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Homework Help: Difference between 2 Langrangians, proof

  1. Apr 23, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Demonstrate that if the Lagrangians [tex]L(q, \dot q ,t)[/tex] and [tex]L'(q, \dot q , t)[/tex] who differ in a total derivative of a function f(q,t) give the same motion equations.
    That is, [tex]L'=L +\frac{d}{dt}f(q,t)[/tex]

    2. Relevant equations
    Euler-Lagrange's equation.


    3. The attempt at a solution
    I tried to use Euler-Lagrange's equation to see if I could reach the same equations for L and L' but without any success. I've checked out in Landau & Lifgarbagez's book. Here is what it more or less says: "If we have [tex]L'=L+\frac{d}{dt}(2a \vec r \cdot \dot \vec r+a \dot r ^2 t)[/tex], we can omit the second term since it's a total derivative with respect to time."
    So according to this book it's obvious while I have to prove it. But I don't understand why it's obvious nor why it's true.
     
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  3. Apr 23, 2010 #2

    Dick

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    It's obvious because the Euler-Lagrange equations come from the variation of the integral of L*dt. A total time derivative in L can only contribute a constant to the integral of L*dt. The variation of a constant is 0.
     
  4. Apr 23, 2010 #3

    fluidistic

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    I thank you for the reply, much appreciated.
    I'm still not getting it. Ok for your 1st and 3rd sentence, but I don't see why the second is true.
    If I understand you, you are implying that [tex]\int _{t_0}^{t_1} \frac{d}{dt} f(q,t) dt=\int _{t_0}^{t_1} d f(q,t)=K[/tex], a constant. Also the integral of d f(q,t) doesn't make sense to me. I'm missing something...
     
  5. Apr 23, 2010 #4

    gabbagabbahey

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    Just use the fundamental theorem of calculus...
     
  6. Apr 23, 2010 #5

    fluidistic

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    That's what I had done first on my draft, but I wasn't sure it was right. So it's worth [tex]f(q,t_2)-f(q, t_1)[/tex]. But I'm not sure q depends on t. If it does then we indeed reach a constant, but if it doesn't, I don't see it.
    Also, I'm not sure I can use the FTC since I have an integral of a differential of a function, not a differential of a variable. I mean I don't even have a "dt" or whatsoever.
     
  7. Apr 23, 2010 #6

    Dick

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    q does depend on t. But in euler-lagrange you fix the endpoints. So q(t1) is fixed and q(t2) is fixed.
     
  8. Apr 23, 2010 #7

    fluidistic

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    Ok I see. If q is constant we're done and if q depends on t we're also done. It cannot depend on anything else I guess.
    Now am I right in post #3? How can I apply the FTC? I don't have a differential of a variable.
     
  9. Apr 23, 2010 #8

    Dick

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    You have integral of d/dt(f(q(t),t)*dt from t1 to t2. That's f(q(t2),t2)-f(q(t1),t1) by FTW as gabbagabbahey said. Everything is fixed at the endpoints. It's a constant.
     
  10. Apr 23, 2010 #9

    fluidistic

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    Thanks I now get it. Those differentials! I missed that I shouldn't have simplified the dt's in the equation [tex]\int _{t_0}^{t_1} \frac{d}{dt} f(q,t) dt=\int _{t_0}^{t_1} d f(q,t)[/tex] and that I should have applied it (the FTC) in the left hand side. Sorry to both for being so slow minded. I'm happy I've understood at least.
    Thanks once again.
     
  11. Apr 24, 2010 #10

    gabbagabbahey

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    On a side note, you can also use the chain rule to show that

    [tex]\frac{\partial L'}{\partial q}-\frac{d}{dt}\frac{\partial L'}{\partial \dot{q}}=\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}[/tex]
     
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