- #26

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

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- #26

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

- #27

PAllen

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The portion of solution you get for one RHow else can one take it?

As above.In what way?

- #28

PAllen

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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 RThen the ether interpretation, based on the same equations, would also mandate that they happen. So it would be falsified too.

- #29

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How would you falsify the GR prediction in this case?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 R^{4}coordinate chart you ended up with, that is 'cut' from a full manifold solution. Two completely different predictions, either of which could be falsified.

You would be, once the ether prediction is not falsified, all the time on the R

- #30

martinbn

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The local solution is not for all of RThe portion of solution you get for one R^{4}coordinate chart is taken to be all of reality. There is no concept of analytically continuing this manifold to a larger manifold.

- #31

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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.Isn't the spin-2 field on flat background supposed to be only the local description?

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But it seems to me that the topology of the universe is a separate assumption. You can do field theory on top of [itex]R \times S^3[/itex] (or whatever) just as well as on top of [itex]R^4[/itex].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.

- #33

PAllen

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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.The local solution is not for all of R^{4}, 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.

IF you insist that the largest R

- #34

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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.You can do field theory on top of ##R \times S^3## (or whatever) just as well as on top of ##R^4##.

- #35

PAllen

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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 SHow would you falsify the GR prediction in this case?

You would be, once the ether prediction is not falsified, all the time on the R^{4}part of the solution.

- #36

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

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 [itex]R \times S^3[/itex] too. Last but not least, a local approximation of an [itex]R \times S^3[/itex] theory would be on [itex]R^4[/itex], and approximations are usually simpler.

That's really hard. Test bodies which cover every single point - because one point of space would be sufficient to SLet'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 S^{3}X R.

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.

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Agree it's hard. There may be a nicer example that is easier.That's really hard. Test bodies which cover every single point - because one point of space would be sufficient to S^{3}X R to [itex]R^4[/itex].

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.

Your approach would predict something different because you want to represent the complete solution in harmonic coordinates on R

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

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

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- #41

martinbn

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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.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 R^{4}coordinate chart you can construct is the whole universe, then you have an alternate theory rather than an interpretation.

- #42

PAllen

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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 R^{4}coordinate chart you can construct is the whole universe, then you have an alternate theory rather than an interpretation.

In the case of the first "if" above, there is no issue with topoplogy. You are handling topology the same say as standard GR that way.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.

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