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Schwarzschild solution radial coordinate and black holes

  1. Nov 18, 2008 #1

    Jonathan Scott

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    When people appear to be getting very confused about the weird nature of black holes, I normally respond with answers based on standard black hole theory, but I sometimes feel I should also call attention to the point that some people now think that the "black hole" solutions to the Schwarzschild equations are quite possibly only the result of an incorrect assumption about the physical meaning of the radial coordinate. Unfortunately, this subject tends to raise such fierce argument that I usually try to forestall questions and provide references at the same time, and even then the result is often further questions, which tend to derail the original subject of the thread. I'm therefore starting a new thread, to act as a reference point and allow some discussion (as long as it can remain scientific).

    I think the best explanation of this situation (and least over-stated) is an article by Salvatore Antoci and Dierck-Ekkehard Liebscher "Reinstating Schwarzschild's Original Manifold and its Singularity", also available at arXiv:gr-qc/0406090.

    These ideas have been forcefully rebutted on-line in a thread "Flogging the Xprint" on sci.physics.research by "T. Essel", and also in a paper by Malcolm MacCallum at arXiv:gr-qc/0608033. However, although in both cases the "rebuttals" appear very authoritative, it appears to me personally that they are based on stating weak arguments very forcefully, which I don't find very convincing, and assuming that any fault with the opposing argument means that their own position is correct.

    I believe that basically the situation is that at present there's no physical theory which allows us to determine the location of the mass point in the Schwarzschild vacuum solution. Schwarzschild assumed it to be at his original r=0, equivalent to R=2GM in the Schwarzschild radial coordinate defined by the proper areas of spheres. Hilbert assumed it to be at R=0 because it seemed to be mathematically valid to do so.

    It was only a few years ago that Leonard S Abrams called attention to this difference of assumptions, which had apparently been ignored since Hilbert "corrected" Schwarzschild's original assumption. Some people have argued that since there are a lot of weird things in black hole theory, Schwarzschild's original assumption must have been correct. I'm inclined to sympathize with this point of view, but it's not a proof. On the other side, the people such as "T. Essel" who argue against them say that most of their arguments against Hilbert's assumption are invalid (which I'd probably agree with), and hence that Hilbert was right (which doesn't actually follow).

    As far as I can see, Antoci and Liebscher's article referenced above gives a more balanced view, giving the background and explaining how the difference arose, and showing that Schwarzschild's original solution does appear to be better in many ways.

    This situation could perhaps be resolved by experiment. There is some evidence in both directions, in that at least one super-massive black hole candidate appears to have a significant intrinsic magnetic field, which is not consistent with black hole theory, but objects which could according to black hole theory be near the borderline between neutron stars or black holes appear to show differences in X-ray emissions suggesting that there is some threshold being crossed (although Abhas Mitra has published papers saying that this threshold doesn't necessarily imply a black hole, but could be some other form of phase change or similar).
     
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  3. Nov 18, 2008 #2
    I don't understand this phrase...I thought black holes could have electric charge and might spin...hence it never occured to me a magnetic field would be prohibited...
     
    Last edited: Nov 18, 2008
  4. Nov 18, 2008 #3

    Jonathan Scott

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    Normal bodies can have strong magnetic fields caused by aligned current loops and magnetic moments of particles, requiring no overall charge.

    If black holes have charge and spin, they can have a magnetic field too, but it is negligible, because it can only be due to the circulation of the overall charge. If a black hole had a significant amount of charge, then it would attract particles with the opposite charge more strongly, which would soon neutralize that charge.
     
  5. Nov 18, 2008 #4

    Hurkyl

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    No it can't. The behavior of the universe in the R > 2GM/c² region is not affected by questions of whether there exists an R <= 2GM/c² region, and if so what goes on in that region.
     
  6. Nov 18, 2008 #5

    Hurkyl

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    Honestly, the whole discussion seems doomed from the start, since its two most prominent points are irrelevant!

    Schwarzschild, Einstein, Hilbert -- none of them were oracles. They did have great insights that lead to the development of GR (and other subjects) -- but their insight does not supersede the decades of development that has been made since their time. e.g. whatever opinions Scharzschild and Hilbert may have had on where the edge of the universe lies are don't really matter today.

    The most important guiding principle behind general relativity is that coordinates have no physical significance; they are merely a system for assigning labels to events in space-time. And so a discussion that focuses on matters of coordinates in general relativity cannot be said to be discussing anything physically meaningful at all!
     
  7. Nov 19, 2008 #6

    Jonathan Scott

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    This is incorrect, because if there is no R <= 2GM/c2 region within the vacuum solution then the central mass must reside outside that region, which prevents the "horizon" from being reached. This also for example allows super-massive objects to have strong intrinsic magnetic fields.

    (Also, even in standard black hole theory, that statement is not completely correct, because the amount of mass, charge and angular momentum inside the region affects what happens outside).

    Have you actually looked at the Antoci/Liebscher paper? This is not just a matter of using alternative coordinates; it is a matter of where the mass point is physically located.

    After studying the usual (Hilbert) interpretation of the radial coordinate, based on his explicit assumption that the central point could be assumed to be at R=0 "without loss of generality", and seeing pictures like Flamm's paraboloid, regions of space-time, Kruskal coordinates and so on it is admittedly difficult to reset one's thinking to understand the alternative view of Schwarzschild's original solution. (I find it also quite shocking even to consider the possibility that all the brilliant minds who have been trying to understand black hole solutions might have been wasting their time on a purely mathematical artefact).

    It is of course true that assuming the vacuum solution extends down to R=0 causes no mathematical loss of generality and has no effect whatsoever on the nature of the solution outside R=2GM. However, if, as Schwarzschild effectively assumed, the origin is actually at R=2GM, that means that it is impossible to reach that within the vacuum solution because of reaching the surface of the mass.
     
  8. Nov 19, 2008 #7
    The reason we have to ask questions of you is because what you are saying does not match the references. Furthermore, and a separate issue, is that the main reference you give is mistaking defects in coordinate choices for actual singularities... this is like claiming the singularities that show up in the metric for a Rindler observer (a uniformly accelerating observer in flat spacetime) are real singularities of the geometry. You claim in a previous thread you understand the Rindler case, but if you did, then we would be done here. The mistakes in conclusions are almost exactly analogous to the Rindler case.


    So let's try to agree on some common ground topics starting from your reference: arXiv:gr-qc/0406090

    Do you agree with the following? (if not, please point out where and why)

    1] Schwarzschild and Hilbert's solutions differ only by the choice of: integration constant rho, and valid range of the radial coordinate.

    2] As can be seen from equations A.10, A.11, A.12, the integration constant rho serves as a coordinate offset controlling which spherical shell is labelled "r=0". The choice of rho is merely selecting a coordinate system choice.

    3] Hilbert's solution has a single point singularity at the origin R=0. Schwarzchild's "origin" at r=0, is not a single point. At r=0, theta and phi can still be varied to specify physically separate points on the manifold. What Schwarzschild considers a singularity is actually a spherical shell.

    4] Therefore, Schwarzschild's choice of rho yields a coordinate choice in which the original coordinate transformation he started with is invalid. It claims x=0,y=0,z=0 is not a single line in spacetime, but a SHEET.

    5] The only choice of rho that fixes this problem is the choice that Hilbert's solution effectively used.


    -- now to your comment: "This situation could perhaps be resolved by experiment."

    6] The two solutions are equivalent outside the Schwarzschild radius. Neither solution allows information at or "below" the Schwarzschild radius to causally affect any event in spacetime outside the horizon. Therefore this situation can only be resolved by appealling to mathematical means.

    7] Consider a collapsing solid sphere constrained to always have uniform density. Schwarzschild's solution would require an abrupt change from a smooth spacetime to one containing a cylindrical hole of finite size. This transition would require the mass at the center to suddenly move ... faster than the speed of light ... from the center to the spherical shell at the event horizon. This violates causality.

    8] Hilbert's solution does not have this problem, and calculations using GR show the evolution of a star into a blackhole show a point singularity as in Hilbert's solution.



    The history behind this problem I found to be interesting as I was not aware of it. Thank you for sharing. But all the mathematical reasons above still leave me convinced Hilbert's solution is correct, and the conclusions of that paper are misleading/incorrect.
     
    Last edited: Nov 19, 2008
  9. Nov 19, 2008 #8

    Jonathan Scott

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    It's not a matter of whether Schwarzschild or Hilbert was "better" in some way, nor purely a matter of coordinates; the same issue can be illustrated using any spherical coordinate system. The point is merely that comparison of their different forms of the same solution illustrates that different assumptions have different physical consequences, and it is only by the use of such supplementary assumptions (which also affect the way in which the external and interior solutions join) that the current standard theory has been defined, so it is important to question the basis for these assumptions and compare alternatives.

    Again, I recommend reading the Antoci/Liebscher paper. It would at least be good to see some criticism of things that they are actually saying, rather than of things that they are NOT saying. They are also more articulate than I am on the subject and better qualified to defend it!
     
  10. Nov 19, 2008 #9

    Hurkyl

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    Wrong, and there is a patently obvious reason -- the Schwarzschild solution is exactly the same as the exterior region in any of the other solutions.

    As you say, this is a vacuum solution, which means there is no matter anywhere in space-time! So there is certainly not any matter located outside the event horizon. :tongue:

    (Hopefully someone more knowledgeable will correct me if the following paragraph is wrong)
    In fact, the notion of mass is notoriously subtle in general relativity. In general, it doesn't even make sense to select a region of space and ask "how much mass is located inside?" Black holes are, as I understand it, one of the few cases (the only case?) where such a statement can be made -- and even then, all you can say is that the region 'inside' the event horizon contains some mass. By the way, I put 'inside' in scare quotes because the event horizon doesn't actually enclose a region of space; the 'inside' goes off 'to infinity' just like the 'outside' does. To wit, if you make a substitution like [itex]R \mapsto 1/(R - 2GM/c^2)[/itex], you get another vacuum solution with the event horizon located at coordinate infinity.


    No it doesn't. All information about the mass, charge, and angular momentum can be read off from the external gravitational field.
     
  11. Nov 19, 2008 #10

    Jonathan Scott

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    True, assuming it went in there in the first place, which can only happen if that region exists.
     
  12. Nov 19, 2008 #11

    Hurkyl

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    Why would you think that?

    I know of nothing in the formulation of GR that prevent matter from 'falling off the edge of the universe' and thus simply vanishing from existence. And when there's a Schwarzschild black hole, that will happen in finite (proper) time -- it doesn't matter whether you put the edge at the event horizon, or somewhere 'inside'.

    (And for that matter, there is nothing that says matter cannot spontaneously appear from an edge, which would happen in a white hole)
     
    Last edited: Nov 19, 2008
  13. Nov 19, 2008 #12

    Jonathan Scott

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    Thank you very much for taking the time to look at the paper and comment on it.

    I understand your points 1] to 5], and I would at least also agree that Hilbert's view is self-consistent. However, I believe that points 3] to 5] are only valid IF you already make Hilbert's assumption. In Schwarzschild's initial coordinate system used for the solution calculation, r=0 really is a single point, at which he assumes the mass point to be located, so the vacuum solution only applies to r>0 and simply cannot reach r=0 for physical reasons. It is of course true that if you make the different assumption that the solution can be physically extended to r=0, then space at that "point" has a finite area and has all the well-known characteristics of the event horizon.

    One could also certainly say that Schwarzschild's choice was perhaps arbitrary and may have been incorrect. However, as far as I can see his original solution was self-consistent and, like Hilbert's, fully consistent with the Einstein Field Equations.

    In response to 6], note that although the vacuum solution is unaffected by the different assumptions, there is a big difference in that if Schwarzschild was right, a super-massive object will consist of ordinary matter instead of having an event horizon. Although this has very little effect on the shape of space outside the object (or outside the equivalent distance from the event horizon), it does mean for example that the object could have a huge intrinsic magnetic field and a well-defined surface (at a high redshift) even though according to black hole theory it could have neither.

    I don't understand your points 7] and 8], but as there is no event horizon, collapse or singularity in the original Schwarzschild solution I don't think these points are valid.
     
  14. Nov 19, 2008 #13

    Jonathan Scott

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    By "vacuum solution" I was referring to the Schwarzschild exterior solution for the vacuum outside a central mass. I believe this terminology is perfectly normal.

    I can't follow the rest of what you are saying, but I'm hoping you may be somewhat enlightened by something in my response to JustinLevy's comments. :smile:
     
  15. Nov 20, 2008 #14
    No, this is demonstrateably false using Schwarzschild's solution.

    Look at the solution given in equation A.14 in the paper you linked. Now calculate the spacetime interval of the path (r(s),theta(s),phi(s),t(s)) where
    s goes from 0 to 2pi
    r(s)=0
    theta(s) = 0
    phi(s) = s
    t(s)=0

    Lo and behold, the spacetime interval is non-zero. r=0 is demonstrateably NOT a single point in spacetime. Therefore his choice of rho makes his initial coordinate transformation INVALID. Chosing rho to make the coordinate transformation valid yields Hilbert's solution.

    Hilbert's solution is self-consistent. Schwarzschild's solution is not.

    Yes, r=0 is not vaccuum according to his assumption. Are you saying this allows you to claim the metric has a different value at r=0? The limit as r->0 for that path is completely defined, AND doesn't diverge, AND is the same as evaluating the metric for r=0. There is no leeway for you to claim this is not the spacetime path length for that path.


    No, again this seems to be a personal claim of yours.

    Using Schwarzschild's solution, do you agree a free falling test particle released at finite r, will reach r=0 in finite proper time?

    In Schwarzschild's assumption all the mass is at his r=0 coordinate. The vaccum solution r>0 is IDENTICAL to Hilberts, and since nothing at r<=0 can give information to the spacetime at r>0 (if special relativity is correct and physics has local poincare symmetry) ... the physics MUST be the same in the r>0 region.

    That is incorrect for it would require information at r=0 to escape to r>0 regions. This is not possible.

    Those points relied on understanding that r=0 is not actually a point in space (a line in spacetime), but actually a sheet in spacetime. A spherical ball of uniform gas of particles interacting only under gravity will collapse to Schwarzchild's r=0 in finite time... this transition would require superluminal speeds if you claim there is no spacetime at r<0, because the particles in the center would suddenly have to be at r=0 when the radius of the sphere of gas reaches the Schwarzschild radius.


    Another way of looking at this is consider a spherical shell of mass M, and proper length circumference 2 pi R. Solve for the metric outside of the shell. You will get Hilbert's solution, which is unique. Schwartschild's assumption is that R must be >= 2 G M / c^2 because otherwise the components of the metric diverge, and this divergence shows there is a physical singularity at those points. The over-arching view of the paper you linked is to try to convince us that indeed this divergence is physical, and thus cannot be empty space, but must be where the mass is located. They come up with many things to try to convince us, but they still require a leap of faith to get to their conclusion.

    Consider the Rindler coordinates. They describe a flat spacetime. There is a divergence in the components of the metric at one point. Similarly, a particle would need infinite proper-acceleration to stay stationary in these coordinates at the Rindler event horizon. This is a geometric quantity. But this doesn't let you take that last leap and pretend you can conclude there is therefore a physical singularity in spacetime at that location ... In this case we KNOW it to be false since spacetime is flat. You can consider Rindler coordinates a counter-example to their logic. Their conclusion is not justified.
     
    Last edited: Nov 20, 2008
  16. Nov 20, 2008 #15

    Jonathan Scott

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    Schwarzschild's "mass point" solution described in this paper is assumed from the start to contain an idealized mass point at the origin, for simplicity of calculation. Newtonian gravity is often described in the same way, given that we know that the gravitational effect outside a sphere in Newtonian theory is the same as if the mass were concentrated at its centre. That is obviously just a limit, and the idea of being able to actually reach that point with continuity from the exterior solution would require the additional unphysical assumption that the mass density was infinite.

    It is certainly true that as the mass radius decreases and its density increases, the proper surface area tends to a limit, but however small the physical radius of that mass might be in practice, there is no reason to suppose that the vacuum solution could reach the middle of it. Instead, when you consider the well-known basic properties of Newtonian masses or GR interior solutions it obviously behaves more like a delta function, in that the smaller and denser it gets, the more steeply the shape of space changes inside it, but it continues to have approximately flat space at its "infinitesimal" centre, and the centre remains a single point, not a surface.

    So my answer is yes, the metric is different at r=0, because the metric at the centre of a mass remains flat as it shrinks, and this limit applies rather than the metric at its surface, which never quite reaches the centre.

    In the general case, Schwarzschild does NOT of course assume that the mass is of infinite density located at the origin; this was merely for the purposes of the exterior solution. He followed up the exterior "mass point" solution paper with a corresponding interior solution expressed in terms of the same original coordinate system, which then joins on to the exterior solution at its surface. It is that solution which describes supermassive objects without giving rise to event horizons.
     
  17. Nov 20, 2008 #16

    George Jones

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    This line of thought is promoted by a small group of people that is largely ignored by the physics community. As, such it is clearly not mainstream, and the Physics Forums rules,

    https://www.physicsforums.com/showthread.php?t=5374,

    in part, say,

    Note that in the last quoted sentence the word "or" is used, not "and."

    This decision has been reached after discussion by several Mentors.

    Although Physics Forums Mentors are under no obligation to defend mainstream physics (as such action could result in unending bickering), in this particular case, I would like to put forward the mainstream's case, but, because of work and family commitments, I won't be able to do this until about two to three weeks from now.

    When I have enough time, I will open up this thread for a while, and anyone who has posted (or not) will be invited to participate.

    Until then, posts about this stuff will be treated in the way posts about non-mainstream stuff are usually treated.
     
    Last edited: Nov 20, 2008
  18. Jan 1, 2009 #17

    Jonathan Scott

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    It seems that there is overwhelming experimental evidence that at least some quasars must have significant intrinsic redshifts (as Arp has been suggesting for years). There is also some evidence from a gravitationally lensed quasar suggesting a significant intrinsic magnetic field. These observations are not compatible with the black hole model derived from the standard interpretation of GR.

    This seems to me to be evidence that the standard interpretation of GR is incorrect. However, it appears that the official line is currently that GR theory is so strong that it cannot be countered by experimental evidence, so there must be some additional effect which is not being taken into account (like dark matter and dark energy). In the case of quasars, it is a statistically unlikely evolution of characteristics with time which just happens to duplicate the weird radial distribution implied by their redshifts. GR just seems to be too tough for Occam's razor!

    I have seen some fairly convincing papers (by Salvatore Antoci et al) which suggest that even though Einstein's GR and the original Schwarzschild solution are correct, black holes only arise as a result of a mathematical change made by Hilbert. With Schwarzschild's original solution, quasars could be hyper-massive objects with well-defined surfaces and unlimited intrinsic redshift, probably spinning at relativistic speeds, at least when first formed. There would then be little need for any special time evolution of the quasar population, as most of them would fall into the same range of properties.

    I'm aware that a few quasars have been shown to be at or near their redshift distances by various means. However, according to Arp's plausible observations (ignoring his implausible theories), only the youngest and most active have significant intrinsic redshifts. These appear to decrease with time, and it appears that old quasars probably evolve into galaxies with redshifts approximately corresponding to distance as usual.
     
  19. Jan 1, 2009 #18

    Hurkyl

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    Re: Quasars and Cosmology

    Schwarzschild's solution has a hole too. The difference is that Schwarzschild's hole is formed by cutting out what we usually the event horizon, along with everything inside, and the 'boundary' of the hole has the geometry of a sphere of finite, nonzero area.

    (I put 'boundary' in quotes, because I'm really referring to the region of space-time surrounding the hole)

    (I said 'geometry' not 'coordinate representation'. The hole is not part of space-time, so a faux-coordinate chart for the region surrounding the hole can 'fill it in' with whatever contractible shape it wants to use, because the 'filled in' portion isn't meant to have any relationship with space-time)


    GR has no explicit prohobition against putting holes whever you want in space-time. e.g. observations are perfectly consistent with there being a hole in space-time where we think the moon is, but the hole simply emits light and stuff to give the appearance of the moon. But as you can imagine, physicists generally avoid introducing holes gratuitously, and only do it when it's really necessary.
     
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