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Magnetic field from vector potential function using tensor notation

  1. Sep 23, 2013 #1
    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[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)

    2. Relevant 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]

    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)[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.
     
    Last edited: Sep 23, 2013
  2. jcsd
  3. Sep 23, 2013 #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.
     
  4. Sep 23, 2013 #3
    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?
     
  5. Sep 23, 2013 #4

    TSny

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    Yes. You should be able to reduce ∂[itex]_{m}[/itex]r[itex]_{m}[/itex] to a number.

    Not quite. It's simpler than you wrote.
     
  6. Sep 23, 2013 #5
    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?
     
  7. Sep 23, 2013 #6

    tiny-tim

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    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:
     
  8. Sep 23, 2013 #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"?
     
  9. Sep 23, 2013 #8
    oh... Okay then let me try again:

    since r is x i + y j + z k when we take the divergence we get 3?
     
  10. Sep 23, 2013 #9

    TSny

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    Yes. Good.
     
  11. Sep 23, 2013 #10
    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?
     
  12. Sep 23, 2013 #11

    tiny-tim

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

    which equals … ? :smile:
     
  13. Sep 23, 2013 #12
    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?
     
  14. Sep 23, 2013 #13

    tiny-tim

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    hold it there!

    so (for fixed m and n) ∂xn/∂xm = … ? :smile:
     
  15. Sep 23, 2013 #14
    Right we don't have a repeated index

    that would then = 0 because you are looking at different components correct?
     
  16. Sep 23, 2013 #15

    tiny-tim

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    ∂xn/∂xm = δmn :wink:
     
  17. Sep 23, 2013 #16
    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!
     
  18. Sep 23, 2013 #17

    tiny-tim

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    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:)
     
  19. Sep 23, 2013 #18
    Oh that was a mistake I knew that but thank you. You have been a wonderful help!
     
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