# 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? 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. 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

The usages of "isotropic" and "rotational invariance" here are not at all standard as applied to a semi-Riemannian space. According to normal usage, the Minkowski metric is isotropic rotationally invariant (as are, e.g., FLRW spacetimes).

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?

Your statement is not true according to standard usage in a semi-Riemannian space. The wolfram.com page that you linked to, which makes this statement, is not about semi-Riemannian spaces, it's about Riemannian spaces.

Spacetime is semi-Riemannian: it has 3 spatial dimensions and 1 time dimension. The term "rotation" is used only for rotations in the 3 spatial dimensions. The Minkowski metric is invariant under those 3-dimensional rotations.

• I guess I should read more about semi-Riemannian spaces. I appreciate the clarifications. Nov 18 '18 at 23:32