Vector potential

  • #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.
 

Answers and Replies

  • #2
Meir Achuz
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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.
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.
 
  • #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.
 
  • #4
223
2
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.
 
  • #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.
 
  • #6
223
2
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.
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
1
0
Why can you change div to div'?
 

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