1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Diagonalized Lagrangian
  1. Dissipative Lagrangian (Replies: 3)

  2. Lagrangian Derivation (Replies: 2)

  3. Lagrangian mechanics? (Replies: 2)

Loading...