Could someone please explain this 4-vector/tensor notation?

[AxiomOfChoice]

If you can answer any of the questions below, your help will be greatly appreciated.

There's an equation from E&M (I believe the definition of the antisymmetric field tensor) that reads:

$$F_{\mu \nu} = \frac{\partial A_\nu}{\partial x^\mu} - \frac{\partial A_\mu}{\partial x^\nu},$$

where (of course) $A$ is the 4-vector potential. I have no idea how to parse this equation...what's going on with taking the partial of $A_\nu$ with respect to $x^\mu$? What's this notation describing, or telling me to do? I just don't see it.

Also (and this pertains to relativistic quantum mechanics), what's meant by something like

$$\partial_\mu [\partial_\nu \Psi(x^\mu)],$$

where (of course) $\Psi(x^\mu)$ is the wave function? I.e., what does it mean to operate on $\partial_\nu$ of something with $\partial_\mu$? Particularly when that something I'm operating on is a function of $x^\mu$? (By the way: In case the notation is unfamiliar to you, $\partial^\mu = \partial / \partial x_\mu$.) Does that turn into something? Can I compactify that? Or do I just have to leave it as $\partial_\mu [\partial_\nu \Psi(x^\mu)]$? Also, would I get an equivalent expression if I interchanged the order of $\partial_\mu$ and $\partial_\nu$; i.e., if I tried to evaluate $\partial_\nu [\partial_\mu \Psi(x^\mu)]$? Or would I get something quite different?

Does all of this boil down to "we need two distinct dummy indices, $\mu$ and $\nu$, to indicate that there shouldn't be any summation going on?"

(P.S. - I would like to thank David J. Griffiths for getting me into the habit of italicizing words I want to emphasize.)

 

[Peeter]

$$F_{\mu \nu} = \frac{\partial A_\nu}{\partial x^\mu} - \frac{\partial A_\mu}{\partial x^\nu},$$

where (of course) $A$ is the 4-vector potential. I have no idea how to parse this equation...what's going on with taking the partial of $A_\nu$ with respect to $x^\mu$?

Given two four vectors, one for the field, and one for the spacetime
position

$$\vec{A} = (A_0, -A_1, -A_2, -A_3)$$
$$\vec{x} = (x^0, x^1, x^2, x^3) = (ct, x, y, z)$$

This says that there's 16 scalar quantities that can be calculated by
taking derivatives. Two examples are:

$$F_{23} = <br /> \frac{\partial A_3}{\partial x^2} - \frac{\partial A_2}{\partial x^3} = <br /> \frac{\partial A_3}{\partial y} - \frac{\partial A_2}{\partial z}$$

$$F_{33} = <br /> \frac{\partial A_3}{\partial x^3} - \frac{\partial A_3}{\partial x^3} = 0$$

For your other question, I'd assume summation over $\mu$ is implied.

 

[AxiomOfChoice]

Thanks a lot for your help!

What if I confront something like

$$\frac{\partial F^{\mu \nu}}{\partial x^\nu} = \partial_\nu F^{\mu \nu}?$$

Does that imply a sum over $\nu$?

 

[nicksauce]

Yes, if an index is repeated as an upper index and a lower index, it implies a sum over that index.