# I Cartesian Coordinates Interpretation in GR?

1. Jan 9, 2016

### clinden

What is the physical interpretation of Cartesian coordinates in GR? Say, e.g., a system centered at the center of a spherical mass. What are x,y, and z physically, i.e., how are they measured?

2. Jan 10, 2016

### Staff: Mentor

Coordinates in GR do not necessarily have any physical meaning. They are just numbers used to label events in spacetime. For coordinates to have any physical meaning, you have to pick them so that they match up with some kind of symmetry of the spacetime. For example, in the case of a spherically symmetric mass, you would use spherical coordinates. But even then there is more than one way of setting up the spherical coordinates as far as the physical meaning of the coordinates goes.

It might help if you gave some more details about what you are trying to figure out.

3. Jan 10, 2016

### HallsofIvy

Perhaps the reason you are asking is that "Cartesian coordinates" in particular can be used only in "flat spaces". And General Relativity is a "geometric" theory using the curvature of space to explain gravity. Space that have curvature are not flat and there do NOT EXIST global Cartesian coordinates.

4. Jan 10, 2016

### clinden

"It might help if you gave some more details about what you are trying to figure out."

I was trying to understand how to physically measure the coordinates if the origin is set at the center of the spherical mass.
I understand if spherical coordinates are set at the origin, the 2 angles are straightforward, bur "r" needs to be measured indirectly as
the circumference of a circle through the point of interest divided by 2 Pi. I was looking for a direct measurement approach, rather than converting in the standard way from the spherical coordinates.

5. Jan 10, 2016

### stevendaryl

Staff Emeritus
Within a small region, say the inside of a small room, you can set up approximate Cartesian coordinates the same way you would in Newtonian spacetime. You pick a point to call the origin. At the origin, you pick three perpendicular directions that you call the "x-axis" the "y-axis" and the "z-axis". Then to establish the coordinates of any other point in that region, you find a way to get from the origin to that point by first moving parallel to the x-axis, then moving parallel to the y-axis, and finally moving parallel to the z-axis. Then you measure the distances traveled along each segment. Or alternatively, you could put a mirror at the point, and you could measure the time for a light signal from a laser to travel from the origin to the mirror and back, and figure out the coordinates from the round-trip time and the orientation of the laser.

What makes this possible is that within a small enough region, you can say approximately that two objects are at rest relative to one another--it just means that the round-trip time for light is approximately constant. You can say that two paths are approximately parallel--it means that corresponding points are a constant distance apart. Etc. Euclidean geometry works approximately in curved spacetime, as long as you restrict consideration to a small enough region.

6. Jan 10, 2016

### Staff: Mentor

I don't think there is one, other than in a small enough region, as stevendaryl describes. There is no guarantee that coordinates can be found in every spacetime such that all of them can be directly measured. Note that in the case of a spherically symmetric mass, the time coordinate $t$ can't be directly measured either, except at infinity; at any finite radius, you have to correct for gravitational time dilation.

7. Mar 12, 2016

### Ilja

This depends on the interpretation. One can have an ether interpretation of the GR equations which contains a flat background and interprets the curved metric as distorted by the ether (the gravitational field). This "distorted" metric may be a solution of the Einstein equations but exist, in this interpretation, on some flat (undistorted) background described by some preferred coordinates.

8. Mar 12, 2016

### stevendaryl

Staff Emeritus
It's sort of interesting: If you think of the metric tensor as just a field on spacetime, then the "shape" of spacetime is almost independent of the metric. But not quite, because certain metrics are only consistent with certain underlying topologies.

9. Mar 12, 2016

### Ilja

No, every metric is consistent with $\mathbb{R}^4$. To cut an arbitrary 4D manifold in such a way that only $\mathbb{R}^4$ remains is trivial, what has to be thrown away is codimension 1. You need something else, like a completeness condition or so, to make the jump from a simple tensor field with (1,3) signature on $\mathbb{R}^4$ to a nontrivial topology.

10. Mar 14, 2016

### pervect

Staff Emeritus
Cartesian coordinates don't exist for the case you describe. The closest thing I've seen to them uses isotropic coordinates. See for instance https://en.wikipedia.org/wiki/Schwarzschild_metric#Alternative coordinates for the metric.

$$\frac{(1-\frac{r_s}{4R})^{2}}{(1+\frac{r_s}{4R})^{2}}{d t}^2 - \left(1+\frac{r_s}{4R}\right)^{4}(dx^2+dy^2+dz^2) \quad R = \sqrt{ x^2 + y^2 + z^2 }$$

A non-technical summary of why isotropic coordinates are preferred over regular Schwarzschild coordinates - it's confusing, and not very physically meaningful, if the speed of light is different in the x,y, and z directions. So you start with isotropic $r, \theta, \phi$ spherical coordinates which has the desired property that the speed of light is the same in all directions, then use the usual trig relationships to convert the spherical coordinates to x,y,z, and you wind up with the above metric.

I should stress again that while the isotropic coordinates are in some sense similar to Cartesian coordinates, they are not Cartesian. True Cartesian coordinates would have a metric that is diagonal(-1,1,1,1). The isotropic coordinates have the metric given in the reference, which you can see is fairly close to being diagonal(-1,1,1,1) when R >> r_s, but this is only an approximation, it's not exact.

Last edited: Mar 14, 2016
11. Mar 15, 2016

### PAllen

Isn't this more than just an interpretation? Certainly, for 'almost all' observables it is the same, as all measurements are affected by 'metric field', such that they match GR. What I am not sure of is how inspiralling BH would be modeled without non-trivial topology. Unless they can, then LIGO can be argued to be an observation which distinguishes the interpretations (and falsifies the flat background). More generally, there are clear differences in prediction, it is just the most of them don't apply to any accessible part of our universe. This is different from the LET interpretation of SR which makes identical predictions in all cases, thus is truly an interpretation. What you describe is a different theory than modern classical GR. That its different predictions are hard to access does not change its nature as a different theory not just an interpretation.

[edit: a concrete example is that to represent a closed FLRW universe, you have to remove a point in the flat bacground theory as compared to the modern GR. This means that the flat background theory predicts there is some possible test body which encounters discontinuity in its history, while the standard theory says there is no such thing.]

Last edited: Mar 15, 2016
12. Mar 15, 2016

### Ilja

I do not have sufficient computational power to model inspiralling BHs. But if this is done, say, in the straightforward way using harmonic coordinates (why not, this is the most common coordinate condition and essentially simplifies the equations) then this would be exactly the same mathematics as in the ether interpretation.

Given that this interpretation excludes nontrivial topologies, and, moreover, also closed causal loops, it can be considered, formally, as a different theory. Even if the equations are identical. In this case, given that it makes additional falsifiable prediction in comparison with the spacetime interpretation, it would be preferable from the point of view of Popper's criterion of empirical power.

13. Mar 15, 2016

### Staff: Mentor

If it turns out that certain current speculations about the effects of quantum corrections are right, and there are no actual event horizons (only apparent horizons), then one actually could model the phenomena we call "black holes" using $R^4$ topology, at least in an asymptotically flat spacetime. (One still couldn't do it in asymptotically de Sitter spacetime, which is the current best candidate for our actual universe, since the topology of de Sitter spacetime is $S^3 \times R$, not $R^4$.)

14. Mar 15, 2016

### PAllen

I don't see this at all. For every prediction difference, you have a falsifiable prediction for both theories. Thus, I see no difference by this criterion. The only difference is which predictions are correct.

15. Mar 15, 2016

### PAllen

Right, but nobody knows how to do this yet, or whether the predictions would be identical in a complete QG theory. It cannot be done in classical GR because the topology inside a horizon is not R4. For example, inside an SC BH, it is S2 X R2.

16. Mar 15, 2016

### Staff: Mentor

This is only true if the horizon is an actual event horizon; it's not true if there's an apparent horizon but no event horizon. In the latter case the topology even inside the apparent horizon can be $R^4$, as long as the spacetime is asymptotically flat, as I said.

17. Mar 15, 2016

### PAllen

Again, that requires quantum corrections, and there is no well founded theory (yet) in which to do such a computation. In classical GR, there are horizons.

18. Mar 16, 2016

### Ilja

No. You have solutions of GR, those with nontrivial topology or closed causal loops, which are valid GR solutions, but do not allow an ether interpretation. Observing that our universe would be described by such a solution would falsify the ether interpretation, but not the spacetime interpretation. Not observing such a solution would leave above interpretations viable. What would falsify the spactime interpretation but leave the ether interpretation valid?

In some sense, this would be done by observing some fundamental violations of the equations, say, by observing the atomic structure of the ether. Then "ether theory" in general would survive, but spacetime would be dead. But this would be, nonetheless, a different theory, not the ether interpretation of the Einstein equations, which would survive only as an approximation.

19. Mar 16, 2016

### Ilja

Why this? AFAIK, the observation favors a globally flat universe. (Of course, this would always allow some measurement uncertainty for the curvature, thus de Sitter would never completely excluded. And there will be always people who for philosophical reasons would prefer a finite universe. But beyond this?)

20. Mar 16, 2016

### stevendaryl

Staff Emeritus
What I assume is the case is that if you tried to model the collapse of a star into a black hole where you treat gravity as a field theory on top of flat $R^4$, then you will be able to reproduce the predictions outside the event horizon, but not inside it. Since nothing inside the event horizon can affect observers outside the event horizon, that would mean (I would think) that the only way to observe a difference in the two theories would be to take a plunge into the black hole.

This brings up an issue with General Relativity that I'm not sure I understand. Mainstream physicists act as if the interior of a black hole is as real as the exterior. In terms of general coordinates, there is nothing special about the event horizon, locally, so there is no physical reason for the manifold to end at the horizon. But is that just an aesthetic choice?

21. Mar 16, 2016

### martinbn

I am not sure I understand the issues with the topology. Isn't the spin-2 field on flat background supposed to be only the local description? Here is an analogy that I hope will clarify my question. Consider curves (smooth) in the plane. Not all of them are graphs of functions, for example a circle is not the graph of a $y=f(x)$. But locally, by the implicit function theorem, all curves can be viewed as graphs of functions, including the circle. The circle's topology is different than that of a line, but there is no problem with the statement above. So what exactly is the problem with space-times with topology different than that of $\mathbb R^4$? Each event will have a neighborhood, where gravity can be described as a field on a flat background?

22. Mar 16, 2016

### PAllen

Not observing such things in a situation that standard GR mandates they happen.

23. Mar 16, 2016

### PAllen

It leads to accepting violation of the principle of equivalence for free fall observers approaching a supermassive BH.

24. Mar 16, 2016

### PAllen

That's true if you accept that the spin 2 theory is local, and is used in conjunction with a global manifold interpretation to arrive at global predictions. Then, it isn't a complete interpretation.

25. Mar 16, 2016

### martinbn

How else can one take it?

In what way?