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Covariant vector differentiation problem with kronecker delta?

  1. Sep 13, 2011 #1
    I'm having trouble understanding the proof/solution below (please see photo, I also wrote out the problem below). I highlighted the part of my problem in red (in the picture attached). Basically I'm not sure what identity they use to get the Kronecker delta after differentiating or whether they use something else to get that part.

    "Since A/[itex]\nu[/itex] is covariant vector, it satisfies


    A/[itex]\nu[/itex] = ([itex]\partial[/itex]x[itex]\rho[/itex][itex]/[/itex][itex]\partial[/itex]x/[itex]\nu[/itex])A[itex]\rho[/itex]

    A/[itex]\rho[/itex] = ([itex]\partial[/itex]x/[itex]\mu[/itex][itex]/[/itex][itex]\partial[/itex]x[itex]\rho[/itex])A/[itex]\mu[/itex]

    This next part gives me trouble with the relation:

    ([itex]\partial[/itex]A[itex]\rho[/itex][itex]/[/itex][itex]\partial[/itex]A/[itex]\nu[/itex]) = ([itex]\partial[/itex]x/[itex]\nu[/itex][itex]/[/itex][itex]\partial[/itex]x[itex]\rho[/itex]) [itex]\delta[/itex][itex]\nu[/itex][itex]\mu[/itex]

    I'm lost as to how the kronecker delta appears. I know that

    ([itex]\partial[/itex]x[itex]\nu[/itex][itex]/[/itex][itex]\partial[/itex]x[itex]\mu[/itex]) = [itex]\delta[/itex][itex]\nu[/itex][itex]\mu[/itex]

    [Mod Note: Over-sized image removed. See attachment. Hoot]
     

    Attached Files:

    Last edited: Sep 13, 2011
  2. jcsd
  3. Sep 13, 2011 #2
    I updated the question without the photo, thanks.
     
  4. Sep 13, 2011 #3

    haushofer

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    One has
    [tex]
    \frac{\partial A_{\rho}}{\partial A'_{\nu}} = \frac{\partial}{\partial A'_{\nu}} \frac{\partial x^{'\mu}}{\partial x^{\rho}} A'_{\mu} = \frac{\partial x^{'\mu}}{\partial x^{\rho}}\frac{\partial A'_{\mu}}{\partial A'_{\nu}}
    [/tex]
    So the functional derivative wrt A goes right to the transformation matrix. The primes on the fields A can then safely be removed; the components of A are a priori linear independent, if you write them in a primed coordinate system or not (the prime is just a label now). This is where the delta comes from:
    [tex]
    \frac{\partial A_{\rho}}{\partial A'_{\nu}} = \frac{\partial x^{'\mu}}{\partial x^{\rho}}\delta^{\nu}_{\mu}
    [/tex]
     
  5. Sep 14, 2011 #4
    What does wrt mean? Written?
    What does priori linear independent mean?

    Are there missing steps that you aren't showing, maybe that would help.

    Forgive me I'm still confused, what math class teaches this in terms of schooling or what theorem are you using. I never really studied this formally. I did take linear algebra once and studied linear independence but not related to calculus and matrices.

    thanks.
     
    Last edited: Sep 14, 2011
  6. Sep 14, 2011 #5

    haushofer

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    wrt means "with respect to".

    "a priori" is maybe confusing; I meant "from the outset". A field without any symmetries and no constraints imposed has linear independent components per definition. So a vector has linear independent components, which is expressed as

    [tex]
    \frac{\partial A_{\rho}}{\partial A_{\nu}} = \delta^{\nu}_{\rho}
    [/tex]

    Similarly, for a tensor T,

    [tex]
    \frac{\partial T_{\mu\nu}}{\partial T_{\rho\sigma}} = \delta^{\rho}_{\mu} \delta^{\sigma}_{\nu}
    [/tex]

    Etc.
    This of course changes whenever T is symmetric or antisymmetric!
     
  7. Sep 14, 2011 #6

    haushofer

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    By the way, these are functional derivatives, so you would like to look at that.
     
  8. Sep 14, 2011 #7
    Why does the subscript for the vectors flip in the kronecker delta?

    thanks.
     
  9. Sep 14, 2011 #8

    haushofer

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    Because you differentiate with respect to A_{\nu}. Compare with an ordinary derivative:

    [tex]
    \partial_{\mu} = \frac{\partial}{\partial x^{\mu}}
    [/tex]
     
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