Another question about tensor derivatives

  • #1
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Main Question or Discussion Point

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]
 
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Answers and Replies

  • #2
George Jones
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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]
Yes.
 
  • #3
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Thanks.
 
  • #4
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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]?
 
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  • #5
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is [itex]A^0[/itex] a valid gauge choice? That's E = 0 isn't it?
 
  • #6
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4
is [itex]A^0[/itex] a valid gauge choice? That's E = 0 isn't it?
Yes, I wonder. Its given as a question in the book.
 
  • #7
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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]

?
 
  • #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.
 
  • #9
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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.
 
  • #10
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4
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?
 
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