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Euler Lagrange derivation in book

  1. Nov 4, 2011 #1

    Can any1 recommend a book that will show the derivation of the Euler-Lagrange equation.
    (I am learning in the context of cosmology ie. to extremise the interval).
    Ideally the derivation would be as simple/fundamental as possible - my maths is not up to scratch!
  2. jcsd
  3. Nov 4, 2011 #2
    The first chapter of Classical Mechanics by Herbert Goldstein has an excellent and very succinct derivation starting from D'Alembert's Principle. (By succinct, I mean that the information density is high and it takes a while to wade through it! However, all the steps are clearly laid out.)

    The second chapter uses variational calculus to derive the equation from Hamilton's Principle. Again, the information density is high, but the treatment is excellent and complete.

    That the book has been constantly in print for 60 years speaks well of its merit.
  4. Nov 5, 2011 #3
    I think maybe I'll take the opportunity to give my intuitive derivation (very non-rigorous and objectionable, but cute), which I don't think you will see anywhere else (at least not in this form), since I came up with it. To save time, I'll be fairly sloppy.

    I'll just do the case of one particle confined to a move in a straight line. You want to choose a path q(t) that is extremal for the functional

    [itex]\int L(q,\dot{q},t) dt [/itex]

    Now, you want to perturb the path by a little bit and the integral doesn't change (to a first order approximation). You can imagine that the path is formed by a bunch of vectors, dq_i, going tail to tip, each one being traversed in time dt. So, imagine you make one of the vectors longer, starting at time, t, so that q gets bigger (make the next vector have the same tip). This makes the integral go up by [itex]\frac{\partial L}{\partial q} [/itex].

    But, you have also changed [itex]\dot{q}[/itex] because the velocity vector is longer. So you add [itex]\frac{\partial L}{\partial \dot{q}} [/itex] at time t and you subtract it at time t + dt.

    So you have three terms that are changing the integral, but you can see that the latter two are just [itex]-\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} [/itex] when dt is small.

    So, there you have it: [itex]\frac{\partial L}{\partial q}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = 0[/itex]

    Quite a beautiful and transparent (perhaps, after some thought) argument, I find, if you are well-prepared enough for it. The case of more particles and dimensions is not significantly different.

    Ah, the wonders of not being rigidly formal all the time...
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