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Proof of something .

  1. Feb 8, 2009 #1
    Proof of something.....

    I posted this in the calculus section...but no reply... I came across this while studying the most general noether's theorem which I will later apply to quantum field theory. So I am posting this here, hoping that some one would have come across this while studying qft

    I was learning noether's theorem and I came up with something that I can't seem to prove.

    Prove -

    [tex] d (\delta x) = \delta (d x)[/tex]

    I was trying to prove the more general expression

    [tex] d (\delta f) = \delta (d f)[/tex]

    where f( x, y, z, t, .....) is a function of n variables.

    If i could prove the first then the second is proved....any help?
  2. jcsd
  3. Feb 8, 2009 #2


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    Re: Proof of something.....

    I have never tried to prove it, I thought it should be taken for granted.

    now to a mathematician, you have to specify what \delta is I think...

    but this is how i think of it:

    [tex] \delta f(x) = f(x+\delta x) - f(x) [/tex]

    [tex] d(\delta f(x)) = d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx [/tex]

    [tex] \delta (df(x)) =
    d(f(x+\delta x) - f(x)) = \dfrac{df(x+\delta x )}{dx}dx - \dfrac{df(x )}{dx}dx [/tex]

    But as I said, this is how I see it, I don't think it is a "proof"
  4. Feb 11, 2009 #3


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    Re: Proof of something.....

    You are right that one must be careful in specifying what [tex] \delta [/tex] is, particularly in the context of Nother's theorem where [tex] \delta L [/tex] is often used to denote the variational derivtaive of the Lagrangian. For what it's worth, on pages 89 and 90 of Anderson's Principles of Relativity Physics , he takes pains to use a bar over [tex] \delta [/tex]to distinguish between the two uses. Furthermore, he states (without proof) the relationship you just derived. Personally, your argument convinces me (but then I'm a lousy mathematician :smile:)
  5. Feb 11, 2009 #4


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    Re: Proof of something.....

    Yeah, most physicsists are louse mathematicians, so am I, but this is the way I convince myself ;-)
  6. Feb 11, 2009 #5
    Re: Proof of something.....

    As they have written, it's necessary to specify what [tex] \delta f [/tex] means, but you also have to specify what [tex] df [/tex] means.

    Usually, but not always, (or, if you prefer, just "sometimes") [tex] \delta f [/tex] means to compute the variation of a definite integral, varying a function to which the integrand function depends. Example, you have the action S:

    [tex]S = \int_{t_1} ^{t_2} L[q(t),q'(t),t]\ dt[/tex]

    If you want to compute the variation of S when you varies the function q(t) for every value of t, without changing the extremes of integration, you will write:

    [tex]\delta S = \delta \int_{t_1} ^{t_2} L[q(t),q'(t),t]\ dt\ =\ \int_{t_1} ^{t_2} \delta L[q(t),q'(t),t]\ dt\ =\ \int_{t_1} ^{t_2}\ [\frac{\partial\ L}{\partial\ q} q'(t)\ +\ \frac{\partial\ L}{\partial\ q'}q''(t)]\ dt[/tex]

    (If you wanted to prove the Euler-Lagrange equation, you have to fix the values [tex]q(t_1)[/tex] and [tex]q(t_2)[/tex] in the evaluation of that integral).

    Now, what does dS mean, in this context? Of course nothing, since it's a number. It has a meaning if, for example, you write:

    [tex]S(t) = \int_{t_1} ^{t} L[q(\tau),q'(\tau),\tau]\ d\tau[/tex]

    But then you evaluate [tex] \delta S [/tex] in a completely different way (I let it to you as exercise).
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