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Levi Civita notations, curls relating to 1/r fields

  • Thread starter NaOH
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  • #1
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This isn't really isn't strictly a homework question; there wouldn't be solutions provided. I know the answer -- if anything it follows from other electrostatic results. I would like to make sure what I have written down as steps are legitimate.

I will also skip ahead some steps because typing everything out would be very time consuming, especially if I have no idea if the codes will turn out right.

Homework Statement


Prove that
[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = 0 [/tex]

[tex]\nabla \times {\bf r} = 0 [/tex]

Homework Equations





The Attempt at a Solution



I begin with
[tex]\nabla \times {\bf r} = 0 [/tex]
Using Levi Civita notations,
[tex]\nabla \times {\bf r} = \epsilon_{ijk} \partial_{j} x_{k}=\epsilon_{ijk} \delta_{jk}[/tex]
Where
[tex] \partial_{j} = \frac{d}{dx_{j}}[/tex]
Here, I reason that for it to have a value, j=k in the kronecker delta, but if that is the case, then the levi civita tensor would be zero, hence, curl of a 1/r is zero.

Next,

[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = \epsilon_{ijk} \partial_{j} (x_{k} (x_{b} x_{b})^{-3/2}) [/tex]
[tex]=\epsilon_{ijk}( (x_{b} x_{b})^{-5/2} \partial_{j}(x_{k}) - 3x_{k}x_{j}(x_{k} (x_{b} x_{b})^{-5/2}[/tex]
[tex]=\frac{1}{r^{3}}\nabla \times {\bf {r}} - \frac{3}{r^5} {\bf {r}} \times {\bf r} = 0-3(0) = 0[/tex]
 
Last edited:

Answers and Replies

  • #2
fzero
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[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = \epsilon_{ijk} \partial_{j} (x_{k} (x_{b} x_{b})^{-3/2}) [/tex]
[tex]=\epsilon_{ijk}( (x_{b} x_{b})^{-5/2} \partial_{j}(x_{k}) - 3x_{k}x_{j}(x_{k} (x_{b} x_{b})^{-5/2}[/tex]
There's a small typo in the 2nd line (the -5/2 exponent in the 1st term should still be a -3/2) but both of your calculations look perfectly fine. I would further nitpick a bit and suggest that when you convert from vector notation to index notation, you stick the free indices on the vector expression so that the indices match on both sides of every equation. So for example, I would write

$$ (\nabla \times \mathbf{r})_i = \epsilon_{ijk} \partial_{j} x_{k}.$$

You seem to have the important ideas sorted out though, so this is more a criticism of form than substance.
 
  • #3
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Thank you for your reply fzero.

My lecturer that introduced these notations to us typed out exactly as I did (maybe without the embarrassing mistake!), so I thought that was the norm. I will include the free indices from now on, makes it clearer for me too.

I am really concerned if what I wrote
[tex] \epsilon_{ijk} \partial_{j} x_{k} = \epsilon_{ijk} \delta_{jk} = 0 [/tex] is a legitimate mathematical statement.
 
  • #4
vanhees71
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Your statement is obviously true, because [itex]\delta_{jk}=0[/itex] for [itex]i \neq j[/itex] and [itex]\epsilon_{ijk}=0[/itex] for [itex]j=k[/itex]. So all terms in the sum [itex]\delta_{jk} \epsilon_{ijk}=0[/itex].

The other thing is that, strictly speaking the notation to equal a vector to one of its components doesn't make any mathematical sense. It is plain wrong! So the correct notation is
[tex](\vec{\nabla} \times \vec{r})_i=\epsilon_{ijk} \partial_j x_k=\epsilon_{ijk} \delta_{jk}=0.[/tex]
Since this holds for all [itex]i \in \{1,2,3 \}[/itex] you have proven that
[tex]\vec{\nabla} \times \vec{r}=0.[/tex]
Here, on the right-hand side stands the zero-vector!
 
  • #5
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Thanks for the reply. I will keep that in mind!
 

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