(Alpha) General relativity is self-inconsistent v2

Discussion in 'Physics & Math' started by zanket, Jun 3, 2007.

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  1. zanket Human Valued Senior Member

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    (The “Alpha” in the title indicates that the Alpha rules apply to this thread.)

    In another thread I showed that general relativity (GR) is self-inconsistent. I think the proof there stands unrefuted. In this thread the lessons I learned from the previous discussion were applied to make a hopefully less contentious and more convincing proof. There’s also a new image below that shows the flaw of GR in a nutshell.

    First some supporting info (or skip to “Now the proof”):

    From pg. 1-19 of Exploring Black Holes by Taylor and Wheeler: “The spacetime arena for special relativity is the free-float (inertial) frame, one in which a free test particle at rest remains at rest and a free test particle in motion continues that motion unchanged. We call a region of spacetime flat if a free-float frame can be set up in it. ... In principle one can set up a latticework of synchronized clocks in a free-float frame. The position and time of any event is then taken to be the location of the nearest lattice clock and the time of the event recorded on that clock. The observer is the collection of all such recording clocks in a given reference frame.”

    From the glossary of Exploring Black Holes:

    • flat spacetime: Region of spacetime in which it is possible to set up a free-float (inertial) reference frame.

    • horizon: One-way surface surrounding a black hole, defined by the property that anything may pass inward through the horizon, but (in the non-quantum description) nothing, not even light, may pass outward.

    • inertial frame (free-float frame): Generally, a reference frame in which a free test particle initially at rest remains at rest. More technically, a reference frame with respect to which relative (tidal) accelerations of test particles can be neglected for the purposes of a given experiment.

    • tidal acceleration [(tidal force)]: Relative acceleration of two free test particles located in different parts of a reference frame.
    From the glossary of Black Holes & Time Warps by Thorne:

    • event: A point in spacetime; that is, a location in space at a specific moment of time. Alternatively, something that happens at a point in spacetime, for example, the explosion of a firecracker.

    • freely falling object: An object on which no forces act except gravity.

    • local inertial reference frame: A reference frame on which no forces except gravity act, that falls freely in response to gravity’s pull, and that is small enough for tidal gravitational accelerations to be negligible inside it.

    • tidal gravity [(tidal force)]: Gravitational accelerations that squeeze objects along some directions and stretch them along others. Tidal gravity produced by the Moon and Sun is responsible for the tides on the Earth’s oceans.

    • Spacetime curvature and tidal gravity [(tidal force)] are different names for the same thing.
    Note that:

    • The spacetime throughout an inertial frame is negligibly curved (all but flat).

    • The definition of an inertial frame allows them to be arbitrarily large (they need be only “small enough”).

    • More than one type of horizon is defined for a black hole. In texts about black holes, “horizon” usually refers to an absolute horizon, as it does here.
    From pg. 2-6 of Exploring Black Holes: “The constant, ever-present "force of gravity" that we experience on Earth is gone, eliminated as we step into a free-float [(inertial)] frame. What remains of "gravity"? Only curvature of spacetime remains. What is this curvature? Nothing but tidal acceleration. Curvature is tidal acceleration and tidal acceleration is curvature.”

    From pg. 98 of Black Holes & Time Warps (the italicized statement is Einstein’s): “In any small, freely falling reference frame anywhere in our real, gravity-endowed Universe, the laws of physics must be the same as they are in an inertial reference frame in an idealized, gravity-free universe. Einstein called this the principle of equivalence, because it asserts that small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity. This assertion, Einstein realized, had an enormously important consequence: It implied that, if we merely give the name "inertial reference frame" to every small, freely falling reference frame in our real, gravity-endowed Universe (for example, to a little laboratory that you carry as you fall over a cliff), then everything that special relativity says about inertial frames in an idealized universe without gravity will automatically also be true in our real Universe. Most importantly, the principle of relativity must be true: All small, inertial (freely falling) reference frames in our real, gravity-endowed Universe must be "created equal"; none can be preferred over any other in the eyes of the laws of physics.”

    GR predicts that an inertial frame can exist everywhere except at the center of a black hole. Here are confirmations:

    • From pg. 2-4 of Exploring Black Holes: “Our old, comfy, free-float (inertial) frame carries us unharmed to the center of a black hole. Well, unharmed almost to the center! ... No one can stop us from observing a black hole from an unpowered spaceship that drifts freely toward the black hole from a great distance, then plunges more and more rapidly toward the center. Over a short time the spaceship constitutes a "capsule of flat spacetime" hurtling through curved spacetime. It is a free-float frame like any other. Special relativity makes extensive use of such frames, and special relativity continues to describe Nature correctly for an astronaut in a local free-float frame, even as she falls through curved spacetime, through the horizon, and into a black hole.” From pg. 2-6: “Confronted by tidal accelerations, how can we define a free-float frame falling into a black hole? At the center of the black hole we cannot; general relativity predicts infinite tidal accelerations there. However, short of the center, [we limit] the space and the time—the region of spacetime!—in which experiments are conducted.” See also the section free-float frame on pg. 2-31.

    • From pg. 21 of Black Holes: A Traveler’s Guide by Pickover: “If you were approaching a 10 solar masses black hole with a radius of 30 kilometers, you would be killed long before you reached the horizon, at an altitude of 400 kilometers. However, you could reach the horizon of a 1,000 solar masses black hole, and even be able to explore the interior of a 10 million solar masses black hole. The tidal forces at the horizon of this gigantic black hole would be weaker than those produced by Earth, which are already impossible for us to feel.”

    • Another online reference: “In a supermassive black hole the tidal forces are weaker, and you could survive well inside the horizon of the black hole before being torn apart.”
    From the Black Holes FAQ: “You can think of the horizon as the place where the escape velocity equals the velocity of light. Outside of the horizon, the escape velocity is less than the speed of light, so if you fire your rockets hard enough, you can give yourself enough energy to get away. But if you find yourself inside the horizon, then no matter how powerful your rockets are, you can't escape. ... The horizon has some very strange geometrical properties. To an observer who is sitting still somewhere far away from the black hole, the horizon seems to be a nice, static, unmoving spherical surface. But once you get close to the horizon, you realize that it has a very large velocity. In fact, it is moving outward at the speed of light! That explains why it is easy to cross the horizon in the inward direction, but impossible to get back out. Since the horizon is moving out at the speed of light, in order to escape back across it, you would have to travel faster than light. You can't go faster than light, and so you can't escape from the black hole.”

    From pg. 2-22 of Exploring Black Holes: “... Einstein predicts that nothing, not even light, can be successfully launched outward from the horizon ... and that light launched outward EXACTLY at the horizon will never increase its radial position by so much as a millimeter.”

    A definition of escape velocity: “In physics, for a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion needs to have to move away indefinitely from the source of the field, as opposed to falling back or staying in an orbit within a bounded distance from the source.”

    Now the proof:

    GR predicts that an inertial frame can fall through the horizon of a black hole (hereafter just “horizon”). Let X be an inertial frame falling through a horizon. By the definition of an inertial frame, the tidal force in X can be neglected for the purposes of a given experiment. The tidal force is synonymous with spacetime curvature, so the spacetime in X is negligibly curved (all but flat). GR says that SR applies in X even as it falls through the horizon. GR predicts that the horizon is at r=2M, where r is the r-coordinate (also known as a reduced circumference) and M is the mass in geometric units. GR predicts that the escape velocity above the horizon is less than c, the speed of light. That is, GR predicts that any object above the horizon can escape to r=infinity.

    At the same time in X, let two free test particles exist in X along the same radius: particle A at r=2M+ε, where ε is arbitrarily small, and particle B at r=2M. Let A be escaping to r=infinity. Let A and B have the same velocity in X (the same velocity as measured in X). SR predicts that two free test particles having the same velocity in an inertial frame are at rest with respect to each other and therefore can share an inertial frame of their own, a frame with respect to which they are both at rest. But GR predicts that A and B cannot be at rest with respect to a common inertial frame; if B shared A’s frame, then B would be passing outward through the horizon to stay at rest with respect to A as A makes its way to r=infinity, in contradiction with the definition of a horizon. Moreover, GR predicts that in B’s frame (or the frame of any free test particle in X that is at or below the horizon), the horizon moves outward at c. Particle A moves outward from the horizon. Then GR predicts that A must recede at a relativistic velocity in B’s frame, whereas SR predicts that A is at rest in B’s frame. Since GR contradicts SR in X (an inertial frame in which GR says that SR applies), GR is proven to be self-inconsistent. The theory contradicts itself.

    [​IMG]

    An argument against this proof, which I call the “not exactly flat” argument, claims that the self-inconsistency is explained away by the fact that the spacetime in X is not perfectly flat (the tidal force in X is not nonexistent). The notion that SR applies in only perfectly flat spacetime (or, put differently, that SR applies in only a zero-sized frame) is refuted by Einstein himself in his above statement of the equivalence principle. If the proof showed just a small difference between what GR and SR predict for B’s frame, then the “not exactly flat” argument might have merit. But the proof shows that the difference between those predictions is the difference between a velocity of zero and a relativistic velocity, a huge difference that is not explained by a negligible tidal force. (Nor can reducing the tidal force in X by, say, choosing a larger black hole, reduce that difference in the slightest.) Those putting forth the “not exactly flat” argument should be prepared to show that the difference is duplicable in the inertial frame of the International Space Station (ISS). After all, the tidal force in X could be less by any degree than the tidal force in the ISS, for a sufficiently large black hole. And the principle of relativity tells us that no inertial frame in our real, gravity-endowed universe can be preferred over any other in the eyes of the laws of physics.

    A variant of the “not exactly flat” argument claims that the escape velocity must be c throughout X, therefore A cannot be escaping to r=infinity. But GR predicts that the escape velocity varies in any nonzero-sized frame. At r=2M+ε, where A initially is, GR predicts that the escape velocity is less than c.

    Another type of argument tries to use the self-inconsistency of GR against the proof of that. For example, the argument may claim that A and B cannot have the same velocity in X, because then B would be passing outward through the horizon, which GR does not allow. But X is an inertial frame in which GR says that SR applies, and nothing about SR suggests that A and B cannot have the same velocity in X. If A and B cannot have the same velocity in X, then GR is self-inconsistent.

    Another argument goes like this: “X is inertial for only an arbitrarily short time, after which A and B cannot be expected to stay at rest with respect to each other”. Not only is the proof designed to work in an arbitrarily short time (if B shared A’s frame for even a moment in that frame, then B would be passing outward through the horizon), but also a frame can remain inertial for an arbitrarily long time in principle. Consider the North Star, which has indicated north on Earth for at least a thousand years. Our Sun and the North Star have remained essentially at rest with respect to each other for at least a thousand years, even as the tidal force imparted on the Milky Way by the Andromeda galaxy has grown as those galaxies have fallen toward each other. The tidal force in X is negligible as it falls through the horizon, and GR predicts that the tidal force in X was even less when it was wholly above the horizon. Then there is no reason—in a self-consistent theory of gravity—why A and B could not remain essentially at rest with respect to each other for an arbitrarily long time as measured in their common frame, as both particles move outward toward r=infinity.

    Some claim that black holes have been observed, so the proof of GR’s self-inconsistency must be wrong. But there is no direct observational evidence of a black hole, and the indirect evidence relies on the validity of GR (as in “if GR is the correct theory of gravity, then the observation indicates a black hole”). The theory has been experimentally tested only in relatively extremely weak gravity far from a theorized horizon. In the strongest-gravity test of GR to date, the minimum r-coordinate is 210000M. GR approximates the correct theory of gravity in weak gravity but not in strong gravity.

    Later I’ll post a solution to GR’s self-inconsistency, a proposed modification to GR that represents a new theory of gravity and one that is fully experimentally confirmed.
     
    Last edited: Jun 8, 2007
  2. James R Just this guy, you know? Staff Member

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    The problem appears to be your assumption that although A and B are specificed to have the same velocity in X, you assume A can escape the black hole while B cannot.
     
  3. zanket Human Valued Senior Member

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    Particle A is given to be escaping, and GR allows that. GR does not allow particle B to escape. B is at the horizon, and GR does not allow anything to pass outward through the horizon; see the definition of a horizon in the OP. I don’t see a problem with the proof.
     
  4. James R Just this guy, you know? Staff Member

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    I questioned your assumption that A and B can have the same velocity in X and yet A can escape but B cannot.

    Can you show that?
     
  5. zanket Human Valued Senior Member

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    I cover that in the OP:

    A and B are just two free test particles in an inertial frame. So you're questioning the assumption that two free test particles can have the same velocity in an inertial frame. Of course they can, regardless how one of the particles is given to be moving.
     
  6. James R Just this guy, you know? Staff Member

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    No. A is restricted to having at least the escape velocity. It cannot have just any velocity.

    You need to show that B can have the same velocity.
     
  7. zanket Human Valued Senior Member

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    B can have the same velocity as A in X because SR allows a free test particle to have the same velocity as another free test particle in an inertial frame. GR says that SR applies in X.
     
  8. James R Just this guy, you know? Staff Member

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    Does it?
     
  9. zanket Human Valued Senior Member

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    Yes. Doesn’t a freely falling object in an inertial frame comprise particles moving all at the same velocity? And can’t the object have any velocity SR allows?
     
  10. James R Just this guy, you know? Staff Member

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    I think your horizon places restrictions on things in the particular frame X.

    So far, you have not shown that it doesn't.
     
  11. zanket Human Valued Senior Member

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    I don't have to show that it doesn't. To show that GR is self-inconsistent, I need only show (and do show) that GR contradicts SR in X. Look at the principle of relativity stated in the OP; there are no preferred inertial frames in our real, gravity-endowed universe. GR says that SR applies in X. Then GR cannot override SR's predictions in X without being self-inconsistent. For GR to be self-consistent, SR must work the same in X as it does in any other inertial frame.
     
  12. James R Just this guy, you know? Staff Member

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    I agree with that, zanket. Here's the crux:

    I agree with this.

    I don't think this is correct.

    Spacetime inside an event horizon is completely different to spacetime outside it, and you have not shown that B can have a radial outwards velocity when it is inside the horizon. In fact, from my understanding of GR, it cannot - at least not in a frame stationary with respect to the black hole itself, which is the frame you are trying to use for A.
     
  13. zanket Human Valued Senior Member

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    Einstein explicitly says in his statement of the equivalence principle in the OP that X is equivalent to any other inertial frame. There are no preferred inertial frames in our real, gravity-endowed universe, says the relativity principle. Einstein says that “the laws of physics must be the same” in any inertial frame. GR incorporates these principles and predicts that an inertial frame can exist everywhere except at the center of a black hole.

    Then I can treat X like any other inertial frame. Particle A is moving outward from the horizon; its frame is not stationary with respect to the black hole. If B shared A’s frame, then B would be passing outward through the horizon to stay at rest with respect to A as A makes its way to r=infinity. This can be confirmed from within X too: In X, if B had A’s velocity, then B would be passing outward through the horizon to stay at rest with respect to A as A makes its way to r=infinity.

    When you say things like “Spacetime inside an event horizon is completely different to spacetime outside it”, I say that you are falling into the logical trap of trying to use the self-inconsistency of GR against the proof of that. GR predicts that the spacetime throughout X is negligibly curved (all but flat), the same as in any other inertial frame. If GR predicts anything that contradicts SR in X, then GR is self-inconsistent. Particle A’s radial position is increasing, which GR allows. SR allows A to have the same velocity as B in X. Then GR must allow B to increase its radial position too, to stay at rest relative to A, even though B is at the horizon. If GR disallows that then it is self-inconsistent—it contradicts itself. And it does disallow that.

    If you focus just on GR’s predictions (as in “In fact, from my understanding of GR, it cannot”) then you may miss seeing the self-inconsistency I purport. To refute me, you need to show that SR’s and GR’s predictions for X do not conflict.
     
    Last edited: Jun 11, 2007
  14. James R Just this guy, you know? Staff Member

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    No. Einstein does not talk about frames straddling event horizons.

    I'm not so sure. I'm not convinced your frame X is a true inertial frame. You need to establish that it is.

    Is A's motion supposed to be radial? Is it's velocity constant? And if so, constant with respect to what?

    Have you considered how the metric changes inside the horizon, compared to outside?

    I'm not convinced. SR is a special case of GR. You need to establish that X can indeed be an SR inertial frame. Otherwise, GR overrides SR.
     
  15. zanket Human Valued Senior Member

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    He does, with the word in bold: “In any small, freely falling reference frame anywhere in our real, gravity-endowed Universe, the laws of physics must be the same as they are in an inertial reference frame in an idealized, gravity-free universe.”

    X is given to be an inertial frame. GR allows an inertial frame to fall through a horizon. In addition to Einstein’s statement above, I gave three references to support that in the OP; search for “Here are confirmations”. Taylor and Wheeler in particular could hardly be clearer about that. Plus I’ve already convinced you before.

    The OP says “Let A be escaping to r=infinity”. Its motion need not be radial for the proof to hold. SR predicts that the velocity of every free test particle in an inertial frame is constant with respect to the frame. In the OP Taylor and Wheeler say “The spacetime arena for special relativity is the free-float (inertial) frame, one in which a free test particle at rest remains at rest and a free test particle in motion continues that motion unchanged”.

    I did, but needed to do that only to the extent needed to show that GR overrides SR in X. The key difference between what GR predicts for inside vs. outside the horizon in X is incorporated into the proof.

    It’s already established in the OP. If you think there are inertial frames of both an SR and non-SR kind, the principle of relativity stated in the OP refutes you. There is only one kind: the SR kind.

    Here’s a preview of my next version of this proof:

    [​IMG]

    In the future I probably won’t be talking about velocities, multiple particles, or frames other than X—too messy.

    GR’s prediction of black holes and their central singularities has always been absurd. They never became less absurd; people just accepted the idea over time. Now it is apparent that in the correct theory of gravity the escape velocity must always be less than c.
     
    Last edited: Jun 12, 2007
  16. RJBeery Natural Philosopher Valued Senior Member

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    If Zanket is still here to defend this view, here are my initial thoughts:
    If I read this correctly, you are saying that A and B's distance from the center of the black hole is arbitrarily small by an amount of ε. Unless I'm confused they would therefore be experiencing a different amount of gravitational pull (albeit arbitrarily small) from the black hole. This means that they cannot both:
    1) share the same velocity and
    2) remain at rest with respect to each other

    Another problem I have with black holes is that, if you accept the math that shows a velocity of c is required to escape the horizon, then the same math shows that velocities > c are required to avoid the center once you have passed the horizon. To me, this suggests that we do not fully understand black holes yet so it may be unfair to use our limited knowledge of them to claim contradictions in existing theories. I personally like the idea that once critical mass has been reached, no matter is able to make it past the horizon. It just keeps traveling very slowly towards it "forever".
     
  17. zanket Human Valued Senior Member

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    Wow, a blast from the past. Yeah I'm back on sciforums after being absent for many moons while taking steps to avoid being Bushwacked.

    2M is the radius where the black hole’s horizon is, so that's not what I'm saying there, unless you meant to say the difference between A and B's distance from the center of the black hole is arbitrarily small.

    Your points #1 and #2 apply to any inertial frame in our real, gravity-endowed universe. Even a speck of dust in a galaxy far, far away from us will make #1 and #2 true in an inertial frame in our Solar System. Two free test particles in an inertial frame can still have the same velocity with respect to that frame to some number of significant digits. Yet in X, which is an inertial frame falling through the horizon of the black hole, particles A and B cannot have the same velocity to any number of significant digits--those particles are moving in opposite directions according to GR. In X, a free test particle located in a certain part of the frame (namely, below the horizon) cannot possibly be at rest relative to a particular other possible free test particle in the frame (namely, one escaping to infinity), to any number of significant digits; this law of physics does not apply in an inertial frame that is wholly above a horizon, which contradicts GR's own equivalence principle that says that the laws of physics are the same in all inertial frames.

    Who says we have limited knowledge? GR makes clear predictions, which are well publicized. If a book on GR shows a contradiction of the theory in plain sight, however subtly, that’s a problem with the book at least. I have seen dozens of books & other publications that contain the contradiction noted above, so it’s safe to say this is a problem with the theory and not just with dozens of very smart authors. And it is certainly fair to say that there is a contradiction in a theory when the theory makes contradictory predictions.

    GR makes clear (albeit sometimes contradictory) predictions. On this point GR predicts that an object can fall through the horizon as the object itself observes, but no object above the horizon can observe that (if only because nothing, not even light, escapes the black hole). An observer hovering above the black hole would observe the falling object to ever more closely & slowly approach, but never reach, the horizon.

    A common objection to my claim of GR’s self-inconsistency is that GR has been well mathematically vetted, so that any self-contradiction would have been discovered long ago. This objection fails because the equivalence principle is an idea expressed in a non-mathematical language (e.g. English). It is the non-mathematical expression that is the core of the theory. Any mathematical expression of it is a translation.
     
  18. AlphaNumeric Fully ionized Moderator

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    The equivalence principle in what you've read might be worded in English, but that doesn't mean it always is or that the immediate consequences of it are not expressed algebraicly. Furthermore, if this was something which lead through to an inconsistency, you should be able to find that inconsistency algebraicly. If algebraic expression X relies on inconsistent wordy expression Y, then X is inconsistent. So where's any algebra?

    This thread is more than a year old and you previously claimed to have a contradiction discussed before. Why are neither now published? It's easy to come onto a forum, even one with the number of actual physicist students as this one, and pick a topic of discussion which noone is particularly well versed in and proclaim you've got some new/critical/irrefutable result. Frames passing through the horizon are precisely the kind of thing a lot of people are a bit fuzzy on, I know I am and even when I was doing relativity a lot, I was still a bit off on it. But the times I asked people much more in the know than me they always instantly had an answer and an explaination.

    Your example involves a particle at r=2m and another at r=2m+e. Immediately alarm bells go off in my head because while you can construct inertial frames iin this situation, you can only do them locally, not of arbitrary size and you'll find that in any region larger than a point you get differences from SR, since space-time curvature is non-zero. You seem to be trying to construct arbitrarily large frames, which are not going to be true for anything more than an instant (so motion of the particles cannot be considered).

    You have remained firmly in the qualitative, not the quantitative, which allows you to nicely avoid doing actual details. Can you not consider a Schwarzchild metric system and then consider test particle motion within it in a way which shows this inconsistency? Do you even know how?
     
  19. RJBeery Natural Philosopher Valued Senior Member

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    Wait a minute Zanket, I don't understand but I want to. Please explain how your black hole scenario is different from the following:

    Consider 2 rockets, A and B, with rocket A an arbitrarily small amount further from the center of the Earth than B. Each rocket is accelerating at 9.0 m/s/s above the Earth's surface at precisely the altitude such that rocket A "escapes to infinity" while B is unable to overcome Earth's gravity, and subsequently falls.

    If this is equivalent to your black hole scenario then I would say that there is nothing contradictory about it. My gut tells me that you aren't applying the definition of inertial frame correctly because an inertial frame should not have a non-uniform field of gravity applied to it. It should be obvious that applying different forces to two objects which are initially at rest wrt each other will cause them not to remain so.
     
  20. Guest254 Valued Senior Member

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    Perhaps it's been too long since I did any real-world GR, but it seems to me the obvious issue is your assertion that the particles should remain at the same velocity within your chosen frame. They are feeling different tidal forces - no matter how small their radial seperation, i.e. how ever small you choose your \(\epsilon\), there will always be a difference.

    With respect, I very much doubt you're in a position to make such bold statements. Sure, if you're Robert Geroch then I don't mind you commenting on the the human race's understanding of black holes, but otherwise, it seems like a throw away comment made to give the impression of wisom. I find it far more likely you don't understand much about GR, and therein lies the problem. This is certainly not meant as an insult - GR is horrendously complicated, even for maths and physics graduates.
     
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