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For anyone who does not beleive in black holes, we what happens to neutron stars then

  1. May 25, 2010 #1
    If black holes are not real, just wondering? By the way they may or may not exist. But since some people don't beleive they exist and some do, then if you DO NOT , how do you explain what happens to neutron stars that are 20 -25 the size of the sun?

    typo in the title, it should read: " what happens to black holes" the typo is where it says "we", how that got there is not clear but , anyways.
    Last edited: May 26, 2010
  2. jcsd
  3. May 25, 2010 #2
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    the evidence for BH are clear, one should believe in them.
  4. May 25, 2010 #3


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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    There are no neutron stars that are 20-25 times the size of the sun.
  5. May 25, 2010 #4
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Please, do tell. 1) What evidence distinguishes a black hole from collapsing matter that will become a blackhole in infinite time where we may either measure time as cosmological time or the time on the clocks of us observerse here on Earth? I'm getting very upset with those who evidently pick up their physics leasons from the producers of the science channel without further ado. 2) What electromagnetic evidence or otherwise can distinguish a massive object, where the bulk of the matter occupies a thin layer on the causual side of critical radius, per the overlayed Minkowskian coordinate system of the Earthly observer, from a black hole?
  6. May 25, 2010 #5
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    pick up any astrophysics text book dude
  7. May 25, 2010 #6


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    Nothing worth nitpicking a definition over.

    We've had the discussion before: that black holes exist is not in doubt. What their exact nature is is still being researched.....kinda like with every other phenomena in science.
  8. May 25, 2010 #7
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    You can investigate this using well defined alternative model. See e.g. http://arxiv.org/abs/gr-qc/0609024" [Broken].
    Last edited by a moderator: May 4, 2017
  9. May 25, 2010 #8

    Jonathan Scott

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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    I think it is quite likely that black holes arise only as the result of a plausible but incorrect mathematical assumption on the part of Hilbert, who modified Schwarzschild's original assumption about the boundary conditions on the radial coordinate. That is, Einstein's field equations are correct but the assumption that the Schwarzschild radial coordinate can actually pass r = 2Gm/c2 is physically incorrect, as this is the location of the (admittedly extremely hypothetical and unphysical) "point mass" in Schwarzschild's original solution.

    A paper supporting this point of view was written by Marcel Brillouin in 1923, and this idea was resurrected by Leonard S Abrams and by Salvatore Antoci and others recently (including Stephen J Crothers, who has an unfortunate tendency to overstate the case and find fault in everything else as well). You can easily find them on Google if you are interested. English translations of some of the key historical papers have been made available on the ArXiv to help clarify the situation.

    However, for reasons which puzzle me greatly, this is a topic which supporters of standard GR (as reformulated by Hilbert) consider heretical - a most unscientific concept - and most arguments on the subject seem to consist of being rude about the opposition. There may be a sound scientific argument for the standard GR position, and I would welcome such clarification, and I have for example seen papers claiming to refute these ideas specifically, including one by Malcolm MacCallum at arXiv:gr-qc/0608033. However, in each case it appears to me that the argument has missed a key point and has in some way implicitly assumed a result which depends on the point it is trying to prove. As far as I can see, GR does not specify the boundary conditions and hence it is possible that neither position can be proven, but Schwarzschild's original position seems more physically plausible, even though Hilbert's is more mathematically compelling.

    If black holes do not occur for this reason, then extremely dense masses would still have exactly the same gravitational field outside their surface as predicted by standard GR at present, but the effective location of the center of the mass as mapped in terms of the exterior coordinate system would be at a point whose Schwarzschild radial coordinate effectively approaches the Schwarzschild radius as the density increases towards infinity. (This model works anyway, as can be confirmed by imagining the mass as being a thin hollow sphere and poking a radial ruler in through a hole). Note also that as the mass would still be made of normal matter, it could have an huge intrinsic magnetic field, and there is some evidence supporting the idea that some quasars have such intrinsic fields (although there are also theories of how a classical black hole could have the "frozen" remains of such a field anyway).

    I also suspect that GR may not be totally accurate anyway in many cases, such as on the galactic and cosmological scale, and indeed down at the quantum scale, but from the experimental evidence it does seem extremely accurate for the simple case of a spherically symmetrical static central mass, as in this case, even if this has not yet been proven to be totally accurate for very strong fields. There is of course also the possibility that Einstein's Field Equations are not quite right in this case anyway. However, even if they are exactly right (at least locally), I'm still not convinced that they lead to black holes, although I hope I'm still open to being convinced if the right evidence comes along.
  10. May 25, 2010 #9
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Please provide a reference to substantiate this claim. The arguments I have been presented, in this forum, have not been well informed. Using the word 'exist', without qualification lends further doubt; we are comparing time intervals, one to another, where the relation between the two is not well behaved. 'Existence' should be used in a relative way, if at all.
  11. May 25, 2010 #10
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    I registered specifically for this thread. Would I be correct in assuming that you believe a black hole never finishes collapsing? I can see how you would view a singularity in this fashion, but to me a black hole is defined by its massive nature and event horizon. Do you believe that these massive bodies could theoretically be directly observed and would not be within an event horizon? I don't understand how that squares with observations of galactic nuclei, or even stellar mass black holes.

    I don't believe that many people believe that there is a point of infinite density and zero volume at the heart of a black hole, but that doesn't mean they are behind an event horizon, making that speculation academic and semantic in the absence of quantum gravity.
  12. May 25, 2010 #11


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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Your request is not applicable. There isn't anything to substantiate any more than there is to prove your name is "Phrak". And besides which, we've already had plenty of discussions on the issue of how well black holes are understood (which is a completely separate issue from whether they exist). There is no need to rehash. The problem, as before, is that you're mixing theory with observation. A "black hole" is a colloquial name given to an observed object (like "Phrak"). The name itself includes no claims about the exact nature of that object. A theory exists to describe the object, but whether the theory is correct or not has no bearing on whether the object exists. It exists.

    Consider the phenomena called "Venus". Does it exist? What about when the ancients viewed it and gave it the name "Venus"? They "theorized" about what it was and were waaaaay off, but today we recognize that they were looking at the same "Venus" we are, they just didn't understand what it was.

    Now with black holes, I suspect of we ranked the properties that define a "black hole" by importance and certainty, we'd probably end up 90+% certain that we understand what a "black hole" is. But even if we ended up being 99% wrong, that's only wrong in the understanding, not in the fact that the object we are trying to understand exists.
    The word "exist" can only be used without qualification. It's definition requires it. It is completely binary: something either exists or it doesn't.
  13. May 25, 2010 #12
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    You keep bringing this up. I am willing to try to help, but as many people have tried before, it is hard to understand what you are not accepting. Please answer these questions to help build a foundation of what is agreed upon personally, and where you personally start to disagree.

    1) True/False? General relativity is a deterministic theory. Given appropriate initial conditions, there is only ONE solution.

    2) Einstein proved that in GR a mass, beyond a certain radius, cannot support itself (ie. no static solution is possible). And furthermore Penrose proved that generically a singularity must form.

    3) Hilbert's solution is a valid solution to the GR equations, if one considers a line singlarity/boundary of spacetime in the center of the event horizon (and this singularity is of zero volume in spacetime). Your suggested solution is a valid static solution only if one considers spacetime to have a boundary (not just a singularity) at the Schwarzschild radius (and this boundary in spacetime is cylindrical with finite area).

    4) Consider a collapsing star. Take a spatial slice of spacetime, and initially there is no singularity. In the simplest cases (full spherical symmetry, dust solution, etc), the unique end solution is what you call "Hilbert's" solution.

    5) Are you claiming, in more general collapsing cases, a non-zero area boundary of spacetime instantly appears? Or are you saying a zero-area boundary (point) shows up in a slice of spacetime, then grows to the Schwarzschild radius? What dynamical equations are you using to claim the movement/growth of a spacetime boundary?

    This is only a guess, but I think the problem you are having is due to the fact that the usual spherical blackhole solution is often (and was historically) derived as a vacuum solution, with the "source" term actually just handled as a boundary condition or interpreting the integration constants in some Newtonian limit. This seems to be leading you to thinking the "source" is a merely a mathematical choice we can argue about. When in reality, GR is a deterministic theory, and while difficult to solve analytically, now-a-days we can solve numerically. Starting from a star and letting it collapse does not yield the cylindrical boundary in spacetime you are claiming. These are not solved as vacuum solutions, as the matter source are considered directly. There is no mathematical "choice" to the solution here. Does this realization help? If not, can you explain better your personal disagreements with these calculations (please don't link me a paper and have me try to guess your understanding)?
  14. May 26, 2010 #13


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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    "a configuration that is a black hole for (almost) all practical purposes, but might be missing the one key ingredient of having a horizon."

    "If the central object is not a black hole, but rather a boson star or something similar, then the inspiraling object will continue to emit long after it would shut off in Kerr (Kesden et al. 2005). This would be a clean and blindingly simple falsification of the central black hole paradigm."
  15. May 26, 2010 #14

    Jonathan Scott

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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    I'm not sure how far this can be taken in this forum, but I'll try to summarize how the alternative viewpoint can be understood in terms of standard theory.

    Although the Schwarzschild external and internal radial coordinates share a common definition, they do not behave in the same way, and the rate of change of this coordinate with respect to proper radius has a discontinuity at the surface of the spherical object, so it is not at all like a physical radius, even with appropriate scale factor.

    Schwarzschild originally assumed a hypothetical point mass at the origin of his solution, defining a conventional radial coordinate in terms of x, y and z. When his original coordinate system is converted to the simplified "Schwarzschild radial coordinate" r (as named by Hilbert), the origin of his original coordinate system corresponds to r=2Gm/c2. If you use the alternative way of defining this coordinate in terms of proper areas of spheres and extend this to the interior, you find that this radial coordinate value refers to the outside surface of the point (which has finite area despite being a point) but the middle of the point has Schwarzschild radial coordinate 0. This doesn't make much sense, so instead of considering something which effectively involves a factor of 0/0, we need to think about limits.

    If we assume the object is slightly larger than a point in Schwarzschild's original model (so its surface is at r > 2Gm/c2), we can see that the radial coordinate decreases through 2Gm/c2 to 0 within the object, and we conclude that the middle of the object is located at r=0, which is therefore that much "further in" than the original, suggesting that Schwarzschild's original model need to be stretched to insert an inner sphere. Hilbert therefore changed the exterior solution model to assume that the origin of that model was at r=0, implying that there is some sort of physical space between r=0 and r=2Gm/c2 which could in theory appear in the exterior solution. Mathematically this also seems more general, because there is a singularity at r=0 but only a coordinate singularity at r=2Gm/c2.

    However, if you look at the scale of the radial coordinate just outside the surface of the object compared with proper radial displacement, and extrapolate that same scale into the middle of the object, you find that in terms of the external radial coordinate, the location of the middle approaches r=2Gm/c2 as the mass shrinks. To avoid the complexities of internal solutions, you can simplify the model to a thin hollow rigid spherical shell whose external field is the same as that of the original mass. Inside that shell we have flat Minkowski space whose proper radius decreases towards the Schwarzschild radius as the external Schwarzschild radial coordinate shrinks towards the same value. Oddly, it seems that if the shell tries to collapse, the spatial coordinates shrink in such a way that its internal radius merely gets a little closer to 2Gm/c2.

    This means that Schwarzschild's original position that the physical origin of the exterior solution is the point where the exterior Schwarzschild coordinate r reaches 2Gm/c2 seems to work perfectly well, and the interior solution, shrunk down very small relative to the exterior coordinate system, still fits between the outside and the inside of the mass without needing to move its origin.

    In answer to your numbered questions:

    1) Yes, I agree GR is deterministic (that seems obvious)

    2) The proof that collapse is unavoidable depends on Hilbert's change of origin, which Einstein appears to have accepted (presumably because Hilbert was clearly the greater mathematician, even though the difference is physics rather than mathematics).

    3) There is no boundary or discontinuity involved in space-time, although the Schwarzschild radial coordinate behaves abruptly in a different way once it reaches a surface.

    4) The standard interior solution is defined by continuity with Hilbert's version of Schwarzschild's exterior solution, and that is what allows collapse.

    5) There is no boundary or discontinuity. The explanation is given above.
  16. May 26, 2010 #15
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Hmm... #1 was a gimme, yet unfortunately that is the only one you agreed with.

    You seem to be misunderstanding something involving the coordinates. So let me try to reformulate some of what I already said in a coordinate independent manner.

    1a) Regardless if you choose to use a coordinate change r' = r + constant, the proper distances as defined by the metric are unchanged. Do you agree with this?

    No, the proof does not depend on your choice of coordinates. Let's rewrite in non-coordinate terms to make this clear. Take a finite sized static spherically symmetric mass in GR. There is a non-zero "proper surface area" limit this object can have. This size limit is LARGER than the "event horizon" size limit. So well before reaching the "event horizon" size, the fate of such an object is sealed ... it must continue collapsing down to zero size.

    So your comments on a static solution with the surface right at or above the event horizon limit are incorrect. Such a static solution in GR is not possible ... unless you artificially impose a boundary of spacetime of non-zero proper area. In which case the mass would run into this boundary in finite proper time (it still would not hover above the boundary).

    Regardless of confusion on coordinate choices for describing a black hole, I really really don't understand how you can wave away solutions that actually start with a non-rotating spherical mass and follow the collapse in. You already agreed the solution is deterministic. As I mentioned, the solution doesn't result in a singularity with a non-zero proper area. And the solutions are not coordinate system dependent (I hope we agree on that).

    You continue to refer to the "Schwarzschild r=0" as a point. This is incorrect. Both mathematically, and physically. If you want to try to claim there is no spacetime existing in "Schwarzschild r<0", fine, and we can discuss that. But there can be NO debate that your "Schwarzschild r=0" is not a point in spacetime.

    5) GR gives Ricci tensor = 0 in vacuum. This is a differential equation for the metric, and thus the solution will have integration constants. The difference between r' = r + constant choice in coordinates has its effect in the metric as a change in choice of the integration constants. To get the final solution we need to fix the integration constant by using the actual source terms (ie. consider the non-vacuum portion of spacetime) or appropriate boundary conditions.

    Since, what you call "Hilbert's solution" is:
    [tex]c^2 {d \tau}^{2} =
    \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \frac{dr^2}{\displaystyle{1-\frac{r_s}{r}}} - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]
    With the constant [tex]r_s = 2 G M / c^2[/tex]. And by Birkhoff's theorem, this is the only solution outside a spherically symmetric mass. So the only thing we can do (and still have a valid GR solution in vacuum outside a spherically symmetric mass) is a change in coordinates.

    Then with a change of variables [tex]\tilde{r} = r - r_s[/tex] in order to have r=0 be on the event horizon, [tex]\tilde{r}=0[/tex] is NOT a point.

    6) Schwarzschild started with x,y,z,t. The origin x=0,y=0,z=0,t=0 is a single point in spacetime. However he chose an integration constant which made r=0, theta=0 have a non-zero proper length from r=0, theta=pi. These are therefore clearly different spacetime points both mathematically (the two labels refer to distinct spacetime points) and physically (they are causally distinct).

    So if you wish to use this coordinate system, you either need to impose a spacetime boundary at r=0, and admit r=0 is not a point. Or consider values r<0. You seem to instead be trying to choose r=0 is a point and there is no boundary ... this is wrong on multiple levels, and is not an available choice.
    Last edited: May 26, 2010
  17. May 26, 2010 #16

    Jonathan Scott

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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    I totally agree that Schwarzschild "point mass" solution is not realistic (and even less so than in Newtonian gravity), and that a point cannot meaningfully have a finite proper area.

    However, Schwarzschild's "point mass" makes perfect sense as an idealized limit case of a mass of finite size being shrunk down towards a point without actually reaching it.

    I also agree that mathematically, the exterior solution doesn't have a boundary at the Schwarzschild radius.

    However, the exterior solution stops at the surface of the mass, and with Schwarzschild's original assumption combined with a not-quite-point mass, that surface remains outside the point as the mass shrinks. The extrapolated limit value of the exterior Schwarzschild radial coordinate at the middle of the mass (which can be formalized by poking a ruler calibrated in those units through the mass and assuming it to be hollow) tends to the Schwarzschild radius as the mass shrinks, suggesting that the middle can be considered to be physically located at a fixed point, at least as far as the exterior coordinate is concerned (which is of course the one used in the exterior solution).

    There are of course other odd things that happen; as the mass shrinks down, it tends towards a limit proper size with a limit proper area. You'd think that for example if it were hollow you could lower the shell inward, but since that makes everything inside the shell shrink in approximately the same proportion, that doesn't work in the expected way. I don't claim to understand all the implications of this viewpoint, but I'll maintain that they are less weird than black holes.

    I think the main source of confusion is the tendency to think of the Schwarzschild radial coordinate as being like a physical radius despite the well-known fact that it changes its nature very abruptly at the boundary between the exterior and interior solutions.
  18. May 27, 2010 #17


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    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    The issue of whether a 'true' singularity can form in a collapse event may be debatable. Formation of an event horizon is not debatable.
  19. May 27, 2010 #18
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Schwartzshild's solution is valid for any spherically symmetric distribution of matter in the region outside of that mass. It's similar to the fact that the gravitational field of a spherically symmetric mass outside the mass looks exactly like that of a point mass in Newton's gravity.
  20. May 27, 2010 #19
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    As already explained, there is no such asymptotic limit case. Before a mass collapses to the "event horizon size", it is already past the static limit. Beyond this, the spherical mass CANNOT prevent further collapse and will collapse to a point. Are you denying this?

    Let's make this more clear:
    1) Do you agree there is a non-zero surface area minimum size limit for a static spherical mass M?

    2) Do you agree with Birkhoff's theorem, that the unique vacuum solution outside a spherical mass distribution is (written here in coordinate form):
    This is the only solution. If you want to write it in coordinate form, the only other solutions are mere coordinate transformations from this.

    3) Do you agree if we look at the Schwarzschild metric in coordinate form for a freefalling observer, it takes a finite proper time to fall from the photon sphere (a location outside the event horizon) to the singularity?

    4) Do you agree that a spherical mass collapsing beyond the minimum static limit size, the entire mass will collapse to a singularity in finite proper time? In other words, not only is there a minimum static size limit, but the mass can not "asymptotically approach" a different size (in effect giving a "pseudo"-static size)?

    Please stop bringing up the interior solution. The interior solution is moot to this discussion. The interior solution CANNOT prevent the collapse of the mass to a singularity beyond the static limit.

    The other question which you continue to avoid is:

    Since you already agreed GR is deterministic, and numerical simulations of star collapse show the result is what you call "Hilbert's" solution. Are you claiming the numerical simulations are wrong?

    There are many people that don't believe a singularity will form in reality. But there can be NO DEBATE about the predictions of a mathematical theory. GR is self-consistent. It predicts a collapsing star, if going beyond the static limit, will collapse beyond an event horizon.

    If your intuition conflicts strongly with a blackhole, or a singularity: fine. You are in good company historically, and even in current time on the later point. But please take the time to understand that there can be no debate about what General Relativity says about them in this very simplified and well understood context of a spherical mass collapsing.

    This again is focusing on moot aspects of the interior solution. Call that r whatever you want. In coordinate system independent terms, the proper area of the spherical mass will collapse to zero.

    What you are bringing up is unrelated enough that I'm getting confused. Maybe I'm horribly misunderstanding your point and we are talking past each other.

    5] Are you claiming that:
    • A) a spherical mass collapsing, can never form an event horizon?
    • B) a spherical mass collapsing beyond the static limit will not go to a singularity in finite proper time, but asymptotically approach a finite proper area size (below the event horizon)?
    • C) a spherical mass collapsing beyond the static limit will go to a zero size proper area point, but (in some unclear sense) the "actual" radius will be non-zero?
    • D) Something else entirely?

    While searching around for the simplest gravitational collapse model I could find, I got this from wikipedia:
    Robert Oppenheimer and Hartland Snyder considered a model of a dust cloud, where the dust particles of the cloud were moving radially, towards a single point, and showed that the dust particles could reach the singularity in finite proper time. After passing the limit, Oppenheimer and Snyder noted that light cones were directed inwards, and that no signal could escape outside.
    J.R. Oppenheimer, H. Snyder, "On Continued Gravitational Contraction", Physical Review 56 (1939) p455.

    Can we at least agree in some cases gravitational collapse leads to an event horizon in GR? And that in this particular case, it leads to a zero surface area singularity?
  21. May 27, 2010 #20
    Re: For anyone who does not beleive in black holes, we what happens to neutron stars

    Those who have us believe there have been black holes observed located in galactic centers have a few things to explain.

    1) What evidence distinguishes a black hole from an incipient black hole?

    2) A 25 kilometer radius black hole takes forever to form in cosmological time. Explain how a 50 kilometer objects fits into a pre-inflation universe several order of magnitude smaller?

    3) The casual, observable influence from an event horizon would require forever to reach us. Explain this one.
    Last edited: May 27, 2010
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