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Vector potential

  1. Mar 15, 2007 #1
    Is it at all ethical to prove divA=0 here A is the magnetic vector potential?
    That is, curlA=B?
    Griffiths goes through a systematic development to show that we may make sure that div A=0 to make life easy.But in our exam the question appeared assuming the vector potential's form:[1/4 pi]int{J dV/|r-r'|},prove that divA=0.Is it justified?Howeevr,I know there is another method to reach the form of vector potential where an additional term (grad phi) appears.And we make it zero.Please help.
  2. jcsd
  3. Mar 15, 2007 #2

    Meir Achuz

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    You show that div A=0 for that specific integral by
    1. Take div inide the integral.
    2. Change div to div'.
    3. Integrate by parts, using the divergence theorem.
    4. The surface lntegral -->0, and use div j=0.
    It is an ethical question.
  4. Mar 15, 2007 #3
    Yes!I want to know the ethics.
    Another thing I want to mention.Our professor showed that way in you did.He argued that for localized current distribution the closed surface integral {[(1/|r-r'|)J(r')] dV']=0 over a surface at infinity...
    Why to take so much task? once you enter the div inside the integral,apply divergence theorem at once.The same surface integral you are referring to results.
  5. Mar 15, 2007 #4
    Except you're taking the divergence in unprimed coordinates, while integrating over primed coordinates, so you can't just apply the divergence theorem that quickly.
  6. Mar 15, 2007 #5
    That's right.But that retards me just a little.I will make div=-div' and then will
    apply divergence theorem directly.That will do.But,what about my original question?
    I am asking if I am justified in "proving" divA=0.
  7. Mar 15, 2007 #6
    Sure. The integral equation you have written down presumes a certain choice of gauge, in this case [tex]\nabla.\mathbf{A} = 0[/tex]. Griffiths discusses gauge transformations to some extent, and you are in particular showing that the vector potential of this form satisfies the Coulomb gauge.
  8. Mar 15, 2007 #7
    Thank you.It's clear now.
  9. Oct 27, 2011 #8
    Why can you change div to div'?
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