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Legendre transform

  1. Jul 10, 2006 #1
    Legendre transform....

    If we define a function f(r) with r=x,y,z,.... and its Legnedre transform
    g(p) with [tex] p=p_x ,p_y,p_z,...... [/tex] then we would have the equality:

    [tex] Df(r)=(Dg(p))^{-1} [/tex] (1) where D is a differential operator..the

    problem is..what happens when g(p)=0?...(this problem is usually found in several Hamiltonians of relativity) then (1) makes no sense since a 0 matrix would have no inverse..how do you define Legendre transform then...:frown: :frown:
     
  2. jcsd
  3. Jul 10, 2006 #2

    HallsofIvy

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    I don't see any problem with "defining the Legendre transform", just defining its inverse- which does not necessarily exist. What kind of functions, f, and p, give g(p)= 0? Why does that not have an inverse?
     
  4. Jul 10, 2006 #3
    The problem of Zero Legendre-transform arises for example in Quantum Field theory...let be the "Lagrangian density":

    [tex] L= \sqrt (-g)R [/tex] where g is the determinant of the "metric" and R is Ricci scalar?...of course if we use a "dot" to indicate derivative respect to time we have that:

    [tex] L=L(g_ab , \dot g_ab,x,y,z) [/tex] now we definte the "momenta"..

    [tex] \pi _ab =\frac{ \partial L}{\partial \dot g_ab) [/tex]

    then the Legendre transform for Quantum gravity is defined by:

    [tex] H=\pi _ab \dot g_ab -L [/tex] where the metric is given by:

    [tex] g_ab =N(t)dt^{2}-g_ ij dx^{i}dx^{j} [/tex] i,j=1,2,3 (einstein sum convention)

    where N(t) is somehow a "lapse" function with a physical meaning so you get the "constraint"..

    [tex] H=0 [/tex]
     
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