Magnetic field from vector potential function using tensor notation

1. Sep 23, 2013

thatguy14

1. The problem statement, all variables and given/known data
We will see (in Chap. 5) that the magnetic field can be derived from a vector potential function as
follows:
B = ∇×A
Show that, in the special case of a uniform magnetic field B$_{0}$ , one possible
vector potential function is A = $\frac{1}{2}$B$_{0}$×r

MUST USE TENSOR NOTATIONm also B0 is constant (uniform magnetic field)

2. Relevant equations

ε$_{ijk}$ε$_{klm}$ = δ$_{il}$δ$_{jm}$ - δ$_{im}$δ$_{jl}$

3. The attempt at a solution
I have tried a bunch of different things but I am missing something near the end.
Here is what I have

B = (∇×A)$_{i}$
B = ε$_{ijk}$∂$_{j}$A$_{k}$
B = ε$_{ijk}$∂$_{j}$ ($\frac{1}{2}$ε$_{klm}$B$_{0l}$r$_{m}$)
B = $\frac{1}{2}$ε$_{ijk}$ε$_{klm}$∂$_{j}$B$_{0l}$r$_{m}$)

where ε$_{ijk}$ε$_{klm}$ = δ$_{il}$δ$_{jm}$ - δ$_{im}$δ$_{jl}$

So B = $\frac{1}{2}$[δ$_{il}$δ$_{jm}$ - δ$_{im}$δ$_{jl}$]∂$_{j}$B$_{0l}$r$_{m}$

Changing indicies gives (noting that the derivative of constant = 0 and using the product rule)

B = $\frac{1}{2}$[B$_{0i}$∂$_{m}$r$_{m}$ - B$_{0l}$∂$_{l}$r$_{i}$]

And that's where I am stuck. What comes next? I am assuming that with the last term there, l and i have to be equal (because if they aren't then it equals 0) and I think I have to introduct the krockner delta somewhere but I am unsure. Any help would be greatly appreciated.

Last edited: Sep 23, 2013
2. Sep 23, 2013

TSny

In the expression $\partial_m r_m$ you are summing over $m$. Consider one term of that sum, say $\partial_2 r_2$. Can you see what that is equal to?

You are right that you can express $\partial_l r_i$ in terms of a Kronecker delta.

3. Sep 23, 2013

thatguy14

Would ∂$_{m}$r$_{m}$ just be equal to the divergence of r i.e. ∇°r?

and for expressing it as a krockner delta is it just ∂$_{l}$r$_{i}$ = δ$_{li}$∂r?

4. Sep 23, 2013

TSny

Yes. You should be able to reduce ∂$_{m}$r$_{m}$ to a number.

Not quite. It's simpler than you wrote.

5. Sep 23, 2013

thatguy14

For the first thing, its just 1 right? And that's because the derivative of r with respect to r is just 1?

For the second... l has to equal i is what I am going to go with so does that mean that that portion is equal to 0? Or is that incorrect?

6. Sep 23, 2013

tiny-tim

no, you can't differentiate wrt a vector

(and if you mean |r|, that isn't in your equation)

instead of ∂/∂r, try ∂/∂x or ∂/∂y or ∂/∂z

7. Sep 23, 2013

TSny

What does the symbol $\partial_m$ stand for? It is a derivative with respect to something. What is the "something"?

8. Sep 23, 2013

thatguy14

oh... Okay then let me try again:

since r is x i + y j + z k when we take the divergence we get 3?

9. Sep 23, 2013

TSny

Yes. Good.

10. Sep 23, 2013

thatguy14

Okay awesome (rather silly of me though)! Now for the second portion I am still lost. I am going to take a guess that it just equals 1 but I can't justify it to myself. Is that correct? and if it is why is it correct?

11. Sep 23, 2013

tiny-tim

if you get confused, remember you can always write ∂mrn as ∂xn/∂xm

which equals … ?

12. Sep 23, 2013

thatguy14

well it would equal 1 only if the indices are equal so then is it equal to 3 also? (because of the y and z components)?

edit: looking only at the x component the indicies are summed to 3 so for the x component is it equal to 3?

13. Sep 23, 2013

tiny-tim

hold it there!

so (for fixed m and n) ∂xn/∂xm = … ?

14. Sep 23, 2013

thatguy14

Right we don't have a repeated index

that would then = 0 because you are looking at different components correct?

15. Sep 23, 2013

tiny-tim

∂xn/∂xm = δmn

16. Sep 23, 2013

thatguy14

hm. I can see how that works now but do you think you can explain that a little further if possible?

Also that means then I have this left with the appropriate substituitons:
$\frac{1}{2}$[3B$_{0i}$ - B$_{0l}$δ$_{li}$]

so then δ$_{li}$] = 1 only if l = i or 0 if l ≠ i

changing indicies

$\frac{1}{2}$[3B$_{0i}$ - B$_{0i}$δ(1)]

so then B = B$_{0}$

and that's it correct?

Thank you by the way guys for your help. I am very glad you didn't just give me the answers and helped me along even if it was a little silly in hindsight!

17. Sep 23, 2013

tiny-tim

once you've used the δ to change the indices, the δ disappears

B$_{0l}$δ$_{li}$ = B0i

(oh, and i forgot to mention: your "B =" on the LHS of all your equations should have been "Bi = " )

18. Sep 23, 2013

thatguy14

Oh that was a mistake I knew that but thank you. You have been a wonderful help!