Any coordinate system in GR?
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From: I.Vecchi on 10 Sep 2006 08:49 Daryl McCullough wrote: > I.Vecchi says... .... > >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. Space-time events (or points as you call them) correspond to measurement outcomes relative to an observer. Coordinates encode measurement outcomes relative to an observer's reference frame. >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. Any physical event (including space-time events) must be measured by an observer. A physical event corresponds to a measurement outcome. The question here is what measurement outcome corresponds to r inside the horizon. In order words, for r<2m, what does it mean to say that A is at r=r_0? What measurement outcome corresponds to such a statement? > > >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? See my parallel reply to Steve Carlip. Cheers, IV
From: I.Vecchi on 10 Sep 2006 09:05 Daryl McCullough wrote: > 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 geodesic equation encodes relationships between measurement outcomes. I am saying that those measurement outcomes are not defined inside the horizon. > > >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. In order to measure t a stationary observer has only to look at his clock. Beyond the horizon however there is no way to measure r. Cheers, IV
From: Daryl McCullough on 10 Sep 2006 09:13 I.Vecchi says... >> Yes, that's what coordinates are. They are just labels for >> spacetime points. > >Space-time events (or points as you call them) correspond to >measurement outcomes relative to an observer. No, they really don't. >Any physical event (including space-time events) must be measured by >an observer. No, that's not true. As I said, coordinates are just *labels*. They don't need to correspond to anything measurable. Figuring out what is going on in the universe is a difficult inverse problem. We take the observations, and we try to work backwards to what caused those observations. The model of the universe that we use is *not* determined by our observations, but is only *constrained* by observations. Observations can prove a model to be wrong, but observations cannot prove a model right. Coordinates are part of our *model*. Observations can show that our model is wrong, but the model is not uniquely determined by the observations. >> 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? > >See my parallel reply to Steve Carlip. I didn't see you address this point. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 10 Sep 2006 09:19 I.Vecchi says... >The geodesic equation encodes relationships between measurement >outcomes. I am saying that those measurement outcomes are not defined >inside the horizon. Yes, and I'm saying that you're wrong. An infalling observer can perfectly well keep track of his current value of r. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 10 Sep 2006 09:16
I.Vecchi says... >> 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? > >See my parallel reply to Steve Carlip. I looked over your reply again, and I certainly didn't see anything addressing this point. Here are your comments: 1. 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. 2. 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. 3. I am saying that "proper measurements" of r in the interior domain are physically impossible. 4. As I wrote, since no observer can hover "at constant r" in the interior domain, r is not a measurable quantity. None of those address the question of what it means for a coordinate to be measurable in a realistic universe in which there are no "stationary" observers. -- Daryl McCullough Ithaca, NY |