# Proper time in General Relativity and change of coordinates

Let $$M$$ be the spacetime manifold and let us consider a local coordinate system \begin{align} \varphi_i:\,U_i&\subset M\to \varphi_i(U_i)\subset \mathbb R^n, \end{align} which associates $$p\in U_i\to \varphi_i(p)\equiv x_i^\mu.$$

This is a way of describing the spacetime event $$p$$ with a set of numbers: a different local coordinate system $$\varphi_j$$ defined on $$U_j$$ describes the same event with different numbers $$x'^\mu$$, which must be related to $$x_i^\mu$$ on the intersection $$U_i\cap U_j$$ by $$x'^\mu=\varphi_j\circ\varphi^{-1}_i(x^\mu)$$.

As far as I understand, local coordinate systems are just observers, who see the same event from different points of view and describe them with different numbers.

Given a pseudo Riemannian metric $$g$$, one can express it wrt two different coordinate system as \begin{align}g&=g_{\mu\nu}(x)dx^\mu dx^\nu\\ &=g'_{\mu\nu}(x')dx'^\mu dx'^\nu,\end{align} from which is it possible to derive the transformation rule of the coefficients $$g_{\mu\nu}.$$

It is often said that the proper time of a moving particle is given by an OBSERVER that sits on the particle, which sees the particle itself at rest. Thus $$d\tau^2=g=g_{\mu\nu}(x)dx^\mu dx^\nu.$$ Is proper time a coordinate system, since it is an observer?

1. If so, LHS should be related to the RHS through some change of coordinates which however seems quite singular, since it does not look invertible. Am I wrong?
2. Furthermore, I struggle making sense of an observer (ie a coordinate systems) that changes point by point as we follow the trajectory of the particle (because point by point the spatial coordinates describing the particle are zero), in the context of differential geometry.
In other words, is the reference frame sitting on the moving particle a single coordinate system?

I would like to understand the above points from a mathematical point of view.

The relation between observers and coordinates isn't quite so simple, unfortunately. This is true in special relativity for inertial observers (to some degree), but in a more general case, the relationship between the two is more subtle.

An observer is, generally speaking, a future-oriented timelike curve $$\gamma$$. Sometimes we also give them additional attributes to reflect how an actual physical apparatus might work : an onboard clock $$h$$ (this is because the time an observer measures may not necessarily be the proper time), which is some monotone increasing function from points of the curve to $$\mathbb{R}$$ :

$$h : \mathrm{Im}(\gamma) \to \mathbb{R}$$

This is used mostly if we're thinking about actual experiments, usually $$h$$ will simply reflect the proper time. An observer may also have a local frame $$e_a$$, which is three linearly independant spacelike directions, so that we can do measurements of directions such as incident angles and such.

Now as an observer is a simple curve, it cannot be equivalent to a coordinate system, because an observer cannot really measure anything beyond its immediate surroundings. A simple reason why not is the following : consider a coordinate system adapted to an observer, and then perform a diffeomorphism on it that is the identity around the observer but not outside of its region. From the point of view of the observer, there will be no difference.

So in what way can we associate an observer to a coordinate system? The simplest way in which this is usually done is that your observer can be a line of constant spatial coordinates, and its coordinate in the timelike direction can be equivalent to the proper time (or its onboard clock, if you so choose). In addition, if our local frame isn't too crazy, we can also make it so that the local frame is oriented along the same direction as the coordinate basis. If the observer is a geodesic, those are the Fermi coordinates. There are more general processes you can use for more arbitrary observers, but those are the most common ones.

Fermi coordinates and other coordinates of their types (such as radar coordinates) are always local in nature, and somewhat arbitrary. They will usually somewhat agree near the observer, but may start diverging the further away you go, and there is usually a point beyond which they stop being valid altogether (such as in the presence of a cut locus).

It is possible to have a better method to construct coordinates "physically" if you have an observer going through every point of space, but that is a more complex process.

On the flipside, can you get an observer the other way around, starting with some coordinates? The answer at first is going to be obviously no. A classic example is the null coordinates

$$ds^2 = -dtdx$$

The constant lines of those coordinates are all null curves, and therefore not observers, even though this is just Minkowski space. If one of your coordinate is a timelike curve, though, then yes, you can simply trace a line along that coordinate (at spatial coordinates zero, for extra fun), put the tetrads as the local frame, and this will indeed be the observer of that coordinate system at that point. But beware that this curve is neither guaranteed to be a geodesic, nor even of proper time. It's a common trick in general relativity that if your metric is stationary, you can always just perform a rather simple coordinate transformation

$$t' = f(t)$$

which will not change anything but the timelike component of the metric, in which case, if your curve was originally going along as proper time, it will no longer do so.