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Diagonalized Lagrangian

  1. Oct 23, 2008 #1
    My Graduate Mechanics text (Walecka and Fetter) says in the chapter on Small Oscillations, that the Modal Matrix Diagonalizes the Lagrangian

    [tex] L = \sum_{\sigma=1}^\infty (\dot{\zeta_\sigma}^2 - \omega_\sigma \zeta_\sigma^2) [/tex]

    where [tex]\zeta[/tex] are the normal coordinates related to the original coordinates [tex]\eta[/tex] via the modal matrix.

    But as far as I understand, the Lagrangian is simply a number, not a matrix. So what does the diagonalization of the Lagrangian mean.
     
  2. jcsd
  3. Oct 24, 2008 #2

    dx

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    The Lagrangian is not a number, it's a function. In the case of small oscillations, it happens to be a quadratic form, and it can be written as

    [tex] L = Q^{T}AQ [/tex],

    Where [tex] Q = (q_1, q_2, ..., q_s, \dot{q}_1, \dot{q}_2, ..., \dot{q}_s) [/tex]. When the [tex] q_i [/tex] are normal coordinates, [tex] A [/tex] becomes a diagonal matrix.
     
  4. Oct 24, 2008 #3
    Sorry for saying that its a number but I still didn't get it. And I think I basically want to know how can Lagrangian the function be transformed into Lagrangian the matrix. e.g. we have [tex] 2L = \dot{\eta}^T m \dot{\eta} - \eta^T v \eta [/tex]. But the [tex] \eta [/tex] are essentially column vectors, giving me a function in the form of a 1x1 matrix after multiplication. So what does it mean to write the Lagrangian in the form of a matrix
     
  5. Oct 24, 2008 #4

    dx

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    Well, its not exactly a matrix. But matrices are not the only things that can be diagonalized. Quadratic forms for example can be diagonalized. If you have some random set of generalized coordinates and velocities [tex] X = (x_i, \dot{x}_i) [/tex], then the Lagrangian for small oscillations is of the form

    [tex] L = \frac{1}{2}\sum (m_{ik} \dot{x_i} \dot{x_k} - k_{ik} {x_i} {x_k}) [/tex].

    This can be written in matrix notation as [tex] L = X^{T}AX [/tex], where A is in general not diagonal. This is the modern way of representing quadratic forms. It turns out that there is a specific set of coordinates [tex] Q_i [/tex] in terms of which the Lagrangian simplifies into the form

    [tex] L = \frac{1}{2}\sum_{\alpha} (\dot{Q_{\alpha}^2} - {\omega_{\alpha}^2} {Q_{\alpha}^2}) [/tex].

    When you write this in the form [tex] Q^{T}AQ [/tex], A will be a diagonal matrix.
     
    Last edited: Oct 24, 2008
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