Is It Justified to Prove divA=0 for Vector Potential?

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Discussion Overview

The discussion revolves around the justification of proving that the divergence of the magnetic vector potential \( \mathbf{A} \) is zero, specifically in the context of an exam question that assumes a particular form of the vector potential. Participants explore the implications of this assumption and the ethical considerations surrounding the proof.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the ethics of proving \( \text{div} \, \mathbf{A} = 0 \) given the assumptions made in the exam question regarding the vector potential's form.
  • Others suggest that using the divergence theorem and integrating by parts can demonstrate that \( \text{div} \, \mathbf{A} = 0 \) for the specific integral provided.
  • A participant mentions that their professor showed a method involving a closed surface integral at infinity, which leads to a similar conclusion about the divergence being zero.
  • One participant points out a potential issue with applying the divergence theorem due to the use of unprimed coordinates while integrating over primed coordinates.
  • Another participant proposes that changing \( \text{div} \) to \( \text{div}' \) could simplify the application of the divergence theorem.
  • There is a mention of gauge choices, specifically the Coulomb gauge, and how the integral form presumes this choice.
  • A participant expresses a desire for clarity on the justification of proving \( \text{div} \, \mathbf{A} = 0 \) in this context.
  • One participant asks why it is permissible to change \( \text{div} \) to \( \text{div}' \), indicating a need for further explanation of this step.

Areas of Agreement / Disagreement

Participants express differing views on the justification of proving \( \text{div} \, \mathbf{A} = 0 \). While some provide methods to support the proof, others raise concerns about the assumptions involved and the ethical implications of the proof process. The discussion remains unresolved regarding the justification of the proof.

Contextual Notes

Participants highlight limitations related to the assumptions made in the proof, particularly concerning the choice of gauge and the treatment of coordinates during integration. These factors contribute to the complexity of the discussion.

Kolahal Bhattacharya
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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.
 
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Kolahal Bhattacharya said:
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.
 
Kolahal Bhattacharya said:
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.
 
Thank you.It's clear now.
 
Why can you change div to div'?
 

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