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Homework Help: 4 dimensional Levi Civita problem

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    I have the following relation:


    And I have to express that via delta functions.

    2. Relevant equations

    I tried looking at the wikipedia but I cannot make sense of that :\
    Plus I have once covariant and once contravariant tensor, and I'm summing over 2 indices.

    3. The attempt at a solution
  2. jcsd
  3. Nov 22, 2011 #2


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  4. Nov 22, 2011 #3


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    I think I would start with a few observations like these:

    If [itex]\alpha,\beta,\rho,\tau[/itex] have four different values, both factors in each term are =0.

    If [itex]\alpha,\beta,\rho,\tau[/itex] have exactly three different values, at least one factor in each term is =0.

    What if [itex]\alpha,\beta,\rho,\tau[/itex] are all the same?
    Last edited: Nov 22, 2011
  5. Nov 23, 2011 #4

    The problem is, and I found the answer in Griffiths, introduction to elementary particles, that it says that the result is difference of 2 kronecker deltas, and in wiki rule there seems to be only product, unless they also use Einstein summation convection.

    This two Levi Civita symbols contract in [itex]\mu[/itex] and [itex]\nu[/itex], right, which will leave me with one Levi Civita?

    I'm totally at loss here. It says that [itex]\varepsilon^{\mu\nu\lambda\tau}[/itex] is zero if any two indices are the same, and + or -1 depending on the permutation. So if they are all the same, shouldn't the answer be zero?

    I still fail to see how can that help me with that number 8 rule on wikipedia :\

    I don't understand what they are trying to say...
  6. Nov 23, 2011 #5


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    Look at the Wikipedia identity again. The brackets on the lower indices denotes antisymmetrization. For example,
    They are using it. When you see a Levi-Civita symbol, you should always assume that the summation convention is used, unless they explicitly say that it isn't.

    Yes, if [itex]\alpha,\beta,\rho,\tau[/itex] are all the same, the result is 0. If they take three or four different values, the result is 0. These three observations tell us that the problem is trivial unless [itex]\alpha,\beta,\rho,\tau[/itex] have exactly two different values. The idea is to keep making observations that tell us more and more about the result.

    Do you at least see why the result is 0 when [itex]\alpha,\beta,\rho,\tau[/itex] have 1,3 or 4 different values?
    Last edited: Nov 23, 2011
  7. Nov 23, 2011 #6
    Oh, I wasn't aware of the antisymmetrization! I am learning general theory of relativity where we use only tensor notation, but our professor kinda sucks :\ And we never got to symmetrization, or antisymmetrization :(

    I'm not quite sure about that last thing you asked :\

    You mean only one Levi-Civita? Or two summed?
  8. Nov 23, 2011 #7


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    If you see any kind of expression that involves at least one Levi-Civita symbol, you can assume that the summation convention is used, unless they're telling you that it's not.

    I forgot a factor when I was talking about antisymmetrization. I think the standard definition includes a factor of 1/n! on the right, like this: [tex]T_{[i_1\cdots i_n]}=\frac{1}{n!}\left(T_{i_1\cdots i_n}+\cdots\right).[/tex] So in particular, [tex]\delta^i_{[j}\delta^n_{m]}=\frac{1}{2}\left(\delta^i_n\delta^j_m-\delta^i_m\delta^n_j\right).[/tex]

    It's not at all hard to prove that [itex]\varepsilon^{\mu\nu\alpha\beta}\varepsilon_{\mu\nu \rho\tau}=0[/itex] when [itex]\alpha,\beta,\rho,\tau[/itex] has 1,3 or 4 unique values, so you should give that another try. If you think it looks like it can't be true, then post your argument so I can tell you what you're doing wrong.
  9. Nov 23, 2011 #8
    I will try it, I have test in nuclear physics on Friday, so this will have to wait a bit :D But thanks for the help :)
  10. Nov 23, 2011 #9


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    OK then. I expect to see your solution here on Friday night then. :wink:
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