Legendre transform of $S(E,V)$ to obtain $F(T,V)$

Question

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?

Answer 1

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 $F = E - TS$.

Answer 2

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.