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Deriving the electromagnetic field strength tensor

  1. Mar 28, 2015 #1
    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.
     
  2. jcsd
  3. Mar 28, 2015 #2
    ## \delta^{i}{}_l\delta^{j}{}_m \partial^lA^m= \partial^iA^j##

    etc.
     
  4. Mar 28, 2015 #3
    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
     
  5. Mar 28, 2015 #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.
     
  6. Mar 28, 2015 #5
    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)
     
  7. Mar 28, 2015 #6

    Orodruin

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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.
     
  8. Mar 28, 2015 #7
    hmm, kind of got that, ok thanks, will work on this some more.

    Thanks for all the replies.
     
  9. Mar 28, 2015 #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
    $$\vec{B}=\vec{\nabla} \times \vec{A}.$$
     
  10. Mar 28, 2015 #9
    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
     
  11. Mar 29, 2015 #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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