Any coordinate system in GR?
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From: I.Vecchi on 7 Sep 2006 02:13 Daryl McCullough wrote: .... > You have to split spacetime up into a collection of overlapping > "patches" such that each patch can be described using a coordinate > system for which the metric components are nonsingular. The patches > have to describe an *open* subset of points (this insures that > overlapping patches overlap in a 4-D *region*). If two patches > overlap, then there must be a 2-way mapping function > between the two coordinate systems that is valid (differentiable, > nonsingular) in the region of overlap. > > So following these guidelines, we can pick one patch to cover > the region infinity > r > 2m, > infinity > t > -infinity. > > The Schwarzchild coordinates are perfectly adequate for describing > this region. But they don't cover the points on r=2m. So you have > to come up with a second patch that covers this region. But since > patches have to be *open* sets, you need a patch covering a little > bit on both sides of r=2m. > > One possibility is to use a Kruskal patch. You have coordinates > R, T with the metric > > 32m^3/u exp(-u/2m) (-dT^2 + dR^2) > > where u is an implicit function of R and T given by > > T^2 - R^2 = (1-u/2m) exp(u/2m) > > Using the mapping > > r_schwarzchild = u(R,T) > t_schwarzchild = 4m arctan(T/R) > > we find that the metrics agree in the region of their overlap, > and that the Kruskal patch covers the region r=2m, as well. > > You claim that choosing a Kruskal patch is arbitrary. Then > show me a different choice: Show me another patch that extends > Schwarzchild spacetime past r=2m which (1) agrees with the > Schwarzchild patch in the region r > 2m, and (2) *disagrees* > with the Kruskal patch in the region r < 2m. I don't think > that there is such a patch. So the Kruskal extension is *unique*, > not arbitrary. You did not use the KS chart to show that an infalling observer reaches the singularity at r=0 in finite proper time. Your argument, as well as any relevant argument I have spotted so far, relies on the interior Schwarzschild chart. I have a problem with that. Both the interior and the exterior Schwarzschild chart correspond to stationary observers (i.e. they map measurement outcomes of space-time events performed by observers that are stationary). For the exterior chart this poses no problem since stationary observers are well defined. However, in the interior domain, according to the current theory, there are no stationary observers. This implies that the interior Scwharzschild chart and the interior soultion defined on it are meaningless, since they refer to measurements that cannot be performed . The t and r coordinate and the metric for r<2m do not correspond to any physically measurable/meaningful quantity. They are just blots on a sheet of paper. I see two mutually exclusive possibilities here. a) It is true that there are no stationary observers in the interior domain. In this case the interior Schwarzschild solution is unphysical and hence meaningless. It should be simply discarded. b) There are stationary obsevers in the interior domain. The arguments purporting to show that there are no stationary observers in the domain of the interior chart are flawed ( e.g. they relie on conservation of quantities that become unobservable/unphysical across the horizon). The Schwarzschild interior solution is valid but it has been grossly misinterpreted. Cheers and thanks for your patient feedback, IV
From: Daryl McCullough on 8 Sep 2006 08:54 I.Vecchi says... >You did not use the KS chart to show that an infalling observer reaches >the singularity at r=0 in finite proper time. Well, the argument goes through the same for Kruskal coordinates. The metric (or line element, as Tom Roberts insists) for KS coordinates is: ds^2 = = f(r) (dT^2 - dR^2) + angular part where f(r) = 32m^3/r exp(-r/2m) and where r is an implicit function of R and T defined by R^2 - T^2 = (r/2m - 1) exp(r/2m) The geodesic equations of motion for radial motion (dtheta = dphi = 0) are 2 d/ds(f(r) (dT/ds)) = df/dr @r/@T ((dT/ds)^2 - (dR/ds)^2) 2 d/ds(f(r) (dR/ds)) = df/dr @r/@R ((dT/ds)^2 - (dR/ds)^2) with the constraint f(r) ((dT/ds)^2 - (dR/ds)^2) = 1 These equations can be solved for T and R as a function of proper time s, and you can compute the proper time at which r(T,S) = 0. >Both the interior and the exterior Schwarzschild chart correspond to >stationary observers (i.e. they map measurement outcomes of space-time >events performed by observers that are stationary). To be more precise, according to relativity, there is no absolute notion of "stationary". What's special about the Schwarzchild chart is that for any observer following a worldline dr/ds = 0, the metric is unchanging. Of course, inside the event horizon (r < 2m), it is impossible to have such a worldline (it represents faster-than-light travel there). >For the exterior chart this poses no problem since stationary observers >are well defined. However, in the interior domain, according to the >current theory, there are no stationary observers. This implies that >the interior Scwharzschild chart and the interior soultion defined on >it are meaningless, since they refer to measurements that cannot be >performed. How does it follow that they are meaningless? Coordinates are useful in *analyzing* a physical situation. They rarely correspond to anything that is locally observable. For example, in good old Euclidean space, we set up coordinates x,y,z. But for an alien observer that is billions of miles from Earth, and who has never heard of Earth, can he measure his value of x? No, he can't. He can set up his own local coordinates, and if the time ever comes when that alien meets up with an Earthling, the two can figure out mappings between their coordinate systems. The same is true of Schwarzchild coordinates. There is no way an observer inside the event horizon can directly measure the Schwarzchild coordinates r and t. However, he can certainly set up his own local coordinate system, and *we* can relate that coordinate system to the coordinate system of an external observer. >The t and r coordinate and the metric for r<2m do not >correspond to any physically measurable/meaningful quantity. >They are just blots on a sheet of paper. Yes, that's what coordinates are. They are just labels for spacetime points. They are nothing physically meaningful. If I call a distant star "Polaris", that has no physical meaning---it is just a label. Similarly, if I identify a spacetime event with coordinates (r,t,theta,phi), that's just a label---it has no physical significance other than for the purposes of saying which event you are talking about. >a) It is true that there are no stationary observers in the interior >domain. In this case the interior Schwarzschild solution is unphysical >and hence meaningless. It should be simply discarded. How does that make any sense? The only meaningful, coordinate-independent way to say that an observer is "stationary" is to say that he is following a worldline such that the metric is constant along that worldline. In the *real* universe, there *are* no stationary observers by that criterion. So does that mean that *all* coordinate systems in the real universe are unphysical and meaningless? Alternatively, you can let your notion of "stationary" be coordinate-dependent, and just say that an observer is "stationary" in coordinate system x^u if x^0 is timelike, and x^1, x^2, and x^3 are all constant for that observer. In that case, we can perfectly well have "stationary" observers in the interior. Just let x^0 = -r, x^1 = t, x^2 = theta, x^3 = phi. Inside the event horizon, it is possible to have a timelike worldline with t, theta, and phi all constant. >b) There are stationary obsevers in the interior domain. The arguments >purporting to show that there are no stationary observers in the domain >of the interior chart are flawed ( e.g. they relie on conservation of >quantities that become unobservable/unphysical across the horizon). In the Kruskal coordinates, nothing traumatic happens at the horizon, so there is no reason for conserved quantities to stop being conserved as the observer crosses the horizon. >The Schwarzschild interior solution is valid but it has been grossly >misinterpreted. That's probably true, but I think that it has been studied so extensively that most of the misconceptions about it have been uncovered. -- Daryl McCullough Ithaca, NY
From: carlip-nospam on 8 Sep 2006 16:33 In sci.physics I.Vecchi <tttito(a)gmail.com> wrote: [...] > Both the interior and the exterior Schwarzschild chart correspond to > stationary observers (i.e. they map measurement outcomes of space-time > events performed by observers that are stationary). Why do you say this? It's true for the exterior chart, in the sense that (1) there's a timelike Killing vector (i.e., a timelike "direction" in which the metric does not change), and (2) the Schwarzschild time coordinate is adapted to the Killing vector, so that an observer at fixed spatial coordinates does not see any change in the metric. In the interior, there is no timelike Killing vector -- the Killing vector that was timelike in the exterior is spacelike in the interior. That means that there is a *spatial* direction in which the metric does not change. Schwarzschild coordinates are still adapted to this Killing vector, but this no longer has anything to do with being stationary. Rather, an interior observer at a fixed *time* r sees the interior as being *spatially* unchanged in the *spatial* direction t. > For the exterior chart this poses no problem since stationary observers > are well defined. However, in the interior domain, according to the > current theory, there are no stationary observers. This implies that > the interior Scwharzschild chart and the interior soultion defined on > it are meaningless, since they refer to measurements that cannot be > performed. The t and r coordinate and the metric for r<2m do not > correspond to any physically measurable/meaningful quantity. That's not true. One just has to perform the proper measurements. I assume you don't object to the standard coordinates in FRW cosmology, even though the Universe is not stationary; these coordinates are chosen by the observation of homogeneity, and the requirement that the spatial slices at constant time be homogeneous. In the same way, the interior Schwarzschild metric has a "homogeneous" spatial direction, which can be used to pick out the coordinate t (which is now a spatial coordinate). The r coordinate is a time coordinate; it has the physical meaning that at constant r, the spatial direction t is foliated by spheres of constant area 4\pi r^2. (You should, in fact, view the interior as a "cosmology" -- it is a particular case of a Kantowski- Sachs cosmological spacetime.) Steve Carlip
From: I.Vecchi on 10 Sep 2006 06:52 carlip-nospam(a)physics.ucdavis.edu wrote: > In sci.physics I.Vecchi <tttito(a)gmail.com> wrote: > > [...] > > Both the interior and the exterior Schwarzschild chart correspond to > > stationary observers (i.e. they map measurement outcomes of space-time > > events performed by observers that are stationary). > > Why do you say this? > > It's true for the exterior chart, in the sense that (1) there's a > timelike Killing vector (i.e., a timelike "direction" in which the > metric does not change), and (2) the Schwarzschild time coordinate > is adapted to the Killing vector, so that an observer at fixed > spatial coordinates does not see any change in the metric. A stationary (hence accelerated) observer (say, on a space-ship) can determine her r coordinate by measuring her proper acceleration a, weighing objects or herself with a dynamometer. Normalising everything in sight a= -m/(r^2*sqrt(1-2m/r)) for r>2m. > In the interior, there is no timelike Killing vector -- the Killing > vector that was timelike in the exterior is spacelike in the interior. > That means that there is a *spatial* direction in which the metric > does not change. Schwarzschild coordinates are still adapted to > this Killing vector, but this no longer has anything to do with being > stationary. Rather, an interior observer at a fixed *time* r sees the > interior as being *spatially* unchanged in the *spatial* direction t. The problem arises because there is no such thing as an interior observer at fixed r. There are no stationary observers in the interior domain, hence the above measurement of r is impossible. > > For the exterior chart this poses no problem since stationary observers > > are well defined. However, in the interior domain, according to the > > current theory, there are no stationary observers. This implies that > > the interior Scwharzschild chart and the interior soultion defined on > > it are meaningless, since they refer to measurements that cannot be > > performed. The t and r coordinate and the metric for r<2m do not > > correspond to any physically measurable/meaningful quantity. > > That's not true. One just has to perform the proper measurements. I am saying that "proper measurements" of r in the interior domain are physically impossible. > I assume you don't object to the standard coordinates in FRW cosmology, > even though the Universe is not stationary; these coordinates are > chosen by the observation of homogeneity, and the requirement that > the spatial slices at constant time be homogeneous. In the same way, > the interior Schwarzschild metric has a "homogeneous" spatial direction, > which can be used to pick out the coordinate t (which is now a spatial > coordinate). The r coordinate is a time coordinate; it has the physical > meaning that at constant r, the spatial direction t is foliated by > spheres of constant area 4\pi r^2. (You should, in fact, view the > interior as a "cosmology" -- it is a particular case of a Kantowski- > Sachs cosmological spacetime.) As I wrote, since no observer can hover "at constant r" in the interior domain, r is not a measurable quantity. Cheers, IV
From: Daryl McCullough on 10 Sep 2006 08:04
I.Vecchi says... >A stationary (hence accelerated) observer (say, on a space-ship) can >determine her r coordinate by measuring her proper acceleration a, >weighing objects or herself with a dynamometer. Normalising everything >in sight a= -m/(r^2*sqrt(1-2m/r)) for r>2m. Well, anyone who has fallen from the exterior to the interior can calculate his r coordinate by integrating the geodesic equations. >The problem arises because there is no such thing as an interior >observer at fixed r. Why is that a "problem"? There are no external observers at fixed t, either. As I said, r can be computed in the interior by integrating the geodesic equations. Alternatively, I believe that r can also be measured by computing the full Riemann curvature tensor R_ijkl. The Einstein tensor G_ij is zero everywhere, but I don't believe that the Riemann curvature tensor is. >I am saying that "proper measurements" of r in the interior domain are >physically impossible. That's not true. >As I wrote, since no observer can hover "at constant r" in the interior >domain, r is not a measurable quantity. That doesn't follow. Nobody can hover at constant t in the exterior, either. -- Daryl McCullough Ithaca, NY |