1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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]

    etc.
     
  4. Feb 14, 2012 #3
    Tnx. Do I get something with Legendre transforms if I defined
    [tex]U=U(T,P)[/tex]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook