Problem in deriving Ampere's law

  • Thread starter dingo_d
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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

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

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

[tex]\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'[/tex]

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

I end up with:

[tex]\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)[/tex]

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

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

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

But I don't get what I need :\

What am I doing wrong?
 

Answers and Replies

  • #2
vela
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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:

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

I was following Jacksons steps and it said integration by parts... so when I take derivation, I'll get [tex]\nabla j(r')=du[/tex]?
 
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  • #4
vela
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Integration by parts is just applying the product rule to rewrite the integrand. In this case, the integrand becomes

[tex]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')[/tex]

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.
 
  • #5
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I did not know that :D

Thank you!!!
 
  • #6
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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'| )
 
  • #7
vela
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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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