Transformation of some formula

  • Thread starter Vermax
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  • #1
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There is a transformation:

[tex]\vec{E}\times\left(\nabla\times\vec{E}\right)=
\vec{k}_{i}\epsilon_{ijk}E_j\left(\nabla\times\vec{E}\right)_k=
\vec{k}_{i}\epsilon_{jk}E_j\epsilon_{klm}\nabla_l E_m=
\vec{k}_i\epsilon_{ijk}\epsilon_{klm}E_j\nabla_l E_m
=\vec{k}_i\left(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\right)E_j\nabla_l E_m= [/tex]
[tex]=\vec{k}_i\delta_{il}\delta_{jm}E_j\nabla_l E_m-\vec{k}_i\delta_{im}\delta_{jl}E_j\nabla_l E_m=\vec{k}_i jE_j\nabla_i E_j-\vec{k}_i E_j\nabla_j E_i={{1}\over{2}}\nabla E^2-\left(\vec{E}\nabla\right)\vec{E} [/tex]

I think it is connected somehow to matrixes, but I do not get it at all. The way how it is done looks quite nice so maybe it is worth to learn this method. Can someone just help me with understanding it?




PS. Sorry if I posted it in a wrong forum, i just do not know where I am supposed to write it...
 

Answers and Replies

  • #2
Cyosis
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With "I do not get it at all" do you mean that you don't understand a single step? It would be nice to narrow it down as much as possible.

Just to check what you know or don't know:

1)Do you know what the Levi-Civita tensor, [itex]\epsilon_{ijk}[/itex]?
2)Do you know the Kronecker delta symbol, [itex]\delta_{ij}[/itex]?
3)Do you know how to write vectors/cross products in components, index notation?
 
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  • #3
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Just I am not too familiar with this Einstein's notation. Starting from the very first part of this equation:
[tex]
\vec{E}\times\left(\nabla\times\vec{E}\right)=
\vec{k}_{i}\epsilon_{ijk}E_j\left(\nabla\times\vec {E}\right)_k [/tex]

So cross product can be written as:
[tex]\vec{u} \times \vec{v}= u^j v^k (\mathbf{e}_j \times \mathbf{e}_k ) = u^j v^k\epsilon^i_{jk} \mathbf{e}_i [/tex]

I assume we do not care about lower or upper position of indexes? If yes, that would be ok I think.


Answering your questions:
1) yes, theoretically, but I have never used them so think I have problems with using them "fluently"
2) yes, on a decent level I think
3) same as first



But now:
[tex]\vec{k}_{i}\epsilon_{jk}E_j\epsilon_{klm}\nabla_l E_m=
\vec{k}_i\epsilon_{ijk}\epsilon_{klm}E_j\nabla_l E_m [/tex]
I do not know how this rotation turn into gradient and how to play with this indexes while changing their positions.
 
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  • #4
Cyosis
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Well in your LHS you lost and index, which I now see you missed in your original equation as well? Is this a typo or intended? Secondly k is a unit vector so it's better to write it with a hat on top of it, or as e_i.

Switching the Levi-Civita symbol and E_j is not a problem at all, realise that these are just numbers inside of a sum.

To make it a bit more clear (hopefully) I will use normal summation instead of Einstein summation:

[tex]
\begin{align*}
\vec{a} \times (\vec{b} \times \vec{c}) & = \sum_{i,j,k} \epsilon_{ijk} \hat{e_i}a_j(\vec{b} \times \vec{c})_k
& = \sum_{i,j,k} \epsilon_{ijk} \hat{e_i}a_j \left(\sum_{l,m} \epsilon_{klm}b_l c_m \right)
& = \sum_{i,j,k} \sum_{l,m} \epsilon_{ijk} \hat{e_i}a_j \epsilon_{klm}b_l c_m
\end{align*}
[/tex]

In the last expression is it clear that you can just rearrange the terms as long as they aren't operators?
 
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  • #5
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Hmm I just copied it from some Wiki-book. If you want to check it by yourself here is the link but it is not in english I am afraid http://pl.wikibooks.org/wiki/Fizyka_matematyczna/Zasady_zachowania_a_twierdzenia_o_właściwościach_pola_elektromagnetycznego [Broken] .

Yes, I must play more with this notation, becouse I think I do not see this changing numbers inside of a sum as I do not see this rot->grad change unfortunately...
 
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  • #6
Cyosis
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That's definitely a mistake from the author. On second thought you mean why the rotation is written as a gradient? I mistook rotation for the swapping around of elements within the sum. So the problem is from the second step to the third step?
 
  • #7
Cyosis
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First convince yourself, by writing down all the components, that the k-th component of a cross product can be written as [itex] (\vec{a} \times \vec{b})_k=\epsilon_{klm} a_lb_m[/itex]. Einstein summation over the indices l and m implied.

The gradient operator is nothing more than [itex](\partial_{x_1},\partial_{x_2},\partial_{x_3})[/itex]. As a result [tex](\nabla \times \vec{a})_k=\epsilon_{klm} \partial_{x_l} a_m=\epsilon_{klm} \nabla_l a_m[/tex]. Differentiation with respect to the l-th coordinate.
 
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  • #8
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I did as you told me to, and this thing:
[itex]
(\vec{a} \times \vec{b})_k=\epsilon_{klm} a_lb_m
[/itex]
Looks for me (almost ofc) like a Laplace determinant expansion with the first versor. It makes more sense now for me.

And rot->grad thing is not so diffcult as it looked like :)

I will try to analyse it one more time and hopefully to the end. Thank you very much for help, especially that such topics are not too easy to explain via forum.
 
  • #9
Cyosis
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No problem. The next step, writing a product of Levi-Civita tensors as they sum of the two times the product of two Kronecker deltas is probably the hardest step.
 
  • #10
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I stucked in the second line.

[tex]
=\vec{k}_i\delta_{il}\delta_{jm}E_j\nabla_l E_m-\vec{k}_i\delta_{im}\delta_{jl}E_j\nabla_l E_m=\vec{k}_i jE_j\nabla_i E_j-\vec{k}_i E_j\nabla_j E_i={{1}\over{2}}\nabla E^2-\left(\vec{E}\nabla\right)\vec{E}
[/tex]

I think:
[tex]\delta_{il}\delta_{jm} = \delta_{im}\delta_{jl} =1 [/tex]
But suddenly:
[tex]\nabla_l E_m=j \nabla_i E_j [/tex]
uh...

Or it is rather:
[tex]\delta_{il}\delta_{jm}\nabla_l E_m=j \nabla_i E_j [/tex]
[tex]\delta_{im}\delta_{jl}\nabla_l E_m=E_j\nabla_j E_i [/tex]
 
  • #11
Cyosis
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That loose j shouldn't be there, again a mistake from the author. The deltas disappear due to contraction. Lets look at the term [itex]\delta_{jm} E_m[/itex]. Now look at the different possibilities. Let [itex]m \neq j[/itex] then [itex]\delta_{jm}=0[/itex], therefore [itex]\delta_{jm} E_m=0[/itex]. Let [itex]m= j[/itex] then [itex]\delta_{jm}=1[/itex], therefore [itex]\delta_{jm} E_m=1*E_m=1*E_j=E_j[/itex].
 
  • #12
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Hehe quite logical!

One last thing I am afraid...
I understand why:
[tex]
\vec{k}_i E_j\nabla_j E_i=\left(\vec{E}\nabla\right)\vec{E} [/tex]

But how:
[tex]
\vec{k}_i E_j\nabla_i E_j={{1}\over{2}}\nabla E^2[/tex]
Why there is 0.5 and where goes the versor k?
 
  • #13
Cyosis
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Lets write the i-th component of the gradient as [itex]\partial_{x_i}[/itex]again.

Using the Einstein summation convention we get:

[tex]
\hat{e_i}E_j \partial_{x_i} E_j & =\frac{1}{2} \hat{e_i} \partial_{x_i} E_j^2 =\frac{1}{2} \nabla E^2
[/tex]

Take a function f(x)^2, then [itex]\partial_x f(x)^2=2 f(x) \partial_{x} f(x)[/itex]. This is nothing more than the chain rule.
 
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  • #14
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Oh I see...
But that was I think the most difficult part.
I am glad, however, that I understad it thanks to you of course :)
 
  • #15
Cyosis
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You're welcome.

Edit: I overlooked something,[itex](\vec{E}\nabla)\vec{E}[/itex] should of course be [itex](\vec{E} \cdot \nabla)\vec{E}[/itex].
 
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  • #16
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Hm yes I think so, and I can trot out that I tried to prove equations from Wiki:
http://en.wikipedia.org/wiki/Vector_calculus_identities
And I can say that I was succesfull in almost every one :) I am still working on some more complicated but it is a matter of time I think.
 
  • #17
Cyosis
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Hehe proving all the vector identities using index notation is an excellent way to get the hang of it. Much easier than writing out all the components.
 
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