Is the formula of temperature $ \frac{1}{T}= \left(\frac{\partial S}{\partial U}\right)_{V,N}$ applicable to all type of ensembles?
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I have seen multiple posts on this page that explained the statistical definition of Temperature as the derivative of the Entropy to the energy:
1T≡(∂S∂U)V,N1T≡(∂S∂U)V,N
Where SS has always been the Boltzmann Entropy,
S=−kBlnΩ(U,V,N)S=−kBlnΩ(U,V,N)
My Question is: Does this Definition also apply to other Ensembles than the micro-canonical Ensemble? For any other Ensemble, can I also calculate ∂S∂⟨U⟩∂S∂⟨U⟩ where ⟨U⟩⟨U⟩ is supposed to be the averaged Energy and S=−kB⟨ln(ρ)⟩S=−kB⟨ln(ρ)⟩ is not any-more the Boltzmann entropy, but instead the Gibbs - Entropy, with ρρ being the probability distribution of the system?
I know that there are ensembles in which temperature is given externally and then defines the state of the system, like the canonical ensembles. But you can still assume for a given set of equilibrium states, that the temperature "changes" with the change of average Energy ⟨U⟩⟨U⟩, and thus my question is up also for those cases
1T≡(∂S∂U)V,N1T≡(∂S∂U)V,N
Where SS has always been the Boltzmann Entropy,
S=−kBlnΩ(U,V,N)S=−kBlnΩ(U,V,N)
My Question is: Does this Definition also apply to other Ensembles than the micro-canonical Ensemble? For any other Ensemble, can I also calculate ∂S∂⟨U⟩∂S∂⟨U⟩ where ⟨U⟩⟨U⟩ is supposed to be the averaged Energy and S=−kB⟨ln(ρ)⟩S=−kB⟨ln(ρ)⟩ is not any-more the Boltzmann entropy, but instead the Gibbs - Entropy, with ρρ being the probability distribution of the system?
I know that there are ensembles in which temperature is given externally and then defines the state of the system, like the canonical ensembles. But you can still assume for a given set of equilibrium states, that the temperature "changes" with the change of average Energy ⟨U⟩⟨U⟩, and thus my question is up also for those cases
kohlisurender:
The Answer is AWSM ^_^ @abhishek
Wlcm
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