicated. Ultimately, the speaker found the answer to their question and the conversation came to a close.

- #1

cryptist

- 121

- 1

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

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- #2

DrClaude

Mentor

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Yes.

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

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

- #3

Jorriss

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Thermodynamics came before statistical mechanics.

- #4

cryptist

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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?

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

Last edited:

- #5

DrClaude

Mentor

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$$

F \equiv U - TS

$$

Entropy you can get from the heat capacity:

$$

C_V \equiv T \left( \frac{\partial S}{\partial T} \right)_V

$$

- #6

cryptist

- 121

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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?

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

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- #7

cryptist

- 121

- 1

DrClaude said:definedas

$$

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.

- #8

cryptist

- 121

- 1

Jorriss said: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.

- #9

cryptist

- 121

- 1

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

- #10

Useful nucleus

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

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

Thermodynamics without partition function is a simplified approach to studying thermodynamic systems without using the partition function, which is a mathematical tool used to calculate the thermodynamic properties of a system. This approach is often used in introductory thermodynamics courses to help students understand the basic principles of thermodynamics.

In traditional thermodynamics, the partition function is used to calculate the thermodynamic properties of a system, such as energy, entropy, and free energy. In contrast, thermodynamics without partition function focuses on the qualitative understanding of thermodynamic concepts without relying on complex mathematical calculations.

One advantage of using thermodynamics without partition function is that it simplifies the study of thermodynamics and allows for a more intuitive understanding of the principles. It also eliminates the need for complex mathematical calculations, making it more accessible to beginners and non-mathematical scientists.

Yes, thermodynamics without partition function is a simplified approach and therefore has limitations. It cannot be used to make precise quantitative predictions about a system's behavior, and it may not be applicable to more complex thermodynamic systems.

No, thermodynamics without partition function is not used in real-world applications. It is mainly used as a teaching tool to introduce students to the basic concepts of thermodynamics before they move on to more advanced methods and applications.

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