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Lorentz gauge

  1. Jun 4, 2008 #1
    1. The problem statement, all variables and given/known data
    The general gauge transformation in electrodynamics is
    [tex]{\bf A}' = {\bf A} + \nabla \lambda[/tex]
    and
    [tex]\phi ' = \phi - {{\partial \lambda}\over{\partial t}}[/tex].
    In the Lorentz gauge, we set
    [tex] \nabla . {\bf A} + {{\partial \phi}\over{\partial t}} = 0 [/tex].

    My question is: Is the Lorentz choice true for the tranformed potentials as well? i.e. it is true that
    [tex]\nabla . {\bf A}' + {{\partial \phi '}\over{\partial t}} = 0[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I'm hoping that it is true. The freedom of the gauge transformations allows us to use either the primed potentials, or the unprimed potentials, without changing the physics (i.e. the E and B fields).
     
  2. jcsd
  3. Jun 4, 2008 #2

    Dick

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    A gauge choice is specific way to chose a definite gauge potential among an infinite number of physically equivalent gauges. Just as there are an infinite number of potentials that give the same force field. But if you fix the value of the potential at a point you fix the potential. So, no, a 'gauge choice' is not 'gauge invariant'. It's the opposite. Are you doing QFT? Or is there a classical use for this as well?
     
  4. Jun 5, 2008 #3
    No I am not doing QFT. I am trying to show that in the lorentz gauge, the function [tex]\lambda[/tex] obeys a homogeneous wave equation.
     
  5. Jun 5, 2008 #4

    Dick

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    lambda doesn't satisfy any equations, it's just the parameter of your gauge transformation. I think you mean that the scalar potential A^0 satisfies a wave equation in the Lorenz gauge. And it's only homogeneous in the vacuum.
     
  6. Jun 6, 2008 #5
    Right. That was the next question, i.e. to show that the scalar and vector potentials both satisfy inhomogeneous wave equations in the lorentz gauge. This I can do using the Maxwell equations.

    But doesn't the "gauge choice" fix the function lambda?
    The original question asks me to show that lambda satisfies a homogeneous wave equation.
     
  7. Jun 6, 2008 #6

    Dick

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    Ooohhhh. I see. You want to assume the primed and unprimed potentials are both in a Lorenz gauge but are gauge transforms of each other. To get the wave equation for the lambda, just take the difference between the two gauge conditions. So specifying that you are in Lorenz gauge isn't enough to fix the gauge potential. You still have the freedom to transform the potential by a function satisfying the homogeneous wave equation and still stay in the Lorenz gauge. BTW, interesting fact. Lorentz and Lorenz are two different people.
     
    Last edited: Jun 6, 2008
  8. Jun 6, 2008 #7
    Ok. Great. I think that gives me what I was looking for. Is this it:?
    [tex]\nabla . {\bf A}' + {{\partial \phi '}\over{\partial t}} = 0[/tex]
    then putting in [tex]{\bf A}' = {\bf A} + \nabla \lambda[/tex] and [tex]\phi ' = \phi - {{\partial \lambda}\over{\partial t}}[/tex]
    we get
    [tex]\nabla . {\bf A} + \nabla^2 \lambda + {{\partial (\phi - {{\partial \lambda}\over{\partial t}})}\over{\partial t}} = 0[/tex]
    Then subtract this equation with
    [tex] \nabla . {\bf A} + {{\partial \phi}\over{\partial t}} = 0 [/tex]
    And arrive at the wave equation for lambda.

    I had never heard of Lorenz. I looked him up. Interesting stuff. Sadly, he passed away just a few weeks ago. May he rest in peace.
    Thanks again.
     
    Last edited: Jun 6, 2008
  9. Jun 6, 2008 #8

    Dick

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    That's it. But that's a different Lorenz. You want Ludwig Lorenz. Died in 1891. This is an old subject. That makes it a little less sad. You're welcome!
     
    Last edited: Jun 6, 2008
  10. Jun 6, 2008 #9
    Great. Thank you very much.
     
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