Please don't answer because U(T,V) don't have S entropy as argument!!!!!!
May I ask a question on thermodynamic? We know that U(V,T) (caloric eq. of state) and S(P,V) (thermodynamic eq of state) can both be derived from thermodynamic potentials (U F G H) and the fundamental relations. However, U(V,T) doesn't hold full thermodynamic info of the system as U(S,V) does, yet S(P,V) also holds full thermodynamic info of the system.
In which step in derivation to get U(T,V) from U(S,V) lost the thermodynamic info? (the derivation is briefly：1. derive U=TdS+ PdV on V, 2. replace the derivative using Maxwell eq. and 3. finally substitute ideal gas eq or van der waal eq)
Why the similar derivation to get S(P,V) retain full thermodynamic info?
Even if we only have U(T,V), can't we get P using ideal gas eq, then calculate the S by designing reversible processes from (P0,V0,T0) to (P',V',T')? If we can still get S, why U(T,V) doesn't have full thermodynamic info?
Natural variables for U: S, V, Ni (for simple systems)
Natural variables for S: U, V, Ni (for simple systems)
T is the partial derivative of U with respect to S, maintaining V and Ni constants:
T=(∂U/∂S)V,Ni
If from the fundamental relationship U=U(S,V,N) you replace S by T just by solving for S in T=T(U,S,V,Ni) and substituting in U=U(S,V,N), then, you loose information because you are replacing a variable with a derivative with respect to that variable.
This problem is worked out with the Lengendre transform. If you search on internet, you may find simple examples: how to change y=f(x) to z=g(p), where p is ∂y/∂x, in the two ways, the incorrect one (calculation of p and plain substitution of p by removing x) and the correct one (calculation of p and applying the Legendre transform, z=g(p)=px-f). In fact, because you do not loose information with the Legendre transform, you may go backwards from z to y, which is not possible with the incorrect way.
Therefore, it is a mathematical "trick".
Applying the Legendre transform to U, with respect to T and S, you get a new thermodynamic potential F=U-TS, the Helmholtz energy, whose natural variables are T, V, and Ni. Beware of the minus sign applied to the Legendre transform (i.e., F is not equal to TS-U, but U-TS).
For a system at constant U, V, and Ni, any possible process will maximize S.
For a system at constant S, V, and Ni, any possible process will minimize U, but not F.
For a system at constant T, V, and Ni, any possible process will minimize F, but not U.
The Legendre transform connecting two thermodynamic potentials parallels the Laplace transform connecting the corresponding partition functions.
The Legendre transform is not only employed in Thermodynamics and Statistical Physics, but also in Classical Mechanics and other fields.
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