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Magnetic field from vector potential function using tensor notation

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Homework Statement


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[itex]_{0}[/itex] , one possible
vector potential function is A = [itex]\frac{1}{2}[/itex]B[itex]_{0}[/itex]×r

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

Homework Equations



ε[itex]_{ijk}[/itex]ε[itex]_{klm}[/itex] = δ[itex]_{il}[/itex]δ[itex]_{jm}[/itex] - δ[itex]_{im}[/itex]δ[itex]_{jl}[/itex]

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)[itex]_{i}[/itex]
B = ε[itex]_{ijk}[/itex]∂[itex]_{j}[/itex]A[itex]_{k}[/itex]
B = ε[itex]_{ijk}[/itex]∂[itex]_{j}[/itex] ([itex]\frac{1}{2}[/itex]ε[itex]_{klm}[/itex]B[itex]_{0l}[/itex]r[itex]_{m}[/itex])
B = [itex]\frac{1}{2}[/itex]ε[itex]_{ijk}[/itex]ε[itex]_{klm}[/itex]∂[itex]_{j}[/itex]B[itex]_{0l}[/itex]r[itex]_{m}[/itex])

where ε[itex]_{ijk}[/itex]ε[itex]_{klm}[/itex] = δ[itex]_{il}[/itex]δ[itex]_{jm}[/itex] - δ[itex]_{im}[/itex]δ[itex]_{jl}[/itex]

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

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

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

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.
 
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Answers and Replies

  • #2
TSny
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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
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Would ∂[itex]_{m}[/itex]r[itex]_{m}[/itex] just be equal to the divergence of r i.e. ∇°r?

and for expressing it as a krockner delta is it just ∂[itex]_{l}[/itex]r[itex]_{i}[/itex] = δ[itex]_{li}[/itex]∂r?
 
  • #4
TSny
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Would ∂[itex]_{m}[/itex]r[itex]_{m}[/itex] just be equal to the divergence of r i.e. ∇°r?
Yes. You should be able to reduce ∂[itex]_{m}[/itex]r[itex]_{m}[/itex] to a number.

and for expressing it as a krockner delta is it just ∂[itex]_{l}[/itex]r[itex]_{i}[/itex] = δ[itex]_{li}[/itex]∂r?
Not quite. It's simpler than you wrote.
 
  • #5
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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
tiny-tim
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For the first thing, its just 1 right? And that's because the derivative of r with respect to r is just 1?
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 :smile:
 
  • #7
TSny
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What does the symbol ##\partial_m## stand for? It is a derivative with respect to something. What is the "something"?
 
  • #8
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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
TSny
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since r is x i + y j + z k when we take the divergence we get 3?
Yes. Good.
 
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  • #10
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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
tiny-tim
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if you get confused, remember you can always write ∂mrn as ∂xn/∂xm

which equals … ? :smile:
 
  • #12
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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
tiny-tim
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well it would equal 1 only if the indices are equal …
hold it there!

so (for fixed m and n) ∂xn/∂xm = … ? :smile:
 
  • #14
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Right we don't have a repeated index

that would then = 0 because you are looking at different components correct?
 
  • #15
tiny-tim
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∂xn/∂xm = δmn :wink:
 
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  • #16
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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:
[itex]\frac{1}{2}[/itex][3B[itex]_{0i}[/itex] - B[itex]_{0l}[/itex]δ[itex]_{li}[/itex]]

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

changing indicies

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

so then B = B[itex]_{0}[/itex]

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
tiny-tim
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… B[itex]_{0l}[/itex]δ[itex]_{li}[/itex]


changing indicies

…B[itex]_{0i}[/itex]δ(1)]
once you've used the δ to change the indices, the δ disappears

B[itex]_{0l}[/itex]δ[itex]_{li}[/itex] = B0i :smile:

(oh, and i forgot to mention: your "B =" on the LHS of all your equations should have been "Bi = " :wink:)
 
  • #18
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Oh that was a mistake I knew that but thank you. You have been a wonderful help!
 

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