Magnetic Potential: Computing the Curl of $\mathbf{A}$

[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?

 

[xboy]

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

 

[latentcorpse]

tried it in cartesian and it got messy-

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

?

 

[xboy]

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 \frac{\mathbf{J(r ^')}}{|\mathbf{r - r^'}|}

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

 

[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'?

 

[xboy]

No, it can't.

 

[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?

 

[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.

 

[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?

 

[xboy]

It does indeed.

 

[latentcorpse]

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

 

[xboy]

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

 

[latentcorpse]

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

 

[xboy]

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