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Question from Lanczos book - genearlized momentum

  1. Apr 8, 2009 #1
    If you have the 4th ed. of Cornelius Lanczos' book on Variational Principles of Mechanics, could you explain to me how he got from eq. 53.4 to 53.5 (page 121). Eq. 53.4 defines the generalized momentum as the derivative of the Lagrangian, L, with respect to the velocities. Then, in eq. 53.5 he refers to the derivatives of the generalized momenta as equaling the derivative of L with respect to the positions. Is this right? This seems so simple yet for some reason (or lack thereof :blushing:) I'm not seeing the math from 53.4 to 53.5.

  2. jcsd
  3. Apr 8, 2009 #2
    I don't think everyone has this book, so could you please write down the equations that he mentions?
  4. Apr 8, 2009 #3


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    It follows from the Euler-Lagrange equations:

    [tex] \frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial v_{\alpha}} = \frac{\partial \mathcal{L}}{\partial x_{\alpha}} .[/tex] ​

    Setting [tex] p_{\alpha} = {\partial{\mathcal{L}}}/{\partial v_{\alpha}} [/tex] in this, you get

    [tex] \frac{d p_{\alpha} }{dt} = \frac{\partial \mathcal{L}}{\partial x_{\alpha}} .[/tex] ​
    Last edited: Apr 8, 2009
  5. Apr 8, 2009 #4
    Thanks dx! Don't know why I didn't seen that.
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