(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 “

From pg. 1-19 of

From the glossary of

From pg. 98 of

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

From pg. 2-22 of

A definition of

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.

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.

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.

*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.

- 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.

*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.”

*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**.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: