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Problem with Einstein notation and Levi-Civita symbol

  1. Sep 6, 2013 #1
    Hello,

    I consider the permutations [itex]\sigma_i[/itex], where [itex]i\in \{1,\ldots,n\}[/itex], of the following kind:
    [itex]\sigma_i[/itex] is obtained by choosing the i-th element from (1,..,n) and by shifting it to the first position; for instance [itex]\sigma_3 = (3,1,2,\ldots,n)[/itex]. The parity of [itex]\sigma_i[/itex] is clearly [itex](-1)^{i-1}[/itex].

    For n=3, I would like to express the following sum based on [itex]\sigma_i[/itex]-permutations in a compact form, using the Levi-Civita [itex]\varepsilon[/itex] and Kronecker [itex]\delta[/itex] symbols:

    [tex]a_{123} - a_{213} + a_{312}[/tex]

    The first index is 1,2,3 so that would be easy to obtain with a repeated index, but I find the remaining two indices difficult to pull out by manipulations with the [itex]\varepsilon[/itex] and [itex]\delta[/itex] symbols.

    Any help?
     
    Last edited: Sep 6, 2013
  2. jcsd
  3. Sep 8, 2013 #2
    Well, first observe that
    εijkaijk = a123-a213+a312-a132+a231-a321.

    You're asking how we could construct a similarly simple formula which kills the last three terms.

    One not-very-pretty way I thought of is that you want to keep all the permutations ijk such that j<k. So you could kill the permutations with j>k by writing

    j+1,kj+2,kijkaijk

    The kroneckers will kill the terms where k<j, giving you what you want.
     
    Last edited: Sep 8, 2013
  4. Sep 8, 2013 #3
    Thanks Jolb,

    killing those three terms is indeed what I am trying to achieve with Einstein notation.
    I see only one little problem with your suggestion: the indices j,k in your formula are repeated three times, which is not in agreement with Einstein notation, am I right?
     
  5. Sep 8, 2013 #4
    Well, my knowledge of Einstein notation comes from physics, and I'd be surprised if any physicist had an issue with the fact that the j's and k's appear three times. It should be pretty clear what that means--since i, j, and k are repeated, then just sum over all possible sets of {i,j,k}.

    Maybe mathematicians have some rigorous way of doing the summation convention that I don't know about, so it's possible that for certain purposes, the expression may be problematic. But I can't think of an example where you would run into a problem using what I wrote... but please let me know what sort of possible problems arise if what I wrote qualifies as an "abuse" of notation.
     
  6. Sep 8, 2013 #5
    I found a couple of old threads in this forum, and more discussions around the net, all saying that it is incorrect to use Einstein notation with more than 2 repeated indices.
    I found the same piece of information from a book I have.

    Though all these sources agreed about the fact that it is "incorrect", unfortunately no one explained *why*.
     
  7. Sep 8, 2013 #6
    Well if you're actually trying to do a calculation rather than some abstract formal math, then you could always just replace what I have written with the actual sum you want a123-a213+a312. If you do some manipulations with the fudgy notation I suggested, you could always go back and plug in the actual sum to check if anything's gone haywire.

    Moreover, you could always just change to normal notation and put the big sigma out front, explicitly stating that you want a sum over {i,j,k}--then you're definitely safe, even for formal math.
     
    Last edited: Sep 8, 2013
  8. Sep 8, 2013 #7

    WannabeNewton

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    Contraction is a mathematical operation; it makes no sense to contract one object or one index with two (or more) other objects or indices simultaneously; such an action would not be well-defined. This is why in Einstein notation one never repeats dummy indices more than twice.
     
  9. Sep 8, 2013 #8
    Thanks WannabeNewton!

    Do you know if it is actually possible at all to express that simple summation using only Levi-Civita's and Kronocker's symbols and in accordance with Einstein notation?
     
  10. Sep 8, 2013 #9

    WannabeNewton

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    I can't really see how you would do it solely using Einstein notation. You would, for example, need proper summation signs to indicate a sum with ##j < k## as jolb said above.
     
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