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Can you help me with Levi Civita Notation, please?

  1. May 14, 2016 #1
    • OP warned about not using the homework template
    I need to prove B.(Gradient . B) - B X(Gradient X B)=Del{i} [B{i}B{j} -1/2 (kroneker delta {ij} B^2]

    where I have used . as the dot product, {} as subscript. Thank you!
     
  2. jcsd
  3. May 14, 2016 #2

    blue_leaf77

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    If I write the left hand side using proper math symbols, is
    $$
    \mathbf{B} (\nabla \cdot \mathbf{B}) - \mathbf{B} \times (\nabla \times \mathbf{B})
    $$
    correct?
     
  4. May 14, 2016 #3
    Hi Blue_leaf77,

    Yes that is correct! Thank you, Im sorry I typed it so poorly I am new to Physics forums.
     
  5. May 14, 2016 #4

    blue_leaf77

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    Then the equality in the original equation does not make sense because the left side is a vector whereas the right side is a scalar.
     
  6. May 14, 2016 #5
    The right is still a vector, I have Bolded the vector quantities B.
     
  7. May 14, 2016 #6
    This is essentially a proof from Jackson. But I need to show it with levi Cevita notation.
     
  8. May 14, 2016 #7

    blue_leaf77

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    No, it's not. For instance you have ##|\mathbf{B}|^2## which is a scalar.
     
  9. May 14, 2016 #8
    I have uploaded the page from Jackson it is equation (6.119) I am trying to prove, however I must use Levi Cevita notation.
     

    Attached Files:

  10. May 14, 2016 #9

    blue_leaf77

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    I see, the left side should be ##[\mathbf{B} (\nabla \cdot \mathbf{B}) - \mathbf{B} \times (\nabla \times \mathbf{B})]_i## which is the i-th component of the vector inside the square bracket and this is a scalar. Start by writing ##\nabla\cdot \mathbf{B}## using Einstein summation notation. Anyway if you have progressed up to any point, just post it here, preferably using LaTex.
     
  11. May 16, 2016 #10

    nrqed

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    You will need to use that the i-th component of a cross product may be written as
    ## (\vec{A} \times \vec{B})_i = \epsilon_{ijk} A_j B_k ##

    and you will need to find the expression for a triple product....hint: what is the following expression equal to?
    ## \sum_i \epsilon_{ijk} \epsilon_{iab} = ? ##
     
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