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I Partial derivative used in Calc of Variation

  1. Sep 11, 2016 #1

    ibkev

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    I'm working through the discussion of calculus of variations in Taylor's Classical Mechanics today. There's a step where partial differentiation is involved that I don't understand.

    Given:
    $$S(\alpha)=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x)\,dx$$

    The goal is to determine ##y(x)## when the derivative of ##S(\alpha)=0## and ensure that ##\alpha=0## at this point.

    So this means:
    $$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \frac {\partial f} {\partial \alpha}\,dx=0$$

    and now here is the step that I don't understand. It's apparently an application of the chain rule but I'm just not seeing it...
    $$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \left( \eta\frac {\partial f} {\partial y}+\eta'\frac {\partial f} {\partial y'} \right)\,dx=0$$

    Suggestions are appreciated! :)
     
    Last edited: Sep 11, 2016
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  3. Sep 11, 2016 #2

    PeroK

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    First, Taylor is being a bit lazy (although in a conventional way) and using ##y## and ##y'## to stand for two different things. In ##\frac {\partial f} {\partial y}##, he's using ##y## as the first argument of the the function ##f## and, likewise, ##y'## as its second argument.

    But, the arguments are not simply ##y## and ##y'## so technically some additional notation is required.

    There's a previous post here for the same question:

    https://www.physicsforums.com/threads/euler-lagrange-derivation-taylor-series.852364/#post-5350620
     
  4. Sep 11, 2016 #3

    Twigg

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    The integrand f is a function of two variables ##y## and ##y'##. The notation used is a little confusing because you end up re-using the same variables ##y## and ##y'## to describe the unperturbed trajectory and its unperturbed velocity.

    To make it a little easier to see what's going on, you can change the notation a bit. Let ##u_{1}## and ##u_{2}## be coordinates for the position and velocity of a trajectory (any trajectory), and reserve the functions ##y(x)## and ##y'(x)## for a trajectory which makes the action integral S stationary. That makes f be a function of ##u_{1}## and ##u_{2}## in general, so it's written ##f(u_{1},u_{2})## in the new notation. That's how the chain rule comes into play. You want to find the trajectory ##(u_{1},u_{2}) = (y,y')##. You get the expression for ##S(\alpha)## by substituting ##(u_{1},u_{2}) = (y + \alpha \eta, y' + \alpha \eta')##. The final expression really says: $$\frac{dS(\alpha)}{d\alpha} = \int_{x_1}^{x_2} (\eta \frac{\partial f}{\partial u_{1}} + \eta' \frac{\partial f}{\partial u_{2}})dx$$
     
  5. Sep 11, 2016 #4

    ibkev

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    Ah ok - I'm starting to make sense of this now. Thanks!

    The thing I find strange about how Taylor does this is that he actually sets up the problem at first similar to the ##(u_1, u_2)## in Twigg's post but calls them ##(Y, Y')## and then deliberately drops them.

    He writes:
    $$\begin{align}
    S(\alpha)&=\int_{x_1}^{x_2} f(Y, Y', x),dx \\
    &=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x),dx
    \end{align}$$
    But then he next writes:
    $$\frac {\partial} {\partial\alpha} f(y+\alpha\eta, y'+\alpha\eta', x) =\eta\frac {\partial f} {\partial y} + \eta'\frac {\partial f} {\partial y'}$$
    Do I understand right that he actually means for that previous equation to say:
    $$\frac {\partial} {\partial\alpha} f(Y, Y' x)=\eta\frac {\partial f} {\partial Y} + \eta'\frac {\partial f} {\partial Y'}$$

    >Taylor is being a bit lazy (although in a conventional way)
    Honestly it seems like it should be an errata, no?
     
    Last edited: Sep 11, 2016
  6. Sep 11, 2016 #5

    Twigg

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    Yeah, this is what he means, but it's also implied that ##Y = y + \alpha \eta## inside the argument of the function f. The variables ##Y## and ##Y'## represent coordinates in a geometric space whose axes are the position and velocity of the trajectory, so the substitution ##(Y,Y') = (y + \alpha \eta, y' + \alpha \eta')## is just a way of parametrizing that space the same way we could parametrize ordinary 3D space with ##\vec{r} = \vec{r_{0}} + \alpha \vec{u}## where ##\vec{u}## is an undetermined unit vector. In that picture, the expression you mentioned and I quoted above is just the directional derivative of f along the "direction" ##(\eta, \eta')## in the ##Y##-##Y'## plane.

    I think Taylor intentionally dropped the Y,Y' notation because he probably didn't want to detract from the important points of this optimization procedure by making the theoretical distinction between y and Y, at the expense of confusing more cautious readers. While his notation is mathematically misleading, it still works and is a convenient shorthand.
     
  7. Sep 12, 2016 #6

    ibkev

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    That's a way of looking at it that I hadn't considered - thanks.

    Incidentally, I checked how the Morin Classical Mechanics text does it and he carries the Y, Y' through to the very end (although he calls them ##(y_\alpha, y_\alpha')## and only then does he replace them with y, y' (albeit magically and without any explanation.)

    After thinking about this, I came up with another way to look at this from the fact that there are TWO goals of our solution not one:
    1. solve ##\frac {dS(\alpha)} {d\alpha}=0## to find the local minimum
    2. require that ##\alpha=0## at that point as well
    So the final step of the derivation recognizes that although we defined ##(Y,Y') = (y + \alpha \eta, y' + \alpha \eta')##, setting ##\alpha=0## at the end as our problem requires, results in ##(Y,Y')=(y,y')##

    Does that seem like a reasonable way to look at it as well?
     
    Last edited: Sep 12, 2016
  8. Sep 12, 2016 #7

    Twigg

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    You are correct, and if you hadn't mentioned it I would've missed a second oversight in Taylor's math. Hamilton's principle does not say $$\frac{dS_{\alpha}}{d\alpha} = 0$$. That would would mean that all functions of the form ##y + \alpha \eta## where ##y## is a trajectory that makes the action integral stationary, are all stationary trajectories. What Hamilton's principle really says is this: $$\frac{dS_{\alpha}}{d\alpha}|_{\alpha = 0} = 0$$. In other words, it says that ##y## is the stationary trajectory, not just any trajectory of the form ##y + \alpha \eta##.
     
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