# Differentiation of metric tensor in new coordinate

I want to understand the explicit meaning of

$$g_{\mu'\nu',\lambda}=0$$

where unprimed coordinates are coordinates of the the original coordinate systems and primed ones are for new coordinate system.

However, I think to really answer this question one must think what exactly is $$g_{\mu\nu}$$. My sense is that (since I don't know the exact mathematical definition of it) it is a rank two tensor such that under coordinate transformation the invariant quadratic form is written as $$g_{\mu\nu}dx^\mu dx^\nu$$ (correct me if I'm wrong or not being precise). It is generally coordinate dependent, i.e. a function that produces 4x4 matrices taking 4 coordinates as input. For a rectilinear coordinate system they are constants, for curved spacetime they are coordinate dependent.

My estimate is that since $$g_{\mu'\nu',\lambda}=(g_{\mu \nu}x^\mu_{,\mu'}x^{\nu}_{,\nu'})_{,\lambda}=0$$, and differentials are not zero in general, $$g_{\mu\nu}$$ is constant. However, I must be getting something like there exists a rectilinear coordinate system which is the primed coordinates in this spacetime.

Can I get help for how to take my steps to this conclusion?

• Actually, it's the covariant derivative of metric, namely, $\nabla_{\lambda} g^{\mu\nu}=0$. And one chooses the connection for covariant derivative such that this is true. See physics.stackexchange.com/questions/47919/…. As for transforming tensors between different coordinate systems, it's a separate question. – Cinaed Simson Oct 17 '19 at 20:51