Cartesian Coordinates Interpretation in GR?

• I
Not observing such things in a situation that standard GR mandates they happen.
Then the ether interpretation, based on the same equations, would also mandate that they happen. So it would be falsified too.

PAllen
2019 Award
How else can one take it?
The portion of solution you get for one R4 coordinate chart is taken to be all of reality. There is no concept of analytically continuing this manifold to a larger manifold.
In what way?
As above.

PAllen
2019 Award
Then the ether interpretation, based on the same equations, would also mandate that they happen. So it would be falsified too.
No it would not. It would predict geodesic incompleteness, e.g. timelike world lines that end in some finite proper time. These would be at the boundary of the R4 coordinate chart you ended up with, that is 'cut' from a full manifold solution. Two completely different predictions, either of which could be falsified.

No it would not. It would predict geodesic incompleteness, e.g. timelike world lines that end in some finite proper time. These would be at the boundary of the R4 coordinate chart you ended up with, that is 'cut' from a full manifold solution. Two completely different predictions, either of which could be falsified.
How would you falsify the GR prediction in this case?
You would be, once the ether prediction is not falsified, all the time on the R4 part of the solution.

martinbn
The portion of solution you get for one R4 coordinate chart is taken to be all of reality. There is no concept of analytically continuing this manifold to a larger manifold.
The local solution is not for all of R4, only for a portion of it. Just like in my analogy, locally the circle can be represented by the graph of a function, the function need not be defined on all of R.

Isn't the spin-2 field on flat background supposed to be only the local description?
I don't think so. One starts on a Minkowski space, which is global. With a linear approximation of a spin 2 field. Which is global too. There is no way one can, during the iteration, obtain something different. One can imagine that the iteration fails in some parts of the Minkowski space, but then we have the result of the iteration only on some part of the Minkowski space, not on something greater.

stevendaryl
Staff Emeritus
I don't think so. One starts on a Minkowski space, which is global. With a linear approximation of a spin 2 field. Which is global too. There is no way one can, during the iteration, obtain something different. One can imagine that the iteration fails in some parts of the Minkowski space, but then we have the result of the iteration only on some part of the Minkowski space, not on something greater.
But it seems to me that the topology of the universe is a separate assumption. You can do field theory on top of $R \times S^3$ (or whatever) just as well as on top of $R^4$.

PAllen
2019 Award
The local solution is not for all of R4, only for a portion of it. Just like in my analogy, locally the circle can be represented by the graph of a function, the function need not be defined on all of R.
We seem to be talking in circles. IF you treat it as local solution whose complete solution must be assembled into some larger manifold whose topology is not determined by the local analysis, you have an interpretation of standard GR.

IF you insist that the largest R4 coordinate chart you can construct is the whole universe, then you have an alternate theory rather than an interpretation.

PeterDonis
Mentor
2019 Award
You can do field theory on top of ##R \times S^3## (or whatever) just as well as on top of ##R^4##.
But you can't put a global Minkowski metric on ##R \times S^3##, can you? There would have to be a coordinate singularity somewhere.

PAllen
2019 Award
How would you falsify the GR prediction in this case?
You would be, once the ether prediction is not falsified, all the time on the R4 part of the solution.
Let's say a bunch of test bodies are sent 'around the closed universe'. If one of is unable to return, you've falsified that the topology is really S3 X R.

But it seems to me that the topology of the universe is a separate assumption. You can do field theory on top of $R \times S^3$ (or whatever) just as well as on top of $R^4$.
Yes, but if you start with SR field theory, you start with $R^4$.

You may ask why the spin 2 iteration approach to obtain the Einstein equations has to start with spin 2 on Minkowski background instead of a spherical one. Hm, I don't know. In the case of the ether interpretation, the situation is different, the harmonic coordinates used there are simply simpler, thus, Occam's razor works. Maybe this can be extended to spin 2 theory on $R \times S^3$ too. Last but not least, a local approximation of an $R \times S^3$ theory would be on $R^4$, and approximations are usually simpler.

Let's say a bunch of test bodies are sent 'around the closed universe'. If one of is unable to return, you've falsified that the topology is really S3 X R.
That's really hard. Test bodies which cover every single point - because one point of space would be sufficient to S3 X R to $R^4$.

And, by the way, why would the ether interpretation predict something different? It is the same equation, the same solution, and the part where above solutions agree covers the whole history of the observer. The argument remains intact.

PAllen
2019 Award
That's really hard. Test bodies which cover every single point - because one point of space would be sufficient to S3 X R to $R^4$.

And, by the way, why would the ether interpretation predict something different? It is the same equation, the same solution, and the part where above solutions agree covers the whole history of the observer. The argument remains intact.
Agree it's hard. There may be a nicer example that is easier.

Your approach would predict something different because you want to represent the complete solution in harmonic coordinates on R4 topology. This, requires, as you note, removing one world line from the complete solution (which requires at least two charts).This missing world line makes a whole class of timelike world lines of the full solution impossible (any that intersect the missing one). In principle this is detectable.

And how you want to find out that particular worldlines have not appeared in the actual universe?

PAllen
2019 Award
And how you want to find out that particular worldlines have not appeared in the actual universe?
I already described a way. You criticized it as hard, but that is not a question of principle. Standard GR says I should never have a problem with one of my test bodies returning. The GR solution minuss a world line predicts that it is possible for one not to return. Thus, ever discovering a non-returning test body falsifies standard GR.

Why would a whole test body not return? The atoms left of the missing point will return, the atoms right of it too, the atomes above and below too. The forces between them remain unchanged too, so your test body would not even look damaged.

martinbn
We seem to be talking in circles.

IF you treat it as local solution whose complete solution must be assembled into some larger manifold whose topology is not determined by the local analysis, you have an interpretation of standard GR.

IF you insist that the largest R4 coordinate chart you can construct is the whole universe, then you have an alternate theory rather than an interpretation.
That wasn't my question. The question was: in the case of your first IF, what is the issue with topology? Your reply was that it wasn't a complete interpretation. I asked in what way.

PAllen