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The topology of spacetimes |
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| Jan3-13, 07:23 PM | #69 |
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The topology of spacetimes |
| Jan3-13, 07:32 PM | #70 |
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| Jan3-13, 07:39 PM | #71 |
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| Jan3-13, 09:37 PM | #72 |
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I mean, we do have an invariant [itex]ds^2[/itex], but this may be positive, negative or null. Then, what is done in physics is that we define our distances simply by [tex] \int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0 [/tex] and this only makes sense if b is inside the lightcone defined by a, so that (by using the proper convention sign [itex]\pm[/itex]) we always get a distance properly defined. This may be a simple trick by using the lightcone, but it can be supported by causality arguments that you want to include in our mode, I think. Does this seems good to you? I have some problems, as I'm thinking of the lightcone as being on the manifold, and not in the tangent space as has been argued. |
| Jan3-13, 10:01 PM | #73 |
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| Jan3-13, 10:05 PM | #74 |
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For example, take the sphere in [itex]\mathbb{R}^3[/itex]. We can endow this with a metric as follows. Take two points on the sphere, draw a straight line through those points and measure the length of the line. So we take the distance on [itex]\mathbb{R}^3[/itex] and restrict it to the sphere. This defines a good distance on the sphere that agrees with the topology. However, this distance is not a very useful one as it relies on the embedding in [itex]\mathbb{R}^3[/itex]. What we want is a distance on the sphere that measures the length of the paths on the sphere. So we don't want a distance that comes from straight lines (which are not on the sphere), but rather a distance that comes from path (=great circles) on the sphere. This distance is a distance coming from a metric tensor and this is much more useful. In the same way, we can endow a distance on a spacetime. But nothing tells us that this distance actually has a physical significance or that it agrees with some metric tensor. |
| Jan3-13, 10:13 PM | #75 |
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It seems to me that there is a meaningful notion of "lightcone" on the manifold as well. (Not sure what the standard terminology is though). This would be the union of all the timelike and null geodesics through the given point. |
| Jan3-13, 10:18 PM | #76 |
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It can be proven (but the proof is not easy by far), that if we have these three topological conditions, then our space is metrizable. So we can always find a metric. Moreover, we can always embed our manifold in [itex]\mathbb{R}^n[/itex]. So even without a smooth structure or a metric tensor, we already have that our manifold is metrizable. Again: the metric of the metric space might not be physical or might not have anything to do with a metric tensor!! If we drop one of the conditions from our list, then the space is not metrizable anymore. For example, if we would define a manifold as just locally euclidean and Hausdorff, then it might not be metrizable (as the long line shows). If we define a manifold as just locally euclidean and second countable, then it might also not be metrizable (as the line with two origins shows). In fact: it might not even be pseudo-metrizable. |
| Jan3-13, 10:31 PM | #77 |
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| Jan3-13, 10:46 PM | #79 |
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| Jan4-13, 12:00 AM | #80 |
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| Jan4-13, 12:05 AM | #81 |
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| Jan4-13, 12:12 AM | #82 |
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| Jan4-13, 04:11 AM | #83 |
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As you notice, in physics the length of the curve is always computed as if the tangent vector at every point where inside the light cone(in the limit at infinity) , regardless of what it is called, i.e. photon's null paths are never considered to have null length. There are no physical examples of spacelike paths so we can leave those out for now. This is the logic thing to do since after all we are working with a smooth manifold that doesn't alter its topology nor its distance function(in the Riemannian manifold case) by the introduction of a pseudoRiemannian metric tensor. The only problem I see is that this seems to be forgotten when applying GR to specific solutions of the EFE. |
| Jan4-13, 04:26 AM | #84 |
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| Jan4-13, 06:11 AM | #85 |
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