- #1
The_Duck
- 1,006
- 108
In special relativity, the metric tensor is invariant under Lorentz transformations:
[tex]\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu g^{\mu \nu} = g^{\alpha \beta}[/tex]
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 [tex]\epsilon^{\mu \nu \sigma \rho}[/tex] and the completely symmetric [tex](g^{\mu \nu} g^{\sigma \rho} + g^{\mu \sigma} g^{\nu \rho} + g^{\mu \rho} g^{\nu \sigma})[/tex]. 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: ..."
[tex]\Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu g^{\mu \nu} = g^{\alpha \beta}[/tex]
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 [tex]\epsilon^{\mu \nu \sigma \rho}[/tex] and the completely symmetric [tex](g^{\mu \nu} g^{\sigma \rho} + g^{\mu \sigma} g^{\nu \rho} + g^{\mu \rho} g^{\nu \sigma})[/tex]. 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: ..."