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Proof that ∇ * A (magnetic vector potential) = 0

  1. Nov 10, 2012 #1
    ∇Δ1. The problem statement, all variables and given/known data

    Show that [itex]\nabla \cdot A = 0 [/itex]

    Where A is formally defined as
    [itex]A(r) = \frac{\mu }{4\pi }\int \frac{J(r')\text{ }}{r} \, dv' [/itex]

    I understand that we can distribute ∇ into the integral, and from there we can do a little bit of algebra to get the terms inside the integrand of the form.

    [itex]\nabla \cdot \left(\frac{J'}{r}\right) + J'\cdot \nabla \left(\frac{1}{r}\right)[/itex]

    Now what happens next is where I get lost. I don't fully understand why we can throw away the first term since ∇ isn't dependent upon J'. I haven't seen anything like this done before and it seems a little fishy. Next griffiths uses some identity to change

    [itex] J'\cdot \nabla \left(\frac{1}{r}\right) = -J'\cdot \nabla '\left(\frac{1}{r}\right) [/itex]

    Then he repeats this whole process again, and throws away the J term. Finally he finishes with stokes theorem to relate the area integral to a line integrand.

    I guess I could just use a little bit of commentary on this proof as it doesn't make much sense to me yet.
     
  2. jcsd
  3. Nov 11, 2012 #2
    First of all ∇ will only operate on those function which depends on r,not on r'.Since in expression of A only 1/r is the one which depends on r,So only this term will be considered and no others.The next one is the usual technique which is used many times.Your 1/r is actually 1/mod(r-r'),you can verify that ∇ acting on r in this will give the same result as -∇' acting on r'.
     
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