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Lagrangian Mass Matrix Question, Pls Help

  1. Feb 9, 2005 #1
    Lagrangian Mass Matrix Question, Pls Help!!

    "Assume that the lagrangian is given in the form

    [tex] L(q_{i}, q_{i}', t) = \sum_{m, n}T_{mn}(q_{i})q_{m}'q_{n}' - V(q_{i})[/tex]

    Show that in order for the principle of least action to hold, [tex]T_{mn}(q_{i})[/tex] has to be positive definite."

    My first intuition is that T has to be positive definite so that the action will be a minimum. However, when I tried to expand the lagragian to the second order, it gets kind of messy and I cannot find a way to eliminate. It is kind of urgent and I hope those who know will kindly help.
     
  2. jcsd
  3. Feb 9, 2005 #2

    dextercioby

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    Since the mass matrix is obviously symmetric (for bosonic coordinates),then there is an orthogonal square matrix which would diagonalize the matrix T.Now,if the diagonal matrix has both negative and positive eigenvalues on the diagonal,when doing the Legendre transformation,the Hamiltonian would not be positively definite,which would mean that,for certain configurations of masses,the total energy could be negative.Therefore,the Lagrangian itself has to be positively definite.

    The same problem appears in QFT,but there we invented the normal product.

    Daniel.
     
  4. Feb 9, 2005 #3
    Thanks. I somehow can convince myself that the mass matrix has to be positive definite. But specifically I need to show that the principle of least action *requires* the mass matrix to be positive definite. So is there a way?

    N.B. By principle of least action I mean, of all the possible paths between two given points, the one that the particle takes is the one of *minimum* (not only extremum) action
     
  5. Feb 9, 2005 #4

    dextercioby

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    It's quite difficult to show that the principle of extremum is in fact a principle of minimum,because,in fact,for nonphysical systems,it's not true.However,i think that it can be shown that the difference between two paths,one of which being the one for extremum (in this case it should be a minimum,but,because of the negative eigenvalues of the T matrix,it could be a maximum as well,or it can be entirely constant),is either positive or negative.You'd have to show that this difference can be negative,which would mean a maximum for stationary path.
    I really don't know how to do it exactly,you may wanna check Goldstein.In Landau & Lifschitz it definitely isn't.
    I gave you the Hamiltonian argument which is actually rock-solid,because the potential is time-independent,therefore the Hamiltonian is the (conserved) energy,which cannot be negative.

    Daniel.
     
  6. Feb 10, 2005 #5
    Are there any other suggestions, pls?
     
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