Deriving the electromagnetic field strength tensor

  • Thread starter rwooduk
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  • #1
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Just one last question today if someone can help. I'm trying to derive the electromagnetic field strength tensor and having a little trouble with (i think) the use of identities, please see below:

YWH2xbX.jpg


I understand the first part to get -Ei, but it's the second line of the next bit I don't understand. I see he wants to get line one to the form he has done so he can use the following identity:

##\varepsilon _{ijk}\varepsilon _{ilm} = \delta _{jl}\delta _{km} - \delta _{jm}\delta _{kl}##

But I'm unsure how he has got there and line 2 to 3 looks a little iffy. I've tried using ##\delta _{ij}\delta _{jk} = \delta _{ik}## but that didn't go very well, and also I've always thought something of the form ##\delta_{i}^{l} ## was the kroneka delta function, so I'm unsure how it would come in here.

If someone can help by putting a couple of inbetween stages in the derivation it would really help my understanding of what is going on and the notation / identities he's used.
 

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  • #2
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## \delta^{i}{}_l\delta^{j}{}_m \partial^lA^m= \partial^iA^j##

etc.
 
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  • #3
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Thanks for the reply, I can see that relation by looking at the difference between the lines, but how did he get the ## \partial^iA^j## term to equal that?
thanks again
 
  • #4
Orodruin
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The sum over ##m## when there is a ##\delta_i^m## is non-zero only when ##i = m## and you can replace the ##m## sum by simply replacing the other ##m## by the ##i## and removing the delta.
 
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  • #5
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The sum over ##m## when there is a ##\delta_i^m## is non-zero only when ##i = m## and you can replace the ##m## sum by simply replacing the other ##m## by the ##i## and removing the delta.

ahh ok i see so j must = m and i must = m so it goes to Am. And likewise i must = l and j must = l therefore it goes to delta l. This is probably a stupid question, but wouldn't you then get zero in the bracket? i.e. (1-1)
 
  • #6
Orodruin
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This is probably a stupid question, but wouldn't you then get zero in the bracket? i.e. (1-1)
No. In one of the terms the i replaces the l and in the other it replaces the m. This gives different terms.
 
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  • #7
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No. In one of the terms the i replaces the l and in the other it replaces the m. This gives different terms.

hmm, kind of got that, ok thanks, will work on this some more.

Thanks for all the replies.
 
  • #8
vanhees71
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Isn't this way too complicated? I usually write
$$F^{ij}=\partial^{i} A^{j} - \partial^{j} A^i=-(\partial_i A^j-\partial_{j} A^{i})=-\epsilon^{ijk} B^{k},$$
where one has to keep in mind that
$$\partial^i=-\partial_i=\frac{\partial}{\partial x^i}$$
and then use that
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
 
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  • #9
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-(\partial_i A^j-\partial_{j} A^{i})=-\epsilon^{ijk} B^{k},$$

do you go from LHS to RHS from memory, or are you saying you used the things you say below it? It's the steps inbetween I'm interested in.

would welcome a simpler 'derivation' of this step though
 
  • #10
vanhees71
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This is simply the definition of the curl operation and the vector potential by ##\vec{B}=\vec{\nabla} \times \vec{A}##.
 
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