# Talk:Electromagnetic four-potential

WikiProject Physics / Relativity  (Rated Start-class, Mid-importance)

## Untitled

Article expanded. To keep both camps happy, I've used both SI and gaussian units. Mpatel

Might be nice to add the formal solution for the 'retarded potentials' in terms of [j] [phi], (eg Bleaney & Bleaney E&M p283) - not sure my ability with the formula generator is up to that.:-) Linuxlad 16:54, 11 Apr 2005 (UTC)

## use of the term four-vector

I got rid of this because four-vectors have to transform like A → ΛA under a Lorentz transform, the EM potential comes up with a derivative term here. This is related to the fact that it's a gauge field, so you can define it however you like. You'd need to fix the gauge completely in one particular way for A to be a four vector. 91.106.44.228 (talk) 19:05, 20 February 2011 (UTC)

If you stick to the Lorenz gauge, it's a four-vector, right? Can we say that?
My understanding was that "A → ΛA" is a valid way to Lorentz-transform, but not the only valid way. You can also do "A → ΛA" followed by a gauge transformation. Is that correct? If so, it seems to me that the term "four-vector" is OK even in general (not just in the Lorenz gauge). Certainly there should be some discussion of this. I agree, we do not want to give the impression "the four-potential in the Coulomb gauge is a four-vector" for example. That would certainly be wrong.
On another topic, I think the idea "The electromagnetic four-potential is built from the magnetic vector potential and the electric potential" is very very important, one of the first things a reader needs to know, in order to relate the electromagnetic four-potential to stuff they are already familiar with. You deleted this from the lead, but I suggest putting it back in some form or another. :-) --Steve (talk) 19:21, 20 February 2011 (UTC)
Hey, my original thought was that it would only be a four-vector in the Lorenz gauge, which seems easy enough to prove. However I'm not 100% sure that even that is correct anymore, I came across some interesting discussions here:
There are quotes from some textbooks that seem to say that you can't add a gauge transformation into the Lorentz transform, but a lot of people disagree. Sorry that I can't say who's correct, so can't help much.
I also think you're right that the relation to the magnetic and electric potential should be put back in. I wanted to emphasize that, really, the four potential is what you define to *be* the electric and magnetic potentials, it's not how you construct the four-potential. But I couldn't think of a way to word it at the time.
Thanks! 91.106.33.90 (talk) 19:14, 24 February 2011 (UTC)
I added a footnote to support the claim that it is a four-vector in the Lorentz gauge. I skimmed the physicsforums links, it seems that the informed participants are in general agreement about this. :-) --Steve (talk) 01:02, 25 February 2011 (UTC)

I've never actually edited anything on here before but this page caused me a great deal of grief when I was briefly convinced that the four-vector was not actually a tensor. It is. The problem people seem to be having (I think) is that most gauge conditions aren't Lorentz invariant. That is; by saying I want to use the Coulomb gauge for example, I am choosing this with respect to a coordinate system. So by performing a Lorentz transformation I will no longer have the Coulomb gauge condition satisfied. This is however not a problem. The Lorenz gauge is however specified in a gauge invariant manner so under Lorentz transformations the Lorenz gauge is preserved. —Preceding unsigned comment added by 130.194.163.203 (talk) 02:52, 6 April 2011 (UTC)

If you can think of a clearer/better way to describe it, go ahead and edit the page. Don't be nervous about editing the article: If you mess anything up, we can clean it up. :-) --Steve (talk) 18:29, 6 April 2011 (UTC)

Ok I've signed up now in case I want to fix anything in the future and I gave my best efforts at fixing it up. Though my grammar isn't always the best and I don't actually have any references to cite so I don't mind if it gets edited beyond recognition. Just that it needs to say it IS a four-vector and transforms as one. Cheers Steve86au (talk) 12:12, 7 April 2011 (UTC)

http://www.physicsforums.com/showpost.php?p=1723974&postcount=13 ? 137.73.5.79 (talk) 17:35, 18 May 2011 (UTC)

I wouldn't generally trust posts on physicsforums.com however I will sit down with a copy of Weinberg soon and have a look at what he is talking about. I've always been under the impression that it's the gauge conditions that aren't defined covariantly because they usually make things nicer in a particular frame but not all. I also would agree it's not actually a 4-vector but technically a pseudo-vector, in that switching from a right-handed to left-handed coordinate system will introduce a negative sign but I felt that was over-complicating things as it's not usually considered. This is just because the cross product is defined in terms of the handedness of the coordinate system. In summary, I'm still sure that if you have picked a well defined gauge then it will transform as a 4-vector. It just won't be gauge covariant. Ok, I decided to check out Weinberg on Google books and the quote on physics forums comes from a quantum discussion so I'm really no use there. The four-potential in question appears to be from a quantum particle and the extra terms that appear are creation and annihilation operators. I'm not sure exactly what's going on there when you're talking about quantum particles, but definitely for classical fields it is. Hopefully this link works, this book uses the 4-potential as an example of a "4-vector" which transforms via Lorentz transformations. http://books.google.com.au/books?id=Ku2o4alas0gC&pg=PA11&dq=electromagnetic+four+potential+transforms+as&hl=en&ei=3QHeTeaxJYemugO9iv3QBQ&sa=X&oi=book_result&ct=result&resnum=5&ved=0CD8Q6AEwBA#v=onepage&q=electromagnetic%20four%20potential%20transforms%20as&f=false Sorry for such a long rant I just kind of kept typing as I was thinking. Steve86au (talk) 07:39, 26 May 2011 (UTC)

## notation for the four-potential

Found it better to use ${\displaystyle \Phi ^{a}}$ rather than ${\displaystyle A^{a}}$ for the four-pot.; ${\displaystyle A^{a}}$ is better reserved for the four-acceleration (cos it's better than ${\displaystyle a^{a}}$). These notations are becoming more standard, see for example, Rindler (1991). --- Mpatel 15:15, 13 Jun 2005 (UTC)

I have to be honest and say that never in my life have I seen ${\displaystyle \Phi ^{a}}$ used for the electromagnetic four-potential, only for other vector fields. I have also never encountered a problem where there is a possibility of confusing the four-potential and four-acceleration. I'd really like to switch them back. --Laura Scudder | Talk 19:23, 13 Jun 2005 (UTC)

While editing the GR pages, some of us preferred to use ${\displaystyle A^{a}}$ rather than ${\displaystyle a^{a}}$ for the four-acceleration - previously, I used ${\displaystyle {\tilde {A}}^{a}}$ for the four-potential, but found it unnecessarily complicated (typographically and verbally). In any case, Rindler is an authority on relativity and the reference I gave to his book should be consulted. ---Mpatel 07:27, 14 Jun 2005 (UTC)

I have to agree with Laura here. ${\displaystyle A^{a}}$ is more standard for the electromagnetic potential. Check out either Jackson's Classical Electrodynamics or Griffith's Introduction to Electrodynamics for example Salsb 18:12, August 15, 2005 (UTC)
Griffiths and Jackson are superseded by Rindler. ${\displaystyle A^{a}}$ used to be standard for 4-potential - no more. If you're only doing e/m, then you can use what you like, but with relativity nowadays, ${\displaystyle \phi ^{a}}$ is more common. ---Mpatel (talk) 15:27, August 16, 2005 (UTC)

## vector?

a Therm noted as ${\displaystyle A^{a}=(\phi ,{\vec {A}}c)}$ isn't a 4-vector - its a quaternionic term. If it would be a vector it would need to get noted as ${\displaystyle {\vec {A^{a}}}=\dots }$. Anyway, the four-potential is afaik quaternionic. -- MovGP0 10:34, 26 May 2006 (UTC)

## E & B fields from 4-potential

The equations for the electric and magnetic fields that result from a 4-potential are needed on this page. I may take a shot at doing the edits myself, if nobody beats me to it.

Pervect 20:55, 22 July 2006 (UTC)

## SI problems?

There appear to be problems with the SI formulation on this and other pages:

See for instance: http://xxx.lanl.gov/abs/physics/0401067 Also, Griffiths "Introduction to Electrdynamics"

We should have in SI units:

${\displaystyle A^{a}=({\frac {\Phi }{c}},{\vec {A}})}$
${\displaystyle A_{a}=(-{\frac {\Phi }{c}},{\vec {A}})}$
This gives ${\displaystyle A^{a}A_{a}=|A|^{2}-\Phi ^{2}}$ in SI units
And, in Gaussian units ${\displaystyle A^{a}A_{a}=|A|^{2}-c^{2}\Phi ^{2}}$

This is the same as http://en.wikipedia.org/wiki/Electromagnetic_tensor in the "derivations" section except for choice of sign conventions. (It's incompatible with earlier statements on the page).

I will let this stand for comment before making any edits on this and/or other affected pages.

Pervect 06:40, 30 July 2006 (UTC)

## Definition of the superscript alpha?

Forgive my ignorance, but could someone define the superscript α? Also, since α does not explicitly appear in the right side of the equation, a discussion of how the equation varies with α might also be helpful. Wikimedes (talk) 19:15, 14 October 2010 (UTC)

Alpha isn't a quantity as such, but a index that can represent the spacetime dimensions 0,1,2,3. It's standard to use greek letters to imply this, see for example http://en.wikipedia.org/wiki/Einstein_notation. :) 91.106.44.228 (talk) 17:25, 20 February 2011 (UTC)
The alpha is implied on the right hand side of the equation where it is equal to zero for the scalar potential and equal to 1,2,3 for the vector potential. Dauto (talk) 18:26, 25 May 2011 (UTC)

## Necessity

I think it would be very helpful to know why is necessary its introduction. Paranoidhuman (talk) 15:25, 25 May 2011 (UTC)

What do you mean? It's an essential part of electromagnetic theory. If you want to understand electromagnetism, than you must introduce the potential. Dauto (talk) 18:11, 25 May 2011 (UTC)
But that's not why. That's just shifting the question to a different part of the sentence : it must be introduced because its necessary for an understanding; but why is it necessary? What is it about an understanding that requires it? That's the question.

Or to put it another way, the article should say

• (a) that it is used to define the EM Field Strength tensor
• (b) that maxwell's equations can be compactly written in terms of that tensor
• (c) creating a lagrangian that involves that tensor, some electron fields, and a co-variant derivative of the four-potential, creates
• (i) The conservation of charge, by Noether's theorem
• (ii) Euler-lagrange equations in the electron fields, that produce the dirac equation, and hence the quantum description of electrons
• (iii) Euler-lagrange equations in the four-potential, that produces half of maxwell's equations
• (d) the other half of maxwell's equations turn up from guage fixing the four-potential
• (e) photons, from quantisation of the EM field strength tensor part of the lagrangian — Preceding unsigned comment added by 94.196.243.2 (talk) 12:29, 13 March 2016 (UTC)

Or in other words, it should mention that however much of a mathematical fiction it might be, the fundamental details of electromagnetism can be written very very simply by using the 4-potential, and that, in a vague mathematical sense, it represents photons themselves. — Preceding unsigned comment added by 94.196.243.2 (talk) 12:27, 13 March 2016 (UTC)

## SI units

${\displaystyle \phi }$ is not a unit, its a variable. What units does it have? What units does A have? And why don't you say that in the article in a section about definition in SI units?

## Sign error?

In the section titled The Lorentz Gauge, the second box (the table following the definition of the D'Alembertian) The SI column has minus signs for the sources, while the Gaussian column has them positive. I don't think that can be right. — Preceding unsigned comment added by 209.129.16.79 (talk) 03:11, 14 April 2017 (UTC)