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Variations, Euler-Lagrange, and Stokes

  1. Sep 10, 2007 #1

    haushofer

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    Hi, I have some questions which I encountered during my thesis-writing, I hope some-one can help me out on this :)

    First, I have some problems interpreting coordinate-transformations ( "active and passive") and the derivation of the Equations of Motion. We have

    [tex]
    S = \int L(\phi, \partial_{\mu} \phi) d^{4}x
    [/tex]

    and

    [tex]
    \delta S = 0
    [/tex]

    , in which the variation of the field is arbitrary. My question is: how exactly is this variation defined? One has 2 options:

    [tex]
    \delta \phi = \phi^{'}(x^{'}) - \phi (x)
    [/tex]
    or
    [tex]
    \delta \phi = \phi^{'}(x) - \phi (x)
    [/tex]

    where the difference lies in the argument. In notes of Aldrovandi and Pereira ( Notes for a classical course on fields ) option 1 is choosen. And in Inverno they say that option 2 is a coordinate transformation. Why is that? I tend to choose for option two, because here you actually change the field; I would say that a scalar quantity is invariant under transformation 1, so here you just state general covariance in stead of obtaining the Equations of Motion. At the other hand, you can always choose x=x'. In the end I want to look at why one is able to commute variations and partial derivatives, so I need the exact definition of the variation of the field.

    Another question concerns Stokes theorem. I understand the theorem totally for n-forms ( where you define the boundary of your region with chains etc ), but why is it also valid for example in the derivation of the Euler-Lagrange equations or tensor densities?

    Many thanks in forward :)

    -edit My TeX-code is not working properly for some reason, but I hope it is clear.
     
    Last edited: Sep 10, 2007
  2. jcsd
  3. Sep 12, 2007 #2

    haushofer

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    No-one? :(
     
  4. Sep 12, 2007 #3

    haushofer

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    Ok, let me put it differently. I know about pull-backs, push-forwards, diffeomorphisms etc. I always regarded a push-forward of a vector as some sort of "generalized coordinatetransformation", but still as passively. Why is a push-forward an active transformation, and a coordinatetransformation just passive?
     
  5. Sep 12, 2007 #4
    I suggest a good textbook -- Analysis, Manifolds and Physics is good. It covers differential geometry over finite and infinite dimensional spaces, which are what you need for the formal treatment of variational calculus.
     
  6. Sep 12, 2007 #5

    haushofer

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    I followed a course on differential geometry, and have the idea that I have enough knowledge of understanding this, but that I'm overlooking something.
     
  7. Sep 15, 2007 #6

    samalkhaiat

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    Last edited: Sep 15, 2007
  8. Sep 21, 2007 #7

    haushofer

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