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Is always a Lagrangian L=T-V ?

  1. Oct 5, 2006 #1
    That's my question..although in more general cases [tex] L=T-V [/tex]

    [tex] H=T+V [/tex] however there're several important exceptions..for example:

    a) Classically (Non relativisitc) the Gravitational "Energy" (=Hamiltonian for a time-independent Potential) is:

    [tex] H=(1/2)\int_{V}\rho (\gra \phi)^{2} [/tex]

    b) Einstein-HIlbert Lagrangian [tex] L=\sqrt (-g) R [/tex] -g is the

    determinant of the metric and R is Ricci scalar.

    Is there always a kind of "transform" so you can always split te Lagrangian into a Kinetic and a potential terms...:confused: :confused:
     
  2. jcsd
  3. Oct 5, 2006 #2
    First of all, it is not sufficient that the hamiltonian have a time-independent potential for it to equal the energy. That's only for the hamiltonian to be conserved in time (actually, it's that [tex]dH/dt = \partial_t L[/tex] that you need). Also, you might want to look up the situation of a charged particle in the magnetic field. the lagrangian is decidedly NOT T - V, but the hamiltonian does equal thte energy of the charged particle (does not include the energy of the field).
     
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