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

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The discussion focuses on deriving Ampere's Law using the curl of the magnetic field and integration by parts. The user struggles with the application of the product rule and the correct assignment of variables during integration. Clarifications indicate that the divergence of the current density j(r') is zero in magnetostatics, leading to the disappearance of certain terms. The surface term from the integration vanishes under the assumption that j(r') is bounded as r' approaches infinity. The conversation emphasizes the importance of correctly applying vector calculus rules in this context.
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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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