Cartesian Coordinates Interpretation in GR?

In summary: In a curved space, say like the Earth, the metric tensor has a lot of extra terms. The extra terms are called "gravitational waves" and they describe the curvature of space. General Relativity is a theory that uses these gravitational waves to explain things like gravity. In a flat space, like the surface of a table, the metric tensor has only a few terms. The few terms are called "gravitational waves" and they describe the curvature of space. But in a curved space, like the Earth, there are a lot of gravitational waves and they describe the curvature of space really well. So General Relativity can explain things like gravity really well
  • #36
stevendaryl said:
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].
Yes, but if you start with SR field theory, you start with [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.

PAllen said:
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 [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.
 
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  • #37
Ilja said:
That's really hard. Test bodies which cover every single point - because one point of space would be sufficient to S3 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.
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.
 
  • #38
And how you want to find out that particular worldlines have not appeared in the actual universe?
 
  • #39
Ilja said:
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.
 
  • #40
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.
 
  • #41
PAllen said:
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.
 
  • #42
PAllen said:
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.
martinbn said:
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.

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.
 

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