Any coordinate system in GR?
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From: Daryl McCullough on 27 Aug 2006 08:38 Koobee Wublee says... >>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 >>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 >> The reason for it this is that the first form of the metric is obtained >> from the second (for which r>0) by means of the coordinate change: > >How do you know which of the two is the first form? And why? If you compute the area of a sphere of radius r and constant t using the top metric, you will find that it is 4 pi (r+K)^2 which implies that the effective radius is r+K, not r. >> ...where you are recycling the letter "r" in the first form where I >> have used "u". Look at the formula r=u+K. It's patently obvious that >> when r changes between 0 and infinity then u changes between -K and >> infinity. Tell me, is this not true? > >No. Are you saying that it's not true that if we start with ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 and change variables to u = r-K, then we obtain ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2 Or are you saying that it's not true that under the variable change u = r-K, the range r=0 to infinity is mapped to the range u=-K to infinity? -- Daryl McCullough Ithaca, NY
From: Sorcerer on 27 Aug 2006 09:05 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ecs3oe0a7t(a)drn.newsguy.com... | Koobee Wublee says... | | >>> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 | >>> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 | | >> The reason for it this is that the first form of the metric is obtained | >> from the second (for which r>0) by means of the coordinate change: | > | >How do you know which of the two is the first form? And why? | | If you compute the area of a sphere of radius r and constant t using | the top metric, Metric. A nonnegative function g(x,y) describing the "distance" between neighboring points for a given set. No metric can contain c or c^2 http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Androcles
From: Daryl McCullough on 27 Aug 2006 09:39 Edward Green says... >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 >gravitaional time dilation? Ed, the usual way of understanding gravitational time dilation is that it is about the *relationship* between two different coordinate systems---the coordinate system of an inertial observer, and the acceleration of a noninertial observer. So it's not really a property of *spacetime*, but of particular coordinates on spacetime. The same effect (gravitational time dilation) takes place inside an accelerating rocket in flat spacetime. Of course, spacetime curvature is relevant because if spacetime is curved, then there *are* no extended inertial coordinate systems, so we're forced to use noninertial coordinates. >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. Let e1 and e2 be two successive ticks of a clock (assuming that you have an old-fashioned clock that ticks, if they make those anymore). There are at least three different ways to compute the "time between ticks": (1) Use the metric, and compute Integral from e1 to e2 of square-root |g_ij dx^i dx^j| along a geodesic path connecting e1 and e2. (2) Look at the elapsed time shown on the clock. (3) Use a particular coordinate system, and use t2 - t1 as the time between ticks (where t2 is the time of e2 in that coordinate system, and t1 is the time of e1). The usual assumption is that methods (1) and (2) will give the same answer. Method (3) can give a wildly different answer, depending on the coordinate system. Two time intervals that are the same "duration" according to methods (1) and (2) might have different durations according to method (3). Now, the usual interpretation of such a discrepancy is to treat the *coordinate* notion of "duration" as not physically meaningful. >How does the formal representation of a metric handle this? 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! But that's *exactly* the modern interpretation of General Relativity: all spacetimes look flat if you look at them locally. You can divide up spacetime into tiny little regions (for example, consider filling spacetime with observers in rocket ships). Each region is approximately like a little piece of flat spacetime. Each region can be described (approximately) using good old inertial cartesian coordinates, and all the laws of physics will (approximately) hold in these coordinates as if there were no gravity, no curved spacetime. Curvature only comes into play when you consider how to glue all these little regions together into a coherent whole. If spacetime is flat, then you can partition it into equal-sized rectangular regions that match neatly on the borders. If spacetime is curved, then the regions will have overlaps or gaps and won't fit together so smoothly. For a lower-dimensional analogy, you can imagine dividing the surface of the Earth into little regions that are say 100 miles by 100 miles. You can draw a map of such a region on a flat piece of paper without worrying about the fact that the Earth is curved. However, if you try to fit all the little maps together, you will find that they *can't* be fit together without overlaps or gaps. >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. No, it really *doesn't* mean that. What a physical clock measures is the *whole* expression ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + r^2 dOmega^2 If the clock travels from the point with coordinates (t1,r1,theta1,phi1) to the point with coordinates (t2,r2,theta2,phi2) then the elapsed time on the clock T will approximately satisfy T^2 = (1 - 2m/r_average) (t2 - t1)^2 - 1/(1 - 2m/r_average) (r2 - r1)^2 - r_average^2 (theta2 - theta1)^2 - r_average^2 sin^2(theta_average) (phi2 - phi1)^2 where r_average = (r2 + r1)/2 and theta_average = (theta2 + theta1)/2. There is no direct way to measure the Schwarzchild coordinate t using clocks. -- Daryl McCullough Ithaca, NY
From: Tom Roberts on 27 Aug 2006 12:09 I.Vecchi wrote: > Tom Roberts ha scritto: >> 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. > > What we are discussing is "the uniqueness of the extension over the > horizon" and the relevant selection criteria . Yes. Extension of the exterior region of the Schw. manifold. The complete extension is known, and is called the "Kruskal extension", and is covered by Kruskal-Szerkes coordinates. You should get a GR textbook and learn about them. It astounds me that you keep trying to argue from ignorance of what is already known about this manifold. > What I am saying is that > there are at least two ways to extend the solution over the horizon, > correspoinding to two different physical situations, the black hole and > the white hole. Knowledge of the complete (inextensible) manifold shows that your "two ways" are merely different regions of a single manifold. I have said this several times, and don't know how to say it differently. <shrug> The two sets of E-F coords. merely cover different regions of the (single) manifold; both include the exterior region, and neither includes the "other" exterior region of the manifold. >> The geodesically complete extension of the Schwarzschild charts is >> unique, given by the Kruskal chart. > > The question is whether the requirement of geodesic completeness (which > in this setting is obtained by shifting trouble to infinity) is > appriopriate, i.e. physically relevant. Of course it is relevant! If the manifold is not geodesically complete (for timelike paths), then objects would simply "leave the manifold", which is tantamount to "leaving the universe". As the manifold is intended to be a model of the entire universe, this does not make sense. > I surmise that there are geodesically incomplete extensions, > corresponding to hybrid solutions which may be physically relevant. You think objects can arbitrarily leave the universe? With the arbitrariness being your personal whim (of which incomplete extension you select)? > I am not keen on throwing away interesting extensions/solutions just > because they do not not comply with some arbitrary uniqueness > criterion. The requirement is not necessarily uniqueness (though that is highly desirable), but for the Schw. manifold the RESULT OF EXTENSIVE ANALYSIS is that the extension is unique. Tom Roberts
From: Edward Green on 27 Aug 2006 12:21
Bill Hobba wrote: > "Edward Green" <spamspamspam3(a)netzero.com> wrote in message > news:1156542691.425027.214860(a)75g2000cwc.googlegroups.com... > > You are merely being insulting now. > > If you thought that then I apologize. You have done nothing to deserve > insults. Thank you. > > > > 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. Possibly it should be. If I ask you if a manifold can be twisted, does it mean I can't distinguish mathematical from physical objects, or does it mean I want to know whether any property of the manifold is analogous to twisting a material body? As a courtesy, please assume enough background on my part to suppose the latter. > Have you considered your musings are better suited to a philosophy forum? Sigh. I guess it just comes naturally then. |