Magnetic Potential

  • #1
hi again, I'm trying to show that [itex]\mathbf{B} = \nabla \times \mathbf{A} [/itex] where [itex]\mathbf{A}[/itex] is the magnetic potential given by:

[itex]\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r')}}{|\mathbf{r-r'}|}[/itex]

i deduced that since [itex]\nabla=(\frac{\partial}{\partial{r_1}},\frac{\partial}{\partial{r_2}},\frac{\partial}{\partial{r_3}})[/itex] it only acts on r terms and so

[itex]\nabla \times \mathbf{A}=\frac{\mu_0}{4 \pi} \int dV' \mathbf{J(r')} ( \nabla \times \frac{1}{|\mathbf{r-r'}|} )[/itex]

i can't figure out how to compute that curl - it should be fairly elementary, no?

Answers and Replies

  • #2
You should choose a specific co-ordinate system and work it.
  • #3
tried it in cartesian and it got messy-

[itex]\epsilon_{ijk} \partial_j [{(r-r')_k (r-r')_k}]^{-\frac{1}{2}}[/itex]

  • #4
Oops, that had been bad advice. My apologies, I should have paid more attention.

Here's some better advice : you want to take the curl of this [tex] \frac{\mathbf{J(r ^')}}{|\mathbf{r - r^'}|} [/tex]

This is a vector multiplied by a scalar. So you got to use the product rule for curl in such situations.
  • #5
ok so
[itex]\nabla \times \frac{\mathbf{J(r')}}{\mathbf{|r-r'|}} = \nabla(\frac{1}{\mathbf{|r-r'|}}) \times \mathbf{J(r')} + \frac{1}{\mathbf{|r-r'|}} \nabla \times \mathbf{J(r')}[/itex]

im confused as to how to take the curl of J(r') because [itex]\nabla=\frac{\partial}{\partial{r_i}}[/itex] not [itex]\nabla=\frac{\partial}{\partial{r'_i}}[/itex] so how can the del operator act on a vector function of r'?
  • #6
No, it can't.
  • #7
so the second term is 0

how do i write out the first term though

[itex]\partial_i \left( (r_j-r'_j)(r_j-r'_j) \right)^{-\frac{1}{2}}[/itex] this appears to violate Einstein summation convention?
  • #8
I would suggest forgetting fancy conventions and working it out in some co-ordinate system with all x s and y s, or r s and thetas, whichever you like.
  • #9
yeah wwe're expected to use that notation though.

i did this:

[itex]a_i=r_i-r'_i[/itex] then we have

[itex]\partial_i(a_ja_j)^{-\frac{1}{2}}=-\frac{1}{2}(a_ja_j)^{-\frac{3}{2}}2a_j(\partial_i(r_j-r'_j))=-\frac{1}{(r-r')^3}a_j \partial_i (r_j-r'_j)=-\frac{1}{(r-r')^3}a_j (\delta_{ij}-0)=-\frac{1}{(r-r')^2}[/itex]

which seems like a suitbale answer?
  • #10
It does indeed.
  • #11
actually, shouldn't this be [itex]-\frac{\mathbf{r-r'}}{(r-r')^3}[/itex] as grad should be a vector?
  • #12
It does not act on [tex] r^' [/tex] . Of course, there should be a unit vector along r .
  • #13
but back in post 9 i got [itex]-\frac{a_i}{(r-r')^3}[/itex] where[itex]a_i=(r-r')_i[/itex] suggesting a vector in both directions?
  • #14
Duh, I am caught napping again. Actually the term in your 11th post is correct. Absolutely and completely perfect.

Suggested for: Magnetic Potential