# Vector potential

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.

Meir Achuz
Homework Helper
Gold Member
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.

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.

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.

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.

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 $$\nabla.\mathbf{A} = 0$$. Griffiths discusses gauge transformations to some extent, and you are in particular showing that the vector potential of this form satisfies the Coulomb gauge.

Thank you.It's clear now.

Why can you change div to div'?