In what coordinates does a non-inertial observer make measurements?

Question

This question is related to my answer here: Why is proper time equated to $ds$? I use the $(-,+,+,+)$ signature.

For an observer whose proper time is $\tau$, the line element can be written as $$ds^2=-d\tau^2=g_{\mu\nu}(x)dx^{\mu}dx^{\nu},$$ where $g_{\mu\nu}(x)$ are the metric components at the point $(x^0,x^1,x^2,x^3)$. This is true for any coordinate system.

In the coordinates the observer would use to make any measurements (say $x'^\mu=(x'^0,x'^1,x'^2,x'^3)$), in other words, in the observer's coordinates, the time coordinate must be the proper time of the observer and the spatial coordinates of the observer must be constants (because in their coordinates the observer is always at rest and everything else moves around them). Hence, for the observer we have $x'^\mu=(\tau,x,y,z)$, where $x,y,z$ are constants, and the 4-velocity of the observer in these coordinates is simply $V^\mu=(1,0,0,0)$.

Now, if the observer is an inertial observer, everything makes sense: their coordinates would be the inertial/normal coordinates at the point where the observer is located. These coordinates have the property that, at that point, the metric is the Minkowski metric and the Christoffel symbols vanish. It is also trivially to see that, at that point, the geodesic equation, $$\frac{d V^\mu}{d\tau}+\Gamma^\mu_{\nu \rho} V^\nu V^\rho=0,$$ is satisfied, as it should be for an inertial observer.

My question is, what is the form of the metric in the coordinates used by a non-inertial observer (again at the point where the observer is located). Is it still the Minkowski metric as for the case of an inertial observer? The Christoffel symbols can not vanish in this case because then the geodesic equation would again be satisfied which would mean the observer is actually an inertial observer. So these coordinates would not be the inertial/normal coordinates at the point where the observer is located.

Answer 1

Technically speaking the metric is "the same" (it describes flat spacetime) but because the coordinate system is non-inertial it has a radically different form in those coordinates. So for example if the observer is rotating they might use Born coordinates, or an accelerating observer would use Rindler coordinates. Because the coordinate systems are non-inertial the laws of physics will take more complicated forms (e.g. in those coordinates the speed of light may not be constant).

As an example, the line element in Rindler coordinates is $ds^2 = - (\alpha x)^2 dt^2 + dx^2 + dy^2 + dz^2$, where $\alpha$ is the proper acceleration (which is assumed to be along the $x$ axis).