Another question about tensor derivatives

[maverick280857]

I just wanted to confirm if the following calculation is correct:

If,

F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}

then

\frac{\partial F_{\mu\nu}}{\partial[\partial_{0}A_{\rho}]} = \delta_{\mu}^{0}\delta_{\nu}^{\rho}-\delta_{\nu}^{0}\delta_{\mu}^{\rho}

 

[George Jones]

I just wanted to confirm if the following calculation is correct:

If,

F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}

then

\frac{\partial F_{\mu\nu}}{\partial[\partial_{0}A_{\rho}]} = \delta_{\mu}^{0}\delta_{\nu}^{\rho}-\delta_{\nu}^{0}\delta_{\mu}^{\rho}

Yes.

 

[maverick280857]

Thanks.

 

[maverick280857]

From this, we can show that the momenta conjugate to A_{\mu} treated as fields, is

\Pi^{\mu} = F^{\mu 0}

and hence

\Pi^{0} = F^{00} = 0

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

Edit: Now, if I choose a gauge such that A^{0} = 0. Then \Pi^{\mu} = -\partial^{0}A^{\mu} = \partial_{0}A^{\mu}?

 

[Peeter]

is A^0 a valid gauge choice? That's E = 0 isn't it?

 

[maverick280857]

is A^0 a valid gauge choice? That's E = 0 isn't it?

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

 

[maverick280857]

If \Phi is a scalar field, what does

\dot{\Phi}

denote?

Is it

\partial_{0}\Phi

or

\partial^{0}\Phi

?

 

[Fredrik]

I think \partial_{0}\Phi is the standard, but I wouldn't be surprised if someone uses the other convention.

 

[Peeter]

I'd guess it's always index lower. If working with a -+++ metric you'd have

<br /> \partial^0 \Phi = -\partial_0 \Phi = -\frac{\partial \Phi}{c\partial t}<br />

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.

 

[maverick280857]

Ok, the confusion stems from the computation of the momentum conjugate to \varphi, the Klein Gordon field. The Lagrangian is

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

You can have "two kinds" of conjugate momenta

\partial^{0}\varphi[/itex]<br /> <br /> or<br /> <br /> \partial_{0}\varphi[/itex]&lt;br /&gt; &lt;br /&gt; The first term \Pi\dot{\varphi} of the Hamiltonian (\mathcal{H} = \Pi\dot{\varphi} - \mathcal{L}) should be&lt;br /&gt; &lt;br /&gt; \partial_{0}\varphi\partial^{0}\varphi&lt;br /&gt; &lt;br /&gt; or&lt;br /&gt; &lt;br /&gt; \partial^{0}\varphi\partial_{0}\varphi&lt;br /&gt; &lt;br /&gt; Is there a notational ambiguity here?