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Homework Help: Magnetic Potential

  1. Feb 16, 2009 #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?
     
  2. jcsd
  3. Feb 16, 2009 #2
    You should choose a specific co-ordinate system and work it.
     
  4. Feb 17, 2009 #3
    tried it in cartesian and it got messy-

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

    ???
     
  5. Feb 17, 2009 #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 gotta use the product rule for curl in such situations.
     
  6. Feb 17, 2009 #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'???
     
  7. Feb 17, 2009 #6
    No, it can't.
     
  8. Feb 17, 2009 #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?
     
  9. Feb 17, 2009 #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.
     
  10. Feb 17, 2009 #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???
     
  11. Feb 17, 2009 #10
    It does indeed.
     
  12. Feb 17, 2009 #11
    actually, shouldn't this be [itex]-\frac{\mathbf{r-r'}}{(r-r')^3}[/itex] as grad should be a vector?
     
  13. Feb 17, 2009 #12
    It does not act on [tex] r^' [/tex] . Of course, there should be a unit vector along r .
     
  14. Feb 17, 2009 #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???
     
  15. Feb 17, 2009 #14
    Duh, I am caught napping again. Actually the term in your 11th post is correct. Absolutely and completely perfect.
     
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