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No (Lorentz) Invariant tensor of rank 3?

  1. Oct 11, 2009 #1
    Hi everyone,

    (This isn't a homework problem). How does one show that there is no Lorentz invariant tensor of rank 3 and the only Lorentz invariant tensor of rank 4 is the 4D Levi Civita tensor?

    Thanks in advance.
     
  2. jcsd
  3. Oct 11, 2009 #2
    Let [tex]\epsilon^{\mu\nu\rho\sigma}=-1,+1[/tex] when the indices are in "even" and "odd" order, respectively. Consider the transformation [tex]\epsilon^{\mu\nu\rho\sigma}\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\nu}\Lambda^{\gamma}_{\rho}\Lambda^{\delta}_{\sigma}[/tex]. Writing this out a bit you can show this is the same as writing [tex]- or + det|\Lambda| [/tex], which gives -1 or +1 when the order of indices on epsilon is "even" or "odd", respectively. This verifies that the transformed epsilon is the same.

    Consider instead a rank-3 tensor [tex]T^{\mu\nu\rho}\Lambda^{\alpha}_{\mu}\Lambda^{\beta}_{\nu}\Lambda^{\gamma}_{\rho}[/tex]. No matter what the symmetrization is on the tensor T, you can see that the resulting tensor depends on the transformation parameters in the Lambdas, because only in the case of the rank-d anti-symmetric tensor in d-dimensions can you get that special "det" result...and that, in turn, only worked since the transformations were "special" in the sense of having unit determinant. Such tensors are in fact always proportional to the basic "invariant volume element" you use, e.g., in an integral.
     
  4. Oct 16, 2009 #3
    Okay, but how do you argue that the Levi Civita is the only rank 4-invariant tensor in 4D?
     
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