- #1
"Don't panic!"
- 601
- 8
I understand that in order to preserve the inner product of two four vectors under a change of coordinates [itex]x^{\mu}\rightarrow x^{\mu^{'}}=\Lambda^{\mu^{'}}_{\,\, \nu}x^{\nu}[/itex] the Minkowski metric must transform as [itex]\eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\, \nu^{'}}[/itex], but how does this imply that the metric tensor [itex]\eta_{\mu\nu}[/itex] is Lorentz invariant? I mean, doesn't an arbitrary (0,2) tensor transform [itex]B_{\mu\nu}[/itex] as [itex]B_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\, \mu^{'}}\Lambda^{\beta}_{\,\, \nu^{'}}B_{\alpha\beta}[/itex], but tensors aren't generally invariant under Lorentz transformations, so how is it obvious that the metric tensor is Lorentz invariant?
The reason I ask is that I've been reading up on the cosmological constant problem and it is mentioned that the vacuum energy density contribution to the energy momentum tensor must be of the form [itex]T^{vac}_{\mu\nu}=-\rho^{vac}g_{\mu\nu}[/itex], the reasoning being that the vacuum is Lorentz invariant and the only Lorentz invariant tensor is the Minkowski metric?!
The reason I ask is that I've been reading up on the cosmological constant problem and it is mentioned that the vacuum energy density contribution to the energy momentum tensor must be of the form [itex]T^{vac}_{\mu\nu}=-\rho^{vac}g_{\mu\nu}[/itex], the reasoning being that the vacuum is Lorentz invariant and the only Lorentz invariant tensor is the Minkowski metric?!