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How can I prove that $$\frac{\partial(T \ln Z)}{\partial T}=\ln \Omega,$$ without using the relation $S=k\ln Ω$?

where $Z$ is the Partition Function, $T$ is the Absolute Temperature, $Ω$ is the Statistical Weight of a Macro-state, $S$ is the Entropy of the system.

For the actual problem, I have provided this below Image file

Lusypher
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  • Check this answer – Roger V. Mar 29 '22 at 08:39
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 29 '22 at 08:54
  • @ytlu Ω is the Number of Micro-states belonging to a particular Macro-state. – Lusypher Mar 29 '22 at 09:18
  • I am not sure what particular system you are working in (or what kind of ensemble, I am assuming canonical), but notice that you can re-arrange equation 2.64a into the Helmholtz free energy as, $TS = kT\ln Z + \bar{U}\rightarrow F = \bar{U} -TS = -kT\ln Z$. Now, if you take the partial derivative of $F$ with respect to $F$, you get, $\frac{\partial F}{\partial T} =-k\frac{\partial}{\partial T}(T\ln Z)$, which is equal to $-S$ to give you $S = k\frac{\partial}{\partial T}(T\ln Z)$. If this is to answer the second exercise, I do not think you need the relationship you posted that you wanted. – MathZilla Mar 29 '22 at 16:06
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    @Cassem02 Thank you for this answer. Actually I thought that, if some how $\frac{\partial(T \ln Z)}{\partial T}=\ln \Omega$ can be established, then the relation (2.64b) can make the relation that has to be proved in the first exercise. So I asked about that. BTW thanks again. – Lusypher Mar 29 '22 at 20:58

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You need to be a bit more careful in how you reframe the problem, because 2.64b is not making an assumption about the ensemble, but your equation does (implicitly, equal a priori probabilities).

The general statement,

$$\frac{S}{k_B} = \frac{\partial}{\partial T}\left ( T\ln Z \right ) \tag{1}$$

Is true. Apply the partition function from the canonical ensemble:

$$Z(T)=\sum_i e^{-E_i/k_B T}$$

The partial derivative of temperature on $Z$ will not be affected by the limits of integration on the phase space, so it is fine to bring the derivative inside the sum.

Recall the Helmholtz Free Energy, $$F=U-TS$$

$F$ has a relationship to $Z$,

$$\ln Z= \frac{-F}{k_B T} \tag{2}$$

Putting all of this together gives $(1)$.

Also, you can use information from the Legendre transform to jump straight to the answer:

$$-F(V,T)+U(V,S)=TS$$

Therefore, at constant volume,

$$\frac{\partial F}{\partial T}=-S$$

It's easy to see from here how we get $(1)$ again.

michael b
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