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Another question about tensor derivatives

  1. May 27, 2009 #1
    I just wanted to confirm if the following calculation is correct:

    If,

    [tex]F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}[/tex]

    then

    [tex]\frac{\partial F_{\mu\nu}}{\partial[\partial_{0}A_{\rho}]} = \delta_{\mu}^{0}\delta_{\nu}^{\rho}-\delta_{\nu}^{0}\delta_{\mu}^{\rho}[/tex]
     
    Last edited: May 28, 2009
  2. jcsd
  3. May 28, 2009 #2

    George Jones

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    Yes.
     
  4. May 28, 2009 #3
    Thanks.
     
  5. May 28, 2009 #4
    From this, we can show that the momenta conjugate to [itex]A_{\mu}[/itex] treated as fields, is

    [tex]\Pi^{\mu} = F^{\mu 0}[/tex]

    and hence

    [tex]\Pi^{0} = F^{00} = 0[/tex]

    What is the physical significance of this result, other than the fact that we cannot use this to solve for the [itex]\dot{A}_{\mu}[/itex]'s?

    Edit: Now, if I choose a gauge such that [itex]A^{0} = 0[/itex]. Then [itex]\Pi^{\mu} = -\partial^{0}A^{\mu} = \partial_{0}A^{\mu}[/itex]?
     
    Last edited: May 28, 2009
  6. May 28, 2009 #5
    is [itex]A^0[/itex] a valid gauge choice? That's E = 0 isn't it?
     
  7. May 28, 2009 #6
    Yes, I wonder. Its given as a question in the book.
     
  8. May 29, 2009 #7
    If [itex]\Phi[/itex] is a scalar field, what does

    [tex]\dot{\Phi}[/tex]

    denote?

    Is it

    [tex]\partial_{0}\Phi[/tex]

    or

    [tex]\partial^{0}\Phi[/tex]

    ?
     
  9. May 30, 2009 #8

    Fredrik

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    I think [itex]\partial_{0}\Phi[/itex] is the standard, but I wouldn't be surprised if someone uses the other convention.
     
  10. May 30, 2009 #9
    I'd guess it's always index lower. If working with a -+++ metric you'd have

    [tex]
    \partial^0 \Phi = -\partial_0 \Phi = -\frac{\partial \Phi}{c\partial t}
    [/tex]

    which I don't think makes much sense to use as this Dot operator. You
    could get away with index up for this time derivative only if using a +--- metric.
     
  11. Jun 2, 2009 #10
    Ok, the confusion stems from the computation of the momentum conjugate to [itex]\varphi[/itex], the Klein Gordon field. The Lagrangian is

    [tex]\mathcal{L} = \frac{1}{2}\partial^{\mu}\varphi\partial_{\mu}\varphi -\frac{1}{2}m^2\varphi^2[/tex]

    You can have "two kinds" of conjugate momenta

    [tex]\partial^{0}\varphi[/itex]

    or

    [tex]\partial_{0}\varphi[/itex]

    The first term [itex]\Pi\dot{\varphi}[/itex] of the Hamiltonian ([itex]\mathcal{H} = \Pi\dot{\varphi} - \mathcal{L}[/itex]) should be

    [tex]\partial_{0}\varphi\partial^{0}\varphi[/tex]

    or

    [tex]\partial^{0}\varphi\partial_{0}\varphi[/tex]

    Is there a notational ambiguity here?
     
    Last edited: Jun 3, 2009
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