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Tensor gymnastic in Noether's Current

  1. Jul 15, 2011 #1
    ok, I hope this is the right place to post this, that's my first post here! ^_^
    Well I'm deriving the Noether's Current in terms of an improved energy momentum tensor and by now all the knowledge about tensors algebra and so on came from my own sadly. I'm stuck with a matter that probably reveals anything that is not really clear about tensors..
    I have:
    [itex]\frac{\partial}{\partial(\partial_{\mu}\phi)} \partial_{\rho}\phi\partial^{\mu}\phi[/itex]
    now I would write:
    a) [itex]=\frac{\partial}{\partial(\partial_{\mu}\phi)} (\partial_{\rho}\phi)\eta^{\mu\mu}(\partial_{\mu} \phi )=(\partial_{\rho}\phi)\eta^{\mu\mu}[/itex]

    a') [itex]=\frac{\partial}{\partial(\partial_{\mu}\phi)} \delta^{\mu}_{\rho}(\partial_{\mu}\phi)(\partial^{\mu}\phi)=
    \delta^{\mu}_{\rho}(\partial^{\mu}\phi)=
    \delta^{\mu}_{\rho}\eta^{\mu \mu }(\partial_{\mu} \phi )=
    (\partial_{\rho}\phi)\eta^{\mu\mu}[/itex]

    and I hope they are correct.. but then I thought.. what about this one?? o_O

    b) [itex]=\frac{\partial}{\partial(\partial_{\mu}\phi)} \delta^{\mu}_{\rho}(\partial_{\mu}\phi)\eta^{\mu \mu}(\partial_{\mu}\phi)=
    2(\partial_{\rho}\phi)\eta^{\mu\mu}[/itex]
     
  2. jcsd
  3. Jul 15, 2011 #2

    dextercioby

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    The only advice I can give you is to use different indices (summation vs free), for example

    [tex] \frac{\partial}{\partial \left(\partial_{\mu}\phi\right)} \left(\partial_{\rho}\phi\right) \left(\partial^{\rho}\phi\right) [/tex]
     
    Last edited: Jul 15, 2011
  4. Jul 15, 2011 #3

    haushofer

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    You write the indices in a wrong way; your expression should be

    [itex]\frac{\partial}{\partial(\partial_{\mu}\phi)} ( \partial_{\rho}\phi\partial_{\mu}\phi)[/itex]

    Then you use the algebraic rule

    [itex]
    \frac{\partial \phi(x)}{\partial \phi (y)} = \delta^4(x-y), \ \ \ \frac{\partial}{\partial(\partial_{\mu}\phi(y))} \partial_{\nu}\phi(x) = \delta^{\mu}_{\nu}\delta^4(x-y)
    [/itex]
    which also holds for higher derivatives. Actually, these are functional derivatives, but lets write it as this. With this you can calculate your expression using the product rule, which holds for functional derivatives.
     
  5. Jul 15, 2011 #4
    it depends on if you mean what you wrote or not!
    Notice: The free indices on both sides of the equality should be the same. this
    rules out your a, a', and b!

    [tex]
    \frac{\partial}{\partial(\partial_\mu \phi)} \partial_\rho \phi \partial^\mu \phi
    [/tex]
    is generally considered invalid because the repeated index should occur once
    in the lower and once in the upper position. your [itex]\mu[/itex] is in the upper
    position in both. [the reason physicists prefer this convention is that for special
    relativity we have a slight difference between 4-vectors with upper and lower indices
    and it's only with the upper/lower einstein convention that you get a lorentz invariant
    quantity.]

    Now let's look at a couple of ways we could "fix" what you wrote - and see how
    we'd work out the derivative in each case.
    [tex]\frac{\partial}{\partial \left( \partial_\mu \phi\right)} \partial_\rho\phi \partial_\mu \phi
    = \frac{\partial (\partial_\rho \phi)}{\partial \left( \partial_\mu \phi\right)} \partial_\mu \phi
    + \partial_\rho \phi \frac{\partial (\partial_\mu \phi)}{\partial \left( \partial_\mu \phi\right)}
    [/tex]
    [tex]
    = \delta^\mu_\rho \partial_\mu \phi + \partial_\rho \phi \delta^\mu_\mu
    = (D+1) \partial_\rho \phi
    [/tex]
    where D is the dimension of your vector space.


    if you meant [itex]
    \frac{\partial}{\partial (\partial_\mu \phi)} \partial_\rho \phi \partial^\rho \phi
    [/itex] as dextercioby suggests you get
    [tex]
    \frac{\partial}{\partial (\partial_\mu \phi)} \partial_\rho \phi \partial^\rho \phi
    = \frac{\partial (\partial_\rho \phi)}{\partial (\partial_\mu \phi)} \partial^\rho \phi +
    \partial_\rho \phi \frac{\partial (\partial^\rho \phi)}{\partial (\partial_\mu \phi)}
    [/tex]
    [tex]
    = \delta^\mu_\rho \partial_\rho \phi + \delta^{\rho \mu} \partial_\rho \phi
    = 2 \partial^\mu \phi
    [/tex]
     
  6. Jul 16, 2011 #5
    thanks guys you really clarify me that formula!And now I understand how to move through it!
    the only thing not clear is:
    given

    [itex]K^{\mu}_{\ \ \rho} = \partial_{\rho}\phi \partial^{\mu}\phi [/itex]

    if I have to compute

    [itex]S^{\mu}=\frac{\partial}{\partial(\partial_{\mu} \phi )}K^{\mu}_{\ \ \rho}[/itex]

    I have to change name to index? which way?thanks!
     
  7. Jul 16, 2011 #6

    haushofer

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    I'd go for


    [itex]S^{\rho}=\frac{\partial}{\partial(\partial_{\mu} \phi )}K_{\mu}^{\ \ \rho}[/itex]

    You seem to be confused about


    [itex]\frac{\partial}{\partial(\partial_{\mu} \phi )} = X^{\mu}[/itex]

    This is a tensor with a upper index mu! So you have to contract it with something which has a lower index mu. One free index then remains.
     
  8. Jul 17, 2011 #7
    ok thanks a lot you all! ^_^
     
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