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From Noether's Theorem to Stress-Energy Tensor

  1. May 26, 2009 #1
    Hi

    The following is a standard application of Noether's Theorem given in most books on QFT, in a preliminary section on classical field theory. Reproduced below are steps from the QFT book by Palash and Pal, which I am referring to, having read the same from other books. I have some trouble with a step which I have highlighted in bold. I would be grateful if someone could clarify my doubts. Thanks in advance!

    While I understand how the last integral has been written, by employing Gauss's Law, I don't understand how

    [tex]\int_{\Omega'-\Omega}d^{4}x \mathcal{L}(\Phi'^{A},\partial_{\mu}'\Phi'^{A}) = \int_{\partial\Omega}dS_{\lambda} \delta x^{\lambda}\mathcal{L}(\Phi^{A},\partial_{\mu}\Phi^{A})[/tex]

    Can someone please explain this step? Where did the [itex]\delta x^{\lambda}[/itex] come from?

    Thanks.

    EDIT - I think I got it: [itex]d^{4}x = dS_{\lambda}x^{\lambda}[/itex]
     
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2
    Yeah, you pretty much got it with your edit. This entire derivation looks mathematically funny, but I think it's correct.

    I don't think it's trivial to say that d^4x=dS dot delta x; you have to imagine the geometry in your head. Say your coordinate transformation is a stretch in the z-direction only. Then why is the new surface only created with its normal to z? You also create surfaces parallel to the z-axis when you stretch the box in the z-direction, but these are only infinitismal and can be neglected. The z-plane that used to be the boundary before the stretch is the non-infinitismal surface that's being changed by your stretch in the z-direction.
     
  4. May 26, 2009 #3
    Thanks RedX. Some more help is needed in interpreting whats going on in the next couple of steps:

    Whats going on here?
     
  5. May 26, 2009 #4
    Take the function f(x)=x^2. You can change it by adding x^3 to it, and the new function is:

    g(x)=x^2+x^3

    If you label g(x) as f'(x), then: f'(x)-f(x)=[tex]\overline{\delta}}[/tex]=x^3

    But you can also change the function by messing with the argument. So change the argument to x+1:

    h(x)=(x+1)^2

    If you label h(x) as f(x'), then f(x')-f(x)=2x+1

    If you add these two changes, you get the total change [tex]\delta f[/tex].

    So they are just classifying the different types of changes, except they are using infintinismal changes.
     
  6. May 27, 2009 #5
    Shouldn't we have [itex]\partial_{\mu}f'(x)\delta x^{\mu}[/itex] in the last equality?
     
  7. May 27, 2009 #6
    The last quotation you have: "where we have ignored higher terms ... in writing the last equality" explains why. You're talking about infinitismal changes. So [tex]\partial_{\mu}f'(x)\delta x^{\mu} [/tex] is already first order because of the small delta x^mu. So if f'(x)=f(x)+[tex]\overline{\delta}f(x) [/tex], then [tex]\partial_{\mu}f'(x)\delta x^{\mu}= \partial_{\mu}(f(x)+\overline{\delta}f(x))\delta x^{\mu}[/tex], and having two deltas is too small to care about - one is enough. So f(x) can be used instead of f'(x) for that term.
     
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