(The “Alpha” in the title indicates that the Alpha rules (http://www.sciforums.com/showthread.php?t=62837) apply to this thread.)

Elsewhere (http://www.sciforums.com/showthread.php?t=67582) 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 (http://www.sciforums.com/showthread.php?t=67582).

The proof
http://zanket.home.att.net/home_files/GRFlaw3.png

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.

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.

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.
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 an incomplete theory that gives incorrect results when trying to apply its inertial frames where gravity is a factor.
SR is irrelevant here. The OP doesn’t mention it.

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

Black hole accretion disks are in a spiral formation due to their rotation, not a sphere.
Accretion disks are irrelevant here. The OP is about theory, not reality.

Kerr black holes, or charged Kerr-Newman black holes, are the types thought to actually ocurr in the universe, not Schwarzchild black holes.
That doesn’t matter. The OP is about theory, not reality.

Supermassive, rotating Kerr black holes are the model with areas of 'negilgible'
tidal effects along the central axis of rotation, not Schwarzchild black holes.
The tidal forces at the horizon of a Schwarzschild black hole can be negligible. See the supporting info.

zanket,
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.
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.
SR is irrelevant here. The OP doesn’t mention it.
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:
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
You then go on to state that tidal forces can be negligible in such a frame:
The supporting info refutes this; search for “Here are confirmations”. The tidal forces at the horizon can be negligible.
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.

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.
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.

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.
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.

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?
From the Wikipedia entry for “'spaghettification” (http://en.wikipedia.org/wiki/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 (http://casa.colorado.edu/~ajsh/singularity.html): “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.

zanket:

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.

In this case, there MUST be a tension force in the rod, since the part above the horizon is accelerating upwards, while the lower end is apparently in free fall.

By predicting that the rod both can and cannot pass outward through the horizon, GR contradicts itself.

GR predicts that tension forces in the rod will cause it to break in this situation.

In this case, there MUST be a tension force in the rod, since the part above the horizon is accelerating upwards, while the lower end is apparently in free fall.
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.

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

Seriously, no one can find a bona fide problem?

I don't think you've progressed past here

In this context, the problem is that your conditions are not consistent:
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.
If 1 and 3 are true, then 2 must be false.

Seriously, no one can find a bona fide problem?
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.

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.
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.

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.

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

In this context, the problem is that your conditions are not consistent:
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.
If 1 and 3 are true, then 2 must be false.
“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?

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.
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.

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?
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”.

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

I think Pete's point is the most fundamental error in Zanket's argument, ...
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.

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.
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.

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.
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.

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).
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.

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

“Negligible” does not mean “zero”. Precisely how small do you think the tidal force on the rod must be to make 2 false?
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.

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.

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.

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.
It is a given that the rod spans a horizon; no experiment needed for that. The experiment in the OP need not be very sensitive. The experiment tests whether the rod breaks. Would you need a sensitive experiment to tell if your arm got torn off?

If the tidal force does not break the rod, then GR is proven to be self-inconsistent. Then “negligible” in the OP means “too weak to break the rod”. To prove your case that “2 must be false” you must show that the tidal force on the rod must be strong enough to break it. You haven’t done that. I’ve given multiple references to show that the tidal force on the rod can be too weak to break it.

It is a given that the rod spans a horizon; no experiment needed for that. The experiment in the OP need not be very sensitive. The experiment tests whether the rod breaks. Would you need a sensitive experiment to tell if your arm got torn off?

If the tidal force does not break the rod, then GR is proven to be self-inconsistent. Then “negligible” in the OP means “too weak to break the rod”. To prove your case that “2 must be false” you must show that the tidal force on the rod must be strong enough to break it. You haven’t done that. I’ve given multiple references to show that the tidal force on the rod can be too weak to break it.
No, you've given some armwaving arguments based on vague, introductory type references.
You haven't shown that the tidal force can to too weak to break the rod - you haven't attempted to quantify the tidal force at all, nor have you mentioned the tensile strength of the rod.

Unless you can produce a more rigorous argument, then I'll stand by my claim - if the rod spans the horizon, and the top end of the rod exceeds the escape velocity at that point, then the tidal force across the rod is not negligible.

Here's another armwaving argument to demonstrate it:
The escape velocity at the top of the rod must be less than c.
But the escape velocity at the event horizon is equal to c.
A changing escape velocity indicates curving spacetime, and a tidal force.

It follows that if the tidal force is negligible, then the escape velocity at the top of the rod is negligibly less than c.
This implies that the top of the rod can't exceed the escape velocity - if it does (as you specify), then the difference between the escape velocity and c is not negligible, and therefore neither is the tidal force.

Your conditions are inconsistent.

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.
Multiple references above show that the tidal force can be negligible on an object falling through a horizon. As an object falls through a horizon the black hole is in flux; it is growing due to the mass the object adds to it. Then the tidal force at the horizon can be negligible even when the black hole is in flux. (And “negligible” in this case means “too weak to break the rod”.) The black hole grows even when a small object falls into it. In principle there need not be a large movement of mass radially for the black hole to grow such that the horizon covers part of the rod. Any degree of growth will do.

You haven't shown that the tidal force can to too weak to break the rod - you haven't attempted to quantify the tidal force at all, nor have you mentioned the tensile strength of the rod.
I have shown that the tidal force can be too weak to break the rod. I’ve given references to show that a human can survive falling through a horizon. If a human can survive there, then a typical rod need not break there. You can assume that the rod does not have a tensile strength so weak that it gets broken by a tidal force that a human cannot even feel. Mentioning that assumption in the OP would be silly. (Nor did Einstein tell us the tensile strength of the rope that accelerates the chest in his thought experiment showing the principle of equivalence, even though the experiment would be ruined if the rope broke with the slightest tug on it.)

The escape velocity at the top of the rod must be less than c.
But the escape velocity at the event horizon is equal to c.
A changing escape velocity indicates curving spacetime, and a tidal force.
Agreed.

It follows that if the tidal force is negligible, then the escape velocity at the top of the rod is negligibly less than c.
This implies that the top of the rod can't exceed the escape velocity - if it does (as you specify), then the difference between the escape velocity and c is not negligible, and therefore neither is the tidal force.
When something is negligible for the purpose of one application of it, you are not prevented from considering it for the purpose of a different application of it. If the amount of water I have is negligible for the purpose of living for one year, is it necessarily negligible for the purpose of living for one day? Of course not. If the tidal force is too weak to break the rod (i.e. negligible for the purpose of breaking the rod), in no way does that prevent using the tidal force to determine that the top of the rod can exceed the escape velocity—for that purpose the tidal force need not be negligible. Then you haven't shown that “2 must be false” in your post #22.

Since GR’s self-inconsistency is determined by whether or not the tidal force breaks the rod, “negligible” in the OP means “too weak to break the rod”. Suppose I changed the first sentence in the OP to say: “Let a freely falling rod span the horizon of a black hole so large that the tidal force on the rod is too weak to break it.” Then “negligible” would not appear in the OP. Would you still think there’s a problem then? Would you say that if the tidal force is too weak to break the rod then no part of the rod can exceed the escape velocity?

When something is negligible for the purpose of one application of it, you are not prevented from considering it for the purpose of a different application of it.
Right, that's why the argument is a handwaving one. Just like yours.
One situation (a human falling into a black hole) is not necessarily equivalent to another (a rod escaping from a black hole).

Suppose I changed the first sentence in the OP to say: “Let a freely falling rod span the horizon of a black hole so large that the tidal force on the rod is too weak to break it.” Then “negligible” would not appear in the OP. Would you still think there’s a problem then? Would you say that if the tidal force is too weak to break the rod then no part of the rod can exceed the escape velocity?
Certainly.
If the rod spans the horizon, and the top of the rod exceeds escape velocity, then the rod will be broken by tidal forces.

Enough handwaving. This question can only be settled rigorously.
Can you describe a specific situation (event horizon radius, rod length, position, and velocity) such that GR predicts that the rod will not break?

If the rod spans the horizon, and the top of the rod exceeds escape velocity, then the rod will be broken by tidal forces.
You’ve given nothing but your opinion to support that, and GR predicts otherwise. It is actually you who is handwaving here.

Can you describe a specific situation (event horizon radius, rod length, position, and velocity) such that GR predicts that the rod will not break?
See the definition of tidal force in the supporting info. The tidal force in a frame is the relative acceleration of two free test particles on either side of the frame. That’s all there is to the definition. Then the same tidal force must apply to any object in a frame regardless of its length, position, or velocity. The supporting info shows that a human can survive falling through a horizon. The rod can be small enough to fit in a human-sized frame. Then the tidal force on the rod can be too weak to break it.

zanket,
You’ve given nothing but your opinion to support that, and GR predicts otherwise. It is actually you who is handwaving here.
No zanket, you predict otherwise based on your inadequate understanding of GR.
See the definition of tidal force in the supporting info. The tidal force in a frame is the relative acceleration of two free test particles on either side of the frame. That’s all there is to the definition.
No, that's not all there is to the definition of tidal forces. Two test particles, one at the top of the frame and one at the bottom of the frame, will accelerate apart when dropped in a gravitational field. The stronger the gravitational field and the greater the distance between the test particles, the greater the velocity of their acceleration. This is what you ignore in your gedanken and this is what breaks the rod.

In addition to that, a rod cannot 'hover' a fixed distance from the event horizon unless a constant accelerating force is applied to the rod. If the force is applied to the lower end of the rod below the event horizon, the rod can never exceed the speed of light and escape the black hole. If the accelerating force is applied to the top of the rod, such as a long rope, the rod will simply stretch and snap due to the stress. The same thing happens when a man is hung. The man accelerates toward the gravitating body due to gravity. The rope snaps his neck when the vector of that acceleration is changed.

The freely falling frame you used is just that, a freely falling frame due to the acceleration of gravity. From the point of view of two stationary observers sitting on each end of your rod in that frame, the event horizon is approaching each at a constantly increasing, but different, velocity until the event horizon is moving at the speed of light when it touches the observer on the lower end of the rod. During this event, the two observers will be separating in distance because the observer on the bottom end is falling faster than the observer on the top. The two observers are not in the same inertial frame of reference. This is an old problem with inertial frames and the very reason Global inertial frames cannot be set up in a gravity field. You simply cannot set up a Global inertial frame in which both ends of the rod are in the same inertial frame in strong gravity. GR can work around this problem by making inertial frames very tiny in extent. SR cannot handle inertial frames that are infinite in extent in a gravity field. Some have said that sounded the death knell for SR in the past, but by making the inertial frames very tiny, SR was saved. A much greater problem for SR comes from George Smoot's discovery of the anisotropy of the CMB, but a different subject from this thread.

You’ve given nothing but your opinion to support that, and GR predicts otherwise. It is actually you who is handwaving here.
We're both handwaving. The difference is that I admit it.

See the definition of tidal force in the supporting info. The tidal force in a frame is the relative acceleration of two free test particles on either side of the frame. That’s all there is to the definition. Then the same tidal force must apply to any object in a frame regardless of its length, position, or velocity. The supporting info shows that a human can survive falling through a horizon. The rod can be small enough to fit in a human-sized frame. Then the tidal force on the rod can be too weak to break it.
See, that's handwaving. You're not making your assumptions explicit.

For example, consider this story, which will sound very familiar:
A human in a freefall lab makes very careful measurements as a rod passes at very high speed, and note that the rod is 2m long, and moving so fast that it can't be distinguished from light speed. (In fact, the rod was moving at (1 - 10^{-100}) c).

Unbeknown to the human, at the moment that the rod was in their lab, their lab was dropping through the event horizon of a billion solar mass black hole. In fact, at the instant they made their measurement of the rod's length, the precise position of the event horizon coincided with the tail of the rod.

Tidal acceleration across the lab was also being measured, but it wasn't within the precision of the lab instruments.

Now,some things to think about, and questions to be answered:
The human measured the rod to be 2m long.
What is the rod's proper length?

The measured tidal acceleration in the lab was between two points in spacetime with a separation in the lab frame of 2 metres and zero seconds.
What was the separation of those points in the rod's rest frame?

Imagine that some equipment on the rod was also measuring tidal acceleration between two points in spacetime, specifically between the head and tail of the rod at the instant the the event horizon coincided with the tail of the rod.
What would be the spacetime separation between those two points in spacetime in the rod's rest frame?
What would be the spacetime separation between those two points in spacetime in the lab frame?

Would the equipment in the lab and on the rod measure the same tidal acceleration?

No, that's not all there is to the definition of tidal forces.
That is all there is to it. The definition in the supporting info is verbatim from Taylor and Wheeler.

Two test particles, one at the top of the frame and one at the bottom of the frame, will accelerate apart when dropped in a gravitational field. The stronger the gravitational field and the greater the distance between the test particles, the greater the velocity of their acceleration.
Acceleration is defined as increase in velocity. There’s no such thing as a velocity of an increase in velocity. Change to “the greater the degree of their relative acceleration” and then it’s right. And then it agrees with the definition of tidal force that I gave.

This is what you ignore in your gedanken and this is what breaks the rod.
I don’t ignore it. You haven’t shown that the tidal force will break the rod. GR allows the tidal force at a horizon to be too weak to break the rod, and the tidal force was given to be negligible; i.e. too weak to break the rod.

In addition to that, a rod cannot 'hover' a fixed distance from the event horizon unless a constant accelerating force is applied to the rod. If the force is applied to the lower end of the rod below the event horizon, the rod can never exceed the speed of light and escape the black hole. If the accelerating force is applied to the top of the rod, such as a long rope, the rod will simply stretch and snap due to the stress. The same thing happens when a man is hung. The man accelerates toward the gravitating body due to gravity. The rope snaps his neck when the vector of that acceleration is changed.
None of this refutes the OP, or even applies to it. Nothing in the OP is noninertially accelerating. Nothing in the OP is hovering.

From the point of view of two stationary observers sitting on each end of your rod in that frame, the event horizon is approaching each at a constantly increasing, but different, velocity until the event horizon is moving at the speed of light when it touches the observer on the lower end of the rod.
The horizon is not approaching in the upper observer’s frame; rather it’s receding. The rod is given to span the horizon. The part of the rod that is above the horizon is given to be escaping to r=infinity, so it’s moving away from the horizon. The lower observer is below the horizon, so if the horizon is approaching in that observer’s frame then GR is self-inconsistent, because GR predicts that all observers who are below a horizon must fall.

The two observers are not in the same inertial frame of reference. This is an old problem with inertial frames and the very reason Global inertial frames cannot be set up in a gravity field. You simply cannot set up a Global inertial frame in which both ends of the rod are in the same inertial frame in strong gravity. GR can work around this problem by making inertial frames very tiny in extent. SR cannot handle inertial frames that are infinite in extent in a gravity field. Some have said that sounded the death knell for SR in the past, but by making the inertial frames very tiny, SR was saved. A much greater problem for SR comes from George Smoot's discovery of the anisotropy of the CMB, but a different subject from this thread.
None of this refutes the OP, or even applies to it. There is no inertial frame in the OP, nor need there be. The OP does not mention SR, even implicitly. The tidal force on the rod is negligible (i.e. too weak to break the rod), not nonexistent.

See, that's handwaving. You're not making your assumptions explicit.
When I say “The rod can be small enough to fit in a human-sized frame”, I meant its proper length, as in “the rod can be small enough to fit in a human-sized frame when it’s at rest relative to that frame”. I agree that I could have been more specific.

Now,some things to think about, and questions to be answered:
The human measured the rod to be 2m long.
What is the rod's proper length?
I won’t calculate, but it would be a lot longer than 2m. I’ll guess one light-second long.

The measured tidal acceleration in the lab was between two points in spacetime with a separation in the lab frame of 2 metres and zero seconds.
What was the separation of those points in the rod's rest frame?
A lot smaller than 2m and greater than zero seconds.

Imagine that some equipment on the rod was also measuring tidal acceleration between two points in spacetime, specifically between the head and tail of the rod at the instant the the event horizon coincided with the tail of the rod.
What would be the spacetime separation between those two points in spacetime in the rod's rest frame?
One light-second and zero seconds.

What would be the spacetime separation between those two points in spacetime in the lab frame?
2m and greater than zero seconds.

Would the equipment in the lab and on the rod measure the same tidal acceleration?
No, but both tidal forces measured could be less than any arbitrarily small value, for a sufficiently large black hole (perhaps larger than a billion solar masses).

You haven’t shown a problem with the OP. Given that the tidal force on a human falling through a horizon can be arbitrarily small, the tidal force on a rod with a proper length of 2m or less and that spans a horizon need not break the rod.

Consider the relativistic rocket equations (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html). These are SR equations of motion for a rocket. The site handles the issue of the tidal force with “We assume that the stars are essentially at rest in this frame”, which is saying “we assume that the spacetime in the lab is essentially flat”, or—since spacetime curvature is synonymous with tidal force—“we assume that the tidal force in the lab is essentially zero”. Notice that the author does not mention that the tidal force becomes significant in the rocket’s rest frame; he should mention that if you are right that velocity necessarily makes a significant difference. In the rocket’s frame two galaxies could be just one meter apart. But if the stars are essentially at rest in the lab frame, they can be essentially at rest relative to one another in the crew’s frame. Let the rocket achieve a velocity of (1 - 10^{-100}) c relative to the stars when it shuts down its engine. Let the crew choose two far-apart stars to measure the tidal force. The measurement can show that the tidal force is essentially zero. That’s obvious, because if the stars were exactly at rest in the lab frame, they would be exactly at rest relative to one another in the crew’s frame, and there would no relative acceleration between them—zero tidal force—in either frame.

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.
I’ve been thinking more about this.

Another way to attain the initial conditions in the OP is to have the whole rod be escaping to r=infinity as it moves through a group of stars falling toward one another to eventually form a black hole. The horizon won’t exist until enough mass exists within a certain volume. Let the horizon form along the rod.

Suppose the tidal force at the horizon is strong enough to kill a human there. The proof in the OP is safe, because the tensile strength of the rod can be arbitrarily high in principle. To refute the OP you would need to show that the tidal force on the rod would always be infinitely high.

Replace the stars with hydrogen atoms. Enough hydrogen in a certain volume will create a black hole, even when each hydrogen atom is at least one light year from its nearest neighbor. (The larger the black hole to be created, the less density of mass is required; I can prove that if you want). Then it’s easy to see that the tidal force on the rod can be arbitrarily small even as the horizon forms along it.

zanket,
That is all there is to it. The definition in the supporting info is verbatim from Taylor and Wheeler.
I quoted you, zanket. I repeat the quote: "See the definition of tidal force in the supporting info. The tidal force in a frame is the relative acceleration of two free test particles on either side of the frame. That’s all there is to the definition." To begin with, acceleration is a vector measurement. You only considered the 'relative acceleration' of two test particles in a plane perpendicular to the gravitational acceleration which is directed downward. You ignored the difference in the relative velocities of two particles at the top and bottom of the frame. That is what stretches the rod in a free fall frame of reference (note: not 'free float').
Acceleration is defined as increase in velocity. There’s no such thing as a velocity of an increase in velocity. Change to “the greater the degree of their relative acceleration” and then it’s right. And then it agrees with the definition of tidal force that I gave.
Acceleration is a time varying increase or decrease in relative velocity. The top and bottom test particles will have different velocities relative to the event horizon, based on each particle's time and distance from the event horizon, which leads to a tidal force in the particles frame. If the distance separating the two test particles is small enough, tidal forces could theoretically be ignored at a large black hole's event horizon. However, you cannot have a rod one lightsecond long and ignore tidal forces.
I don’t ignore it. You haven’t shown that the tidal force will break the rod. GR allows the tidal force at a horizon to be too weak to break the rod, and the tidal force was given to be negligible; i.e. too weak to break the rod.
GR allows the tidal force at a horizon be too weak to break a very short rod. You are challenging GR, so it is your responsibility to define the length of your rod and the exact tidal force experienced by that rod in order to prove the tidal force is 'negligible'.
None of this refutes the OP, or even applies to it. Nothing in the OP is noninertially accelerating. Nothing in the OP is hovering.
Of course not, you started your gedanken at a point after a noninertial acceleration was applied to the rod to reverse its direction and velocity relative to the event horizon. And, yes, an object in free fall is accelerating, regardless of its direction of relative motion. An object in free fall must pass through at least a short period of 'hovering' when its vector of motion relative to the event horizon is changed from 'toward' to 'escaping from'. A 'force' need not be felt in a freely falling frame, but the relative velocity must be changing with time and distance or else its not a freely falling frame of reference. A free falling object will increase relative velocity at a constant rate in direct relation to the gravitational acceleration of a given mass. A particle moving away (escaping) from the mass will decrease relative velocity at the same constant rate. A particle in an SR-type inertial frame of reference (free float) does not change velocity relative to anything, it is a constant-velocity frame. In contrast, a freely falling frame is a constantly accelerating frame, with varying relative velocities.
The horizon is not approaching in the upper observer’s frame; rather it’s receding.
If the horizon did not approach the rod at some point in a physically accurate gedanken, part of the rod could not be below the event horizon.
The part of the rod that is above the horizon is given to be escaping to r=infinity, so it’s moving away from the horizon.
And that motion required a period of non-inertial acceleration and huge g-forces, regardless of whether you ignore that aspect in your OP. It is un-physical to state objects fall away from gravitating masses. SR ignores periods of acceleration when switching between free float frames, but GR does not ignore acceleration in freely falling frames. Again, was this acceleration applied to the top or the bottom of the rod?

In case you haven't noticed, your rod had to experience millions of g's before it started 'escaping' the black hole. If the bottom of the rod was below the event horizon already, the acceleration had to be applied to the part of the rod above the event horizon. The g-forces of acceleration would break the rod, allowing the bottom portion to fall toward the singularity, while the top of the rod escaped the black hole. The rod was already broken when you began your gedanken.

No, but both tidal forces measured could be less than any arbitrarily small value, for a sufficiently large black hole (perhaps larger than a billion solar masses)
Now you begin chasing your tail.
Look at your diagram for a moment:
http://zanket.home.att.net/home_files/GRFlaw3.png
Place the top of the rod at r = top
Now, if you make the hole larger, the escape velocity is higher at top.
The escape velocity is higher, the rod has to be moving faster.
The rod is moving faster, there is more length contraction.
There is more length contraction, the proper length of the rod must be longer in order to reach the event horizon.
If the proper length of the rod is longer, then the tidal acceleration across the rod in the rod's rest frame is greater.
So, you have to make the hole larger to reduce the tidal acceleration...

Yes, this is a handwaving argument. Like I said, the only way to resolve this is to make it rigorous.

So, here's a specific question for you:
For what hole mass, rod length, rod speed, and rod position does GR predict that the bottom end of the rod will escape from the horizon?

Given that the tidal force on a human falling through a horizon can be arbitrarily small, the tidal force on a rod with a proper length of 2m or less and that spans a horizon need not break the rod.
Certainly... as long as the rod is falling into the horizon like the human. But you're talking about a different situation.
I suggest that if the top of the rod is moving fast enough to escape the horizon, then the bottom of the rod will not reach the horizon unless the rod is being stretched.

Consider the relativistic rocket equations (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html). These are SR equations of motion for a rocket. The site handles the issue of the tidal force with “We assume that the stars are essentially at rest in this frame”, which is saying “we assume that the spacetime in the lab is essentially flat”, or—since spacetime curvature is synonymous with tidal force—“we assume that the tidal force in the lab is essentially zero”. Notice that the author does not mention that the tidal force becomes significant in the rocket’s rest frame;
Not relevant. John Baez is specifically discussing the predictions of SR in a gravity-free environment. If you asked him whether tidal forces would eventually become significant in the rocket's frame, he might say yes... or he might laugh and say "not before the rocket is vaporized by background radiation"!

To begin with, acceleration is a vector measurement. You only considered the 'relative acceleration' of two test particles in a plane perpendicular to the gravitational acceleration which is directed downward. You ignored the difference in the relative velocities of two particles at the top and bottom of the frame. That is what stretches the rod in a free fall frame of reference (note: not 'free float').
The definition of tidal force in the supporting info does not restrict you from considering two free test particles at the top and bottom of the frame. It does not require you to consider two free test particles only in a “plane perpendicular to the gravitational acceleration which is directed downward”.

Acceleration is a time varying increase or decrease in relative velocity. The top and bottom test particles will have different velocities relative to the event horizon, based on each particle's time and distance from the event horizon, which leads to a tidal force in the particles frame. If the distance separating the two test particles is small enough, tidal forces could theoretically be ignored at a large black hole's event horizon. However, you cannot have a rod one lightsecond long and ignore tidal forces.
“Small enough” is arbitrary, and could be one light second long in principle. The tidal force from one end to the other of a rod that is one light second long could be far less than the tidal force on your body now, for a sufficiently large black hole. In any case the OP does not restrict the size of the rod. It could be microscopic.

GR allows the tidal force at a horizon be too weak to break a very short rod. You are challenging GR, so it is your responsibility to define the length of your rod and the exact tidal force experienced by that rod in order to prove the tidal force is 'negligible'.
GR allows the tidal force at a horizon be too weak to break an arbitrarily long rod, for a sufficiently large black hole. The OP does not restrict the size of the rod, so it can always be small enough that the tidal force on it is too weak to break it, for any given black hole. The OP does not restrict the size of the black hole, so it can always be large enough that the tidal force on the rod is too weak to break it, for any given rod. Then defining the exact length of the rod or the size of the black hole is superfluous.

In Einstein’s thought experiment showing relativity of simultaneity (http://www.bartleby.net/173/9.html), he was not required to define the exact size of the train and the passenger in the train, to show that the train was large enough to fit the passenger. Instead the reader can assume that that the passenger fits in the train, since he made it a given that the train has a passenger and he did not restrict the size of the train or the passenger. Likewise in the OP the reader can assume that the rod is small enough that the tidal force on it is negligible, because it is given that the tidal force on the rod is negligible and there is no restriction on the size of the rod or the black hole.

An object in free fall must pass through at least a short period of 'hovering' when its vector of motion relative to the event horizon is changed from 'toward' to 'escaping from'.
Nowhere in the OP is it given that the rod ever moved inward toward the horizon. Nor need it ever do that. In post 31 (http://www.sciforums.com/showpost.php?p=1496618&postcount=31) I show how the initial conditions in the OP can be met without the rod ever noninertially accelerating.

A 'force' need not be felt in a freely falling frame, but the relative velocity must be changing with time and distance or else its not a freely falling frame of reference. A free falling object will increase relative velocity at a constant rate in direct relation to the gravitational acceleration of a given mass. A particle moving away (escaping) from the mass will decrease relative velocity at the same constant rate. A particle in an SR-type inertial frame of reference (free float) does not change velocity relative to anything, it is a constant-velocity frame. In contrast, a freely falling frame is a constantly accelerating frame, with varying relative velocities.
I don’t see any problem of the OP specified here.

If the horizon did not approach the rod at some point in a physically accurate gedanken, part of the rod could not be below the event horizon.
No, the horizon could have formed along the rod as it was escaping to r=infinity. This covers the rest of your post.

zanket,
Nowhere in the OP is it given that the rod ever moved inward toward the horizon. Nor need it ever do that. In post 31 I show how the initial conditions in the OP can be met without the rod ever noninertially accelerating.
.....
No, the horizon could have formed along the rod as it was escaping to r=infinity. This covers the rest of your post.
If an event horizon formed midway along the rod while the rod was moving at near the speed of light relative to the horizon (escape velocity), the part of the rod below the event horizon would be prevented from moving upward through the event horizon. The part of the rod escaping to infinity above the event horizon could continue to do so, after the rod broke into two pieces. Again, the rod must separate (break).

zanket,

If an event horizon formed midway along the rod while the rod was moving at near the speed of light relative to the horizon (escape velocity), the part of the rod below the event horizon would be prevented from moving upward through the event horizon. The part of the rod escaping to infinity above the event horizon could continue to do so, after the rod broke into two pieces. Again, the rod must separate (break).

Hi 2inq,
That's exactly half zanket's point - GR predicts that the rod must break (or stretch).
The other half (which I disagree with) is that for a large enough hole, GR doesn't predict any force which would break (or stretch) the rod.

Taken together, these two halves would make an inconsistency.

Hi 2inq,
That's exactly half zanket's point - GR predicts that the rod must break (or stretch).
The other half (which I disagree with) is that for a large enough hole, GR doesn't predict any force which would break (or stretch) the rod.

Taken together, these two halves would make an inconsistency.
Pete, assume we have two bowling balls. Connect the two balls with a thin piece of thread. Drop one ball down a deep hole while simultaneously pitching the other ball upwards. Both bowling balls are in freely falling frames, but with different directions of inertial motion. Are you claiming GR predicts no inertial force to break the string? 'Tidal forces' are a sidetrack and not the reason the string, or the rod, breaks in zanket's particular gedanken. In zanket's example, the event horizon would form somewhere along the string in between the balls. The part of the rod below the event horizon would be immediately accelerated toward the singularity at the speed of light, while the part of the rod outside the event horizon would be travelling at above escape velocity in the opposite direction. Unless the rod were massless, the rod would experience tremendous kinetic energy and inertial forces in opposite directions, breaking or stretching the rod. If the rod were unbreakable, the gravitational acceleration below the event horizon would be impossible to overcome, dragging the entire rod through the event horizon.

Tidal forces are a separate mechanism. A freely falling rod will break due to tidal forces even if it drops unhindered into a black hole. The size and mass of the black hole and the length of the rod determines whether the break happens above or below the event horizon.

I do agree with zanket that the equivalence principle is false, but not for the same reasons. Freely falling frames of reference are not equivalent to SR inertial frames. There are no true inertial frames in a gravitational field. The freely falling frames can be made small enough that the non-inertial effects may not be easily measurable, but freely falling frames are still non-inertial frames.

Pete, assume we have two bowling balls. Connect the two balls with a thin piece of thread. Drop one ball down a deep hole while simultaneously pitching the other ball upwards. Both bowling balls are in freely falling frames, but with different directions of inertial motion. Are you claiming GR predicts no inertial force to break the string?
Of course not. Your example is not relevant to the thread, in which zanket has specified that all parts of the rod are at rest w.r.t. each other.
We all agree that GR predicts that such a rod can't exist if it crosses the event horizon and is moving outward, but zanket is suggesting that GR also predicts that such a rod can exist.

'Tidal forces' are a sidetrack and not the reason the string, or the rod, breaks in zanket's particular gedanken. In zanket's example, the event horizon would form somewhere along the string in between the balls. The part of the rod below the event horizon would be immediately accelerated toward the singularity at the speed of light, while the part of the rod outside the event horizon would be travelling at above escape velocity in the opposite direction.
Yes, that's what tidal forces means.
The difference in acceleration between the two ends of the rod is called "tidal acceleration".
The resulting tension in the rod is thus a tidal force.

Pete,
Of course not. Your example is not relevant to the thread, in which zanket has specified that all parts of the rod are at rest w.r.t. each other.
zanket did not specify that all parts of the rod are at rest w.r.t. each other. He continually stated the part of the rod above the event horizon was escaping to infinity. He only stated the lower end of the rod was below the event horizon.

zanket did not specify that all parts of the rod are at rest w.r.t. each other. He continually stated the part of the rod above the event horizon was escaping to infinity. He only stated the lower end of the rod was below the event horizon.
I apologize - he didn't directly specify that... at least not in this thread.
However, that was certainly his intention. More formally, he intends that there exists some inertial frame in which all parts of the rod have the same velocity.

Place the top of the rod at r = top
Now, if you make the hole larger, the escape velocity is higher at top.
The escape velocity is higher, the rod has to be moving faster.
The rod is moving faster, there is more length contraction.
There is more length contraction, the proper length of the rod must be longer in order to reach the event horizon.
If the proper length of the rod is longer, then the tidal acceleration across the rod in the rod's rest frame is greater.
So, you have to make the hole larger to reduce the tidal acceleration...

Yes, this is a handwaving argument. Like I said, the only way to resolve this is to make it rigorous.
I’m not clear on what you mean by your first statement. If top is wherever the top of the rod is, then there’s no problem for any given rod, regardless of its size. It’s easier to see that in the rod’s frame. The rod is given to be spanning the horizon, so it need not reach the horizon; it’s already there. The black hole is given to be so large that the tidal force on the rod is negligible. The velocity of the rod at top, relative to the horizon, is whatever it needs to be to satisfy the given that the part of the rod that is above the horizon is escaping to r=infinity, which is a velocity less than c. There’s no problem with that scenario, so neither will there be a problem in a frame that is moving relative to the rod, wherein the rod is length-contracted.

If top is wherever the top of the rod is, then I think that observing the rod from a frame that is moving relative to the rod is tripping you up, because you are neglecting that the distance such observer measures between r-coordinates is also length-contracted in that frame. For example, in the muon’s frame in the muon experiment, a confirmation of SR, the height of the Earth’s atmosphere is length-contracted to about 5% of what a lab on the ground measures, and that’s how the muon has enough time to reach the ground before it decays.

If top is some specific r-coordinate, then you’re manufacturing a problem by adding a new condition, which shows no problem with the OP.

For what hole mass, rod length, rod speed, and rod position does GR predict that the bottom end of the rod will escape from the horizon?
Specifying any of those would be superfluous, just like Einstein needn’t have specified the size of the train and the passenger, to show that the passenger could fit in the train (see post 35). I request that you first show relevancy by showing, in a way that I cannot refute, how my omitting those shows a problem with the OP.

Certainly... as long as the rod is falling into the horizon like the human. But you're talking about a different situation.
It’s not different in a way that shows that the tidal force on a 2m long rod in the OP would be different than the tidal force on the human. The human could measure the tidal force by measuring the relative acceleration of two free test particles each at rest relative to the rod. Or the particles could be moving in opposite directions each close to c. Or one could be at rest relative to the human and the other moving at close to c, in any direction. The human could measure the tidal force in her frame by simply reading the value off a gauge affixed to the rod that directly measures the tidal strain on the rod; the definition of tidal force—namely that it’s independent of the velocity of the particles—shows that.

You haven’t made your case that tidal force is dependent on velocity. Tidal force, synonymous with spacetime curvature, is dependent only on the mass of the body and the distance from the body. GR’s spacetime curvature factor, incorporated into the Schwarzschild metric, is:

spacetime curvature factor = sqrt(1 – (2M / r))

Also consider that the tensile strength of the rod can be arbitrarily high in principle. So the OP is refuted only if it can be shown that the tidal force on the rod is always infinitely high, so that the rod always breaks as GR predicts it should.

I suggest that if the top of the rod is moving fast enough to escape the horizon, then the bottom of the rod will not reach the horizon unless the rod is being stretched.
How would it being stretched help the bottom of it reach the horizon?

Not relevant. John Baez is specifically discussing the predictions of SR in a gravity-free environment.
Look again. It’s discussing the predictions of SR in our real, gravity-endowed universe; e.g. “For distances bigger than about a thousand million light years, the formulas given here are inadequate because the universe is expanding.” A while ago I confirmed with the latest author, Don Koks, that the statement “We assume that the stars are essentially at rest in this frame” is for the express purpose of letting the tidal force be negligible.

If you asked him whether tidal forces would eventually become significant in the rocket's frame, he might say yes... or he might laugh and say "not before the rocket is vaporized by background radiation"!
If the tidal force on the rocket eventually became significant then the equations would be invalid above some velocity v. So why does the author neglect to mention that crippling fact? The site does mention the “intense bombardment of cosmic rays and other particles”.

Think Pete: Let the rocket achieve a velocity close to c relative to the Earth when it shuts down its engine. If the tidal force on the rocket is then greater than the tidal force on the larger Earth, then “relativity” is a misnomer. One of the first things relativity teaches—even Galileo’s version—is that movement is relative. If when two identical objects pass nearby each other the tidal forces on them significantly differ, then movement is an intrinsic property, a property of an object that does not depend on any other object. An object could determine whether it’s moving by measuring the tidal strain on itself. Doesn’t that sound silly? Do you think that when two identical objects pass nearby each other the tidal forces on them can significantly differ?

If an event horizon formed midway along the rod while the rod was moving at near the speed of light relative to the horizon (escape velocity), the part of the rod below the event horizon would be prevented from moving upward through the event horizon. The part of the rod escaping to infinity above the event horizon could continue to do so, after the rod broke into two pieces. Again, the rod must separate (break).
I second Pete's reply in post 37 (http://www.sciforums.com/showpost.php?p=1499059&postcount=37). I couldn't have said it better.

'Tidal forces' are a sidetrack and not the reason the string, or the rod, breaks in zanket's particular gedanken.
The tidal force is the only force that GR features that can break a freely falling object like the rod. Tidal force is synonymous with spacetime curvature, the supporting info notes.

Since the tidal force on the rod is given to be negligible (not strong enough to break it), as GR allows, and the tidal force is the only force GR features that can break the rod, then what breaks it?

I apologize - he didn't directly specify that... at least not in this thread.
However, that was certainly his intention.
That is my intention for half of my point that you noted in post 37 (http://www.sciforums.com/showpost.php?p=1499059&postcount=37). Except that the rod can stretch somewhat. The OP shows that GR predicts both that the rod need not break (although it can stretch somewhat) and must break.

More formally, he intends that there exists some inertial frame in which all parts of the rod have the same velocity.
The OP does not depend on an inertial frame or imply one (on purpose, to forestall the erroneous and popular “not exactly flat” argument noted in the supporting info). The “negligible” tidal force in the OP refers to a tidal force that is too weak to break the rod, which can be a tidal force strong enough to rip apart a human, since the rod’s tensile strength can be arbitrarily large in principle.

Next time I’ll be more specific on what “negligible” means, if only because an argument has already been floated that boils down to this: “Since the tidal force is negligible, the frame is inertial and SR applies, but the spacetime is not exactly flat, so the frame isn’t inertial and SR doesn’t apply.” :eek:

I’m not clear on what you mean by your first statement. If top is wherever the top of the rod is, then there’s no problem for any given rod, regardless of its size. It’s easier to see that in the rod’s frame. The rod is given to be spanning the horizon, so it need not reach the horizon; it’s already there. The black hole is given to be so large that the tidal force on the rod is negligible. The velocity of the rod at top, relative to the horizon, is whatever it needs to be to satisfy the given that the part of the rod that is above the horizon is escaping to r=infinity, which is a velocity less than c. There’s no problem with that scenario, ...
I suggest that there is, as I've said more than once. I suggest that the three conditions are contradictory.

...so neither will there be a problem in a frame that is moving relative to the rod, wherein the rod is length-contracted.

If top is wherever the top of the rod is, then I think that observing the rod from a frame that is moving relative to the rod is tripping you up, because you are neglecting that the distance such observer measures between r-coordinates is also length-contracted in that frame.
Yes, this is certainly a problem - I don't have a great grasp of the metric. Like I said, it's a handwaving argument, and a proper treatment is required to resolve it properly.

If top is some specific r-coordinate, then you’re manufacturing a problem by adding a new condition, which shows no problem with the OP.
Yes, that is an artificial condition, an artifice to show the tail-chasing situation of your prior post.

More fundamentally, the objection is this:
Your OP relies on references saying that the tidal acceleration across the event horizon can be negligible for an infalling reference frame.
But, you acknowledge that the tidal acceleration across the rod in the rod frame is not necessarily the same as the tidal acceleration across the rod in the falling lab frame.
Therefore you now need to show that the tidal acceleration across the event horizon can be negligible for an escaping reference frame.

My tail-chasing argument was a hand-waving argument to show that you can't just set that condition - I suggest that perhaps it is not possible for the tidal acceleration across the event horizon to be negligible for an escaping reference frame. (My terminology is getting sloppy... let me know if I'm not making sense!)

Specifying any of those would be superfluous, just like Einstein needn’t have specified the size of the train and the passenger, to show that the passenger could fit in the train (see post 35).
That story is a handwaving argument in a popular text. It's not rigorous, and would need to be supported by a proper treatment if there were a dispute over its validity.

I request that you first show relevancy by showing, in a way that I cannot refute, how my omitting those shows a problem with the OP.
Your OP is a handwaving argument, that is not convincing to me.
I have provided an opposing handwaving argument that is not convincing to you.
We are at a stalemate. I haven't proven GR to be consistent in the given scenario. You haven't proven GR to be inconsistent in the given scenario. We can proceed no further without a proper treatment.

Specific quantities aren't necessary - you can use variables if you prefer.
And don't forget this rule, by the way:
# The interpretation of the rules is at the discretion of the moderator. Suggestions regarding interpretation may be made, and the moderator will consider those suggestions, but the final say is the moderator's alone.

You haven’t made your case that tidal force is dependent on velocity.
*Ahem*:
Would the equipment in the lab and on the rod measure the same tidal acceleration?
No, but...
The "but" is irrelevant. You acknowledged that the measured tidal acceleration between the two ends of the rod in this situation is frame dependent.

Also consider that the tensile strength of the rod can be arbitrarily high in principle. So the OP is refuted only if it can be shown that the tidal force on the rod is always infinitely high, so that the rod always breaks as GR predicts it should.
Conversely, GR is only shown to be inconsistent if it can be shown that the tidal force across an escaping, horizon-spanning rod is finite.
The burden of proof is on you, zanket.
Not so - the tidal force need only be strong enough to pull the escaping end across the horizon into the hole.

How would it being stretched help the bottom of it reach the horizon?
If it is stretched enough that the bottom end is falling in while the top end isn't.

The OP does not depend on an inertial frame or imply one (on purpose, to forestall the erroneous and popular “not exactly flat” argument noted in the supporting info).
Ignoring a useful frame doesn't make it go away. And the "not exactly flat" argument is precisely how you can have the top of the rod at escape velocity, so I can see why it makes you uncomfortable.
Also, your OP relies on citations that GR allows negligible tidal forces in an infalling inertial frame.

The “negligible” tidal force in the OP refers to a tidal force that is too weak to break the rod, which can be a tidal force strong enough to rip apart a human, since the rod’s tensile strength can be arbitrarily large in principle.
Careful... if the rod's tensile strength is arbitrarily large, than a large enough tidal force would pull the escaping end across the horizon into the hole without breaking the rod. To be negligible, the tidal force must also be too weak to stop the rod from escaping.
I'll have to edit my previous post.

Next time I’ll be more specific on what “negligible” means,
Like I said - a proper treatment is required.

Hello all

Isn't it true that the rate of change of a tidal gradient is dependent on the size of the object causing the gradient? So a larger black hole would have a smaller tidal gradient for a given length of rod?

:)

Isn't it true that the rate of change of a tidal gradient is dependent on the size of the object causing the gradient? So a larger black hole would have a smaller tidal gradient for a given length of rod?
Yes.

If when two identical objects pass nearby each other the tidal forces on them significantly differ, then movement is an intrinsic property, a property of an object that does not depend on any other object.

No, the different tidal forces are caused by movement relative to whatever mass is causing the gravity in question, not any kind of abstract or absolute "movement." Your silly conclusions only result if you neglect to include the third massive object that is producing the gravitational field in the first place.


An object could determine whether it’s moving by measuring the tidal strain on itself. Doesn’t that sound silly?

No, not when you consider that all you'd be determining is the object's motion relative to the gravitating object. You can measure this in any number of ways, and it's not at all surprising that it would be reflected in the tidal forces.


Do you think that when two identical objects pass nearby each other the tidal forces on them can significantly differ?

Certainly, as long as their velocities relative to the gravitating object are substantially different. In this example, the lab is approaching the singularity at almost the speed of light, while the rod is moving away at almost the speed of light, so I'd expect radically different tidal forces in each frame.

Yes, this is certainly a problem - I don't have a great grasp of the metric. Like I said, it's a handwaving argument, and a proper treatment is required to resolve it properly.
You just need to show a problem logically. I showed that there need be no problem in the rod’s frame, in which case there is no problem in anyone else’s frame either. GR is not so strange that a rod that spans the horizon in its own frame does not span the horizon in some other frame. If you take into account the fact that the distance between r-coordinates is length contracted in the frame in which the rod is moving, then your paradox goes away.

Yes, that is an artificial condition, an artifice to show the tail-chasing situation of your prior post.
Well you can’t show a problem with the OP, or any other thought experiment, by adding a condition to it. That’s like disproving Einstein’s relativity of simultaneity thought experiment by starting with “Let the train derail ...”

But, you acknowledge that the tidal acceleration across the rod in the rod frame is not necessarily the same as the tidal acceleration across the rod in the falling lab frame.
Here “across the rod in the falling lab frame” just means “in a region the rod fits in when it’s length contracted”. That doesn’t mean that the tidal force the rod feels is affected by its velocity, as you imply. More on this below.

That story is a handwaving argument in a popular text. It's not rigorous, and would need to be supported by a proper treatment if there were a dispute over its validity.
I disagree. It’s a rigorous argument. An argument is rigorous if the validity of each step and the connections between the steps is explicitly made clear in such a way that the result follows with certainty (definition from MathWorld (http://mathworld.wolfram.com/Rigorous.html)). Then you can prove it’s not rigorous only by showing ambiguity. The passenger is given to be in the train, so it’s a certainty that the train is large enough to fit the passenger. Einstein needn’t have specified the size of the train or the passenger to make his argument rigorous. What is your definition of rigorous?

Your OP is a handwaving argument, that is not convincing to me.
You haven’t shown that it’s not rigorous. You’ve not shown any ambiguity. You’ve simply disputed one of GR’s predictions upon which the argument is based.

We are at a stalemate. I haven't proven GR to be consistent in the given scenario. You haven't proven GR to be inconsistent in the given scenario. We can proceed no further without a proper treatment.
Actually what you’ve done is request that I prove a negative. You’re asking me to prove that the tidal force is not dependent on velocity. I’ve shown you GR’s equation that predicts spacetime curvature (tidal force), and velocity is not an input. But you want me to prove that GR does not have some caveat or exception to that equation. It’s logically impossible for anyone to do that. Beyond that equation the best I can show you is that relativity would be nonsensical if you were right.

Specific quantities aren't necessary - you can use variables if you prefer.
Later I’ll post an equation from one of my books for the tidal force an object would feel given only M, r, and the length of the object. But it won’t stop you from asking me to prove that the equation is not dependent on velocity; i.e. prove a negative. You could still insist that the equation applies to only an infalling object, say. You could insist that the equation doesn’t apply to rods. Could I prove you wrong?

You acknowledged that the measured tidal acceleration between the two ends of the rod in this situation is frame dependent.
You had the lab measure the tidal force “between two points in spacetime with a separation in the lab frame of 2 metres and zero seconds”. That’s not measuring the tidal force the rod feels. Many rods, having various proper lengths (including much more than 2m) and velocities relative to the lab, could fit in that 2m region. The tidal force the lab measures in that 2m region would not necessarily be the same as the tidal forces that those rods feel. The rods could feel various tidal forces since their proper lengths can vary. The only tidal force that matters in the OP is the tidal force that the rod feels. Only a tidal force that the rod feels can break it. Your example with the lab and the rod didn’t show anything amiss with the OP.

Not so - the tidal force need only be strong enough to pull the escaping end across the horizon into the hole.
The tidal force can’t do that. GR predicts that an object above the horizon can escape to r=infinity regardless of the tidal force at the horizon or on the object. The tidal force stretches and squeezes an object in place; it doesn’t pull it in a particular direction. There are high tides on both sides of the Earth even when both the Sun and the Moon are on the same side of the Earth.

Ignoring a useful frame doesn't make it go away.
Nevertheless the OP does not rely on SR or an inertial frame.

And the "not exactly flat" argument is precisely how you can have the top of the rod at escape velocity, so I can see why it makes you uncomfortable.
The part of the rod that is above the horizon can escape to r=infinity regardless of the tidal force (spacetime curvature) on it. GR does not require that the tidal force on an escaping object be below some value. The escape velocity at the top of the rod could be 0.5c if the rod had sufficient tensile strength.

Also, your OP relies on citations that GR allows negligible tidal forces in an infalling inertial frame.
That doesn’t mean that the tidal force can be negligible only in infalling frames. Those are just common examples. “Negligible” in the OP means a tidal force that is too weak to break the rod. The tidal force can be arbitrarily high in principle, since the rod's tensile strength can be arbitrarily high.

Careful... if the rod's tensile strength is arbitrarily large, than a large enough tidal force would pull the escaping end across the horizon into the hole without breaking the rod. To be negligible, the tidal force must also be too weak to stop the rod from escaping
I covered this above.

Like I said - a proper treatment is required.
It seems that your definition of “proper treatment” includes proving a negative. I’ve given a proper treatment by showing you GR’s equation for spacetime curvature (tidal force), and noting that velocity is not an input. I’ve noted that the definition of tidal force does not mention velocity, and the two particles used to measure the tidal force can each have any velocity in the observer’s frame. That should have been good enough.

No, the different tidal forces are caused by movement relative to whatever mass is causing the gravity in question, not any kind of abstract or absolute "movement."
Not according to GR. According to GR, tidal forces are independent of movement. Hence a difference in tidal forces is also independent of movement. In GR the tidal force an object feels is dependent only on the mass of the body that is “causing the gravity in question”, the size of the object and the object’s distance from the body.

Your silly conclusions only result if you neglect to include the third massive object that is producing the gravitational field in the first place.
No, I didn’t neglect that. You can assume that at least one body is imparting a tidal force on the two identical objects.

No, not when you consider that all you'd be determining is the object's motion relative to the gravitating object. You can measure this in any number of ways, and it's not at all surprising that it would be reflected in the tidal forces.
Movement is not reflected in the tidal force according to GR. In change of tidal force yes.

Certainly, as long as their velocities relative to the gravitating object are substantially different. In this example, the lab is approaching the singularity at almost the speed of light, while the rod is moving away at almost the speed of light, so I'd expect radically different tidal forces in each frame.
Can you support that with some equation from GR? The theory disagrees with you.

Taylor and Wheeler say that the Schwarzschild metric is a complete description of empty spacetime around a Schwarzschild object. The metric differs from the SR’s metric for flat spacetime by only one equation incorporated into it:

spacetime curvature factor = sqrt(1 - (2M / r))

Spacetime curvature is synonymous with tidal force. Then the tidal forces on the identical objects should be identical when they pass nearby each other, because the r’s (on either end of the objects) are the same and M is the same, and there are no other inputs to the equation. For Schwarzschild geometry, GR has nothing more to say about tidal force than this equation.

zanket, all you need to do is produce an unambiguous situation in which GR predicts that the rod will escape the event horizon, or in which GR predicts that the tidal force on the rod is negligible. Pick some numbers, apply the equations, give the result.

You have already acknowledge that tidal force on the rod is dependent on the reference frame. Like you said, "The tidal force the lab measures in that 2m region would not necessarily be the same as the tidal forces that those rods feel.[/quote]
So we agree, do we not, that the tidal force along the rod in the rod frame might be significantly larger that in the in-falling lab frame?
So it doesn't matter that we agree that the tidal force in the in-falling lab frame can be negligible - you need to support your assertion that it is possible for the tidal force in the rod frame to be negligible.

Anyway, I'm done with handwaving. I agree with your cited definition of rigorous, and note that ambiguity is the opposite of certainty.

I require an answer to this direct and relevant question:
For what hole mass, rod length, rod speed, and rod position does GR predict that the tidal acceleration along the rod in the rod frame is small enough that the rod can escape the event horizon?

Spacetime curvature is synonymous with tidal force. Then the tidal forces on the identical objects should be identical when they pass nearby each other, because the r’s (on either end of the objects) are the same and M is the same, and there are no other inputs to the equation. For Schwarzschild geometry, GR has nothing more to say about tidal force than this equation.

Armwaving again.

In Kip Thorne's excellent book (the source of this assertion), he says:
Thus, Einstein and Newton, with their very different viewpoints on the nature of space and time, give very different names to the agent that causes the crossing. Einstein calls it spacetime curvature; Newton calls it tidal gravity. But there is just one agent acting. Therefore, spacetime curvature and tidal gravity must be precisely the same thing, expressed in different languages.

Professor Thorne is eloquently describing a fundamental connection between alternative explanations for tidal accelerations... and that's all.

He does not intend that the two things are numerically equivalent.
He does not say or imply that tidal acceleration (or force) is equal to the difference in the Schwarzschild metric's spacetime curvature factor.

And regardless, between which two events would you compare the curvature factor? Simultaneous events at different locations (ie each end of the rod at some instant)?

Not according to GR. According to GR, tidal forces are independent of movement. Hence a difference in tidal forces is also independent of movement. In GR the tidal force an object feels is dependent only on the mass of the body that is “causing the gravity in question”, the size of the object and the object’s distance from the body.

No, the motion of the frame in question relative to the gravitator also matters, particuarly when the speed becomes relativistic. All you've done is spit out a bunch of unsupported assertions, and I'm not convinced by any of them. Pete has already attempted to get you to follow the math behind this, and you even admitted that the tidal force was frame-dependent, so I don't know what you're trying to accomplish with this stuff...

Taylor and Wheeler say that the Schwarzschild metric is a complete description of empty spacetime around a Schwarzschild object. The metric differs from the SR’s metric for flat spacetime by only one equation incorporated into it:

spacetime curvature factor = sqrt(1 - (2M / r))

Spacetime curvature is synonymous with tidal force. Then the tidal forces on the identical objects should be identical when they pass nearby each other, because the r’s (on either end of the objects) are the same and M is the same, and there are no other inputs to the equation. For Schwarzschild geometry, GR has nothing more to say about tidal force than this equation.

No, spacetime curvature is not synonymous with tidal force. Spacetime curvature is the underlying *cause* of tidal forces, but that doesn't mean that the terms are interchangeable. "Spacetime curvature" refers to the curvature of a theoretical 4-dimensional manifold, while "tidal force" refers to differences in a (measurable) frame force throughout a lab (i.e., some finite region of spacetime). They are two distinct ideas, applicable to distinct contexts, that happen to be related. It's certainly true that spacetime curvature doesn't depend on the motion of the rod (supposing its mass is negligible compared to the black hole that is), but that doesn't mean that the tidal forces induced by said curvature are independent of the rod's motion.

zanket, all you need to do is produce an unambiguous situation in which GR predicts that the rod will escape the event horizon, or in which GR predicts that the tidal force on the rod is negligible. Pick some numbers, apply the equations, give the result.
Do you see any numbers or equations in Einstein’s relativity of simultaneity thought experiment? Only logic was necessary there. You’re asking for more than a rigorous proof. The OP is already unambiguous. It is based on predictions of GR that you dispute without good reason; that doesn’t show a flaw of it.

So we agree, do we not, that the tidal force along the rod in the rod frame might be significantly larger that in the in-falling lab frame?
Yes.

So it doesn't matter that we agree that the tidal force in the in-falling lab frame can be negligible - you need to support your assertion that it is possible for the tidal force in the rod frame to be negligible.
That has already been supported in spades. The only reason the tidal force along the rod in the rod frame might be significantly larger than that in the in-falling lab frame is because the rod might be significantly longer than 2m. So just give the rod a proper length of 2m. Or—as is given in the OP—make the black hole large enough that the tidal force on the rod (regardless of its proper length) is negligible.

I require an answer to this direct and relevant question:
For what hole mass, rod length, rod speed, and rod position does GR predict that the tidal acceleration along the rod in the rod frame is small enough that the rod can escape the event horizon?
I say it’s overkill, but here is a valid answer:

A mass of 10 million solar masses.
A rod proper length of 1 m.
Any rod speed.
Any rod position.
Any rod color.

What about my question about the relativistic rocket site? Why do you think the author neglected to mention the crippling fact that tidal force will rip apart the rocket when it reaches some velocity?

He does not intend that the two things are numerically equivalent.
You’re really disputing that “precisely the same thing” does not mean they are the same thing? Really?

He does not say or imply that tidal acceleration (or force) is equal to the difference in the Schwarzschild metric's spacetime curvature factor.
In the supporting info Taylor and Wheeler agree with Thorne, saying “Curvature is tidal acceleration and tidal acceleration is curvature.” Then a difference in spacetime curvature is a difference in tidal force.

And regardless, between which two events would you compare the curvature factor? Simultaneous events at different locations (ie each end of the rod at some instant)?
What matters to the OP is the tidal force the rod feels. So let equipment on the rod measure the relative acceleration of two free test particles that are at either end of the rod at the same time in the rod’s frame. The curvature factor tells us at a glance that there is a black hole large enough that the tidal force the rod measures is arbitrarily small, regardless of the rod's speed, position, or color.

You've degenerated into being argumentative, so this is no longer an Alpha thread.

Thanks for the education, but I don't think it's productive to continue now.

You say that two things that are “precisely the same thing” are not numerically equivalent, and I'm the one being argumentative?

Is logic forbidden on sciforums now?

Look at the quote.
Exactly what two things does he say are "precisely the same thing"?
Tidal acceleration and the Schwarzschild metric's spacetime curvature factor?

No. He says that "tidal gravity" and "spacetime curvature" are precisely the same thing. Unfortunately, these are qualitative rather than quantitative concepts. It does not follow that any given quantitative representations of those things will be equivalent.

Yes, you are being argumentative. I think that you are too close to your argument to consider it objectively. That's perfectly understandable, but it makes productive discussion difficult, if not impossible.

What matters to the OP is the tidal force the rod feels. So let equipment on the rod measure the relative acceleration of two free test particles that are at either end of the rod at the same time in the rod’s frame. The curvature factor tells us at a glance that there is a black hole large enough that the tidal force the rod measures is arbitrarily small, regardless of the rod's speed, position, or color.
You're glancing too quickly. Perhaps if you plugged in some numbers to check?
A quick way to determine that something odd is happening is to consider a rod with one end touching the event horizon, and the other at some distance outside. Give the rod a velocity of zero or greater away from the horizon.
The Schwarzschild metric tells us that in the rod's reference frame, the length of the rod is infinite.

So straight away, your flippant response:
I say it’s overkill, but here is a valid answer:

A mass of 10 million solar masses.
A rod proper length of 1 m.
Any rod speed.
Any rod position.
Any rod color.
Is clearly not valid.

No. He says that "tidal gravity" and "spacetime curvature" are precisely the same thing. Unfortunately, these are qualitative rather than quantitative concepts. It does not follow that any given quantitative representations of those things will be equivalent.
He is saying that the two terms are synonyms for the same thing. He makes that ultra clear with “But there is just one agent acting.” If there is only one agent, then there is only one thing, with two names.

Yes, you are being argumentative.
I think you’re confusing being logical with being argumentative. You want to see numbers and equations, i.e. quantitative concepts, when those are not required, just like they aren’t required in Einstein’s thought experiments. Do you know that math is just one form of logic? Math is no more valid than any other form of logic.

A quick way to determine that something odd is happening is to consider a rod with one end touching the event horizon, and the other at some distance outside. Give the rod a velocity of zero or greater away from the horizon.
The Schwarzschild metric tells us that in the rod's reference frame, the length of the rod is infinite.
That’s a common misconception. The directly measured radial distance between the horizon and some higher point is finite. Taylor and Wheeler prove that in Exploring Black Holes.

So straight away, your flippant response:
The response is based on an example given in the supporting info (search for “10 million”). I included “color” to emphasize that anything that is not an input to the spacetime curvature factor will not affect the tidal force the rod feels.

He is saying that the two terms are synonyms for the same thing. He makes that ultra clear with “But there is just one agent acting.” If there is only one agent, then there is only one thing, with two names.
What two terms?
You are asserting that "tidal gravity is synonymous with spacetime curvature" means exactly the same as "the tidal acceleration between two events is exactly equal to the difference in the Schwarzschild metric's spacetime curvature factor between the same two events". You're going from ambiguous qualitative terms to specific quantitative terms without justification, and that is not logical.

I agree that tidal gravity is the same as spacetime curvature.
Now you need to show that tidal acceleration is quantitatively equal to the difference in the Schwarzschild spacetime curvature factor between two events.

And I still maintain that it not necessarily relevant, because you're comparing different events anyway.
That has already been supported in spades. The only reason the tidal force along the rod in the rod frame might be significantly larger than that in the in-falling lab frame is because the rod might be significantly longer than 2m. So just give the rod a proper length of 2m. Or—as is given in the OP—make the black hole large enough that the tidal force on the rod (regardless of its proper length) is negligible.
You haven't shown that this is possible.
You have shown (and I agree) that the tidal acceleration for a horizon-spanning infalling reference frame can be arbitrarily small.
You haven't shown the same for an escaping reference frame.
You can't simply make the hole larger, because this has other effects - making the hole larger means increasing either the escape velocity or the r-coordinate of the top of the rod.

I think you’re confusing being logical with being argumentative. You want to see numbers and equations, i.e. quantitative concepts, when those are not required, just like they aren’t required in Einstein’s thought experiments.
Einstein's thought experiments are handwaving popular descriptions. They are supported by rigorous quantitative descriptions.


That’s a common misconception. The directly measured radial distance between the horizon and some higher point is finite. Taylor and Wheeler prove that in Exploring Black Holes.
For all reference frames, or only for an infalling reference frame?

The response...
Is flippant. You threw out some numbers without attempting to determine if they met the requirements.

That’s a common misconception. The directly measured radial distance between the horizon and some higher point is finite. Taylor and Wheeler prove that in Exploring Black Holes.

I've plugged some numbers and found that you're right - thanks.
I've also found that you can't go one millimeter beyond the event horizon or distance becomes complex... which is a problem for your rod.

No, the motion of the frame in question relative to the gravitator also matters, particuarly when the speed becomes relativistic. All you've done is spit out a bunch of unsupported assertions, and I'm not convinced by any of them.
Your assertion in your first statement is unsupported. Can you support it? I don’t see any support anywhere else. I also see that it contradicts the equivalence principle; see my response to Pete below. I’m not convinced.

Pete has already attempted to get you to follow the math behind this, and you even admitted that the tidal force was frame-dependent, so I don't know what you're trying to accomplish with this stuff...
Pete’s claim that the math he requests is required is unfounded. He hasn’t proven that such math is required. I admitted only that the tidal force is proper-length dependent, which doesn’t refute the OP.

No, spacetime curvature is not synonymous with tidal force.
Then you disagree with all of Taylor, Thorne, and Wheeler in the supporting info.

Spacetime curvature is the underlying *cause* of tidal forces, but that doesn't mean that the terms are interchangeable.
Thorne says there is “only one agent acting”, therefore “they must be precisely the same thing, expressed in different languages”. He could not have made it clearer that they are synonymous. Taylor and Wheeler agree, saying “[tidal force] is curvature, curvature is [tidal force]”.

It's certainly true that spacetime curvature doesn't depend on the motion of the rod (supposing its mass is negligible compared to the black hole that is), but that doesn't mean that the tidal forces induced by said curvature are independent of the rod's motion.
Prove it. The Schwarzschild metric differs from the metric for flat spacetime only by the spacetime curvature factor equation I gave you. Then everything GR has to say about either spacetime curvature or tidal force (regardless whether you agree that they are synonymous) is encapsulated into that equation. The only inputs to that equation are r and M. And if you were right, the equivalence principle as stated by Einstein in the OP would be wrong, making GR self-inconsistent since the principle is a postulate of it. You can’t refute the OP by showing a self-inconsistency of GR. You can’t use a self-inconsistency of GR against a proof of that—that’s illogical. I’m not convinced by your unsupported claim.

You are asserting that "tidal gravity is synonymous with spacetime curvature" means exactly the same as "the tidal acceleration between two events is exactly equal to the difference in the Schwarzschild metric's spacetime curvature factor between the same two events".
I haven’t asserted this. ...

I agree that tidal gravity is the same as spacetime curvature.
... This I have asserted.

Now you need to show that tidal acceleration is quantitatively equal to the difference in the Schwarzschild spacetime curvature factor between two events.
The Schwarzschild metric differs from the metric for flat spacetime only by the spacetime curvature factor equation I gave you. Then everything GR has to say about either spacetime curvature or tidal force is predicted by that equation. Since there is only one agent acting, as Thorne notes, there can be no difference between spacetime curvature and tidal force at some level (they might be presented differently). The equation can rightly be called a tidal force factor equation.

And I still maintain that it not necessarily relevant, because you're comparing different events anyway.
You can forget all about the concept of spacetime curvature and focus on tidal force. Just call the spacetime curvature factor equation a tidal force factor equation. You can choose any one among a set of synonyms and ignore the rest. I will do that myself, starting now.

You haven't shown that this is possible.
You have shown (and I agree) that the tidal acceleration for a horizon-spanning infalling reference frame can be arbitrarily small.
You haven't shown the same for an escaping reference frame.
I have shown that this is possible, by ruling out velocity as a determiner of tidal force. Only the r-coordinate and the mass is an input to the tidal force factor equation. An equation is not dependent on something that is not an input to it.

There is another way to tell that the tidal force is independent of velocity. Look at the equivalence principle as stated by Einstein in the supporting info. If you were right then the equivalence principle would be wrong. If you were right then the laws of physics in some small, freely falling reference frames would greatly differ than those in an inertial reference frame in an idealized, gravity-free universe. The equivalence principle is a postulate of GR and no, Einstein’s statement of the principle in the supporting info is not handwaving. If you were right that the tidal force is dependent on velocity then GR would be self-inconsistent since it would contradict the equivalence principle. All the experimental tests of SR to date would be suspect since we can’t be sure that the lab was not escaping from a black hole somewhere at a relativistic velocity; agreement with SR would not indicate that one way or the other. You can’t refute the OP by showing a self-inconsistency of GR. You can’t use a self-inconsistency of GR against a proof of that—that’s illogical.

You can't simply make the hole larger, because this has other effects - making the hole larger means increasing either the escape velocity or the r-coordinate of the top of the rod.
Neither of those effects would affect the validity of the OP, so they are moot points. For example, if the escape velocity at the top of the rod was 1-(10^-70)c, or if the top of the rod was at r=10^70M, the OP would be fine. A higher escape velocity at the top of the rod is no problem. The tidal force on the rod needs to be infinite for the OP to be refuted. The tidal force on the rod can be made smaller by making the black hole larger.

Einstein's thought experiments are handwaving popular descriptions.
Prove they are handwaving, or quit saying this. I say they are rigorous proofs, according to the definition of rigorous. If you want to call them handwaving, then prove it by showing something ambiguous about them.

Is flippant. You threw out some numbers without attempting to determine if they met the requirements.
No flippancy. You didn’t ask for anything more than the numbers. I am confident they meet the requirements because the supporting info tells me that a human can survive falling through a 10 million solar masses black hole, a 1m rod is smaller than a human, and the tidal force is not dependent on velocity.

I've also found that you can't go one millimeter beyond the event horizon or distance becomes complex... which is a problem for your rod.
If that was a real problem then it would likewise be a real problem for anything passing inward through the horizon. But the supporting info shows there’s no real problem there. It’s a “coordinate singularity” problem, which only looks like a real problem until you switch to Gullstrand-Painlevé coordinates, a.k.a. “rain frame” coordinates, a.k.a. the river model. Even an observer below the horizon can determine the r-coordinate she’s passing; Taylor and Wheeler show how in Exploring Black Holes.

Your assertion in your first statement is unsupported.

You haven't proven that I need to support it. It's funny how you keep trying to shift the burden of proof onto others. This is a thread that *you* started in order to convince others of *your* proposition. The burden is unambiguously on *you*. The fact that you refuse to understand the various criticisms that have been posted doesn't convince anyone (other than you) that you're correct. You're not fooling anybody with these silly rhetorical games.

Then you disagree with all of Taylor, Thorne, and Wheeler in the supporting info.

Or rather, I disagree with your misunderstanding of Taylor, Thorne and Wheeler. Not that I'd have any qualms about disagreeing with that reference directly. It's not exactly what I'd consider authoritative.


He could not have made it clearer that they are synonymous.

Do you even understand what "synonymous" means? It's a very strong condition. It means that you can take any sentence that uses (as opposed to mentions) a particular term, replace every instance of the term with its synonym, and the sentence will have exactly the same meaning. That is not the case with "spacetime curvature" and "tidal force." For example, consider the following sentence:

"The tidal force is a secondary effect of the force of gravity"

This sentence does not make sense if you replace "tidal force" with "spacetime curvature," as spacetime curvature is not an effect at all, but rather a theoretical explanation for said effect. Just because spacetime curvature is the sole theoretical cause of tidal forces does not make the terms interchangable. Falling is strictly a manifestation of gravity, but that doesn't mean that "falling" is synonymous with "gravity."

The Schwarzschild metric differs from the metric for flat spacetime only by the spacetime curvature factor equation I gave you.

So what? You might as well argue that nothing is frame-dependent in flat spacetime, since the Minkowski metric doesn't depend on velocity. Just because you don't understand enough about the application of the Schwarzchild metric to see how tidal forces come out to be frame-dependent doesn't prove anything.

I haven’t asserted this. ...
You've based you're argument on it.

... This I have asserted.
And I agree. Note that the terms are qualitative, not quantitative.
Note that "tidal gravity" is ambiguous... is it "tidal acceleration", "tidal force", or something else?

The Schwarzschild metric differs from the metric for flat spacetime only by the spacetime curvature factor equation I gave you. Then everything GR has to say about either spacetime curvature or tidal force is predicted by that equation. Since there is only one agent acting, as Thorne notes, there can be no difference between spacetime curvature and tidal force at some level (they might be presented differently). The equation can rightly be called a tidal force factor equation.
This is not logical.
You are conflating "tidal gravity" (an ambiguous qualitative term) with "tidal force (a specific quantitative term).
You are also conflating different things simply because they have similar labels. The (1-2M/r) quantity in the schwarzschild metric is given the label "spacetime curvature factor"... but it's only a label. You could give it any label you want ("tidal gravity factor", for example), but that doesn't change what it is, and it does not make it equal to a different quantity with a similar label, such as tidal acceleration. Don't get caught in the semantics, focus on the quantities.

If you were right then the laws of physics in some small, freely falling reference frames would greatly differ than those in an inertial reference frame in an idealized, gravity-free universe.
If a reference frame contains a coordinate singularity, then I don't think that frame can be considered inertial.

Neither of those effects would affect the validity of the OP, so they are moot points. For example, if the escape velocity at the top of the rod was 1-(10^-70)c, or if the top of the rod was at r=10^70M, the OP would be fine.
Please support that assertion.
Show that they way that the variables affect each other allows a finite length rod to be both escaping and crossing the horizon.

Prove they are handwaving, or quit saying this. I say they are rigorous proofs, according to the definition of rigorous. If you want to call them handwaving, then prove it by showing something ambiguous about them.
You've done that yourself, by identifying some assumptions. The assumptions might be reasonable and obvious... but they're still assumptions.

No flippancy. You didn’t ask for anything more than the numbers. I am confident they meet the requirements because the supporting info tells me that a human can survive falling through a 10 million solar masses black hole, a 1m rod is smaller than a human, and the tidal force is not dependent on velocity.
You beg the question.

If that was a real problem then it would likewise be a real problem for anything passing inward through the horizon. But the supporting info shows there’s no real problem there.
Right. In the rod frame coordinates, there is a coordinate singularity.
For an infalling frame there is none.
It’s a “coordinate singularity” problem, which only looks like a real problem until you switch to Gullstrand-Painlevé coordinates, a.k.a. “rain frame” coordinates, a.k.a. the rive