# Concept of coordinate reference frame in GR

1. Nov 7, 2013

### EventHorizon91

In SR there is a whole family of so called inertial observers that are defined as those observers that move at relative constant speed with respect to one another, whose descriptions of nature are all equivalent and whose spacetime coordinate are related by Lorentz transformations i.e. those transformations that leave the interval between any two events invariant. A frame of reference is naturally associated with an observer: we can imagine the observer at the origin and a frame of synchronized clocks and rigid rods that fills all of space, comoving with the observer.

In GR spacetime is some general manifold and so it may be impossible to set up coordinates everywhere. Locally, however, we can find coordinates that make the neighbourhood of an event diffeomorphic to R4. Different coordinate charts are possible, as long as they are smoothly related to each other. The physics doesn't depend on the chart, we can choose whatever coordinates we like. Also, these coordinates are not always easy to understand physically. For example, the t coordinate not always corresponds to the time measured by a real clock. The interpretation depends on the specific form the metric takes in the coordinates chosen.

This is the abstract talk. Now, where is the observer in GR? Is the observer associated with a chart? But then how are observers related 'kinematically'? By that I mean: in SR there is a clear physical parameter, the relative velocity, that enters into the coordinates transformation between two observers, but in GR how can we 'translate' the description of one observer in the description of another? For example, a Schwarzchild BH. In the usual Schwarzchild coordinates t is the time measured by an observer at infinity, so these are the coordinates 'natural' for him. Another observer is freely falling towards the BH. Whate are the coordinates natural for him? How can I, distant observer, know what he experiences or measures? Are these two observers'equivalent'? One lives practically in Minkowski space and the other is freely falling, so they should be. Or not? And what happens at the horizon? To sum it up, in SR we can answer this sort of question because we know explicitly how our coordinates are related to the coordinates of another observer and these coordinates are directly related to physical observables, like time intervals or space lenghts, but in GR?

2. Nov 7, 2013

### The_Duck

In GR I think the only sense in which observers, i.e. actual things moving around in spacetime, are naturally associated with coordinates is that if you pick a point on an observer's worldline you can construct a coordinate system for an infinitesimal neighborhood of that point in which the observer is at rest and the metric is the Minkowski metric. In that neighborhood, those are the "natural" coordinates for that observer to use. If two observers' paths cross at some event, then the "natural" coordinate systems that they can construct centered on that event are related to each other--within that infinitesimal neighborhood--by the familiar Lorentz transformations.

But in GR, if two observers are separated, there is no simple relationship between their experiences. As an extreme example, if one observer falls past a horizon then they can experience things that can never be observed by anyone outside the horizon.

If the observer is relying on his eyes, for instance, just calculate trajectories of light rays and figure out which ones hit his eyes and at what time (his proper time). This lets you calculate what he will see as a function of his proper time.

What do you mean, "equivalent"?

3. Nov 7, 2013

### PAllen

There is a slightly broader sense in which you can do this in GR. You can find coordinates covering a world tube encompassing an arbitrary world line such that the metric is Minkowski all along the chosen central world line. If and only if, the world line is inertial, the connection components can also be made to vanish along the central world line; otherwise the connections components (being dependent on derivatives of the metric) cannot be made vanish along the whole world line, even though the metric is Minkowski along it. Such coordinates are Fermi-Normal coordinates.

What you can achieve at any single event on any world line is a Minkowski metric and vanishing connection, with the world line's 4-velocity being 0 in those coordinates at the chosen event. These represent a momentarily co-moving inertial frame.

Last edited: Nov 7, 2013
4. Nov 7, 2013

### EventHorizon91

By equivalent I mean equivalent in the sense of the EP, the two observer cannot say if they are falling in a gravitational field or drifting at constant speed through empty space. Along this line of thought: Neglecting tidal forces, the falling observer crosses the horizon unaware of everything? So, the two observers are not equivalent in the SR sense i.e. two observers always agree about what physical events happen or do not happen (because for the distant observer the other observer never crosses the horizon in a finite time).

Also, if I start thinking about what the two observers actually see in this case I get immediately confused. Imagine a massive star begins collapsing to form a BH. The first observer is located near the surface of the star and is also falling towards the gravitational radius of the star with the same velocity as the external star gases. He is freely falling, so he notices nothing (and continues seeing the star, until eventually he hits the singularity?). For the distant observer using Schwarzchild coordinates an infinite coordinate time is needed to the first observer to cross the horizon and, since his metric is really minkowskian at infinity, also the real time of his clock should be infinite. So he continues seeing the observer falling and the surface of the star collapsing forever, so the BH is not really black? But shouldn't the photons from the falling observer and collapsing star be infinitely redshifted near the horizon, so the distant observer really sees nothing?

5. Nov 7, 2013

### The_Duck

This page has a nice discussion of the experiences of an observer who falls into a black hole and an observer who remains outside it, and also whether black holes are black.

6. Nov 7, 2013

### WannabeNewton

Say I have an observer $O$ with 4-velocity $u$. At any event $p$ on the observer's worldline, I can find a local Lorentz frame $\{e_{\mu}\}$, where $e_0 = u$, and then transport this frame along the entire worldline of the observer via Fermi-Walker transport (orthonormality of the local Lorentz frame is preserved and the frame will be non-rotating according to the Fermi-Walker transport equation). We don't have to pick a non-rotating frame but I'm just choosing the simplest of frame transport equations for the sake of discussion. Physically what this corresponds to is the observer carrying mutually orthogonal meter sticks ($e_1,e_2,e_3$) which are gyrostabilized, and a clock ($e_0$).

If another observer, $O'$, passes by $O$ at some event $p$ coincident on both their worldlines then I can read off the 3-velocity $\vec{v}$ of $O'$ relative to $O$ at $p$ by using the fact that the 4-velocity $\tilde{u}$ of $O'$ is given by $\tilde{u} = \tilde{u^{\mu}}e_{\mu} = \gamma(1,\vec{v})$ just like in special relativity (the same equation holds in GR because we are working in a local Lorentz frame at a single event). Other quantities such as the energy and 3-momentum of $O'$ relative to $O$ also follow suit from the exact same SR equations. Furthermore, if I have the components of a tensor in the local Lorentz frame of $O$ then I can figure out the components of the tensor in the local Lorentz frame of $O'$ by performing a local Lorentz boost at the event $p$ just like in SR, the only difference being the Lorentz boost has to be performed from one observer to another when their worldlines intersect.

Finally, I can use the Fermi-Walker transported local Lorentz frame of $O$ to define a coordinate system that is always comoving with $O$ simply by applying the exponential map. This is a physical coordinate system corresponding to a laboratory that is constructed by $O$ by forming a lattice of meter sticks and clocks in a neighborhood of $O$ using his/her original three meter sticks and clock (the $\{e_{\mu}\}$) and then carrying this lattice along the worldline of $O$. Such coordinates are usually called Fermi-Normal coordinates: http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&amp;page=articlesu18.html [Broken]
For the case of a freely falling observer, it corresponds to a set of locally inertial coordinates along the entire worldline of the observer.

Also we've restricted ourselves to kinematics between two observers but we can generalize this to entire time-like congruences of observers and, using the same local Lorentz frame apparatus, define kinematical quantities such as the expansion, shear, and twist (i.e. vorticity) of the congruence.

Hope that helps!

Last edited by a moderator: May 6, 2017
7. Nov 7, 2013

### pervect

Staff Emeritus
The concept of "an observer" is not an absolute necessity. For instance , Misner writes in "Precis of General Relativity", http://arxiv.org/abs/gr-qc/9508043

My personal choice to replace the concept of an observer in GR (and one which I believe is in relatively wide use) is to use Fermi-Normal coordinates along a given worldline.

Given a general metric, one can find fermi-normal coordinates around any worldline
(see for instance http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&amp;page=articlesu18.html [Broken]). Howver, these coordinates wont, in general, cover all of space-time, just a narrow tube around the worldline. So it's not a perfect replacement for "an observer".

A paper on the "physical meaning" of Fermi Normal coordinates is http://arxiv.org/abs/gr-qc/9402010 which agrees with my personal views is quoted below:

One can also in principle find the transformation between the fermi-normal coordinates between two different "observers" following two different wordlines.

H Nikolic does this in http://arxiv.org/abs/gr-qc/9909035, as part of a larger agenda

I've never used the particular method / equations that Nikolic describes - I can say that even for the simple case of a uniformly accelerating observer, it's a real chore to try and come up with the transformation equations (I don't think I ever got a closed-form expression).

Last edited by a moderator: May 6, 2017
8. Nov 7, 2013

### WannabeNewton

Also see if this helps: https://www.physicsforums.com/showpost.php?p=4559456&postcount=10