Any coordinate system in GR?
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From: JanPB on 27 Aug 2006 04:05 Koobee Wublee wrote: > JanPB wrote: > > Koobee Wublee wrote: > >> ** 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 > >> > >> K is an integration constant chosen to fit Newtonian result. It is > >> > >> ** K = 2 G M / c^2 > >> > >> How do you know which of the two is the first form? And why? > > > > If one starts - as is usual - with a spherically symmetric > > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric > > metric can be locally written in the two forms we had before (not the > > two forms you wrote above - the two forms with the sign switch). One > > solves the Einstein equations which describe a metric (uniquely, > > because the equations happen to be ODEs in this case) described by > > components which happen to blow up at r=2m in that basis. This is your > > second equation. > > Of course, r >=0. This is spherically symmetric polar coordinate. Yes, although there is nothing magical about 0. It's just a convention everyone uses by analogy with the planar polar coordinates. One can assign the number 42 to the centre of symmetry. > > If one started instead with a spherically symmetric > > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number) > > then you'd end up with the solution that looks like your first > > equation. Same thing, different chart. > > Is there any reason for you to avoid my simple question? I though I'd just answered it. I assume it was "How do you know which of the two is the first form?" Did you mean something else? I assumed you meant "how does one know to which form r>0 applies to?". > > > > ...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 sure? This is aritmetic now. One more time: assuming r=u+K if r > > has values between 0 and infinity then u has values between what and > > what? > > Very sure. In both equations, r >= 0. No. The first equation follows from the second when you change the coordinates as r=u+K. So for clarity you should not use the same letter "r" there. I used "u". So if in the second equation the range of r is 0 through infinity it means that the range of u=r-K is -K through infinity. Elementary. -- Jan Bielawski
From: I.Vecchi on 27 Aug 2006 05:05 Tom Roberts ha scritto: > I.Vecchi wrote: .... > > 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. What we are discussing is "the uniqueness of the extension over the horizon" and the relevant selection criteria . 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. > It is quite clear that the universe we inhabit is not > described by the Schw. manifold. Yes, as I said, it is a physically irrelevant mathematical construct, whose physical irrelevance reflects the irrelevance of the criteria it fulfills. > >>> 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. You may call it as you like. > > > 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. 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. I surmise that there are geodesically incomplete extensions, corresponding to hybrid solutions which may be physically relevant. I am not keen on throwing away interesting extensions/solutions just because they do not not comply with some arbitrary uniqueness criterion. Cheers. IV
From: Sorcerer on 27 Aug 2006 08:03 "JanPB" <filmart(a)gmail.com> wrote in message news:1156665939.840854.177660(a)m79g2000cwm.googlegroups.com... | Koobee Wublee wrote: | > JanPB wrote: | > > Koobee Wublee wrote: | > >> ** 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 | > >> | > >> K is an integration constant chosen to fit Newtonian result. It is | > >> | > >> ** K = 2 G M / c^2 | > >> | > >> How do you know which of the two is the first form? And why? | > > | > > If one starts - as is usual - with a spherically symmetric | > > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric | > > metric can be locally written in the two forms we had before (not the | > > two forms you wrote above - the two forms with the sign switch). One | > > solves the Einstein equations which describe a metric (uniquely, | > > because the equations happen to be ODEs in this case) described by | > > components which happen to blow up at r=2m in that basis. This is your | > > second equation. | > | > Of course, r >=0. This is spherically symmetric polar coordinate. | | Yes, although there is nothing magical about 0. It's just a convention | everyone uses by analogy with the planar polar coordinates. One can | assign the number 42 to the centre of symmetry. | | > > If one started instead with a spherically symmetric | > > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number) | > > then you'd end up with the solution that looks like your first | > > equation. Same thing, different chart. | > | > Is there any reason for you to avoid my simple question? | | I though I'd just answered it. I assume it was "How do you know which | of the two is the first form?" Did you mean something else? I assumed | you meant "how does one know to which form r>0 applies to?". | | > > > > ...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 sure? This is aritmetic now. One more time: assuming r=u+K if r | > > has values between 0 and infinity then u has values between what and | > > what? | > | > Very sure. In both equations, r >= 0. | | No. Yes, fuckhead. r>=0. c= 0/0. The problem in both equations is c which is the velocity of light from A to A in time t'A-tA. http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Androcles
From: Daryl McCullough on 27 Aug 2006 08:02 I.Vecchi says... >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. The answer is definitely "yes". Consider an observer in freefall near the event horizon. Using his local coordinates, there is *nothing* to prevent him from reaching and passing the event horizon in a finite amount of proper time, because for him, spacetime near the event horizon is approximately *flat*. Anything other than demanding geodesic completeness would violate the equivalence principle, I think. >I surmise that there are geodesically incomplete extensions, >corresponding to hybrid solutions which may be physically relevant. > >I am not keen on throwing away interesting extensions/solutions just >because they do not not comply with some arbitrary uniqueness >criterion. I don't think that there is anything arbitrary about the criteria. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Aug 2006 08:24
Koobee Wublee says... >JanPB wrote: >> Koobee Wublee wrote: >>> ** 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 >Of course, r >=0. This is spherically symmetric polar coordinate. What about the first metric implies that the range of r is between 0 and infinity? Are you thinking that the use of the name "r" implies such a range? You do realize, don't you, that the first metric, with range r > -K is exactly the same as the second metric, with range r > 0? Or, the other way around, the first metric with range r > 0 is exactly the same as the second metric with range r > K>? -- Daryl McCullough Ithaca, NY |