General relativity is self-inconsistent v3

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

Elsewhere I showed that general relativity (GR) is self-inconsistent. The proof there stands unrefuted. In this thread the lessons I learned from previous discussions were applied to hopefully improve the proof. For supporting info see the other thread.

The proof

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Let a freely falling rod span the horizon of a black hole so large that the tidal force on the rod is negligible. Let the part of the rod that is above the horizon be escaping to r=infinity. By the definition of a horizon, nothing can pass outward through it. But GR must allow the rod to pass outward through the horizon, since the theory features no force other than the tidal force that can break it. By predicting that the rod both can and cannot pass outward through the horizon, GR contradicts itself.​

 

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Zanket, I have read your gedankins before and you seem to make the same errors in all them. First you state the rod is 'freely falling', then you state the outer end of the rod is escaping to infinity. The outer end of the rod must accelerate to escape the black hole, it is not an inertial frame in free fall, thus SR-type inertial frames do not apply. SR is an incomplete theory that gives incorrect results when trying to apply its inertial frames where gravity is a factor. Second you state that the black hole is so large that tidal forces are 'negligible'. At what location are those tidal forces negligible? They certainly are not at the event horizon. Black hole accretion disks are in a spiral formation due to their rotation, not a sphere. The event horizon is the inner region of the accretion disk where tidal forces are great, even in supermassive black holes. It is thought possible for a particle, or rod, to pass through the center of a supermassize Kerr black hole along the axis of rotation, thereby bypassing the accretion disk and the tidal forces at the event horizon. Kerr black holes do not have a central 'infinity', but instead have a thin ring-shaped infinity immediately below the circular accretion disk's event horizon. Kerr black holes, or charged Kerr-Newman black holes, are the types thought to actually ocurr in the universe, not Schwarzchild black holes. Supermassive, rotating Kerr black holes are the model with areas of 'negilgible'
tidal effects along the central axis of rotation, not Schwarzchild black holes.

 

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No, the escape velocity above the horizon is less than c, and escape velocity applies to freely falling objects. Look at the definition of escape velocity in the supporting info. Noninertial acceleration is not required to escape to r=infinity from anywhere above the horizon.

SR is irrelevant here. The OP doesn’t mention it.

The supporting info refutes this; search for “Here are confirmations”. The tidal forces at the horizon can be negligible.

Accretion disks are irrelevant here. The OP is about theory, not reality.

That doesn’t matter. The OP is about theory, not reality.

The tidal forces at the horizon of a Schwarzschild black hole can be negligible. See the supporting info.

 

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zanket,

In your gedanken, the rod was gravitationally accelerating toward the event horizon of the black hole. I know what escape velocity is. The rod must change the direction of motion (its vector) before it can escape the black hole. That demands an acceleration to do so.

Yes, your OP mentions SR indirectly by your link and reference to the earlier thread. You used the definition of a SR flat spacetime inertial frame to set up your gedanken. Quote:

You then go on to state that tidal forces can be negligible in such a frame:

These are the errors that I was posting about. A free-falling frame in a black hole's gravitational field is NOT an SR-type inertial frame. First, if you re-read the definition of an SR inertial frame you posted, you will see this quote: and a free test particle in motion continues that motion unchanged. You seem to interpret that as meaning a particle in gravitational free-fall is an SR inertial frame. Not so. The test particle is accelerating as it moves nearer the event horizon, not 'continuing that motion unchanged'. The particle increases its velocity relative to the event horizon.
Tidal forces are huge factor near the event horizon of any black hole, regardless of the black hole's size. Where do you think the oft-used description of an object undergoing 'spaghettification' comes from? It's from the immense tidal forces near a black hole's event horizon. Tidal forces increase at an inverse cube ratio toward the gravitating body, unlike gravity's inverse square. Gravity measures the same at the event horizon of any black hole, regardless of the comparative sizes of the black holes. Likewise, tidal forces measure the same at the event horizon of any black hole, regardless of size. This is precisely the location where the escape velocity of light is exceeded, regardless of the black hole's mass. It is the supermassive, rotating Kerr black holes that have an area along the rotational axis where it is possible to pass through the black hole. Its like the center of the Earth where you feel no gravitational forces. In the Kerr black hole, all the mass is concentrated in a ring near the event horizon. The same is not true for the Schwarzchild model of a black hole which predicts a central infinity.

 

The OP does not say that the rod was gravitationally accelerating toward the horizon. The only information given about the rod’s motion is that the part of it that is above the horizon is escaping to r=infinity.

Not all of the supporting info in the other thread applies to this thread. The thought experiment in this thread does not rely on SR.

From the Wikipedia entry for “'spaghettification”: “The point at which these tidal forces become fatal depends on the size of the black hole. For a very large black hole, such as those found at the center of galaxies, this point will lie well inside the event horizon, so an astronaut may cross the event horizon without noticing any squashing and pulling whatsoever (although it's only a matter of time, because once inside an event horizon, it is impossible to avoid falling towards the center).”

From the supporting info, another 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.”

The tidal forces at the horizon of a Schwarzschild black hole can be negligible, for a sufficiently large black hole. You could have crossed a horizon while reading this sentence.

 

The part above the horizon is not accelerating. The whole rod is freely falling, as given. "Freely falling" is synonymous with "in free fall"; see the definition of "freely falling object" in the supporting info.

 

Well that was easy. Where do I collect my Nobel?

Seriously, no one can find a bona fide problem?

 

How about that the intial conditions of the gedanken are invalid? The only way a rod can be spanning the event horizon and travelling upwards with a velocity v is if it came from below the horizon, which is impossible. Coming from outside the horizon you could have the rod go into orbit with the bottom of the rod "kissing" the horizon, but then the rod would have no velocity radially outward and your problem goes away.

 

[post=1324276]I don't think you've progressed past here[/post]

In this context, the problem is that your conditions are not consistent:

1. Let a freely falling rod span the horizon of a black hole...
2. ...so large that the tidal force on the rod is negligible.
3. Let the part of the rod that is above the horizon be escaping to r=infinity.

If 1 and 3 are true, then 2 must be false.

 

Well I don't know what GR actually says, but it could easily be something like the event horizon already moving at c in any reference frame in free fall into a black hole - which would make your scenario impossible.

Another possibility (although somehow I doubt this) is that it's only the concept of a black hole that is incorrect. To illustrate: Newton's theory of gravity allows for the existence of objects from which the escape velocity is greater than c (in fact, I believe the idea that such "black holes" might exist even predates General Relativity). If I then define the "black hole" as the volume within which the escape velocity greater than the speed of light, I can claim that the "event horizon" of such a black hole is a point of no return. Obviously the scenario from the OP applies here, but what does it actually prove? That the whole of Newtonian gravity is internally inconsistent, or that there's a problem with my claim that the event horizon is a point of no return?

Personally, I imagine if physicists claim event horizons in GR are (almost) true points of no return, then there's probably some reason for it.

 

The whole rod could have been above the horizon and escaping to r=infinity when the black hole grew to cover part of the rod. So there’s no problem with the initial conditions of the thought experiment.

 

I think Pete's point is the most fundamental error in Zanket's argument, but there's another one I think is worth mentioning. Zanket states that GR forbids any object from passing outwards through the horizon, but this is not really true (or, at least, not in the way he interprets it). What GR actually predicts is that no distant observer will ever see any object come out through the horizon. However, as was demonstrated in gory detail by Andrew Gray in a side-thread, this does not mean that the object can't pass out of the horizon in its own proper time. It's just that this process takes an infinite amount of time from the perspective of distant observers, and so happens "after the end of time" (i.e., never). Thus, there is no inconsistency between the rod's view (curvature is negligible, and the rod cruises right out past the horizon) and the distant observer's view (the rod never comes out of the horizon).

Note that the situation is exactly the same for infalling objects: from the object's point of view, it cruises right past the horizon and continues on towards the singularity. But from the distant observer's point of view, the object slows down as it nears the horizon, and never actually crosses it. You could just as well argue that this well-known fact proves GR is inconsistent, and you'd be just as wrong.

 

In that case I doubt very much whether the tidal forces near the horizon are small.

 

“Negligible” does not mean “zero”. Precisely how small do you think the tidal force on the rod must be to make 2 false? What evidence can you give for that exact value being predicted by GR?

 

The rod is not given to be freely falling into a black hole. It is freely falling, and the part of it that is above the horizon is moving away from the black hole. In the supporting info “freely falling object” is defined as “an object on which no forces act except gravity”. A ball thrown upward is freely falling as its altitude increases, neglecting air friction.

It proves neither. The OP is about GR, not Newtonian mechanics. The OP relies on GR’s prediction that the horizon is a point of no return. In the supporting info “horizon” is defined as “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”.

The reason is that GR predicts that they are, and not “almost”. GR’s math supports the definition of “horizon”.

 

Pete hasn’t shown an error. Escape velocity in GR is less than c above the horizon of any Schwarzschild black hole, regardless how small the tidal force is at its horizon.

GR predicts that nothing can pass outward through the horizon; that’s why no distant observer will see it happen. Distant observers can in principle see what happens just above the horizon (the photons from there reach them in a finite time on their clocks). In the supporting info “horizon” is defined as “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”. GR’s math supports that definition, and the definition makes no exception to allow an observer to see itself pass outward through a horizon.

GR predicts otherwise. Andrew did not refute hundreds of physicists who have used GR’s math to show that no object can pass outward through a horizon in its own proper time. The basis for his claim was a white hole, not a black hole. The OP is about a black hole.

If the rod could pass outward through the horizon in its own proper time, then according to GR it could subsequently reach the point where the distant observer is, in which case the observer would have a hard time explaining how the rod got there without passing outward through the horizon. In principle the observer can be at that distant location for all time, waiting for the rod to pass by. GR does not feature multiple independent realms of time above a horizon, such that the rod can reach the distant observer’s location after forever on that observer’s clock.

Note that when you piggyback on other people’s claims, you need to support them yourself.

 

Can you support your doubt with evidence? The tidal force at the horizon is a function of only the mass of the black hole, and not whether the black hole grew. The black hole could have been supermassive before it grew to cover part of the rod.

 

That depends on the sensitivity of your experiment.
If your experiment is sensitive enough to locate the event horizon somewhere along the rod, then the experiment is sensitive enough for the tidal force to be significant.

 

I am under the impression that there are momentum density terms (the stress part) in the stress energy tensor. On a more practical scale the momentum density field of the Earth's rotation causes frame dragging. Now if the event horizon shifts there must be a large movement of mass radially. I'm guessing but it would surprise me greatly if this didn't cause some serious tidal forces near the horizon. Call it a strong suspicion that the predition that certain static black holes have negligable tidal forces won't hold for a dynamic black hole in flux.