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The thermodynamic free energy is defined by $F=U-TS$ with $U,T,S$ being the internal energy, temperature and entropy respectively.

I have also seen another formula for the free energy, $F=-T \log{Z}$ where $Z=\int \cal{D} \phi e^{-I[\phi]}$ is the partition function.

First of all, I'm not sure the $Z$ in the log is the same as the one I've defined in the path integral. Can someone confirm this please? Secondly, how can I show these two definitions of free energy are equivalent?

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  • $\begingroup$ Both of your expressions are correct. The connection between them is through U = -d(lnZ)/dbeta, where beta=1/(kT). en.wikipedia.org/wiki/… I can make an answer from this but I'd like to write something about the specific form of the path integral, i.e. what the I[phi] is. Could you specify what kind of system you are looking at? $\endgroup$ Jan 25, 2016 at 20:00
  • $\begingroup$ @Numrok Thanks. It is the action for a black hole spacetime. $\endgroup$
    – user11128
    Jan 25, 2016 at 20:12
  • $\begingroup$ I'll pass on writing an answer in that case, since I don't know enough about that kind of problem. Hope my comment helped though. $\endgroup$ Jan 25, 2016 at 20:55

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The weight for a specific state is $w(\phi)=e^{-E/T+F/T}\equiv e^{-I(\phi)+F/T}$. Due to the definition of the entropy $S$, we have $S=-\int D\phi w\ln w=\frac{1}{T}(\int D\phi wE-F\int D\phi w)=\frac{1}{T}(U-F)$. Then we have $F=U-TS$.

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  • $\begingroup$ Why is $S=-\int D \phi w \ln{w}$? $\endgroup$
    – user11128
    Jan 26, 2016 at 9:07
  • $\begingroup$ @user11128 As what I have said, it is just the definition of the entropy (or the so-called Boltzmann H function). $\endgroup$
    – Wen Chern
    Jan 26, 2016 at 9:29
  • $\begingroup$ I don't think this is fully correct. Firstly there is no need to employ a second definition of the entropy, since we have the partition function and it contains all the information. Secondly the Boltzmann H is a very questionable quantity. It is correct if you don't consider interparticle attractions of your gas, if you are (which you really need to, otherwise you have no pressure) you should be using Gibbs H instead. Thirdly if we are pedantic the entropy in F=U-TS is the third kind of entropy, also called the "experimental entropy". They are equivalent in many contexts though. $\endgroup$ Jan 26, 2016 at 10:02
  • $\begingroup$ @Numrok How would you recommend demonstrating this then? Thanks. $\endgroup$
    – user11128
    Jan 26, 2016 at 11:13
  • $\begingroup$ @user11128 I think I actually slightly misunderstood your question and Wen Chern's answer. I thought the question was for the relation between the equations and not for a derivation of the one from the other (which I didnt even know existed). Clearly Wen Chern has suggested such a derivation. I honestly just can't really tell if it is correct since I don't know enough about it, so apologies, I will withdraw my criticism. $\endgroup$ Jan 29, 2016 at 17:26

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