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Hamiltonian as Legendre transformation?

  1. Jun 4, 2010 #1
    The definition of a Legendre transformation given on the Wikipedia page http://en.wikipedia.org/wiki/Legendre_transformation is: given a function f(x), the Legendre transform f*(p) is

    [tex]f^*(p)=\max_x\left(xp-f(x)\right)[/tex]

    Two questions: what does [tex]\max_x[/tex] mean here? And why is it not (explicitly?) included in the definition of the Hamiltonian

    [tex]H(q,p)=p\dot{q}-L(q,\dot{q})[/tex]

    if the Hamiltonian is a Legendre transformation?
     
  2. jcsd
  3. Jun 4, 2010 #2
    I get it. If we want to maximize

    [tex]g(x)=xp-f(x)[/tex]


    then we set [tex]g'(x)=0[/tex] which is the same as putting [tex]p=f'(x)[/tex]. In mechanics this amounts to

    [tex]p=\frac{\partial L}{\partial\dot{q}}[/tex]

    Well.. thanks to anyone who read and at least thought about replying. :-)
     
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