What Am I Doing Wrong in Deriving Ampere's Law?

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Homework Help Overview

The discussion revolves around deriving Ampere's Law within the context of magnetostatics, specifically referencing the treatment in Jackson's textbook. The original poster encounters difficulties in applying integration by parts to a specific expression involving the magnetic field B and the current density j.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply integration by parts but is unsure about the correct assignment of variables and the resulting expressions. Some participants question the application of the product rule and suggest clarifying the roles of the functions involved in the integration.

Discussion Status

Participants are actively engaging with the original poster's confusion, providing insights into the integration process and the product rule. There is a recognition of the complexities involved in handling vector functions and divergence in this context. The discussion is ongoing, with some participants offering clarifications while others express their own uncertainties.

Contextual Notes

There is mention of assumptions in magnetostatics, particularly regarding the divergence of the current density j, and the treatment of surface terms in the integration process. The original poster's reference to Jackson's steps indicates a reliance on established methods that may require further unpacking in this discussion.

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Homework Statement




I'm going through Jackson a bit, reading on Magnetostatics, and I came into a bump.

I'm looking at

\nabla\times B=\frac{1}{c}\nabla\times\nabla\times\int\frac{j(r')}{|r-r'|}d^3r'

I expand that using 'BAC-CAB' rule and I get:

\nabla\times B=\frac{1}{c}\nabla\int j(r')\cdot\nabla\left(\frac{1}{|r-r'|}\right)d^3r'-\frac{1}{c}\int j(r')\nabla^2\left(\frac{1}{|r-r'|}\right)d^3r'

So after changing the \nabla into \nabla ' and using the fact that \nabla^2\left(\frac{1}{|r-r'|}\right)=-4\pi\delta(r-r')

I end up with:

\nabla\times B=-\frac{1}{c}\nabla\int j(r')\cdot\nabla '\left(\frac{1}{|r-r'|}\right)d^3r'+\frac{4\pi}{c}j(r)

And here it says that the first part after integration by parts becomes:

\nabla\times B=\frac{1}{c}\nabla\int \frac{\nabla '\cdot j(r')}{|r-r'|}\right)d^3r'+\frac{4\pi}{c}j(r)

I tried integration by parts like this:
j(r')d^3r'=dv\Rightarrow j(r')=v and \nabla '\left(\frac{1}{|r-r'|}\right)=u\Rightarrow \nabla^2'\left(\frac{1}{|r-r'|}\right)d^3r'=du

But I don't get what I need :\

What am I doing wrong?
 
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You have u and v backwards. Also, when you integrate j(r'), you don't get j(r').

It's probably easier to see using the product rule for the divergence:

\nabla\cdot(\phi \mathbf{F}) = (\nabla\phi)\cdot \mathbf{F} + \phi(\nabla\cdot\mathbf{F})
 
oh so j(r')=u and \nabla '\left(\frac{1}{|r-r '|}\right)d^3r'=dv?

I was following Jacksons steps and it said integration by parts... so when I take derivation, I'll get \nabla j(r')=du?
 
Last edited:
Integration by parts is just applying the product rule to rewrite the integrand. In this case, the integrand becomes

j(r')\cdot\nabla'\left(\frac{1}{|r-r'|}\right) = \nabla'\cdot\left(\frac{j(r')}{|r-r'|}\right) - \frac{1}{|r-r'|}\nabla'\cdot j(r')

With vector functions, saying u=this and dv=that gets kind of confusing, so it's better to just use the relevant product rule directly.
 
I did not know that :D

Thank you!
 
There is still something I don't understand ... ( I know that I join this topic a bit lately )

Using the product rule, two terms come out. According to Jackson, the first disappears and the second is zero because the divergence of J is zero in magnetostatics. I understand this point, but what about the first term ?(I mean the one with J(r')/|r-r'| )
 
That's what's called a surface term. It vanishes because you assume J is bounded so J(r')/|r-r'| goes to 0 as r' goes to infinity.
 

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