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What is the variation

  1. Dec 16, 2007 #1
    I always like to do the variational calculations in rigor way for example like this

    0 = D_{\alpha} \Big(\int dt\; L(x(t)+\alpha y(t))\Big)\Big|_{\alpha=0} = \cdots

    because this way I understand what is happening. However in literature I keep seeing the quantity

    \delta x(t)

    being used most of the time. What does this delta mean really? Does it have a rigor meaning? It seems to be same kind of mystical* quantity as the [itex]df[/itex], but this time an... infinite dimensional differential?

    *: mystical in the way, that even if the rigor meaning exists, it is not easily available, and the concept is usually used in non-rigor way.
  2. jcsd
  3. Dec 16, 2007 #2

    Ben Niehoff

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    According to my copy of Goldstein's Classical Mechanics (3rd ed., p.38), the variation is defined as:

    [tex]\delta y = \left \frac{\partial y}{\partial \alpha} \right|_{\alpha = 0} d\alpha[/tex]


    [tex]y(x,\alpha) = y(x,0) + \alpha \eta(x)[/tex]

    for some arbitrary function [itex]\eta[/itex].

    I share your sentiment that the mathematics of physics ought to be done in a rigorous way...for my own edification, I went through the variational calculus bits of Goldstein's and proved them all myself. It's actually not too hard, despite the fact that function space is infinite-dimensional.
  4. Dec 16, 2007 #3
    Now [itex]\delta y[/itex] depends on the chosen [itex]\eta[/itex]. If a notation [itex]\delta y_{\eta}[/itex] was given, would that be analogous to [itex]dx_i[/itex] where i=1,2,3?
  5. Dec 16, 2007 #4

    Ben Niehoff

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    No, because [itex]\eta(x)[/itex] is not really an "index". [itex]\eta(x)[/itex] is just an arbitrary function; it is simply left undefined exactly what sort of function it is. You never "choose" it. When you work through the details of the math, you get to a point where the definition "[itex]\eta(x)[/itex] is arbitrary" can be used to advance the proof. Specifically, you will get to a point where you have

    [tex]\int_{x_1}^{x_2} M(x)\eta(x)\, dx = 0[/tex]

    which, if it is true for all functions [itex]\eta(x)[/itex], must therefore imply

    [tex]M(x) = 0[/tex]

    And, setting M(x)=0 allows you to derive the Euler equations (which are generic to the calculus of variations, and have nothing specific to do with the Lagrangian formalism).
  6. Dec 16, 2007 #5
    There was a mistake with my analogy. In three dimension we can take a gradient of a function with expression [itex]u\cdot\nabla f[/itex], where [itex]u=(u_1,u_2,u_3)[/itex] is some vector. The mapping [itex]x\mapsto \eta(x)[/itex] should be considered analogous to the mapping [itex]i\mapsto u_i[/itex]. So [itex]\eta[/itex] alone, is analogous to [itex]u[/itex], and not to the indexes.

    Anyway there are several things that don't make sense to me. What is the [itex]d\alpha[/itex]?

    If [itex]\eta[/itex] is arbitrary, then [itex]\delta y[/itex], whatever it is, at least seems to be arbitrary too.

    What did the "arbitrary" mean? I would like to have some kind of parameter interpretation for it. For example we might say that x is arbitrary in f(x), but we don't really want to mean an arbitrary number f(x), but instead a mapping [itex]x\mapsto f(x)[/itex].
    Last edited: Dec 16, 2007
  7. Dec 16, 2007 #6

    Ben Niehoff

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    [itex]\alpha[/itex] is just a parameter, so [itex]d\alpha[/itex] is just the (ordinary) differential of [itex]\alpha[/itex]. It ends up going under an integral sign.

    [itex]\eta[/itex] is arbitrary in the same sense that a constant of integration is arbitrary:

    [tex]\int f(x) dx = F(x) + C[/tex]

    Here, C is arbitrary, meaning that you can choose any number to be C, and the equation is true.

    The only difference is that [itex]\eta(x)[/itex] is a function. So, "arbitrary" in this case means that [itex]\eta(x)[/itex] can be any function. (Actually, there is a slight restriction, in that [itex]\eta(x_1) = 0[/itex] and [itex]\eta(x_2) = 0[/itex], but [itex]\eta(x)[/itex] can take on any value whatsoever at other points).
  8. Dec 16, 2007 #7
    But we are not integrating over alpha anywhere in the calculation.
  9. Dec 16, 2007 #8
    This started very confusingly. It could be smarter to not try to understand everything that got already said....
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