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Equation of Motion from Lagrangian - Zee QTF in a Nutshell

  1. May 29, 2014 #1
    I need help understanding how to apply the formula for converting a Lagrangian to an equation of motion in this following specific application.
    On page 4 of Zee's QTF in a Nutshell, he gives a Lagrangian (equation 1). In the following sentence he gives the corresponding equation of motion. I know the formula for converting a Lagrangian, but I don't get/understand Zee's answer. I have searched books and online without finding a similar worked out example.
    Help is most appreciated. (I hope the formulas show below show up correctly.)
    Thanks, Richard
    (1) L = 1/2(∑am(dqa/dt)2 - ∑a,bkabqaqb - ......)

    equation of motion => m(d2qa/dt2)a = - ∑bKabqb
  2. jcsd
  3. May 29, 2014 #2
    Maybe you can show your attempt to go from the given Lagrangian to some equations of motion? It would be helpful to see where exactly you get stuck or go wrong.
  4. May 29, 2014 #3
  5. May 29, 2014 #4


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    Just apply the Euler Lagrange equations. That is always the standard way of going from a Lagrangian to an equation of motion:

    $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}=0$$
  6. May 29, 2014 #5
    Here is what I had done that didn't work.
    Chosing a particular point mass which I still called qa, gets rid of the summations over index a. Now I applied the Lagrangian to equation of motion formula.

    d/dt(∂L/∂q[itex]^{}.[/itex]a) - ∂L/∂qa = 0

    The first term LHS comes out correct as mq[itex]^{}..[/itex]a

    for the second term LHS I get 1/2∑bkabqb

    The 1/2 multiplier from the Lagrangian didn't go away. This doesn't match the equation of motion in the book.
  7. May 29, 2014 #6
    My previous post has the time derivative dot and the "a" subscript reversed. Oops.
  8. May 29, 2014 #7


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    The second term has that 1/2 because the sum is over all values of a and b, and the summand is symmetric (assuming that k is symmetric).

    Maybe it's easier to see if you write out the terms assuming only 2 q's q1 and q2:

    You will get a Lagrangian which looks like this:


    Since k should be symmetric (it is, after all, defined to be the spring constant between 2 masses, and so how could it not be symmetric?), you get 2 identical terms at the end there.
  9. May 29, 2014 #8
    So are you Ok now?
  10. May 30, 2014 #9
    Thanks for all the good input. I followed the advice to write out a few terms in the series and to note that the K terms with the indexes swapped were identical. Then I was able to see how the 1/2 went away in the equation of motion. From there I could see how to generalize the answer to more point masses. Thanks again!
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