The metric tensor for a flat spatial manifold gives us length on object, or separation between two space points. Similarly, $g_{\mu \lambda} dx{^\mu} dx{^\lambda}$ gives separation between two space time points, or two events. But I don't understand the relative (-) sign for time coordinate in metric tensor. Please provide some insights on the topology of metric tensor and length element.
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$\begingroup$ This is way too short for an answer, but the negative sign is because "distance" in spacetime, is not physical distance you can measure with rods, but an abstract "distance" between two events. This abstract distance need not have the usual properties of distance. Why we took that as length element is based on special relativity. If the metric had (++++) signature, then the linear isometry group of spacetime would be $\mathrm{O}(4)$, group of 4d rotations. However from electrodynamics and SR you can see that... $\endgroup$– Bence RacskóCommented Jul 18, 2015 at 7:17
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$\begingroup$ ...the group of linear isometries for spacetime is the Lorentz group, and $\mathbb{R}^4$ with the (-+++) metric has the Lorentz group as its group of linear isometries. $\endgroup$– Bence RacskóCommented Jul 18, 2015 at 7:18
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3$\begingroup$ The minus sign is what characterises Lorentzian manifolds. You really need to go off and read around the subject of Lorenztian manifolds. As it stands your question is too broad to be usefully answered here. $\endgroup$– John RennieCommented Jul 18, 2015 at 7:19
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$\begingroup$ @Uldreth I understand that its some abstract distance, some scalar with a lot of relevance in general relativity.But I feel like missing some interpretation. $\endgroup$– seeking_infinityCommented Jul 18, 2015 at 7:26
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$\begingroup$ It isn't so much a distance as a "proper time". Imagine you have a parallel-mirror light clock. If it's motionless, the light path is straight up & down, and the proper time is in essence the number of reflections. If you move it fast the light path zigzags, and the number of reflections and so the proper time reduces. If you move it at c the number of reflections is zero, and you have a light-like interval s²=0. But the light-path length is not zero, and the events are not simultaneous. $\endgroup$– John DuffieldCommented Jul 18, 2015 at 12:24
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The metric tensor is unrelated to the topology of spacetime, as it does not actually qualify as a metric in the topological sense : it is not positive definite, nor does it apply to spacetime points (at least not directly : you can always find the length between two spacetime points by integrating the tangent vector of the path between them). The topology of spacetime, like any manifold, is determined by the basic metric of $\mathbb{R}^n$. That is why when you speak of the neighbourhood of a spacetime point, you still refer to a little ball around it, and not to the entire neighbourhood of the light cone, even though the distance between a point and the surface of any part of its lightcone is always 0 according to the metric tensor.
That out of the way, the meaning of the signature of the metric tensor.
The - of the metric tensor is here for the causal structure of spacetime. If your spacetime is able to have a (-+++) signature (or (+---) signature, which is almost identical), then you can define a vector field on it that is everywhere timelike ($(x,x) < 0$) and, if no topological shenanigans are involved, future directed ($x_0 > 0$). This roughly corresponds to the direction of time.
Locally, you can see this through the light cone. The light cone divides the spacetime around a point p in several regions. One is just the point itself. Then you have :
- the chronological future, all the points that can be reached via a future-directed timelike curve
- the chronological past, all the points that can be reached via a past-directed timelike curve
- the future light cone, all the points that can be reached via a future-directed lightlike curve
- the past light cone, all the points that can be reached via a past-directed lightlike curve
- And the rest, which is all points reachable via a spacelike curve
Without the metric such as it is, there is no way to separate time in a way to construct any causality. You could make arbitrary turns around in spacetime, go faster than light or go back where you came.
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$\begingroup$ I have been studying topology from last two days only. I thing I want to clarify is, the light cones we consider, do they have meaning in the tangent plane of manifold? Am I making any sense? $\endgroup$ Commented Jul 18, 2015 at 7:48
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$\begingroup$ The tangent plane is where the tangent vectors are located, hence you can't really speak of the light cone in there (although you will have the same kind of structure in it, since it is ~ Minkowski space, they correspond to timelike vectors, lightlike vectors and spacelike vectors). $\endgroup$– SlereahCommented Jul 18, 2015 at 8:23