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Coulomb Gauge

  1. Nov 28, 2009 #1
    Coulomb gauge fixes gauge by setting div(A)=0.
    What has it to do with adding a gradient to A and subtract a partial time derivative from V?
     
  2. jcsd
  3. Nov 28, 2009 #2
    Hello intervoxel
    What is Coulomb gauge?
    greetings Janm
     
  4. Nov 29, 2009 #3
    Hi,

    A gauge transformation in the electromagnetic field is the folowing pair of equations:

    A' = A + grad(lambda),

    V' = V - drond lambda / drond t,

    where V is the scalar potential and A the vector potential. Lambda can be any function.

    This has the nice property of leaving the fields E and B unnafected. It is used to simplify
    expressions derived from Maxwell`s equations.

    The two most common gauge transformations are the Coulomb and Lorentz gauges.
     
  5. Nov 29, 2009 #4
    Hello intervoxel
    So you have large scale fields based on the scalar potential A and vector potential V.
    The elctric field based on the scalar potential and the magnetic field based on the vector potential is it not?
    ... and now a gauge is a local changement of overall field if I understand right.
    Is Lambda a scalar function and what is drond?
    greetings Janm
     
  6. Nov 30, 2009 #5
    drond means partial differential.

    lambda is a scalar function.

    The theory says that the EM field can be derived from two potential functions through the relations:

    E = -grad(V)-drond A / drond t
    B = curl(A)

    We may then have many pairs of potential functions generate the same EM field.

    Gauge transformations extend this set even further.

    Oh, come on folks, let's return to my original question: What div(A)=0 has to do with gauge transformation?
     
  7. Nov 30, 2009 #6
    Hello Intervoxel
    The chapter inhomogeneous wave equation speaks of Lorentzcondition:

    divA+mu*eps*drondphi/drondt+mu*sig*phi=0,

    which gives the wave equation in free space.

    divA+(1/c^2)*drondphi/drondt=0

    and continues with introduction of the d'Alenbertian and after that retarded potentials, so what is your problem?
    greetings Janm
     
  8. Nov 30, 2009 #7
    Now I see, JANm, thank you.

    The Lorentz condition is a statement of the law of conservation of charge -- It imposes a constraint between A and V. If I select a V invariant in time I get Coulomb gauge:

    A' = A + grad(V)
    V' = V - 0
     
    Last edited: Nov 30, 2009
  9. Dec 1, 2009 #8

    Born2bwire

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    You would not want to explicitly set the scalar potential to be time invariant. The primary reason being that it restricts the physical phenomenon that can occur. For example, under the Coulomb gauge we can derive the expression that
    [tex]\frac{1}{c^2}\nabla\frac{\partial^2 \Phi}{\partial t^2} = \mu_0\mathbf{J}_t[/tex]
    where J_t here is the irrotational current, the curl free portion of the currents. Thus, by fixing a time invariant scalar potential you are forcing the condition that there can be no irrotational currents, among other things I am sure.

    The reason for confusion is that I believe you are still trying to satisfy the Lorenz condition, which is the defining property that gives rise to the Lorenz gauge. The Lorenz guage requires that the Lorenz condition be satisfied,
    [tex]\nabla\cdot\mathbf{A}+\frac{1}{c^2}\frac{\partial \Phi}{\partial t} = 0[/tex]
    This gives rise to the two decoupled inhomogeneous wave equations for \Phi and A.
    The Coulomb gauge, however, satisfies the inhomogeneous wave equation,
    [tex]\nabla^2\mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J} + \frac{1}{c^2}\nabla \frac{\partial \Phi}{\partial t}[/tex]
    We can see that by decomposing the current into irrotational and solenoidal components we can decouple the above wave equation as previously stated above. While this is very similar to what arises in the Lorenz gauge, it allows a different set of freedoms in the choice of our potentials. You can find a detailed explanation of this in Jackson's text.
     
  10. Dec 2, 2009 #9
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