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Indicial notation proof

  1. Sep 14, 2010 #1
    Hello, I need to prove the following:

    [tex]\nabla\times\nabla a[/tex] = 0

    where a is a scalar function of x1, x2, x3 and 0 is the zero vector
     
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  3. Sep 14, 2010 #2

    vela

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    Do you know how to express the cross product using the Levi-Civita symbol?
     
  4. Sep 14, 2010 #3
    oh yes of course

    this is how far i get:

    [tex] \epsilon_{ijk} \nabla_{j} ( \nabla a)_{k} [/tex]

    now if i do try to keep going i have to take the del operator to the del times a in parantheses. Assuming this is where i have to use chain rule i would get a times del of del plus del times (del times a)
    or in written form i get:

    [tex] \epsilon_{ijk} ( a_{k} ( \nabla_{j} \nabla_{k} ) + \nabla_{k} ( \nabla_{j} a_{k} ) ) [/tex]

    not quite sure what del of del is, or del times a for that matter
     
  5. Sep 14, 2010 #4

    vela

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    What is ak? You said a is a scalar function.

    There's no need to use the chain rule. The kth component of ∇a is ∇ka, right?
     
  6. Sep 14, 2010 #5
    I think when i say a is a scalar function, i mean you plug in random values for x1, x2, x3 and you get a scalar.

    so basically a could look something like this:

    [tex] a = x_{1}^2 + x_{2}^2 + 2*x_{3} [/tex]
     
  7. Sep 14, 2010 #6

    vela

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    Right, so what does ak mean? There's no kth component of a.
     
  8. Sep 14, 2010 #7
    oh i dont really know, i guess it doesnt have components.
    thats actually the part where i really dont know how to continue the problem

    can you help?
     
  9. Sep 14, 2010 #8

    vela

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    My point was what you wrote doesn't make sense. You were okay up to here:

    [tex]\epsilon_{ijk} \nabla_j (\nabla a)_k[/tex]

    Use what I said earlier about what the kth component of ∇a is.
     
  10. Sep 14, 2010 #9
    ok so then i have the following:

    [tex] \epsilon_{ijk} ( a_{k} ( \nabla_{j} \nabla_{k} ) + \nabla_{k} ( \nabla_{j} a ) ) [/tex]

    but i still have no idea what [tex] \nabla_{j} \nabla_{k} [/tex] or [tex] \nabla_{j} a [/tex] is?
     
  11. Sep 14, 2010 #10

    vela

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    Can you explain how you got that? It seems like you're just pushing symbols around without any understanding of what they mean.

    Let's clear up your understanding of the notation first. Let B=∇xA. How is this written in indicial notation? Compare that to how you learned to write ∇xA before.
     
  12. Sep 25, 2010 #11
    If we assume that we are working in cartesian coor. system, we can write;
    [tex]\hat{e}_i \epsilon_{ijk} \nabla_j (\nabla a)_k = \hat{e}_i \epsilon_{ijk} \partial_j \partial_k a [/tex]

    here [tex]\nabla_i = \partial_i[/tex]

    then let's write this sum term by term:
    [tex]\sum_{ijk} \hat{e}_i \epsilon_{ijk} \partial_j \partial_k a = \hat{e}_1 \epsilon_{123} \partial_2 \partial_3 a + \hat{e}_1 \epsilon_{132} \partial_3 \partial_2 a + \cdots[/tex]

    [tex]\epsilon_{123} = 1[/tex] and [tex]\epsilon_{132} = -1[/tex]
    It should be easy to see that all term cancels each other.Hope could be helpful.
     
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