Another question about tensor derivatives

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]

 

Thanks.

 

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]?

 

Yes, I wonder. Its given as a question in the book.

 

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]

?

 

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?