Any coordinate system in GR?
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From: carlip-nospam on 17 Sep 2006 17:08 In sci.physics I.Vecchi <tttito(a)gmail.com> wrote: > carlip-nospam(a)physics.ucdavis.edu wrote: > 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. This is true *if* she already knows that the metric is the Schwarzschild metric, and she knows that she's stationary, and she knows the value of m. >> 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. "The above measurement" may be impossible. But if the observer knows that the metric is of Schwarzschild form, and she knows the mass m, she will have no difficulty in measuring r. All she will need to do is to measure the tidal forces in a freely falling laboratory -- that is, the relative accelerations of two (say) radially separated test masses. There's actually a deep issue here. As people who work in quantum gravity are painfully aware, virtually all observables in general relativity -- all quantities that can be genuinely measured -- are nonlocal. The radial coordinate r for the exterior Schwarzschild solution, for example, is defined in terms of areas of two-spheres, which are highly nonlocal. In your alternative proposal, your observer must know that she is "at rest" (requiring a comparison to distant objects -- you can't decide this by a local measurement); she must weigh something (also not a local operation); she must determine the value of m (not at all easy if she doesn't already know the value of r at some point); and she must know that the metric is the Schwarzschild metric, a very highly nonlocal problem. What this means is that people who do real observations involving GR, in any context that goes beyond a weak field approximation with preferred "background" Minkowski coordinates, have to be careful to first translate all observations into invariant -- that is, coordinate independent -- form. For example, you don't use radar times to planets to determine "position coordinates" of Venus; instead, you write down a fairly general form of the metric with a bunch of free parameters, compute the (observable, coordinate-independent, and nonlocal) proper time for the round trip of a radar signal from Earth to Venus as a function of (arbitrary) coordinates and unknown parameters, and then fit everything. Steve Carlip
From: Ahmed Ouahi, Architect on 17 Sep 2006 17:45 However, along that matter, do also start by a taking a just a number, especially a fraction among the 0 zero, and the 1 one, and a just double it, furthermore, it have to be along which, you would drop the integer part, especially the part to a left, along the decimal point, and then a just it would be repeated that process. Therefore, as a most of the numbers would be irrational, that process would a just a produce an infinite sequences of a numbers, as it would be a magical for any observer, a definitely as a matter a fact. -- Ahmed Ouahi, Architect Best Regards! <carlip-nospam(a)physics.ucdavis.edu> wrote in message news:eekdha$ram$1(a)skeeter.ucdavis.edu... > In sci.physics I.Vecchi <tttito(a)gmail.com> wrote: > > > carlip-nospam(a)physics.ucdavis.edu wrote: > > > 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. > > This is true *if* she already knows that the metric is the Schwarzschild > metric, and she knows that she's stationary, and she knows the value of m. > > >> 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. > > "The above measurement" may be impossible. But if the observer knows > that the metric is of Schwarzschild form, and she knows the mass m, > she will have no difficulty in measuring r. All she will need to do > is to measure the tidal forces in a freely falling laboratory -- that > is, the relative accelerations of two (say) radially separated test > masses. > > There's actually a deep issue here. As people who work in quantum gravity > are painfully aware, virtually all observables in general relativity -- > all quantities that can be genuinely measured -- are nonlocal. The radial > coordinate r for the exterior Schwarzschild solution, for example, is > defined in terms of areas of two-spheres, which are highly nonlocal. In > your alternative proposal, your observer must know that she is "at rest" > (requiring a comparison to distant objects -- you can't decide this by a > local measurement); she must weigh something (also not a local operation); > she must determine the value of m (not at all easy if she doesn't already > know the value of r at some point); and she must know that the metric is > the Schwarzschild metric, a very highly nonlocal problem. > > What this means is that people who do real observations involving GR, in > any context that goes beyond a weak field approximation with preferred > "background" Minkowski coordinates, have to be careful to first translate > all observations into invariant -- that is, coordinate independent -- form. > For example, you don't use radar times to planets to determine "position > coordinates" of Venus; instead, you write down a fairly general form > of the metric with a bunch of free parameters, compute the (observable, > coordinate-independent, and nonlocal) proper time for the round trip of > a radar signal from Earth to Venus as a function of (arbitrary) coordinates > and unknown parameters, and then fit everything. > > Steve Carlip
From: I.Vecchi on 19 Sep 2006 05:01 carlip-nospam(a)physics.ucdavis.edu wrote: > In sci.physics I.Vecchi <tttito(a)gmail.com> wrote: > > > carlip-nospam(a)physics.ucdavis.edu wrote: > > > 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. > > This is true *if* she already knows that the metric is the Schwarzschild > metric, and she knows that she's stationary, and she knows the value of m. I am assuming in the exterior domain there are stationary observers. In that case, if the observer is changing her r, she will measure changes in her weight, since getting farther or closer to the horizon will make her lighter or heavier respectively. This is a local process, i.e, it can be implemented in the neighbourhood of a point. Stationary observers populating the exterior domain may define the metric through appropriate measurements and information exchange. It is still unclear to me how such a process can be implemented if there are no stationary observers, i.e. observers that cannot hover at a point. I read a post on spr mentioning infalling "Frolov observers", so I surmise that the maybe the problem is not entirely trivial. > > >> 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. > > "The above measurement" may be impossible. But if the observer knows > that the metric is of Schwarzschild form, and she knows the mass m, > she will have no difficulty in measuring r. All she will need to do > is to measure the tidal forces in a freely falling laboratory -- that > is, the relative accelerations of two (say) radially separated test > masses. > > There's actually a deep issue here. Yes, a deep and crucial one, which is essentially the one Wigner is raising. > As people who work in quantum gravity > are painfully aware, virtually all observables in general relativity -- > all quantities that can be genuinely measured -- are nonlocal. The radial > coordinate r for the exterior Schwarzschild solution, for example, is > defined in terms of areas of two-spheres, which are highly nonlocal. In > your alternative proposal, your observer must know that she is "at rest" > (requiring a comparison to distant objects -- you can't decide this by a > local measurement); she must weigh something (also not a local operation); > she must determine the value of m (not at all easy if she doesn't already > know the value of r at some point); and she must know that the metric is > the Schwarzschild metric, a very highly nonlocal problem. > > What this means is that people who do real observations involving GR, in > any context that goes beyond a weak field approximation with preferred > "background" Minkowski coordinates, have to be careful to first translate > all observations into invariant -- that is, coordinate independent -- form. > For example, you don't use radar times to planets to determine "position > coordinates" of Venus; instead, you write down a fairly general form > of the metric with a bunch of free parameters, compute the (observable, > coordinate-independent, and nonlocal) proper time for the round trip of > a radar signal from Earth to Venus as a function of (arbitrary) coordinates > and unknown parameters, and then fit everything. You may remember that we had an exchange about this ([1]). In this setting I still regard the distinction between quantum and classical GR as spurious. Your argument appears to confirm my point. There is no gtr without a space-time measurement model. Beside the general issue of measurabilty , however, I was raising a more specific point. I have been claiming that an external observer switching to free fall cannot calculate the proper time it will take him to reach the singularity. Darryl McCullough engaged in an instructive exchange with me about this, then wrote that the radial parameter in the inner domain is an ARBITRARY constant which has to be SET to 2m. I still do not see any physical reason for this. Actually I regard it as a vindication of my previous claims that the argument (which appears similar to what one finds in mainstream textbooks) is circular. In [2] I replied to Darryl who wrote >> the solution will have an >> arbitrary constant in it. That constant is determined by setting >> the maximum value of r in the interior region to be 2m. I wrote > By setting it? > And why precisely 2m? Why not 3m? What about infinity? > What is the physical criterion that motivates this choice? > By the way, isn't this closely related to "making some assumptions on > the time it will take to reach the singularity at r=0" or "the > matching at the horizon appears arbitrary and it constitutes an 'ad > hoc' assumption" or "your argument assumes that to an infalling > observer the inner domain appears as a sphere of radius 2m", as I > wrote? There appears to be no reason why an infalling observer should comply with arbitrary parameter fixing. Cheers, IV [1] http://groups.google.com/group/sci.physics.research/browse_frm/thread/9512d809de0c6ec9/ [2] http://groups.google.com/group/sci.physics.relativity/msg/f6c757e545f6fe26
From: Daryl McCullough on 19 Sep 2006 10:45 I.Vecchi says... >There appears to be no reason why an infalling observer should comply >with arbitrary parameter fixing. As Steve Carlip says, the value of r internally can be uniquely determined by measuring *local* tidal forces. So it is incorrect to say that r is somehow more measurable outside the event horizon than inside. -- Daryl McCullough Ithaca, NY
From: jmfbahciv on 20 Sep 2006 06:12
In article <eekdha$ram$1(a)skeeter.ucdavis.edu>, carlip-nospam(a)physics.ucdavis.edu wrote: <snip> >What this means is that people who do real observations involving GR, in >any context that goes beyond a weak field approximation with preferred >"background" Minkowski coordinates, have to be careful to first translate >all observations into invariant -- that is, coordinate independent -- form. >For example, you don't use radar times to planets to determine "position >coordinates" of Venus; instead, you write down a fairly general form >of the metric with a bunch of free parameters, compute the (observable, >coordinate-independent, and nonlocal) proper time for the round trip of >a radar signal from Earth to Venus as a function of (arbitrary) coordinates >and unknown parameters, and then fit everything. Ouch. You must be glad that computers exist. /BAH |