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Is Susskind's derivation Euler-La Grange rigorous?

  1. Sep 26, 2012 #1
    Beginning at 31:03 Dr. Susskind presents an intuitively very satisfying derivation of the Euler-La Grange equation(s). But, I'm not convinced it is rigorous. It seems his choice of variation is not the only possible choice for the neighborhood he selected.




    The reason this matters to me is because, in his derivation velocity and position seem to be implicitly coupled. My understanding of the Euler-La Grange equations is that position and velocity are independent variables. Lemons Section 2.2 gives a more abstract and symbolic derivation. I understand that to say that: given any variation in position, there are an infinite number of variations in velocity possible, and vis-versa.

    Can these two derivations be shown to be equivalent?
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Sep 27, 2012 #2
    IMO That might be the worst explanation i've ever seen. Although,
    it has a nice feel, and the ideas are solid. This is the technique for
    path integrals of Feynman.

    No, it's not rigorous. He's passing from discrete to continuous
    very slipshod-illy. It can be made rigorous provided you can give
    a good description how to pass to and from the continuum.
     
  4. Sep 27, 2012 #3

    vanhees71

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    It should be clear that in the Lagrange version of Hamilton's principle the position-space trajectories are varied, not position-velocity space trajectories. Since time is not varied by definition, you have the connection
    [tex]\delta \dot{q} =\frac{\mathrm{d}}{\mathrm{d} t} \delta q.[/tex]
    Further the endpoints of the position-space trajectory are fixed.

    I've not the time to watch lectures on the internet. So I can't say, whether Suskind's derivation is rigorous or not.

    BTW: Hamilton's principle in the Hamilton formulation is extended compared to the Lagrange version, and there the variation is wrt. phase-space trajectories, i.e., position-space and conjugate-momentum variables are varied independently. Still only the position-space variables are fixed at the boundary, the momenta are free.

    Indeed you get a deeper physics understanding of why these variational principles work as they do is from Feynman's path-integral formulation of quantum mechanics, but that's of course not necessary to understand classical analytical mechanics.
     
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