Legendre Transforms: U=U(S,V) vs U(V,P)

[matematikuvol]

When people do Legendre transforms they suppose that U=U(S,V). But you can see in some books that heat is defined by:
dQ=(\frac{\partial U}{\partial P})_{V}dP+[(\frac{\partial U}{\partial V})_P+P]dV

So they supposed obviously that U=U(V,P).

In some books you can that internal energy is function of T,P, and in some books function of V,T. Why then in definition of Legendre transforms of thermodynamics potential we use U=U(S,V). Tnx for the answer.

 

[vanhees71]

You can, of course, express any thermodynamic potential by any pair of quantities you like, but there are "natural" ones. E.g. for the internal energy, U, the natural variables are S and V, because of the fundamental laws of thermodynamics:

\mathrm{d} U=T \mathrm{d} S-p \mathrm{d} V.

Now you can define other potentials to have other "natural" independent variables by Legendre transformations. E.g. the free energy trades S for T via:

F=U-T S.

Taking the total differential gives

\mathrm{d} F=\mathrm{d} U - T \mathrm{d} S-S \mathrm{d} T=-S \mathrm{d} T-p \mathrm{d} V,

etc.

 

[matematikuvol]

Tnx. Do I get something with Legendre transforms if I defined
U=U(T,P)?