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Euler-Lagrange equation derivation

  1. Jun 26, 2011 #1
    I'm trying to understand the derivation of the Euler-Lagrange equation from the classical action. This has been my main source so far. The issue I'm having is proving the following equivalence:

    [tex]
    \int_{t_1}^{t_2} [L(x_{true} + \varepsilon, \dot{x}_{true} + \dot{\varepsilon},t) - L(x_{true}, \dot{x}_{true},t)] \mathrm{d}t = \int_{t_1}^{t_2} (\varepsilon \frac{\partial L}{\partial x} + \dot{\varepsilon} \frac{\partial L}{\partial \dot{x}}) \mathrm{d}t
    [/tex]

    I understand the idea behind their equivalence intuitively, The derivative of a function is the change in that function, and I see how on the left side there is a representation of a small change in the lagrangian, but I'm having a hard time proving this to myself mathematically and I'd like some help.

    I understand all the other steps shown in the derivation.

    Thanks to anyone that responds.
     
  2. jcsd
  3. Jun 27, 2011 #2
    You should ignore the integral, and just focus on showing that the things under the integrals are equal. If you have a function f(x), then by definition of the derivative:

    [f(x+h)-f(x)]/h=f'(x)

    when h goes to zero.

    Therefore f(x+h)-f(x)=h*f'(x)

    This is for one variable, and the generalization to two variables is:

    f(x+h,y+g)-f(x,y)=h*Dxf+g*Dyf

    where Dx and Dy are the partial derivatives in the x-direction and y-direction respectively.

    So just let y be x dot, and f be your Lagrangian, and you get the result.
     
  4. Jun 27, 2011 #3
    Ok, that makes sense. I understand how it works with just one independent variable, but I did not realize the generalization to two variables. Thanks for that.
     
    Last edited: Jun 27, 2011
  5. Jun 27, 2011 #4
    O, I didn't know that you know how it works with one variable. In that case:

    f(x+h,y+g)-f(x,y)=[f(x+h,y+g)-f(x,y+g)]+[f(x,y+g)-f(x,y)]=
    h*Dxf(x,y+g)+g*Dyf(x,y)

    which you get just from the 1-variable definition of the derivative.

    Now the key is that applying the one variable definition of the derivative again:

    h*Dxf(x,y+g)=h*Dxf(x,y)+g*h*DxDyf(x,y)

    and if g and h are really small, just keep the first term.

    So f(x+h,y+g)-f(x,y)=h*Dxf(x,y+g)+g*Dyf(x,y)=h*Dxf(x,y)+g*Dyf(x,y)
     
  6. Jun 27, 2011 #5
    Thanks for clearing that up for me. That's a clever trick to apply the one variable definition again. To be honest, it's bit unsettling to me that we disregard a term based on g and h being very small. However, I do realize in the derivation on the wikipedia page I linked that that term would be very small.
     
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