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Sloppy Derivation of Euler-Lagrange.

  1. Jan 13, 2016 #1
    Just wondering how much validity there is to this derivation, or if it's just a convenient coincidence that this works.

    We have a Lagrangian dependent on position and velocity: [tex] \mathcal{L} (x, \dot{x}) [/tex]
    Let's say now that we've perturbed the system a bit so we now have:
    [tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) [/tex]
    But the physics can't change from a perturbation, therefore:
    [tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) [/tex]

    We can expand the perturbed Lagrangian to first order for both position and velocity:
    [tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) + \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} [/tex]

    This implies that:
    [tex] \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} = 0 [/tex]
    which I will call "equation A".

    [tex] \delta \dot{x} = \frac{d}{dt} \delta x [/tex]
    we can use the product rule:
    [tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x + \frac{\partial \mathcal{L} }{\partial \dot{x}} \delta \dot{x} [/tex]

    which we can plug into equation A and get:
    [tex] \left( \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} \right) \delta x + \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = 0 [/tex]

    You can see the Euler-Lagrange equation in there. All that's left to eliminate is the term:
    [tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) [/tex]
    You can argue that this term is 0, because complete time derivative terms can be integrated over t, and the endpoints can be chosen to have dx(endpoint) = 0.

    So then you are left with:
    [tex] \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} = 0 [/tex]

    Anyone with the patience to read this, what are your thoughts? Thank you.
  2. jcsd
  3. Jan 14, 2016 #2
    You have most of the steps of the derivation of the Euler-Lagrange equations (modulo defining an action and stating an extremization principle), but the statement which is clearly unfounded is ##\mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x})##. This hugely restricts the set of Lagrangians you can consider. In fact it eliminates the Lagrangians we almost always consider which are quadratic in ##\dot{x}## and a non-constant in ##x##.
  4. Jan 14, 2016 #3


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    The point is that you want to find stationary points of the action, i.e., to solve a variational problem. Then all these steps can be made rigorous. Of course, the Lagrangian is not invariant under arbitrary shifts of the orbit in configuration space. This would imply that the Lagrangian is a constant.
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