# Isotropy of the Minkowski metric

An isotropic tensor has the same components in all rotated coordinate systems.

Scalars or tensors of rank zero are isotropic. Tensors of rank one or vectors are not isotropic. The only isotropic rank-2 tensor is the Kronecker delta.

How could one show or prove that the Minkowski metric $$\eta_{ab}$$ is not isotropic?

Could it be shown that the roatational invariance

$$\eta_{ij}= R_{im} R_{jn} \eta_{mn}= \eta'_{ij}$$

does not hold for the Minkowski metric?

• I don't understand this topic well. If the only rank-2 isotropic tensor is the Kronecker delta or a multiple of the Kronecker delta, how can the Minkowski metric be isotropic? – George Pa1 Nov 18 '18 at 23:07
• I saw this question a couple of days ago in an online chat room about physics. I did some research about this subject, then I asked the question here. I have some knowledge about tensors but regretfully it's limited. – George Pa1 Nov 18 '18 at 23:18
• I see, sorry if my tone came off as hostile. It smelled like homework. – user4552 Nov 18 '18 at 23:20