Any coordinate system in GR?
in [Math]
|
Prev: Prime numbers
Next: Am I a crank?
From: LEJ Brouwer on 4 Sep 2006 17:33 Tom Roberts wrote: > LEJ Brouwer wrote: > > [The Kruskal extension] > > is the unique _maximal_ extension (I would still appreciate a > > reference for that. > > Hawking and Ellis is the best reference I know for this. > > > > Anyway, > > this is not the case I am interested in. I am interested specifically > > in whether 'non-maximal' extensions are possible of the _exterior_ > > Schwarzschild solution > > Given that the maximal extension is known, all other extensions are > simply submanifolds of it. So take the Kruskal manifold and cut away any > parts you wish, and you'll still have an extension of the exterior > region (as long as you don't cut part of the exterior away (:-)). That's one of the reasons I would like to see a proof - I would like to know exactly how the event horizon is treated, and in particular if it needs to be treated in a special way. The Kruskal diagram event horizon is very odd, as it is a big fat 'X' shape with pointy edges, which I imagine would complicate the analysis. > You seem to be thinking/hoping/dreaming that there is some other > solution -- as long a smoothness and geodesic connectivity are required > there isn't (though I'm not certain what all caveats apply). There is nothing wrong with that. And yes, I believe that the usual rules must be broken at the event horizon. The event horizon is a very strange place after all. > Cutting away parts of the complete manifold is, of course, unphysical -- > it make mathematical sense and is well defined, but physically it makes > no sense to model the universe with a manifold from which objects can > simply disappear! So geodesic completeness for timelike geodesics is > required on physical grounds; the only exception is at a curvature > singularity where it is clear that theory (GR) breaks down. Objects don't just disappear from my manifold - but they *do* disappear from yours. In my model there are no singularities where GR breaks down, just weird edges where odd things happen. Maybe it's just a matter of taste, but as I told Jan, my model predicts electrodynamics, MOND, quantum theory and the standard model gauge group (and a bunch of other things, but let's take one thing at a time, here). > Tom Roberts - Sabbir.
From: JanPB on 4 Sep 2006 17:53 I.Vecchi wrote: > JanPB ha scritto: > > > I.Vecchi wrote: > > > > > > Consider the following alternative scenario. Take the complete Kruskal > > > solution and paste together the horizons of the white and of the black > > > hole. This is still a well-behaved extension. > > > > I don't see why the metric should remain smooth (at tne horizon) after > > such regluing. > > I would say it does, also based on stuff I've read on the web, but I > may be wrong. . I think the metric would not be smooth for the following reason. If I understand you correctly you want to join together the horizon between I and II with the horizon between I and III - correct? \ II / \ / IV\/ I /\ / \ / III\ ....and similarly the IV-II horizon with the IV-III horizon. Let's look e.g. at the gluing of the I-II with the I-III. You can pick any gluing you want as long as the result is a smooth manifold. The claim is that no matter what the gluing, the metric won't be smooth at the horizon. The metric is: ds^2 = (32m^3/r) exp(-r/2m) (-dT^2 + dX^2) + r^2 dOmega^2 ....where r = r(T,X) is a function given implicitly by: (1 - r/2m) exp(r/2m) = T^2 - X^2 Look at the graph of r(T,X) shown in terms of the RHS expression T^2-X^2: http://www.mastersofcinema.org/jan/rTX.pdf (I plotted it by a brute force calculation in Java.) The 1 on the horizontal axis represents T^2-X^2=1, i.e. the black hole singularity. 0 represents T^2-X^2=0, i.e. the horizon, etc. Now you can imagine the actual surface graph of r(T,X) over the K-S diagram: r is constant on the hyperbolae T^2-X^2=const., beginning with height 0 at the BH singularity, reaching level 2m at the horizon T=+/-X and continuing higher in the exterior. Now you can see the smoothness problem in the glued version of the manifold: r(T,X) has the slope pointing down as you cross from region I to II but as you emerge (in the glued version) from III back to I again, the slop reverses sharply as it's rising again. This sharpness is the problem: the graph I showed in the file rTX.pdf has a nonzero slope at T^2-X^2=0, so the gluing would result in a nondifferentiable V-shaped curve (imagine erasing the right hand portion of the magenta curve corresponding to T^2-X^2>0 and replacing it with the mirror image of the left portion). If the slope of r at 0 was zero, we'd have no problem (at least no problem with r). > > > The infalling > > > observers, raising his eyes after checking on his clock that he's > > > passed the horizon of the black hole, would just find himself gushing > > > out of a white hole horizon. After browsing around I think this is > > > called an Einstein-Rosen bridge. > > > > The Einstein-Rosen bridge is a spacelike surface (really a 3D manifold) > > corresponding to the lines T=const. (horizontal) on the K-S diagram. As > > you can see from the K-S it can be traversed only if you move faster > > than light. > > I'll try to work that out. From what I read, as well as from my wobbly > intuition, the Einstein-Rosen solution is well-behaved at horizon. What do you mean by the Einstein-Rosen solution? > I > would say that such a space-like surface (photons staying put on the > horizon) is there in any case and it is spanned by the K-S charts > (which , as far as I understand, get smoothly pasted together in the > Einstein-Rosen case), but that may reflect my current inadequate > understanding of the problem. Hm, not following you here. > > > I surmise this may be the same as a > > > blackhole were the singularity at r=0 is blown up to infinity > > > (together with tau_0) through a coordinate change for the interior > > > domain 2m<r<0. Repeating the above question, how do we know that this > > > is not the "right" chart (i.e, the chart corresponding to space-time > > > measurements of an infalling observer)? > > > > Not sure what you mean... Proper measurements don't change when > > coordinates change. > > True, but I still have doubts about (or don't understand, as you > prefer) the argument fixing an infalling observer's proper arrival time > at the singularity at r=0 . That's Darryl's argument that "you can > *compute* r from your proper time s", where it's still not clear to to > me whether the proper time is uniquely defined (this is the doubt that > has been motivating all my posts). So I still suspect that the > Einstein-Rosen solution may actually be described by what is regarded > as the interior solution, short of additional assumptions . > > Essentially, as far as I understand, you are saying that Einstein-Rosen > solutions (aka wormholes) are not smooth at the horizon. I'll study > this a bit more, gather references and then possibly state my point > again. They are smooth spacelike submanifolds. Their definition AFAICT is not 100% strict but pretty much the idea is that any family of spacelike submanifolds (smooth) that behave like the slices T=const. on the K-S diagram are "the Einstein-Rosen bridge". By themselves they are not solutions to any PDE that I know of (which is not saying much). -- Jan Bielawski
From: I.Vecchi on 5 Sep 2006 07:57 Daryl McCullough ha scritto: > I.Vecchi says... > > >Yes, but this assumes that the parameter r corresponds to to the the > >radius of the black hole, right? When you write that tau_0 "depends on > >the initial conditons" you are making some assumptions on the time it > >will take to reach the singularity at r=0. > > No, not really. Let me switch from tau to s, because s is > easier to write. The geodesic equation implies the following > constraints on the motion > > 1. (1-2m/r) (dt/ds)^2 - 1/(1-2m/r) (dr/ds)^2 = 1 > 2. (1-2m/r) (dt/ds) = some constant k Objection 1. Both equations may hold for r!=2m, otherwise they are meaningless as t becomes infinite. I also do not understand what may be the meaning of dt/ds and hence of the second equation inside the horizon, where t becomes space-like. > > Using 2, we can simplify 1 to get > > k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1 > > or > > 1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0 Again, only if r>2m. There is no reason to assume that this holds on both sides of the horizon. The matching at the horizon appears arbitrary and it constitutes an "ad hoc" assumption. I think that it entails assuming that the radius seen from inside the blackhole matches the radius as seen from the outside. > > To compute k, you need to know the initial conditions. > In particular, let's suppose that the observer is initially > at rest according to the Schwarzchild coordinates. That > means that when s=0, we have: dr/ds = 0. objection 2. This means that observer has yet to switch his rockets off, right? I see another issue here. When he switches the rockets off there is an abrupt change in dr/ds. How does it get encoded in the equation, if at all? The problem here is that you must establish a map between the time of a stationary (hence accelerated) observer and the time of an inertial observer in free-fall. In order to establish such a correspondence you must track proper time as it diverges from the time of a stationary observer. I'd rather plot dr/ds as a Heaviside function. > In that case, letting r0 be the value of r when s=0 > Then we have, at proper time s=0: > > (1-2m/r0) (dt/ds)^2 = 1 > > from equation 1 and > > (1-2m/r0) (dt/ds) = k > > from equation 2. These two together imply > > k = square-root(1-2m/r0) > > So our equation 1' above becomes > > (dr/ds)^2 + 2m/r + 2m/r0 = 0 > > This equation gives us r as a function of s and r0. > > >I don't think that an infalling observer can measure the > >inner radius, since in order to do that he would have to reach the > >singularity at r=0, where the metric blows up for good. > > Yes, you're right that r is not directly measurable. However, > using the above equation, you can *compute* r from your proper > time s. > > >More in general, apparently your argument assumes that to an > >infalling observer the inner domain appears as a sphere of radius r, > > No, I didn't assume that. I just assumed that r and t are two > coordinate such that the metric expressed in terms of r and t > is > > ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + angular part > > >Consider the following alternative scenario. Take the complete Kruskal > >solution and paste together the horizons of the white and of the black > >hole. > > I'm not sure what this means. Look at the Kruskal diagram as > shown in > http://io.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes_files/image011.gif > > In that picture, the Kruskal "time" coordinate is v, whose axis is shown > oriented vertically. The Kruskal "space" coordinate is u, whose axis is > shown oriented horizontally. > > In terms of compass directions: The black hole singularity is to the > north, the white hole singularity is to the south, the black hole > exterior is to the east, and the mirror exterior is to the west. > The black hole event horizon is the V-shaped boundary of the northern > quadrant. It runs diagonally from the northwest corner to the center > (v=0, u=0), and then turns a right angle and runs diagonally to the > northeast corner. > > The white hole event horizon is the inverted V-shaped boundary of > the southern quadrant. It runs from diagonally from the southwest > corner to the center, and then diagonally from the center to the > southeast corner. > > So which portions of the horizon are you proposing to paste together? I am folding and pasting the line r=2m symmetrically with respect to the origin, i.e, along the red circle in http://casa.colorado.edu/~ajsh/wormbig_gif.html . See below. .. > > >This is still a well-behaved extension. The infalling > >observers, raising his eyes after checking on his clock that he's > >passed the horizon of the black hole, would just find himself gushing > >out of a white hole horizon. After browsing around I think this is > >called an Einstein-Rosen bridge. > > As I understand it, the bridge connects the exterior > in the east to the mirror exterior in the west. Yes. Einstein-Rosen originally suggested casting the metric into the coordinate rho^2=r-2m and considering only real values of rho, with rho>0 corresponding to the black hole exterior and rho<0 corresponding to the white hole exterior. They are connected at rho=0 where the determinant of the metric vanishes, i.e. on the white and black horizons that are thus identified. I read that the Einstein-Rosen construct is considered erroneous/arbitrary, but, as I explained already, it's not clear to me yet that similar objections can't be raised for the Schwarzschild interior solution. > I don't > see how it is accurate to describe someone crossing the > bridge as "gushing out of the white hole horizon". Well, you cross r=2m and you are on the other side. IV
From: I.Vecchi on 5 Sep 2006 09:28 I.Vecchi ha scritto: > Daryl McCullough ha scritto: > > > I.Vecchi says... > > > > objection 2. > > This means that observer has yet to switch his rockets off, right? > I see another issue here. When he switches the rockets off there is an > abrupt change in dr/ds. How does it get encoded in the equation, if at > all? The problem here is that you must establish a map between the time > of a stationary (hence accelerated) observer and the time of an > inertial observer in free-fall. In order to establish such a > correspondence you must track proper time as it diverges from the time > of a stationary observer. I'd rather plot dr/ds as a Heaviside > function. Forget objection 2. It's dr^2/ds^2 that changes abruptly. IV
From: Daryl McCullough on 5 Sep 2006 10:55
I.Vecchi says... > >Daryl McCullough ha scritto: >> No, not really. Let me switch from tau to s, because s is >> easier to write. The geodesic equation implies the following >> constraints on the motion >> >> 1. (1-2m/r) (dt/ds)^2 - 1/(1-2m/r) (dr/ds)^2 = 1 >> 2. (1-2m/r) (dt/ds) = some constant k > >Objection 1. >Both equations may hold for r!=2m, otherwise they are meaningless as t >becomes infinite. Right, they are valid in the region r > 2m. They are *also* valid in the region r < 2m. >I also do not understand what may be the meaning of dt/ds and >hence of the second equation inside the horizon, where t >becomes space-like. The meaning of dt/ds is the rate of change of the coordinate t as a function of proper time s. The fact that t is spacelike makes dt/ds *less* mysterious, it would seem to me. dt/ds is just proper velocity along the t-axis. It's no more mysterious than dx/ds or dy/ds in Minkowsky spacetime. >> Using 2, we can simplify 1 to get >> >> k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1 >> >> or >> >> 1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0 > >Again, only if r>2m. There is no reason to assume that this holds on >both sides of the horizon. There is nothing in the derivation that makes any assumption about the sign of (1-2m/r), so the derivation is just as valid when r < 2m as when r > 2m. >The matching at the horizon appears arbitrary and it constitutes >an "ad hoc" assumption. There is nothing arbitrary about it. It's the unique answer, as far as I see. To put together a Lorentzian spacetime manifold, you need to do the following: You have to split spacetime up into a collection of overlapping "patches" such that each patch can be described using a coordinate system for which the metric components are nonsingular. The patches have to describe an *open* subset of points (this insures that overlapping patches overlap in a 4-D *region*). If two patches overlap, then there must be a 2-way mapping function between the two coordinate systems that is valid (differentiable, nonsingular) in the region of overlap. So following these guidelines, we can pick one patch to cover the region infinity > r > 2m, infinity > t > -infinity. The Schwarzchild coordinates are perfectly adequate for describing this region. But they don't cover the points on r=2m. So you have to come up with a second patch that covers this region. But since patches have to be *open* sets, you need a patch covering a little bit on both sides of r=2m. One possibility is to use a Kruskal patch. You have coordinates R, T with the metric 32m^3/u exp(-u/2m) (-dT^2 + dR^2) where u is an implicit function of R and T given by T^2 - R^2 = (1-u/2m) exp(u/2m) Using the mapping r_schwarzchild = u(R,T) t_schwarzchild = 4m arctan(T/R) we find that the metrics agree in the region of their overlap, and that the Kruskal patch covers the region r=2m, as well. You claim that choosing a Kruskal patch is arbitrary. Then show me a different choice: Show me another patch that extends Schwarzchild spacetime past r=2m which (1) agrees with the Schwarzchild patch in the region r > 2m, and (2) *disagrees* with the Kruskal patch in the region r < 2m. I don't think that there is such a patch. So the Kruskal extension is *unique*, not arbitrary. >I think that it entails assuming that the radius seen from >inside the blackhole matches the radius as seen from the outside. No, it doesn't make any assumptions about what the infalling observer *sees*. >> To compute k, you need to know the initial conditions. >> In particular, let's suppose that the observer is initially >> at rest according to the Schwarzchild coordinates. That >> means that when s=0, we have: dr/ds = 0. > >objection 2. > >This means that observer has yet to switch his rockets off, right? It's a geodesic. The particle is in freefall. All motion is without the help of rockets. I'm assuming an initial condition in which the test particle has dr/ds = 0. >> So which portions of the horizon are you proposing to paste together? > >I am folding and pasting the line r=2m symmetrically with respect to >the origin, i.e, along the red circle in >http://casa.colorado.edu/~ajsh/wormbig_gif.html . See below. I don't understand what that diagram means. However, if you identify the two regions labelled r=2m, then you are identifying the black hole with the white hole. That doesn't affect the event horizon. -- Daryl McCullough Ithaca, NY |