Can four-potential be uniquely defined?

  • Thread starter USeptim
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In summary, different four potentials can generate the same electromagnetic fields, with A'=A+(e/c)∇χ and \phi'=\phi+(∂χ/∂t) being two examples. In electrostatics, the potential is well-defined except for a constant that must be the same throughout space. However, in electrodynamics, the gauge conditions are more complex and there is no unique way to define exact potentials. Instead, gauge-fixing procedures, such as the Lorentz gauge, are used. The Lienard-Wiechert potentials give the potential generated by a moving charge at every point in space, and can be used to determine the exact potential by integrating the potential caused by all charges and including a
  • #1
USeptim
98
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Different four potentials can generate the same electromagnetic fields. Specifically the following degrees of freedom yield the same EM fields:

A'=A+(e/c)∇χ
[itex]\phi[/itex]'=[itex]\phi[/itex]+(∂χ/∂t)

Where A es the vector potential, [itex]\phi[/itex] is the scalar potential and ∇χ is any scalar field.

In the electrostatic, potential is well defined except for a constant that has to be the same on all the space. Besides, for a continuous charge distribution, the potential energy of the charges is the same than the energy stored in the field. So it looks like the potential can be uniquely defined in this special case.

In electrodynamics however, the gauge conditions are more complex and as far as I have seen, there is no document or textbook stating how to define exact potentials. Instead of that, “gauge-fixing” procedures are used, for example de Lorentz gauge: [itex]∂_{\alpha}[/itex][itex]A^{\alpha}[/itex]=0.

However Lienard-Wiechert potentials show the potential generated by a moving charge on every point of the space, I understand this is the “effect” of the moving charge on the point. So I think it should be possible to determine the exact potential by integrating the potential caused by all the charges at δ( -[itex]t_{0}[/itex]-t)|[itex]x_0{}[/itex] - x|) plus a term corresponding to the boundary radiated fields.

Is there anything that may be wrong in this assumption?

NOTE: [itex]t_{0}[/itex] and [itex]t_{0}[/itex] is the space-time point when we measure the potential and t and x are the position of the charges that cause effect on this point.
 
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  • #2
The Lienard-Wiechert potentials are the retarded Green's function solutions to Maxwell's equations in the Lorenz gauge, ##\square A^{\mu} = \frac{4\pi}{c} j^{\mu}## and ##\partial_{\mu}A^{\mu} = 0##, so they already assume a choice of gauge. There is simply no unique choice of 4-potential for a given electromagnetic field configuration.
 
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  • #3
Thanks WannabeNewton.

So if we calculate the potential integrating the Lienard-Wiechert potentials we are implicitly using the Lorentz gauge which is as arbitrary as any other.Sergio
 
  • #4
Yes indeed.

P.S. is your name short for Uriel Septim, from Elder Scrolls IV: Oblivion?
 
  • #5
Just that the OP learns: it's the Lorenz not Lorentz gauge.
 
  • #6
WannabeNewton said:
Yes indeed.

P.S. is your name short for Uriel Septim, from Elder Scrolls IV: Oblivion?


Yes it is!, :)
 
  • #7
I've written my posting in real LaTeX, because it's easier to write the typographically a bit complicated equations (see attached pdf).

BTW: It's Lorenz gauge, because the retarded solution of the equations for the four-potential in this gauge goes back to the Danish physicist Ludwig Lorenz. He was quite a while earlier than the Dutch physicist Hendrik Antoon Lorentz, who finally fully analyzed the meaning of what we call "gauge symmetry" nowadays. In the older literature usually they name the Lorenz gauge after Lorentz which is a bit unjust. That's why usually nowadays people quote Lorenz for historical justice.

http://arxiv.org/abs/hep-ph/0012061
http://dx.doi.org/10.1103/RevModPhys.73.663
 

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  • coulomb-lorenz-gauge-trafo.pdf
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  • #8
In fact, if we apply a boost to the electrostatic potential (V,0,0,0), we get the Lienard-Wiechert potential so it is somehow the “extension” to the electrostatic potential.

However, the electrostatic potential is a clear example of Lorenz gauge since the four components of [itex]∂_{\alpha}[/itex][itex]A^{\alpha}[/itex]=0 are equal to zero! Of course the boosted electrostatic potential meets also the Lorenz gauge.

So this way to calculate the potentials can be considered the “natural” way but not the only one.

I read a QFT book when it was said (or at least I understood… I have a reduced knowledge on this matter) that choosing a gauge or another did not change the properties of the system but I hardly believe how a system can evolve the same way if different four potentials are applied at each point of the wave function.
 
  • #9
A boost cannot give you the retarded potential of arbitrarily moving charges but only for a uniformly moving charges. Particularly this will give you no radiation, which you only get when the charge is accelerated.

The gauge symmetry is at the heart of the standard model of elementary particles, but to have a chance to understand it, you must have a good understanding of classical electrodynamics first!
 
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  • #10
USeptim said:
Yes it is!, :)

Haha I love that game so much. I dropped a good chunk of my childhood on it.

USeptim said:
I read a QFT book when it was said (or at least I understood… I have a reduced knowledge on this matter) that choosing a gauge or another did not change the properties of the system but I hardly believe how a system can evolve the same way if different four potentials are applied at each point of the wave function.

Can you believe that the interaction of a charge with an electromagnetic field doesn't care about how you orient your coordinate system?
 
  • #11
vanhees71 said:
A boost cannot give you the retarded potential of arbitrarily moving charges but only for a uniformly moving charges. Particularly this will give you no radiation, which you only get when the charge is accelerated.

Hi vanhees71,

So, the potential exerted by a charge depends not only of the reference frame where it is in the retarded time.

What other factor we have then? Just charge's acceleration in the retarded time? or is there any other? I have not found in any textbook how can potentials be calculated under acceleration. If you have some reference I would be glad to know about it.


Sergio
 
  • #12
WannabeNewton said:
Can you believe that the interaction of a charge with an electromagnetic field doesn't care about how you orient your coordinate system?

I also spent quite time with Oblivion :).

While it's evident that interactions does not depend on rotations, I cannot see so clear that the gauge symmetry will not affect the interaction. Of course if the system has symmetry about some variable, its results are not affected by changes in that variable, therefore my caveat is that I have not understood how the EM gauge becomes a symmetry in QFT. I will read about that.
 
  • #13
USeptim said:
Hi vanhees71,

So, the potential exerted by a charge depends not only of the reference frame where it is in the retarded time.

What other factor we have then? Just charge's acceleration in the retarded time? or is there any other? I have not found in any textbook how can potentials be calculated under acceleration. If you have some reference I would be glad to know about it.


Sergio

I don't understand your first paragraph. Have you had a look at my pdf file from my first posting? There I work in a fixed inertial reference frame and show explicity, how you find the potentials in the two most common gauges from Maxwell's equations. The point is gauge invariance, which leads to the same (retarded) electromagnetic fields at given charge-current distributions for different but gauge covariant potentials.

Look for Lienard-Wiechert potentials. These give the retarded potentials (fulfilling the Lorenz gauge condition) for the em. field of an arbitrarily moving point charge.

You can calculate them yourself by using the charge-current distribution of a point particle. Let [itex]\vec{y}(t)[/itex] it's trajectory as function of (coordinate) time. Everything is, of course, in a fixed inertial frame. Don't mix the problem of Lorentz transformations with your problem. Start to understand everything working in a fixed inertial frame! The charge-current distribution for a moving point charge reads
[tex]\rho(t,\vec{x})=q \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(t,\vec{x})=q \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)].[/tex]
The four-current is, as usual, [itex](j^{\mu})=(c \rho,\vec{j})[/itex], and it's really a four-tensor since you can write the above expressions in manifestly covariant form, using an arbitrary parameter [itex]\lambda[/itex] for the world line of the particle:
[tex]j^{\mu}(x)=q \int_{\mathbb{R}} \mathrm{d} \lambda \frac{\mathrm{d} y^{\mu}(\lambda)}{\mathrm{d} \lambda} \delta^{(4)}[x^{\mu}-y^{\mu}(\lambda)].[/tex]
It's also independent on the choice of the parameter. You come back to the first form in terms of the coordinate time, if you use [itex]\lambda=t[/itex].

This you plug into the retarded-potential solutions and do the integrals, using the [itex]\delta[/itex] distribution. This leads to the Lienard-Wiechert retarded potentials.
 
  • #14
vanhees71 said:
I don't understand your first paragraph. Have you had a look at my pdf file from my first posting? There I work in a fixed inertial reference frame and show explicity, how you find the potentials in the two most common gauges from Maxwell's equations. The point is gauge invariance, which leads to the same (retarded) electromagnetic fields at given charge-current distributions for different but gauge covariant potentials.

Hello vanhees71,

I have read now your pdf, you show the scalar "gauge field" necessary to connect Coulomb and Lorenz, not Lorentz :), potentials. You speak also about Jefimenko equations which calculate the EM field including the effect of charge acceleration which I didn't know.

What I was asking however was how to retrieve the potentials and not the electromagnetic field, however maybe this question lacks of sense since different gauge potentials are equivalent.

I think I will learn a bit more about the mathematical properties of the pointcaré group and the nature of gauge fields so that I can understand how the gauge symmetry works.



Sergio
 

1. What is four-potential?

Four-potential, also known as the electromagnetic four-vector potential, is a mathematical quantity used in electromagnetism to describe the potential energy of an electric and magnetic field. It is composed of four components, three for the electric field and one for the magnetic field.

2. Why is it important to uniquely define four-potential?

Uniquely defining four-potential is crucial in order to accurately calculate and predict the behavior of electromagnetic fields. Without a unique definition, there may be inconsistencies and errors in calculations, leading to inaccurate results.

3. How is four-potential uniquely defined?

Four-potential is uniquely defined by specifying its boundary conditions, which are the values of the potential at the boundaries of a given system. These conditions must be consistent with the physical properties of the system in order to ensure a unique solution.

4. Can four-potential have multiple solutions?

No, four-potential must have a unique solution in order for it to accurately describe the behavior of electromagnetic fields. If there are multiple solutions, it means that the boundary conditions were not properly specified or there are inconsistencies in the physical properties of the system.

5. What happens if four-potential is not uniquely defined?

If four-potential is not uniquely defined, it may lead to inconsistencies in calculations and predictions of electromagnetic fields. This can result in errors and inaccuracies in scientific experiments and technological applications that rely on the accurate understanding of electromagnetic fields.

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