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Legendre transforms

  1. Feb 14, 2012 #1
    When people do Legendre transforms they suppose that [tex]U=U(S,V)[/tex]. But you can see in some books that heat is defined by:
    [tex]dQ=(\frac{\partial U}{\partial P})_{V}dP+[(\frac{\partial U}{\partial V})_P+P]dV[/tex]

    So they supposed obviously that [tex]U=U(V,P)[/tex].

    In some books you can that internal energy is function of [tex]T,P[/tex], and in some books function of [tex]V,T[/tex]. Why then in definition of Legendre transforms of thermodynamics potential we use [tex]U=U(S,V)[/tex]. Tnx for the answer.
  2. jcsd
  3. Feb 14, 2012 #2


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    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, [itex]U[/itex], the natural variables are [itex]S[/itex] and [itex]V[/itex], because of the fundamental laws of thermodynamics:

    [tex]\mathrm{d} U=T \mathrm{d} S-p \mathrm{d} V.[/tex]

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

    [tex]F=U-T S.[/tex]

    Taking the total differential gives

    [tex]\mathrm{d} F=\mathrm{d} U - T \mathrm{d} S-S \mathrm{d} T=-S \mathrm{d} T-p \mathrm{d} V,[/tex]

  4. Feb 14, 2012 #3
    Tnx. Do I get something with Legendre transforms if I defined
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