Re: [Phys-L] four-vector potential ... or not

[John Denker <jsd@av8n.com>]

On 10/19/2015 10:35 AM, I wrote:

> Here's something amusing:  Take a look at
>   https://en.wikipedia.org/wiki/Electromagnetic_four-potential
>
> Do you notice anything wrong?

The problem is, the whole article is devoted to something that
does not actually exist.

Contrary to what it says in the article, the object
     A  :=  [ϕ/c, Ax, Ay, Az]

is not a four-vector.  It has four components, but that is not
sufficient to make it a well-behaved 4-vector.  It does not 
behave properly under Lorentz transformations.

This is not tragic, because the potentials are not directly observable.
The only thing that matters is the difference between two potentials, 
and that turns out to be well behaved, for the following reason:

Loosely speaking, if you start out with a vector potential in a certain 
gauge and then change to a different reference frame, you get a vector
potential with the same physical meaning /in some other screwy gauge/.
If you try to calculate A by evaluating it in one frame and then boosting
it into another frame, you will almost certainly get the wrong value for 
A.  However, when you compute any physical observable, the gauge drops 
out, so you might end up with the right physics.

In particular, the key equation is OK:

	F  =  dA           
 	   =  ∇∧A
 
The electromagnetic field F is a well-behaved bivector.  The exterior 
derivative on the RHS annihilates any and all gauge fields.

This partially explains why the Liénard-Wiechert potentials are usually
written in an inelegant, non-covariant, non-spacetimey form.