So, using the legendre transform: $$g=f-x\left ( \frac{df}{dx} \right )_y$$ I tried to obtain the Helmholtz free energy from Entropy and I got: $$F(T,V)=S(E,V)-E\left ( \frac{\partial S}{\partial E} \right )_V$$$$F(T,V)=S-\frac{E}{T}$$ which definitely is not a correct expression for the Helmholtz free energy. What am I doing wrong in here?
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asked Apr 29, 2020 at 8:25
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$\begingroup$ Related : A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein. $\endgroup$– VoulkosCommented Apr 29, 2020 at 8:29
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1$\begingroup$ Look up Massieu, Planck and Massieu-Planck functions $\endgroup$– SmallPieceOfBreadCommented Apr 17, 2022 at 17:25
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2 Answers
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Helmholtz free energy is a Legendre transformation of $E(S,V)$ and not $S(E,V)$. You will get $F(T,V)=E(S,V)-S\left ( \frac{\partial E}{\partial S} \right )_V$ and $\left ( \frac{\partial E}{\partial S} \right )_V = \left ( \frac{\partial S}{\partial E} \right )_V^{-1} = \frac{1}{T}^{-1} = T$. Then $F$ becomes $F = E - TS$.
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$\begingroup$ Yeah, I know that F is a Legendre transform of and E and can derive from E, but why can't we use it for S as well and derive $F(T,V)$ from $S(E,V)$? $\endgroup$ Commented Apr 29, 2020 at 8:33
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$\begingroup$ @RoshanShrestha Then I would suggest looking at this post. You cannot directly go from S to F, only by inverting it to E first. As you tried it, you are creating a new variable which is the Legendre transform of entropy but (I think) it is not really related to F directly. Only in the sense that $-T\cdot(S-\frac{E}{T}) = E - TS = F$. $\endgroup$ Commented Apr 29, 2020 at 8:36
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$\begingroup$ Legendre transforms of S are Massieu functions, and they are quite similar to the usual potentials. Precisely because they are so similar, people have not bothered to work too hard in terms of them. After all, you get similar physics. I mean, studying $F = E - TS$ will get a similar result as studying $-F/T = S - E/T$ $\endgroup$ Commented Apr 28, 2023 at 6:20
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My guess is that by doing a Legendre transform to $S(E, V)$ you obtain a different thermodynamic potential $\hat{F}(T, V)$, one that is different from $F(T,V)$. You can try using this new thermodynamic potential to express the equations of motion of an ideal gas and see if it will do the job. If all equations of motion can come out of this potential, then it's fine to use it I guess, maybe we did it with E(S,V) instead of S(E,V) for historical reasons.
