Curved space-time VS change of coordinates in Minkowski space

Question

I'm looking for a rather intuitive explanation (or some references) of the difference between the metric of a curved space-time and the metric of non-inertial frames.

Consider an inertial reference frame (RF) with coordinates $\bar x^\mu$, in flat spacetime $\eta_{\mu \nu}$ (Minkowski metric).

1. If I have well understood, on one hand, I can go to an accelerated RF by change of coordinates $x^\mu(\bar x)$. The metric is given by: $$\tag{1}g_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}$$

2. On the other hand, I know that a curved space-time with metric $q_{\mu \nu}$ cannot be transformed to Minkowski $\eta_{\mu \nu}$ by coordinate transformation. In other words there does NOT exist any coordinate $x^\mu(\bar x)$ such that (in the whole coordinate patch): $$\tag{2}q_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}\qquad \leftarrow \text{(does not exists in curved space)}$$

So far, everything is more or less ok... But my question is:

1. What is the difference between $q_{\mu \nu}$ and $g_{\mu \nu}$? I mean, in both cases a particle would "feel" some fictitious forces (in which I include the weight force due to the equivalence principle).

2. What physical situation can $q_{\mu \nu}$ describe and $g_{\mu \nu}$ cannot?

I additionally know that by change of coordinates $q_{\mu \nu}$ is locally Minkowski. But still, I can't see clearly the difference.

Answer 1

Gravity is a gauge theory. Gauge transformations are diffeomorphisms (coordinate changes) described by your equations. Therefore, the space of all possible metrics (the moduli space) is the quotient of the space of all $g_{\mu \nu}$ over these coordinate changes.

So your $g_{\mu \nu}$ can be set to $\eta_{\mu \nu}$ by some coordinate transformation. It means that they belong to the same equivalence class.

On the other hand, $q_{\mu \nu}$ belongs to another equivalence class. It can be seen by computing the Riemann curvature tensor. For any $g_{\mu \nu}$ it should be zero, but not for $q_{\mu \nu}$.

Answer 2

OP's question (v2) seems partially caused by imprecise use of the word local:

1. If OP's eq. (1) holds locally in a neighborhood $U\subseteq M$, then there exists coordinates in $U$ such that the metric $g_{\mu\nu}$ becomes on Minkowski-form in $U$, and then the (Levi-Civita) Riemann curvature tensor $R^{\sigma}{}_{\mu\nu\lambda}$ vanishes in $U$, or eqivalently, the manifold $M$ is by definition flat in $U$. The implications also hold in the opposite direction, after possibly going to a smaller neighborhood $V\subseteq U$.

2. For an arbitrary point $p\in M$ on a Lorentzian manifold $(M,g)$, there exist Riemann normal coordinates in a sufficiently small coordinate neighborhood $U\subseteq M$ of the point $p$ such that the metric $g_{\mu\nu}$ becomes on Minkowski-form with vanishing (Levi-Civita) Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ locally in the point $p$ (but not necessarily in the punctured neighborhood $U\backslash\{p\}$ and the manifold $M$ is not necessarily flat in $U$). In particular, the (Levi-Civita) Riemann curvature tensor $R^{\sigma}{}_{\mu\nu\lambda}$ does not necessarily vanish at $p$.