# Is the Minkowski metric coordinate independent?

Suppose I have some vector $$\mathbf{P} = p^{\mu} e_{\mu}$$.

Now, for a flat spacetime, the contravariant components can be lowered via the Minkowski metric,

$$p_{\mu} = \eta_{\mu \nu} p^{\nu}.$$

My question is, does $$\eta_{\mu \nu} = (-1,1,1,1)$$ always, for all coordinates? My immediate response would be no, only for Cartesian coordinates, as in the Wikipedia example. However, the motivation for this question is from my inability to understand the step in going from eqn B5 to B6 in Kulkarni et al. 2011.

In this paper the author appears to be working in spherical polar coordinates, but makes a transform using the Cartesian metric. Surely instead the metric in spherical coordinates should be used?

• Such "Am I right?" questions are difficult to answer. What except "Yes, you're right" do you expect an answerer to write here in the case where you are right? – ACuriousMind Jul 26 '19 at 13:48
• Well, really I expect to be wrong; that there is somewhere a hole in my understanding and that I am not understanding the linked paper correctly. I then expect an answer to point out this hole? – user1887919 Jul 26 '19 at 13:51
• So you expect answerers to read the linked paper? It would help if you cited the relevant section directly, then. – ACuriousMind Jul 26 '19 at 13:53
• In the paper you refer to, they use a formulation of general relativity called the Tetrad Formalism. Here the metric is always $(-1,1,1,1)$ and information about the curvature of the spacetime is encoded in the basis vectors at each point of the tangent space (which by construction are always orthonormal to one another). – Greg.Paul Jul 26 '19 at 13:54
• @Greg.Paul I lilke your comment but since precision of ideas is crucial here, I think you should not say "at each point of the tangent space" but rather "at each point of the manifold" (because there is not one tangent space but many) – Andrew Steane Jul 26 '19 at 14:19