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Invariance of maxwell's equations under Gauge transformation

  1. Feb 4, 2008 #1
    [SOLVED] invariance of maxwell's equations under Gauge transformation

    1. The problem statement, all variables and given/known data

    Show that the source-free Maxwell equations [tex]\partial_{\mu} F^{\mu\nu}=0[/tex] are left invariant under the local gauge transformation

    [tex] A_{\mu}(x^{\nu})\rightarrow A'_{\mu}(x^{\nu})=A_{\mu}(x^{\nu})+\frac{\partial}{\partial x^{\mu}}\varphi(x^{\nu}) [/tex]

    for an arbitrary scalar function [tex]\varphi(x^{\nu})[/tex]

    2. Relevant equations

    The definition of the field strength tensor: [tex]F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}[/tex]

    3. The attempt at a solution

    This is a course I don't quite have the foreknowledge of. There is no book, so some things I can't quite figure out. The teacher gave the solution on the blackboard already, so that's not the real problem. I wrote it down real good and neat, but I don't get it at all.

    I just don't know what a Gauge transformation is, or what the definition of maxwell's equation in that tensor is, I've only seen it as four equations just yet (both integral and differential form). If someone could just help me understand the problem, or give me some useful links to understand it, I'd appreciate it a lot... or mainly what the bigger part of those symbols mean.
    Last edited: Feb 4, 2008
  2. jcsd
  3. Feb 4, 2008 #2


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    You could just show that field strength tensor itself is invariant under the gauge transformation. If you substitute the gauge transformation into the definition of the field strength tensor this is just a matter of mixed partial derivatives being equal. I.e. d^2f/dxdy=d^f/dydx. Isn't it?
  4. Feb 4, 2008 #3


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    Read the chapter on relativity in Griffith's electrodynamics book. And the chapter about gauge freedom too.

    Short digest:

    Remember that in a any given situation involving electromagnetic fields, the potentials phi and A are not unique. All that matters is that when differentiated according to maxwell's equations, they give the right fields E and B. For instance, adding a constant c1 to phi and a constant vector c2 to A does not change the derivatives, and so phi' = phi + c1 and A' = A + c2 are again valid potentials for the physical situation. Choosing a value for phi and A is called choosing a gauge, and a switch from one gauge to another, such as going from phi and A to phi' and A' above is called a... gauge transformation.

    As it turns out, for any function f(x,t), the gauge transformation phi' = phi - df/dt and A' = A + grad(f) leaves the fields unchanged! (but you gotta change phi and A simultaneously!)

    This is just what the question is asking you to show.
  5. Feb 6, 2008 #4
    Dick: that's right.

    Quasar: thanks, and in my house there are like 5 copies of Griffiths and I my teacher pointed out a specific chapter in Peter Jackson's classical electrodynamics I should read to catch up. So I'm confident it should work out :)
  6. Jun 1, 2011 #5
    Re: [SOLVED] invariance of maxwell's equations under Gauge transformation

    Because im stupid and i dont understand can anybody explain me better how it turns out that if i substitute Aμ' =Αμ +θμF ,F scalar fuction, i get the same Fμν ??
  7. Jun 2, 2011 #6


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    Re: [SOLVED] invariance of maxwell's equations under Gauge transformation

    Well, you need to show that

    [tex] F'_{\mu\nu} \left(A'_{\lambda}\left(A_{\sigma}(x)\right)\right) = F_{\mu\nu} \left(A_{\sigma}\right) [/tex]

    knowing the connection between the potentials A'(A) and how F is defined in terms of A.

    [Edit] The LaTex is apparently not working properly.

    [Later Edit] See post #7.
    Last edited: Jun 2, 2011
  8. Jun 2, 2011 #7


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    Re: [SOLVED] invariance of maxwell's equations under Gauge transformation

    You're missing a \ before the first right.
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