# Can you help me with Levi Civita Notation, please?

1. May 14, 2016

### hellomynameisscottt

• 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. May 14, 2016

### blue_leaf77

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. May 14, 2016

### hellomynameisscottt

Hi Blue_leaf77,

Yes that is correct! Thank you, Im sorry I typed it so poorly I am new to Physics forums.

4. May 14, 2016

### blue_leaf77

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. May 14, 2016

### hellomynameisscottt

The right is still a vector, I have Bolded the vector quantities B.

6. May 14, 2016

### hellomynameisscottt

This is essentially a proof from Jackson. But I need to show it with levi Cevita notation.

7. May 14, 2016

### blue_leaf77

No, it's not. For instance you have $|\mathbf{B}|^2$ which is a scalar.

8. May 14, 2016

### hellomynameisscottt

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. May 14, 2016

### blue_leaf77

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. May 16, 2016

### nrqed

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?
$\sum_i \epsilon_{ijk} \epsilon_{iab} = ?$