# Another question about tensor derivatives

## Main Question or Discussion Point

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

If,

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

then

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

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George Jones
Staff Emeritus
Gold Member
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.

Thanks.

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 $\Pi^{\mu} = -\partial^{0}A^{\mu} = \partial_{0}A^{\mu}$?

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

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.

If $\Phi$ is a scalar field, what does

$$\dot{\Phi}$$

denote?

Is it

$$\partial_{0}\Phi$$

or

$$\partial^{0}\Phi$$

?

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

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

$$\partial^0 \Phi = -\partial_0 \Phi = -\frac{\partial \Phi}{c\partial t}$$

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.

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

$$\partial^{0}\varphi[/itex] or [tex]\partial_{0}\varphi[/itex] The first term $\Pi\dot{\varphi}$ of the Hamiltonian ($\mathcal{H} = \Pi\dot{\varphi} - \mathcal{L}$) should be [tex]\partial_{0}\varphi\partial^{0}\varphi$$

or

$$\partial^{0}\varphi\partial_{0}\varphi$$

Is there a notational ambiguity here?

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