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Can four-potential be uniquely defined?

  1. Aug 15, 2014 #1
    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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  3. Aug 15, 2014 #2

    WannabeNewton

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    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.
     
  4. Aug 15, 2014 #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.

    Best regards,
    Sergio
     
  5. Aug 15, 2014 #4

    WannabeNewton

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    Yes indeed.

    P.S. is your name short for Uriel Septim, from Elder Scrolls IV: Oblivion?
     
  6. Aug 15, 2014 #5

    dextercioby

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    Just that the OP learns: it's the Lorenz not Lorentz gauge.
     
  7. Aug 16, 2014 #6

    Yes it is!, :)
     
  8. Aug 16, 2014 #7

    vanhees71

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    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
     

    Attached Files:

  9. Aug 16, 2014 #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.
     
  10. Aug 16, 2014 #9

    vanhees71

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    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!
     
  11. Aug 16, 2014 #10

    WannabeNewton

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

    Can you believe that the interaction of a charge with an electromagnetic field doesn't care about how you orient your coordinate system?
     
  12. Aug 16, 2014 #11
    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.

    Best regards,
    Sergio
     
  13. Aug 16, 2014 #12
    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.
     
  14. Aug 17, 2014 #13

    vanhees71

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    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.
     
  15. Aug 17, 2014 #14
    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.


    Best regards,
    Sergio
     
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