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Derivation of Euler-Lagrange Equation

  1. Apr 3, 2009 #1
    I am stuck in trying to understand the derivation of the Euler-Lagrange equation. This mathematical move is really bothering me, I can't figure out why it is true.

    [tex]\frac{\partial f(y,y';x)}{\partial\alpha}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha}+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial\alpha}[/tex]

    Why is it not:

    [tex]\frac{\partial f(y,y';x)}{\partial\alpha}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha}+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial\alpha}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial\alpha}[/tex]

    Or better yet:

    [tex]\frac{df(y,y';x)}{d\alpha}=\frac{\partial f}{\partial y}\frac{d y}{d\alpha}+\frac{\partial f}{\partial y'}\frac{d y'}{d\alpha}+\frac{\partial f}{\partial x}\frac{d x}{d\alpha}[/tex]


    Edit: I think I have found some sound reasoning. That y is a function of x and alpha. So the derivative of f w.r.t. alpha does not involve x. In fact x is held constant for that derivative. That is why dy/d(alpha) and dy'/d(alpha) are partials and not total derivatives.
     
    Last edited: Apr 3, 2009
  2. jcsd
  3. Apr 3, 2009 #2
    I don't think you would say that [itex]\partial f/\partial x = 0[/itex]. Instead you are allowing y to vary with alpha (for all and any values of x), but x is not a function of this variation.

    I don't know if you've seen it, but Feynman's lectures (volume II in what he calls the "entertainment" chapter)
    has some nice simple examples of these ideas.
    They cut the complexity out of the picture and treat some simple concrete cases. One of these simplest examples was a pure kinetic Lagrangian density, like the following (he probably included the mass and factor of one half but I've been lazy) :

    [tex]
    S = \int_{a}^b \dot{y}^2 dx
    [/tex]

    Let

    [tex]
    y = \bar{y} + \alpha
    [/tex]

    Here [itex]\bar{y}[/itex] is the optimal solution that you are looking for, and [itex]\alpha(a) = \alpha(b) = 0[/itex] (variation vanishes at the end points).

    Taking derivatives
    [tex]
    \dot{y} &= \dot{\bar{y}} + \dot{\alpha}
    [/tex]

    [tex]
    \begin{align*}
    S
    &= \int_{a}^b \left(\dot{\bar{y}} + \dot{\alpha} \right)^2 dx \\
    &= \int_{a}^b \left(\dot{\bar{y}}^2 + 2 \dot{\alpha}\dot{\bar{y}} + \dot{\alpha}^2 \right) dx \\
    &= \int_{a}^b \left(\bar{\dot{y}}^2 - 2 {\alpha}\ddot{\bar{y}} - {\alpha}\ddot{\alpha} \right) dx \\
    \end{align*}
    [/tex]

    and for the derivative to be zero
    [tex]
    \begin{align*}
    0 &= \left. \frac{dS }{d\alpha} \right\vert_{\alpha = 0} \\
    &= -2 \int_{a}^b \ddot{\bar{y}} dx \\
    \end{align*}
    [/tex]

    Dropping the overbar, you have for the desired solution to the variation

    [tex]
    \ddot{y} = 0
    [/tex]

    (particle moves with constant velocity in absence of force).

    EDIT: In retrospect, the point of my example may not be clear. When x is the dependent variable (y = y(x)), x is not varied, instead it is y that is varied, and I was hoping the example showed why you don't have a [itex]\partial x/\partial \alpha[/itex] as in the second two equations you listed.
     
    Last edited: Apr 3, 2009
  4. Apr 3, 2009 #3
    I think perhaps a better understanding might come from writing
    [tex]
    y = \bar{y} + \alpha \eta(x)
    [/tex]
     
  5. Apr 5, 2009 #4
    Thank you, everyone always seems to come through for me here. I am not sure why I have to be reminded to check the feynman lectures, it should be one of the first things I think of by now.
     
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