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Problem with tensors

  1. Nov 6, 2011 #1
    I'm having alot of problems with tensors. Here is what the professor in class told us in the lecture notes

    In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor
    F[itex]^{\mu\nu}[/itex] = -F[itex]^{\nu\mu}[/itex] is equivalent to an axial Lorentz vector, F[itex]^{\mu\nu}[/itex] = e[itex]^{\mu\nu\lambda}[/itex]F[itex]_{\lambda}[/itex]. Consequently, in 3D
    one can have a massive photon despite unbroken gauge invariance of the electromagnetic
    field A[itex]_{\mu}[/itex]. Indeed, consider the following Lagrangian:

    L = -(1/2)*F[itex]_{\lambda}[/itex]F[itex]^{\lambda}[/itex] + (m/2)*F[itex]_{\lambda}[/itex]A[itex]^{\lambda}[/itex] (6)


    F[itex]_{\lambda}[/itex](x) = (1/2)*[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]F[itex]^{\mu\nu}[/itex] = [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex],

    or in components, F[itex]_{0}[/itex] = -B, F1 = +E[itex]^{2}[/itex], F[itex]_{2}[/itex] = -E[itex]^{1}[/itex].

    In 2+1 dimension, [itex]\epsilon[/itex][itex]^{\alpha\beta\gamma}[/itex][itex]\epsilon[/itex][itex]_{\alpha}[/itex][itex]^{\mu\nu}[/itex] = g[itex]^{\alpha\mu}[/itex]g[itex]^{\beta\nu}[/itex] - g[itex]^{\alpha\nu}[/itex]g[itex]^{\beta\mu}[/itex]

    That last part above may be a typo because I've never seen an epsilon without all of its indices either upstairs or downstairs

    I'm having trouble with two things
    1. Using the last part above, does that mean that
    [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex] = g[itex]^{\lambda\mu}[/itex]g[itex]^{\mu\nu}[/itex] - g[itex]^{\lambda\nu}[/itex]g[itex]^{\mu\mu}[/itex] = g[itex]^{\lambda\mu}[/itex]g[itex]^{\mu\nu}[/itex] - g[itex]^{\lambda\nu}[/itex]?

    If so, would that give [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\nu}[/itex] = [itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex][itex]\partial[/itex][itex]^{\nu}[/itex]A[itex]^{\mu}[/itex] - [itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]^{\lambda}[/itex] ?

    2. But when I tried to write out the Lagrangian, I got

    L = -(1/2)[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\nu}[/itex] + (m/2)[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex]A[itex]^{\lambda}[/itex]

    [itex]\frac{\partial L}{\partial A^{\lambda}}[/itex] = (m/2)[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex] [itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex]

    [itex]\frac{\partial L}{\partial (\partial_{\mu}A_{\nu}) }[/itex] = -[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex] + (m/2)[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]g[itex]^{\mu\nu}[/itex]A[itex]^{\lambda}[/itex]
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2
    I think I can help with [itex]\epsilon_{\alpha\beta\gamma} \epsilon^{\alpha\beta\gamma}[/itex]. The epsilon tensor can only have three values (-1, 0, 1), and we are contracting two of them. So we are going to have pairs of -1s and 1s and 0s begin multiplied and added together. That is
    \epsilon_{\alpha\beta\gamma} \epsilon^{\alpha\beta\gamma}= \epsilon_{000}\epsilon^{000} +...+ \epsilon_{123}\epsilon^{123} +...+ \epsilon_{213}\epsilon^{213}+...[/tex]
    So considering these will be something like (-1)(-1)+(0)(0)+...+(-1)(-1)+... etc... what will the sum be?

    Unless you made a typo... in which case you are refering to the identity that two epsilon tensors contracting an index turn into two permutations of the remaining indecies for the metric, which is like what you wrote. These identities are on the wiki page.

  4. Nov 7, 2011 #3
    is the epsilon that does not have ALL of its indices either upstairs or downstairs a typo? If not, then the usual product of epsilons is of no use since they will now follow the relation the prof gave us in 2+1 dimensions
  5. Nov 7, 2011 #4
    The prof just told me the indices not being ALL up or downstairs is not a typo. I still would like to know if I'm calculating the Euler lagrange equations correctly or not.
  6. Nov 7, 2011 #5


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    There's a basic property of tensor notation that you're overlooking - a term can only have two occurrences of the same index in it, one upstairs and the other downstairs. You need to correct this before anything else can be said about your derivation.
  7. Nov 7, 2011 #6
    so you're saying that to fix the error, I should set F[itex]_{\lambda}[/itex]F[itex]^{\lambda}[/itex] = (1/4)[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]F[itex]^{\mu\nu}[/itex][itex]\epsilon[/itex][itex]^{\lambda\alpha\beta}[/itex]F[itex]_{\alpha\beta}[/itex]?

    If so, that still would not get rid of the m term in the equation I got for [itex]\frac{\partial L}{\partial (\partial_{\mu}A_{\nu})}[/itex], which I think is where the problem is
  8. Nov 7, 2011 #7
    If I'm not mistaken that looks good, and you can write [itex]\epsilon_{\alpha\beta\gamma}\epsilon^{\alpha\delta\lambda}[/itex] as an identity in Kronecker deltas by permuting the indices.
  9. Nov 7, 2011 #8
    I forgot to say that I recalculated

    [itex]\frac{\partial L}{\partial (\partial_{\mu}A_{\nu})}[/itex] = -([itex]\partial^{\mu}[/itex]A[itex]^{\nu}[/itex] - [itex]\partial^{\nu}[/itex]A[itex]^{\mu}[/itex])) + (m/2)[itex]\epsilon[/itex][itex]_{\lambda}[/itex][itex]^{\mu\nu}[/itex]A[itex]^{\lambda}[/itex]
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