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A Chain rule - legendre transformation

  1. Dec 26, 2016 #1

    df=∂f/∂x dx+∂f/∂y dy and ∂f/∂x=p,∂f/∂y=q

    So we get

    df=p dx+q dy

    d(f−qy)=p dx−y dq

    and now, define g.

    g=f−q y
    dg = p dx - y dq

    and then I faced problem.

    ∂g/∂x=p←←←←←←←←←←←←←←← book said like this because we can see g is a function of x and p so that chain rule makes ∂g/∂x=p

    but I wrote directly ∂g/∂x so i can get a result,

    ∂g/∂x=∂f/∂x−y ∂q/∂x−q ∂y∂x=p−y ∂q∂x - 0(y is independent of x)

    I know that variation y is independent of x, but I'm not sure that q is also the independent function of x. what if the func f is xy??? if I set like that,


    it's not same with p!!!

    mathematical methods in the physical sciences, Mary L. Boas 3rd edition page231
  2. jcsd
  3. Dec 26, 2016 #2
    In taking partial derivatives , it is recommended to explicitly show which variable is maintained constant. So you want to calculate (∂g/∂x)|q. There are two ways. The easiest one, is to look at the differential of g and set dq=0 since you are holding q constant. This way you get (∂g/∂x)|q = p. The second is to use your method. Then,

    (∂g/∂x)|q=(∂f/∂x)|q - q (∂y/∂x)|q --------- (1)

    notice that (∂f/∂x)|q ≠ p. Correct is (∂f/∂x)|y = p.

    It is possible to manipulate expression(1) above to reach to the same conclusion that we obtain easily from the differential which is (∂g/∂x)|q = p.

    Also note that there no need to assume that y is independent of x. In fact in the context of applying Legendre Transform to Thermodynamics, the natural variables do depend on each other. For example U=(S,V,N) where U is the internal energy, and we have a dependence of S on N and V.
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