Maxwell Field in General Relativity: Explained

[Einj]

Hello everyone,
I'm studying some applications of AdS/CFT and I came across an expression of the Maxwell field written in the following way:
$$
A=A_t(r)dt+B(r)xdy.
$$
How does this notation work? Is it simply a way of writing the four-vector? If so, why do we use this notation?
Thanks a lot!

 

[PeterDonis]

I came across an expression of the Maxwell field written in the following way:

​

$A=A_t(r)dt+B(r)xdy$.

Are you sure this is transcribed correctly? Can you give the reference for where you got it? It doesn't look right to me; the Maxwell field should be a 2-form, i.e., it should be expressed as a sum of wedge products of the form $dt \wedge dx$, $dy \wedge dz$, etc.

 

[Einj]

yes, it's in here: http://arxiv.org/abs/0903.3246. Equation (56). Thanks!

 

[wabbit]

Looks like the expression is the potential, not the field.

Edit : no idea why he picks this form, didn't read the rest of the paper sorry

 

[Einj]

He calls it field but yes, I pretty sure he means the potential. Does this simply mean that the four-potential have component $A_t(r)$ and $A_y(r,x)=B(r)x$? If so, where does that notation come from?

 

[PeterDonis]

He calls it field but yes, I pretty sure he means the potential.

Yes, he does.

Does this simply mean that the four-potential have component $A_t(r)$ and $A_y(r,x)=B(r)x$?

Yes, although I also think $r = \sqrt{x^2 + y^2 + z^2}$, so any function of $r$ is really a function of $x, y, z$.

where does that notation come from?

It's differential form notation; the 1-form $A$ is expressed in terms of its components as $A_{\mu} dx^{\mu}$, where $dx^{\mu}$ are the basis 1-forms $dt$, $dx$, $dy$, and $dz$. The electromagnetic field itself is then expressed as the 2-form $F = dA$, which in components is $F = \frac{1}{2} F_{\mu \nu} dx^{\mu} \wedge dx^{\nu}$, and $F_{\mu \nu} = \partial_{\mu} A_{\mu} - \partial_{\mu} A_{\nu}$. This notation is often used in field theory.

 

[Einj]

Oh great thank you!