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How to arrive at Lorentz gauge condition?

  1. Oct 23, 2012 #1
    How to arrive at Lorenz gauge condition?


    I know it's used to simplify the 2 partial differential equations of the potentials, but why can we put such a restriction on the potentials? Doesn't that restriction restrict the possible electrical and magnetic fields that can be derived from the potentials?

    Could I make any arbitrary restriction on the potentials? For example could I say that the laplacian of A (A being the magnetic vector potential) is equal to 0?

    I asked my teacher and looked in Jackson's book but they just justify it saying "the potentials are arbitrary".
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 23, 2012 #2
    Two of Maxwell's equations reduce to a (generalized) curl equation:

    $$\nabla \wedge F = 0$$

    (Read that as curl and you basically get the right meaning.) ##F## is the EM tensor and ##\nabla = e^t \frac{1}{c} \partial_t + e^x \partial_x + e^y \partial_y + e^z \partial_z## is just the SR version of del.

    This then suggests that the EM tensor comes from some four-potential ##A## such that:

    $$\nabla \wedge A = F \implies \nabla \wedge \nabla \wedge A = 0$$

    which automatically holds as a result. The latter equation can be rewritten

    $$\nabla^2 A - \nabla(\nabla \cdot A) = 0$$

    Setting ##\nabla \cdot A = 0## generates the Lorenz gauge condition. (Note it is Lorenz, not Lorentz.)

    If your question is more "why do we have this freedom?" then the answer is to go back to the equation ##\nabla \wedge A = F##. You can add any potential ##X## such that ##\nabla \wedge X = 0## without changing the value of ##F## (see that ##\nabla \wedge (A + X) = \nabla \wedge A##).

    So the Lorenz gauge condition is really define ##X## such that ##\nabla \wedge X= 0## and ## \nabla \cdot X = - \nabla \cdot A##, or ##\nabla X = -\nabla \cdot A## for short.

    This then generates a new four-potential ##A' = A + X## that obeys ##\nabla \wedge A' = F## and ##\nabla \cdot A' = 0##.
  4. Oct 24, 2012 #3
    Can you explain that not using tensors please? We still don't use that formalism (it's only an introductory electromagnetism class), we only use vectors. As A I meant the magnetic vector potential.
  5. Oct 24, 2012 #4
    What I wrote is just as valid in 3d as it is for special relativistic tensor fields. The wedge is a curl. Because only the curl is important, you can add any curlless vector field to the potential and not change anything about the electromagnetic field. Because the divergence doesn't matter, you have total freedom to specify the divergence of the potential to your liking. These two freedoms together are the freedoms that give you the ability to do gauge transformations.

    So, the short answer to your questions:

    Because the combination of the two potentials obey a differential equation where only some of their derivatives (this curl I've been talking about) matter, not all lof them.

    This condition is basically the formulae for the EM fields. If you transform ##\varphi \mapsto \varphi'## and ##A \mapsto A'##, then

    $$-\nabla \varphi - \frac{1}{c} \frac{\partial A}{\partial t} = -\nabla \varphi' - \frac{1}{c} \frac{\partial A}{\partial t}, \quad \nabla \times A = \nabla \times A'$$

    These conditions must be kept true regardless of the transformation you try to impose. Doing so guarantees the E and B fields remain the same. So that's the constraint.

    The potentials are already restricted, as above. Setting the Laplacian of the vector potential to zero does not, as far as I can see, break that restriction, so you could.

    The potentials are arbitrary aside from those constraints above.
  6. Oct 25, 2012 #5
    A simple explanation for gauge freedom comes from the fact that there are 8 maxwell eqn. and only 6 unknowns.can you see that
  7. Oct 25, 2012 #6

    Meir Achuz

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    Re: How to arrive at Lorenz gauge condition?

    The use and justification for the Lorenz gauge is given on pp. 337-338 of
    Franklin, "Classical Electromagnetism". It also makes the mistake of "Lorentz" instead of Lorenz, but otherwise is correct and simple.
  8. Oct 30, 2012 #7
    Thank you all, I understand now.

    I'll check that, thanks.
  9. Oct 30, 2012 #8


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    For the record, it would be nice to know which was the first textbook on electromagnetism which made the Lorentz (Dutch) vs Lorenz (Danish) confusion.
    AFAIK even the respectable textbook by Landau & Lifschitz (Vol.3 in their series) makes the error and whole dozens of others...
    The result is that essentially all (I have found only an exception in the 2nd volume of Atkinson & Johnson, where the word 'Lorentz' is changed with 'Lorenz' after the initial print, but not everywhere throughout the text (it's kept Lorentz in the appendix)) books on QFT use Lorentz instead of Lorenz. :(
    The reason is that this condition (of course, in the non-covariant version) was brought to worldwide attention by the Dutchman H.A. Lorentz and his book <Theory of electrons> which states it as formula (14) of Appendix 5, page 239 (2nd edition, Leipzig, B.G. Teubner, 1916).

    P.S. As stated in the <Talk> page on Wikipedia, the 3rd edition of the 'bible' by J.D. Jackons restores justice at textbook level (explanatory note on page 294).
    Last edited: Oct 30, 2012
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