# Legendre transforms

1. Feb 14, 2012

### 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.

2. Feb 14, 2012

### 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.

3. Feb 14, 2012

### matematikuvol

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