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Can you help me with Levi Civita Notation, please?

  • #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!
 

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  • #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?
 
  • #3
Hi Blue_leaf77,

Yes that is correct! Thank you, Im sorry I typed it so poorly I am new to Physics forums.
 
  • #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.
 
  • #5
The right is still a vector, I have Bolded the vector quantities B.
 
  • #6
This is essentially a proof from Jackson. But I need to show it with levi Cevita notation.
 
  • #7
blue_leaf77
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The right is still a vector, I have Bolded the vector quantities B.
No, it's not. For instance you have ##|\mathbf{B}|^2## which is a scalar.
 
  • #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.
 

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  • #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.
 
  • #10
nrqed
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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!
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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