Relativistically Invariant Tensors

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SUMMARY

In the context of special relativity, the metric tensor is established as the unique rank 2 tensor invariant under Lorentz transformations, expressed mathematically as \(\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu g^{\mu \nu} = g^{\alpha \beta}\). The discussion identifies two rank 4 tensors with similar invariance properties: the completely antisymmetric tensor \(\epsilon^{\mu \nu \sigma \rho}\) and the completely symmetric combination \(g^{\mu \nu} g^{\sigma \rho} + g^{\mu \sigma} g^{\nu \rho} + g^{\mu \rho} g^{\nu \sigma}\). The conjecture posits that all rank 4 tensors invariant under Lorentz transformations can be expressed as linear combinations of these two. Furthermore, it is concluded that there are no invariant rank 1 tensors and that odd rank tensors likely do not exist, while even rank tensors can be constructed from the metric tensor.

PREREQUISITES
  • Understanding of Lorentz transformations in special relativity
  • Familiarity with tensor algebra and rank definitions
  • Knowledge of metric tensors and their properties
  • Basic concepts of quantum field theory as referenced in Srednicki's QFT book
NEXT STEPS
  • Study the properties of Lorentz transformations in detail
  • Explore the construction of symmetric and antisymmetric tensors
  • Read chapter 34 of Srednicki's QFT book for insights on invariant symbols
  • Investigate the implications of tensor invariance in physical theories
USEFUL FOR

Physicists, particularly those specializing in theoretical physics, students of quantum field theory, and anyone interested in the mathematical foundations of special relativity and tensor analysis.

The_Duck
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In special relativity, the metric tensor is invariant under Lorentz transformations:

\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu g^{\mu \nu} = g^{\alpha \beta}

Is this the unique rank 2 tensor with this property, up to a scaling factor? How would I go about proving that?

I know that two rank 4 tensors with this property are the completely antisymmetric \epsilon^{\mu \nu \sigma \rho} and the completely symmetric (g^{\mu \nu} g^{\sigma \rho} + g^{\mu \sigma} g^{\nu \rho} + g^{\mu \rho} g^{\nu \sigma}). My guess is that any rank 4 tensor invariant under Lorentz transformations is a linear combination of these two. I haven't been able to prove that, though. Any hints?

What about other ranks? I can see that for even ranks I can go on constructing completely symmetric relativistically invariant tensors from the metric tensor as for rank 4. I can also see that there is no invariant rank 1 tensor. My conjecture is that there are none of odd rank, and that except for the completely antisymmetric rank 4 tensor all such tensors are completely symmetric and built from the metric tensor much like the rank 4 one.

I'm asking about this because I just completed a homework problem which would have taken me about 3 fewer pages if I could have just said, "This completely symmetric tensor is invariant under Lorentz tranformation. Therefore it must have this form: ..."
 
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If g and g' are invariant then g^{-1}g' commutes with all \Lambda and therefore g' is proportional to g.
 
arkajad, thanks, I can see now why there is a unique rank 2 invariant. I've found that chapter 34 of Srednicki's QFT book talks about these "invariant symbols" so I will read that.
 

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