Any coordinate system in GR?
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From: Sorcerer on 26 Aug 2006 15:12 "Edward Green" <spamspamspam3(a)netzero.com> wrote in message news:1156618284.840466.162370(a)p79g2000cwp.googlegroups.com... | Ken, I seem to be subject to a mini-curse: nobody quite recognizes what | I'm asking; and finally, given enough perserverence, I may come up with | answers, still unrecognizable. Not recognizing the question, | respondants are either sympathetic or insulting (or silent) according | to their own personality, but not, from my point of view, on target. | | I asked how the metric -- particularly the explicit coordinate | representation of the metric -- adapts to dilations in spacetime. Edward, I seem to be subject to a mini-curse: nobody quite recognizes what I'm asking; and finally, given enough perserverence, I may come up with answers, still unrecognizable. Not recognizing the question, respondants are either sympathetic or insulting (or silent) according to their own personality, but not, from my point of view, on target. I asked how the egg -- particularly the explicit ovum representation of the bright green flying elephant -- adapts to contours in black holes. Androcles
From: Ken S. Tucker on 26 Aug 2006 18:34 Edward Green wrote: > Ken S. Tucker wrote: [brevity snip] > I asked how the metric -- particularly the explicit coordinate > representation of the metric -- adapts to dilations in spacetime. > Hobba's answer is, there is nothing corresponding to a dilation in > spacetime. I'm not sure if I believe this or not. What about > gravitational time dilation? This phrase tends to indicate that a > second as measured by a standard physical clock recorded over _here_ > may not correspond to a second over _there_. Physical clocks and > rulers, composed of standard arrangement of atoms, give what we might > consider a natural local metric to spacetime. If we have reason to > believe -- without the possibility of direct comparison -- that these > metrics would not agree in different regions, then it seems plausible > to say that there is a relative dilation of spacetime beween the > regions. Yes, I agree. > How does the formal representation of a metric handle this? Well first we need to agree that survey's are done using light, radar and laser's, by astronomers, and astrophysicist's. In the case of Newton, the "metric" was assumed to be equalivalent to a Cartesian CS, and I repeat, light is unaffected in a Cartesian CS, (metric) because the light-ray paths define the axes. The so-called "spacetime-field" is defined by the velocity (direction and speed) of light-rays. Subsequent to SR and QT, conservation of the energy of light needed to be included in the method of surveying astronomical calculations, primarily using the Einstein shift and to a lesser degree the deflection of light. Thus the old Cartesian metric was subtly altered, accounting for the effects mentioned above, and I think an outstanding proof is given by a more accurate description of Mercury's orbit and the deflection of light. The altered metric is different from Cartesian, and changes thoses predictions w.r.t Newton. I've studied your post below, but I'll hold here pending a mutual understanding, you know, it's the kinda of basics I grew up with. Cheers Ken S. Tucker > Is the > change in a time-like variable measured in local seconds prefixed with > a different coefficient in region one than in region two? Or are local > seconds always prefixed with the same constant coefficient in a given > representation of a metric, as are, say, local centimeters. That > doesn't seem right, otherwise all spacetimes would look like flat > spacetime as seen through the lens of the explicit metric! > > Schwarschild's metric, e.g. > > http://en.wikipedia.org/wiki/Schwarzschild_metric > > trivially has the property that the prefix to dt is a function of r. > The "dt" implicitly represents time as measured by a appropriate > standard physical clock. So evidently not all such clocks are equal > with respect to the metric -- time is "stretched" in some regions > relative to others. > > Now, I can rephrase my question. Suppose we replaced "t" with t' = > a(t)t , a( ) a positive increasing function. This can be interpreted > as changing the units of time from "natural clock seconds", and not in > a uniform way. How does the explicit form of the metric handle this? > > In this case, a possible answer seems trivial: we simply insert the > inverse of the positive factor "a(t)"... better, a^-1(t') ... in the > formal metric, and everything is fine. > > This is the point where I see a possible problem. We now have a > function of variables labeled t',r,O (omega) which corresponds to the > same spacetime as a different function of variables labeled t,r,O . > So, what's the problem? We've just changed varables. The problem is, > suppose we could demonstrate a second _different spacetime_ with the > same variable labels t',r,O and the same formal metric. Is this a > problem? > > I can't carry through my threat here, with the Schwarschild metric, but > that was the point of my toy one dimensional manifold -- I tried to > motivate such a possibility. > > OK, Ken, here is my current postulation for an answer: you _can't_ > reconstruct a physical spacetime simply from a set of variable labels > and an explicit form of a metric on such labels: there is an element > missing. That element is a perscription of the relationship between > the local differentials of the variables and standard local physical > processes -- "so many wavelengths of a certain spectral transition", > and so forth. If this perscription is the same everywhere, then it can > simply be stated as "choice of units", as in, centimeters and seconds; > if the perscription varies, then the additional information will be > more complicated. But there is additional structure required to > complete the model. > > Comment: the "standard local clock" may need clarification. > > Comment: my internal pendulum is again swinging towards supposing my > question was close to being well-posed or at least well-posable, and > that therefore those who merely dismissed it as containing elementary > errors would have done better to remain silent, than open their mouth > and remove all doubt. > > > > GR treats all Frames of Reference's (FoR's) equally. > > AE's law, G_uv = T_uv constrains the metrical > > geometry, to solutions of the tensor g_uv satisfying > > that law, that is why G_uv=T_uv is a called a field > > equation. For example a simplistic solution follows > > from G_uv=0 called the Schwarzchild Solution, that > > I'm sure you're aware of, that provides metrics like, > > > > g_00 = 1 - 2m/r , g_rr = 1/g_00 etc. > > > > The reason for CS independence is because g_uv > > is a tensor and can be properly transformed via > > the usual procedure, > > > > g'_ab = (&x^u/&x'_a) (&x^v/&x'_b) g_uv > > > > to any other CS, using light-years, millimetres > > polar, elliptical, cylindrical as you please, and > > in time, seconds or dog-years, meaning the > > physical reality is independent of units. > > Of course. I'm thinking however that the units must be stated, which > is sufficiently trivial as to be invisible, and that if we allow > arbitrary point to point fluctuations in the units, we may need some > additional non-trivial statements to complete the specification, beyond > the explicit form of the metric. > > As always, I could be wrong. > > Thanks, Ken.
From: Ken S. Tucker on 26 Aug 2006 18:50 JanPB wrote: > Ken S. Tucker wrote: > > > > Jan, I think you should study the original paper > > Ed Green cited, there is no such thing as an > > event horizon or Black-Hole's. BTW, Dr. Loinger > > and I discussed this at length. In short, the > > original Schwarzschild Solution has been > > bastardized and mis-understood for simplicity. > > No, the "original" solution is the only one. This follows from basic > ODE theory and the definition of tensor. > > > The bastardized version became popularized > > and embraced by astronomers who now see > > BH's under their beds. > > There is no other version, bastardized or not. How many solutions do > you have to the following ODE: > > f'(x) = f(x) > > ...given the initial condition f(0) = 1 ? I can see f(t) = exp(t). Is > there any other? > > How about another ODE, just slightly more complicated: > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0 > > ...given the initial condition f(1) = 0 ? > > -- > Jan Bielawski What Newton assumed was, r= sqrt(x^2 + y^2 +z^2) but in GR the "real" R = r - m where m =1.47 kms in the case of the Sun. Most treatise begin by defining "r" that way. Schwarzschild emphasized that difference and obtained a singularity only at the very center of a mass, and that would assume an infinite density, which is physically impossible. But it's a good exercise. Ken
From: Bill Hobba on 26 Aug 2006 22:23 "Edward Green" <spamspamspam3(a)netzero.com> wrote in message news:1156542691.425027.214860(a)75g2000cwc.googlegroups.com... > Bill Hobba wrote: > >> "Edward Green" <spamspamspam3(a)netzero.com> wrote in message >> news:1156222614.472337.73970(a)p79g2000cwp.googlegroups.com... >> > Bill Hobba wrote: >> > >> >> "Edward Green" <spamspamspam3(a)netzero.com> wrote ... >> > >> > <stretching the manifold vs. compressing the coordinate system> >> > >> >> First why is one unphysical and the other not? >> > >> > Well, that depends on whether it is "physical" to stretch the manifold. >> >> Since it is not a material then obviously it is a 'word salad's with no >> meaning like saying stew smells like isomorphism. > > You are merely being insulting now. If you thought that then I apologize. You have done nothing to deserve insults. > > Obviously a manifold is a mathematical abstraction. Obviously any word > normally applied to material objects tentatively when applied to a > mathematical abstraction should be understood to refer to a putative > property of the abstraction analogous to the named material property. That is not so obvious to me. Have you considered your musings are better suited to a philosophy forum? Thanks Bill > I could equally well scoff that since a manifold is not a material then > it makes no sense assign "curvature" to it. And yet sense can be so > made, as you know. > > Of course, in the continuing verbal self-defense your minimax strategy > makes necessary (proving me maximally stupid based on minimal > evidence), I hasten to add that I am _not_ arguing that since a meaning > can be assigned to "curvature" meaning must also be assignable to the > strain or "stretching" of a manifold. I am merely indicating that such > a putative identification is not the silly category error you imply. > > In fact, the idea will have meaning to the extent we assign meaning to > it. Modeling the rubber band by a one dimensional manifold, we will > find it natural to assign a pointwise property to the model called > "strain". Is this part of the minimal structure of a differential > manifold, or something added? I don't know. Is there something > analagous to strain assigned to the manifold which is an element of the > model of GR? I don't know. > > Since I know that I don't understand the model, I don't know that there > must be a property of the model of GR plausibly described as a local > strain state of spacetime. I am not confident however that there is no > such property is possible merely because Bill Hobba asserts it, because > Bill Hobba denigrates me merely for describing a putative property via > a word applicable to material objects, as if such alone were sufficient > proof of idiocy, when Bill Hobba in fact knows that mathematics freely > borrows terms from the quotidian world. > > Your counter argument is falacious. > >> > In a simple minded analogy, the one dimensional manifold might be >> > represented by a rubber band: stretching the rubber band is a >> > physically distinct situation from compressing the coordinates -- one >> > might be able to determine the state of strain by local measurements, >> > or at least determine the relative strain for two locations. In GR the >> > "stretch" might correspond to a gravitational field vs. flat spacetime. >> > Not _everything_ is an artifact of the coordinate system. >> >> You were on the right track above - stretching is not a property of the >> space-time manifold. > > That's better: a blunt assertion. Maybe you are right, but you will > forgive me if I don't assign your claim 100% Bayesian confidence just > at this moment. You have after all given me no reason at all to > believe you beyond your blunt assertion. > > (Leave off the implicit insults in the interest of brevity. Thanks.) >
From: Tom Roberts on 27 Aug 2006 00:21
I.Vecchi wrote: > Tom Roberts ha scritto: >> It is arbitrary which portion of the horizon you consider, but as both >> are contained in the manifold it is only the analyst who is subject to >> this choice, not the manifold itself (or the physical system for which >> it is a model). > > Are you saying that for every black hole in the universe there is a > corresponding white hole? Not at all! But I am pointing out that in the Schwarzschild manifold there is both a black hole and a white hole. That manifold _IS_ what we are discussing. It is quite clear that the universe we inhabit is not described by the Schw. manifold. >>> It remains a fact that the extension >>> across the horizon is not unique and that each extension yields a >>> distinct physical object. >> Not "physical object" but rather region of the manifold. > > Which corresponds to a physical/observational object. Otherwise we are > talking about nothing. I don't know what you mean by "a physical/observational object". But it IS clear that a region of the manifold is not any sort of "object" at all. <shrug> One might call it an "observational domain" I suppose, but "object" is quite definitely not applicable. > Beside the above, I surmise that there are other space-time extensions > across the horizon corresponding to hybrid white hole/black hole > solutions (*). This would be impossible according to your argument, > right? The geodesically complete extension of the Schwarzschild charts is unique, given by the Kruskal chart. Tom Roberts |