Magnetic Potential

1. Feb 16, 2009

latentcorpse

hi again, i'm trying to show that $\mathbf{B} = \nabla \times \mathbf{A}$ where $\mathbf{A}$ is the magnetic potential given by:

$\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r')}}{|\mathbf{r-r'}|}$

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

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

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

2. Feb 16, 2009

xboy

You should choose a specific co-ordinate system and work it.

3. Feb 17, 2009

latentcorpse

tried it in cartesian and it got messy-

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

???

4. Feb 17, 2009

xboy

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

This is a vector multiplied by a scalar. So you gotta use the product rule for curl in such situations.

5. Feb 17, 2009

latentcorpse

ok so
$\nabla \times \frac{\mathbf{J(r')}}{\mathbf{|r-r'|}} = \nabla(\frac{1}{\mathbf{|r-r'|}}) \times \mathbf{J(r')} + \frac{1}{\mathbf{|r-r'|}} \nabla \times \mathbf{J(r')}$

im confused as to how to take the curl of J(r') because $\nabla=\frac{\partial}{\partial{r_i}}$ not $\nabla=\frac{\partial}{\partial{r'_i}}$ so how can the del operator act on a vector function of r'???

6. Feb 17, 2009

xboy

No, it can't.

7. Feb 17, 2009

latentcorpse

so the second term is 0

how do i write out the first term though

$\partial_i \left( (r_j-r'_j)(r_j-r'_j) \right)^{-\frac{1}{2}}$ this appears to violate Einstein summation convention?

8. Feb 17, 2009

xboy

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. Feb 17, 2009

latentcorpse

yeah wwe're expected to use that notation though.

i did this:

$a_i=r_i-r'_i$ then we have

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

which seems like a suitbale answer???

10. Feb 17, 2009

xboy

It does indeed.

11. Feb 17, 2009

latentcorpse

actually, shouldn't this be $-\frac{\mathbf{r-r'}}{(r-r')^3}$ as grad should be a vector?

12. Feb 17, 2009

xboy

It does not act on $$r^'$$ . Of course, there should be a unit vector along r .

13. Feb 17, 2009

latentcorpse

but back in post 9 i got $-\frac{a_i}{(r-r')^3}$ where$a_i=(r-r')_i$ suggesting a vector in both directions???

14. Feb 17, 2009

xboy

Duh, I am caught napping again. Actually the term in your 11th post is correct. Absolutely and completely perfect.