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

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SUMMARY

The discussion focuses on deriving Ampere's Law from the context of Jackson's "Classical Electrodynamics," specifically in the section on Magnetostatics. The user encounters difficulties in applying the integration by parts technique correctly, particularly in the context of vector calculus and the product rule. Key points include the correct identification of functions for integration by parts and the understanding that certain terms vanish due to the properties of the divergence in magnetostatics. The final expression derived is consistent with the expected form of Ampere's Law.

PREREQUISITES
  • Understanding of vector calculus, particularly the product rule and integration by parts.
  • Familiarity with Ampere's Law and its derivation in magnetostatics.
  • Knowledge of the Dirac delta function and its properties.
  • Proficiency in using Jackson's "Classical Electrodynamics" as a reference for electromagnetic theory.
NEXT STEPS
  • Study the application of the product rule in vector calculus.
  • Review the derivation of Ampere's Law in Jackson's "Classical Electrodynamics."
  • Learn about the properties of the Dirac delta function in the context of electromagnetism.
  • Explore advanced topics in magnetostatics, including boundary conditions and surface integrals.
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Students of electromagnetism, physicists working in theoretical physics, and anyone studying advanced vector calculus in the context of electromagnetic theory.

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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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