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Vector Potential and Zero Divergence

  1. Mar 19, 2013 #1
    I'm trying to understand when a vector field is equal to the curl of a vector potential. Why is it possible that there is always a vector potential with zero divergence?

    Relevent Equation:

    B=∇χA

    I'm trying to understand the proof that the above vector potential A can be one with zero divergence.
     
  2. jcsd
  3. Mar 19, 2013 #2

    WannabeNewton

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    We know that if we perform a gauge transformation [itex]A = A' + \triangledown \xi[/itex], where [itex]\xi[/itex] is an arbitrary scalar field, then both [itex]A'[/itex] and [itex]A[/itex] result in the same observed magnetic field i.e. [itex]B = \triangledown \times A' = \triangledown \times A[/itex] (and of course, as usual, we have to perform the associated gauge transformation of the scalar potential to keep the observed electric field the same).

    Say we are given a vector potential [itex]A'[/itex]. We can find a [itex]\xi[/itex] that solves [itex]\triangledown ^{2}\xi = -\triangledown \cdot A'[/itex]. Performing the gauge transformation [itex]A = A' + \triangledown \xi [/itex] we see that [itex]\triangledown \cdot A = \triangledown \cdot A' + \triangledown ^{2}\xi = 0[/itex] hence we can fix this gauge (again, after performing the associated gauge transformation of the scalar potential) so that we have [itex]B = \triangledown \times A, \triangledown \cdot A = 0[/itex]. This is called the Coulomb gauge.
     
  4. Mar 19, 2013 #3
    Thank You!!!
     
  5. Mar 19, 2013 #4

    WannabeNewton

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    No problem!
     
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