General relativity is self-inconsistent

[andrewgray]

Zanket,

 

Don't forget we are talking about the consistency of GR. I realize that there can be different interpretations of acceleration, but we are talking about GR, and GR's treatment of acceleration. So here we must stick to that.

 

Here is the simple accelerometer given by MTW:

 

 

This is how GR treats and defines acceleration. If the balls go through the holes, the frame is not accelerating. If they don't, then there is acceleration. If we cannot get past this point, then we must digress further.

 

So, the balls would go through the holes in the freefall inertial frame next to the rocket, and the balls would not go through the holes if this accelerometer were attached to the rocket. That is, the rocket is accelerating and the inertial frame is not.

 

Andrew A. Gray

 

P.S. Note that "rotation" is a centripetal acceleration.

[Guest Zanket]

This is how GR treats and defines acceleration. If the balls go through the holes, the frame is not accelerating. If they don't, then there is acceleration. If we cannot get past this point, then we must digress further.

The device tests whether it's in free fall or noninertially accelerating. That doesn't mean that an object in free fall is not gravitationally accelerating.

 

I don't want to go down a blind alley here, so maybe we can cut to the chase. Do you think an object’s ability to increase its radial position is independent of gravitational acceleration? That is, do you think it's effortless to lift a dumbbell? Is that really what you're trying to show? It seems to me that you're trying to show that an object can increase its radial position despite an arbitrarily high gravitational acceleration on it, so you can show that an object can pass outward through r=2M. You're trying to show that it's effortless to lift a dumbbell. Obviously that is not the case.

[andrewgray]

In the meanwhile, if anybodies is able and willing, could they directly address this question, which summarizes my current confusion?

 

Craig,

 

In GR, there is a metric, so typically, the velocity in a frame is not say just [imath]\frac{dr}{dt}[/imath]. For example, in a Schwarzchild frame, V_r is defined as:

 

[math]V_r = \frac{|g_{rr}|^{1/2}}{|g_{tt}|^{1/2}}\frac{dr}{dt} = \frac{1}{\sqrt{1-2M/r}} \frac{1}{\sqrt{1-2M/r}} \frac{dr}{dt} = \frac{1}{(1-2M/r)} \frac{dr}{dt}[/math]

 

Why is this? Well it is best explained by example. In the polar coordinates metric we have

 

[math]-d\tau^2 = -dt^2 + dr^2 + r^2d\theta^2[/math]

 

You know what the tangential velocity is. But using the above formula gives:

 

[math]V_\theta = \frac{|g_{\theta \theta}|^{1/2}}{|g_{tt}|^{1/2}} \frac{d\theta}{dt} = \frac{rd\theta}{dt}[/math]

 

So the acceleration felt by a rocket at constant r (and Zanket, this acceleration is rocket-motor acceleration, not gravitational acceleration according to GR), is the acceleration of the rocket relative to an inertial frame that is momentarily comoving and in freefall right next to the rocket. This inertial frame would be a frame that "moves upwards from below the rocket, peaks right at the rocket, the starts downward" in freefall. To get this acceleration with a metric lurking about, things get a little messy. One must find the relationship between these two frames and then take the curvilinear second derivative of the motion with respect to proper time. I will look into doing this and get back.

 

Andrew A. Gray

[andrewgray]

Zanket, we have plenty of time, it seems to me. It also seems to me that if you are going to prove GR inconsistent, then you must learn GR's definitions.

 

So lets say that you are a prisoner on a spaceship, somewhere inbetween center of the Milky Way and the center of Andromeda. The aliens have their engines off. So you are not "rocket-motor" accelerating. You are given scientific equipment in your cell, and you cannot see outside. Can you tell if you are "gravitationally accelerating" or not?

 

If so, how? And can you tell how much? Can you "feel" the "gravitational acceleration"?

 

 

Andrew A. Gray

 

P.S. I am sorry about the blind alley trip. But it just doesn't seem that I can explain the rest of your logic unless we can agree on common definitions, and I think it important to use GR's. And according to GR, "gravitational acceleration" is an oxymoron, and the question cannot be answered, as it does not make sense according to GR.

[Guest Zanket]

Zanket, we have plenty of time, it seems to me.

Not really. I want to be efficient here. Can you answer my questions above about the dumbbell and where you're going with this?

 

So lets say that you are a prisoner on a spaceship, somewhere inbetween center of the Milky Way and the center of Andromeda. The aliens have their engines off. So you are not "rocket-motor" accelerating. You are given scientific equipment in your cell, and you cannot see outside. Can you tell if you are "gravitationally accelerating" or not?

All observers are subject to gravitational acceleration in our real, gravity-endowed universe. So the answer is yes.

 

If there was a rocket within the cell that is hovering above a body, I could measure the gravitational acceleration at the rocket's r-coordinate for that body in an MCIF of the rocket that I set up. In the MCIF some device I employ would measure the rocket's acceleration. (Then I could know the ratio of r to M, by plugging the rocket's acceleration into GR's equation for gravitational acceleration.) The fact that such rocket could always exist in my cell in principle proves that I can tell that I am gravitationally accelerating.

 

GR says that gravity is not a force that pulls down. That doesn't mean that gravity can't accelerate freely falling objects relative to r-coordinates. It can, in a way that those objects don't feel, but they can measure.

 

And according to GR, "gravitational acceleration" is an oxymoron, and the question cannot be answered, as it does not make sense according to GR.

If there was some problem with the concept, I doubt I'd get 687 educational institution (.edu) hits googling for "general relativity" "gravitational acceleration". That's a lot of professors misusing the term.

[Qfwfq]

In the meanwhile, if anybodies is able and willing, could they directly address this question, which summarizes my current confusion?

You appear to be treating it as you would a first derivative, except that the result would have the reciprocal of the root. I will suppose you mean the second derivative, for which the correct notation would be [math]\frac{d^2r}{dt^2}[/math]. Try again. I was considering to give you a hint to what you probably meant to do, regardless of whether it be useful starting from Newton's potential, but I realize I'm too short on time to be sure of getting it straight. It would take a bit more calculus anyway to address your query.

 

The text excerpts posted in this thread seem to tantalizingly support my idea of the “impossibility” of an object passing an event horizon being due to time dilation ([math] \frac{dt_0}{dt}[/math]), not a local mechanical effect, but without an adequate understanding of the full text, I can’t reach a confident conclusion.

A freefalling body, starting from some arbirary [math]r>r_S[/math], will inevitably fall through the event horizon in a finite proper time.

 

Zanket, inertial frames do not accelerate past the rocket. A frame is "accelerating" if and only if an accelerometer detects acceleration in the frame. In the freefall inertial frame momentarily at rest by the rocket, the accelerometer measures zero. Agreed?

 

Therefore, the inertial frame is not accelerating past the rocket. It is moving past the rocket with constant velocity. Only the rocket is accelerating since its accelerometer is measuring something finite. Agreed?

I disagree. I'm also at work and I've never been able to fully participate here, but I find you saying some things that differ from the basics of GR despite that you seem to know your way around the calculus. In this case you are disregarding the very idea of general covariance, elsewhere you said that phyisics "is only valid in" inertial coordinates and this is not exactly the general principle of relativity. :xx:

 

Zanket I wasn't being coy, you simply fail to understand the distinction I made. You are implicitly supposing that a box crossing the horizon will consequentially straddle it. This would be fine for the border between Switzerland and Liechtenstein but the event horizon is a totally different thing. Given the well known consequences of the Schwarzschild metric, upon crossing the event horizon, the implication you make is a paralogism.

You said:

The main trouble is, first, the meaning of saying that the box straddles the horizon.

 

To any reasonable person, this is suggesting a problem with the idea of something straddling the horizon.

And that is tautological, congratulations, but due to paralogism you did not address it. The statement I did make is relevant, although not indispensable, at least to start a purely logical evaporation of the contradiction you are discussing. If you got confused, you must not have noticed me reminding that for [math]r<r_S[/math] dr becomes timelike.

 

GR says that an inertial frame can fall through a horizon (see supporting info I gave you above). X is given to be such a frame. Then the box can be on both sides of the horizon simultaneously according to clocks throughout X and at rest with respect to it.

You are only repeating the paralogism, aside from any nitpicking on your wording. Let's just make it that a geodesic can cross the horizon, hence a free-falling object, which is related to there being a choice of a locally inertial chart according to which the object is at rest, for each point along the geodesic.

 

If A occurs, then the box cannot be at rest with respect to the escaping particle, in violation of SR, which allows any two objects to be at rest relative to each other in an inertial frame.

Does SR say that the box must be at rest with the esacping particle? In no way SR gives a method to infer that, if the particle can escape, so can the box. It is trivial to infer this if one supposes that the particle can't leave the box and the box doesn't break, but SR doesn't imply this in any way. Without the differential geometry of the Schwarzschild solution, there is no way to say whether the particle or any part of the box can escape.

 

If B occurs, then the frame is not inertial, because the tidal force is the only force that exists in X that could break the box, and the tidal force is negligible in an inertial frame by the definition of an inertial frame. Then if A and B are the only possibilities, GR is self-inconsistent.

First, B can occur in an inertial frame, who says only the tidal force exists? Second, who says the tidal force is negligible? There is no inertial frame near the event horizon, while there are charts that are locally inertial there; this is what the principle of equivalence says, as assumed for GR. This also means that:

If a “locally inertial chart” is different than an inertial frame, then it’s probably overkill for this discussion.

is false because locally inertial charts are definitely essential for discussing the Schwarzschild solution. If you say they are "probably" an overkill then you must get the differential geometry straight, before discussing these topics. By "getting it straight", I do not mean thinking that a space with non-zero intrinsic curvature can be equivalent to a "straight" space.

[Guest Zanket]

Does SR say that the box must be at rest with the esacping particle?

No, but it doesn't have to. SR allows the box to be at rest relative to the escaping particle. GR does not allow that. That's a self-inconsistency.

 

First, B can occur in an inertial frame, who says only the tidal force exists?

No other force is given. Thought experiments need not rule out every eventuality. For example, in his relativity of simultaneity thought experiment, Einstein did not have to rule out the possibility that the train derails. "Who says the train doesn't derail?" would not refute that thought experiment.

 

Second, who says the tidal force is negligible? There is no inertial frame near the event horizon, while there are charts that are locally inertial there; this is what the principle of equivalence says, as assumed for GR.

You disagree with quotes in the OP by Einstein, Taylor, Thorne, and Wheeler. Taylor and Wheeler in particular say that SR works fine in an inertial frame falling through a horizon. The equivalence principle as stated in the OP says that any small freely falling frame anywhere in our real, gravity-endowed universe is equivalent to an inertial frame in an idealized, gravity-free universe. Then clearly X can be an inertial frame in which SR applies. And by the definition of an inertial frame, the tidal force in it is negligible.

 

If you say they are "probably" an overkill then you must get the differential geometry straight, before discussing these topics.

No, the supporting info in the OP already amply supports the proof there. There's no need to obfuscate. Can you show that T&W are wrong about SR working fine in an inertial frame falling through a horizon?

[andrewgray]

GR says that gravity is not a force that pulls down. That doesn't mean that gravity can't accelerate freely falling objects relative to r-coordinates. It can, in a way that those objects don't feel, but they can measure.

 

Zanket, we almost have it here. We are simply arguing strictly about semantics here, but it is important.

 

Of course one can think of things "accelerating" relative to accelerated coordinates. But the point is that you cannot feel "gravitational acceleration" if you are in freefall. And, GR treats freefall as non-acceleration, because all accelerometers measure zero acceleration. If we cannot get past this, then I will have to wave a white flag.

 

So in your view (both Qfwfq and Zanket), in the hyperbolic scenario in flat spacetime, the inertial observers are "accelerating" past the hyperbolic astronaut?

 

 

Andrew A. Gray

 

P.S. Where am I going with this? Well, I want to get into the details of the accelerated time coordinates going to infinity during the crossings.

[Guest Zanket]

Of course one can think of things "accelerating" relative to accelerated coordinates. But the point is that you cannot feel "gravitational acceleration" if you are in freefall. And, GR treats freefall as non-acceleration, because all accelerometers measure zero acceleration.

That’s fine, as long as you don’t think that gravitational acceleration has no effect on an object in free fall just because it can’t feel it.

 

So in your view (both Qfwfq and Zanket), in the hyperbolic scenario in flat spacetime, the inertial observers are "accelerating" past the hyperbolic astronaut?

Sure. Acceleration is just a change in velocity over time relative to something. An inertial observer can measure its acceleration relative to something hyperbolic, in which case the observer and that thing are accelerating relative to each other by definition. It doesn’t matter whether the inertial observer or the hyperbolic thing measures the acceleration.

 

P.S. Where am I going with this? Well, I want to get into the details of the accelerated time coordinates going to infinity during the crossings.

I think that would be a lost cause. To forestall any argument that uses infinity, I’ve been careful to restrict my refutations of your claims to the region between r=2M and r=2M+ε, where ε is zero in the limit. Gravitational acceleration is arbitrarily high above r=2M and below r=2M+ε, preventing any material object from reaching r=2M+ε from r=2M. Then the box in the OP cannot be at rest relative to the particle, making GR self-inconsistent. Go ahead and make your argument, but it’s likely I’ve already refuted it, in which case I would just refer to the existing refutation.

[Guest Zanket]

Without the differential geometry of the Schwarzschild solution, there is no way to say whether the particle or any part of the box can escape.

Missed this one. No, this is just obfuscation. That’s like saying, “Without the differential geometry of the Schwarzschild solution, there is no way to say whether GR predicts black holes”. As if a thousand textbooks saying that GR predicts black holes are immaterial. It is perfectly okay for a thought experiment to refer to widely accepted predictions. The OP lists the relevant predictions from reputable sources.

[Qfwfq]

No, but it doesn't have to. SR allows the box to be at rest relative to the escaping particle. GR does not allow that. That's a self-inconsistency.

It is not a self-inconsistency, you are doing exactly what I warned against, i. e. thinking that a space with non-zero intrinsic curvature can be equivalent to a flat space.

 

No other force is given. Thought experiments need not rule out every eventuality. For example, in his relativity of simultaneity thought experiment, Einstein did not have to rule out the possibility that the train derails. "Who says the train doesn't derail?" would not refute that thought experiment.

No other force is given, in my fictuous example of the Swiss-Liechtenstein border? This was the Gedankenexperiment in question, and I did say "by chopping it off". You can't refute my argument by claiming that SR forbids to use an axe or a chainsaw, even less by saying I needn't rule out every eventuality. To the contrary, and unlike derailing trains, I envisaged it. It was just one branch in the beginning of a logical argument lacking the objection I myself raised and, also, it is believed the tidal force will tear objects to shreads by the time they reach the event horizon. What do all those X-rays come from?

 

You disagree with quotes in the OP by Einstein, Taylor, Thorne, and Wheeler. Taylor and Wheeler in particular say that SR works fine in an inertial frame falling through a horizon. The equivalence principle as stated in the OP says that any small freely falling frame anywhere in our real, gravity-endowed universe is equivalent to an inertial frame in an idealized, gravity-free universe. Then clearly X can be an inertial frame in which SR applies. And by the definition of an inertial frame, the tidal force in it is negligible.

 

No, the supporting info in the OP already amply supports the proof there. There's no need to obfuscate. Can you show that T&W are wrong about SR working fine in an inertial frame falling through a horizon?

Aside from appeal to authority, you are committing the same fallacy as above. How small is "small"? It's a colloquial enough expression when Einstein, in Die Grundlagen, (and others) say "infinitely small" but he explicitly states that his aim is not to be rigourous and refers readers to the sources of differential geometry. To further remove the word infinitely causes you to err.

 

In a proper treatment, the meaning of a chart being locally inertial for a given point of space-time is that certain first-order terms are zero, the dynamics differ from SR by terms of higher order. Can your box be infinitely small? For a given size of it, there can always be fields in which tidal forces are not negligible and, anyway, these depend on intrinsic curvature. The principle of equivalence, as stated for GR, is that you can always choose a chart such that the tangent space is described by SR.

 

Missed this one. No, this is just obfuscation. That’s like saying, “Without the differential geometry of the Schwarzschild solution, there is no way to say whether GR predicts black holes”. As if a thousand textbooks saying that GR predicts black holes are immaterial. It is perfectly okay for a thought experiment to refer to widely accepted predictions. The OP lists the relevant predictions from reputable sources.

Now if that isn't obfuscation! On just what grounds do you think those thousand(!?!!!) textbooks say that GR predicts black holes? Beware of replacing actual reason with appeal to authority. Like, wow! Like someone says that we wouldn't have inferred the universe is expanding without having observed redshift, and you piss him down by pointing out how many folks say that it is! Ipse dixit!

 

So in your view (both Qfwfq and Zanket), in the hyperbolic scenario in flat spacetime, the inertial observers are "accelerating" past the hyperbolic astronaut?

A free-falling object is accelerating according to any observer that isn't free-falling. Where's the problem? I can see the trouble is basically just in your wording, I think this has aggravated the problem in getting across to one that is reading very divulgative text and trying to prove something based on colloquial explanations not properly interpreted.

[Guest Zanket]

It is not a self-inconsistency, you are doing exactly what I warned against, i. e. thinking that a space with non-zero intrinsic curvature can be equivalent to a flat space.

The spacetime in an inertial frame need not be perfectly flat, according to Einstein, Taylor, Thorne, and Wheeler, as shown in the supporting info in the OP. You’re making what I call the “not exactly flat” argument, which I refute in the OP. The fact that the spacetime in X is negligibly curved, not perfectly flat, does not explain any degree of difference between the laws of physics in X and the laws of physics in an inertial frame in perfectly flat spacetime. In his equivalence principle as quoted in the OP, Einstein says that the laws of physics in a frame in which the spacetime is negligibly curved, like X, are the same as they are in an ineritial frame in which the spacetime is perfectly flat. He did not use the word “same” to indicate “different”, like you are implying. Einstein did not redefine the words “same” and “equivalence” to mean the opposite of what the dictionary says they mean. The equivalence principle refutes you. It clearly says the opposite of what you are claiming.

 

No other force is given. Thought experiments need not rule out every eventuality.

You can't refute my argument by claiming that SR forbids to use an axe or a chainsaw, even less by saying I needn't rule out every eventuality.

In my quote I did not say or imply either thing you suggest I did. In my first sentence I'm talking about my thought experiment, the proof in the OP, not yours. The box in the OP need not break because the tidal force is given to be negligible (because X is given to be an inertial frame) and no other force is given in the proof which might break the box. And thought experiments like mine need not add text like "and no one tears apart the box with an axe, or a chainsaw, or a sledgehammer, or a ..." just like Einstein didn't need to add "and the train doesn't derail" to his thought experiment showing relativity of simultaneity.

 

... it is believed the tidal force will tear objects to shreads by the time they reach the event horizon.

The supporting info in the OP refutes this; the tidal force in a region straddling the horizon can be negligible. Search the OP for “Here are confirmations”.

 

What do all those X-rays come from?

The question isn’t relevant here. The OP already shows ample evidence from reputable sources that an inertial frame can fall through a horizon, and the tidal force in an inertial frame is negligible by the definition of an inertial frame.

 

How small is "small"?

This question is answered in the OP by Thorne: “small enough for tidal gravitational accelerations to be negligible inside it”.

 

It's a colloquial enough expression when Einstein, in Die Grundlagen, (and others) say "infinitely small" but he explicitly states that his aim is not to be rigourous and refers readers to the sources of differential geometry. To further remove the word infinitely causes you to err.

Do you really think SR is not experimentally confirmed, since we don’t have infinitely small measuring equipment? If SR is widely accepted to be experimentally confirmed, then you are refuted, and it is. The smaller the inertial frame (or the shorter its duration), the more accurate SR’s predictions are expected to be; that’s all that’s implied by “infinitely small”. Taylor and Wheeler explain this in the supporting info in the OP.

 

Can your box be infinitely small? For a given size of it, there can always be fields in which tidal forces are not negligible and, anyway, these depend on intrinsic curvature. The principle of equivalence, as stated for GR, is that you can always choose a chart such that the tangent space is described by SR.

The equivalence principle is given in the OP as stated by Einstein himself, and it shows that the box need not be infinitely small, just like the measuring equipment used for experimentally confirming SR need not be infinitely small. Taylor, Thorne, and Wheeler also make that clear in the OP.

 

On just what grounds do you think those thousand(!?!!!) textbooks say that GR predicts black holes? Beware of replacing actual reason with appeal to authority. Like, wow!

It is an extremely widely accepted practice in scientific works to refer to the scientific works of others, using their works as a basis for an argument. Just like you did above when you said “it is believed the tidal force will tear objects to shreads ...” Except that you only alluded to a reference, whereas I gave mine.

[Qfwfq]

Ummmh, ahhh, wait, what's that? Cup of :cup:, OK, coming... another :cup:... getting weary, really needed a weekend's rest from all this...

 

It is also an extremely widely accepted practice for masons to lay new tiers on top of the lower ones, that are already stable. Does this mean that, if you later remove the lowest ones, you could expect the upper tiers to remain where they are without alternative support? Zanket if you want to persuade people that GR has an inconsistency then you must think scientifically; bootstrap arguments will get you nowhere.

 

Now, I see you're finally starting to get it about the equivalence principle but you're still in a muddle and, :omg:, you're even saying it refutes me! You mean that what I've been trying to tell you refutes me?

Anyway you still basicaly don't have the concept clear. What you're finally saying is essentially that coordinates might be not inertial but locally inertial (for a given point, I would add), and I in no manner said that:

SR is not experimentally confirmed, since we don’t have infinitely small measuring equipment

so NO, I do NOT think this at all and neither am I making the “not exactly flat” argument, if you don't understand what I've been saying, get the calculus straight before trying to discuss it. But to say:

The fact that the spacetime in X is negligibly curved, not perfectly flat, does not explain any degree of difference between the laws of physics in X and the laws of physics in an inertial frame in perfectly flat spacetime.

shows you are still not there yet, at all, so quit presuming to teach me GR and quit quoting your OP as if quoting Einstein. 'Infinitely small' is an expression used by many, including Einstein and you can find it in Die Grundlagen, just to spare repeating all those epsilons and lessthans; initially you had omitted any manner of specifying and I had only, simply pointed this out. Anyway if that statement were true, what would the Riemann tensor be for? Even the connection [imath]\Gamma_{\mu\nu}^\sigma[/imath] is superfluous in a flat space, a choice of chart is enough to make it identically vanish. Oh, you meant perhaps to say that about the tangent space? Wasn't that what I said?

 

...and the tidal force in an inertial frame is negligible by the definition of an inertial frame.

Oh no! Not "by the definition of an inertial frame", no sir! If it is negligible for a given extended object, of a given shape and size, in a given position in a given field, it is negligible regardless of the chart (and likewise if it isn't negligible). This is true else general covariance schmemeral schmovariamce, choice of chart changes the representation of the same physical thing.

 

Now, obviously, the smaller the box the smaller the tidal force but, for a typical black hole, by the time you reach the event horizon it'll need to be a mighty small box indeed. But this doesn't even matter, you've been battling against one of the two points of my initial argument that are alternatives, IOW they weren't meant to be both necessary in a given case, I included derailing the train... :doh: uhm, breaking the box only to be exhaustive, and I meant it to be regardless of who or what breaks the box, so there's no point saying:

In my first sentence I'm talking about my thought experiment, the proof in the OP, not yours.

especially if you trace back upstream through the replies; this exchange began with your quote of me mentioning my alternative about the box breaking: "First, B can occur in an inertial frame, who says only the tidal force exists?".

 

Now, if you want to make progress and have a reasonable discussion, instead of getting lost and taking us round in circles about the case in which it already seems clear there is no contradiction, can we get on with the case in which the box doesn't break? That is, regardless of whether there are black holes for which this can ocurr and of whether it makes sense to say the box straddles the horizon, things I shall not discuss in this thread. So, the little ball is still on the Swiss side but the box can no longer come back, the guards have claimed that side of it as Princely property, this is irrevokable and makes that side unexportable. Axiom. In what manner does this contradict SR?

[Guest Zanket]

Zanket if you want to persuade people that GR has an inconsistency then you must think scientifically; bootstrap arguments will get you nowhere.

There’s no circular reasoning in the proof. You don't prove otherwise; you just suggest that there is.

 

Now, obviously, the smaller the box the smaller the tidal force but, for a typical black hole, by the time you reach the event horizon it'll need to be a mighty small box indeed.

Or for an atypical black hole it could be a cube 1 light year per side in proper length, with a tidal force in it less than the tidal force on your body now.

 

Now, if you want to make progress and have a reasonable discussion, instead of getting lost and taking us round in circles about the case in which it already seems clear there is no contradiction, can we get on with the case in which the box doesn't break?

There’s a lot of obfuscation in your post. We’re talking about one of the simplest things in GR: an inertial frame. The supporting info in the OP shows that an inertial frame can fall through a horizon. Either you can refute that, which you haven’t, or else there’s no need to mention a Riemann tensor. An inertial fame need not be infinitely small. Either you can refute that, which you haven’t, or else there’s no need to suggest otherwise.

 

... quit quoting your OP as if quoting Einstein

The equivalence principle in the OP is a quote from Einstein. He says “small”, not “infinitely small”. Taylor, Thorne, and Wheeler agree with Einstein there and show how small in their definitions given in the OP. They make it clear that an inertial frame need not be infinitely small, which is good because otherwise SR could not be experimentally confirmed. That said, the proof in the OP works for an arbitrarily small inertial frame X.

 

That is, regardless of whether there are black holes for which this can ocurr and of whether it makes sense to say the box straddles the horizon, things I shall not discuss in this thread. So, the little ball is still on the Swiss side but the box can no longer come back, the guards have claimed that side of it as Princely property, this is irrevokable and makes that side unexportable. Axiom. In what manner does this contradict SR?

It contradicts SR because SR does not require that the box and the particle move relative to each other. In SR you can put any two objects anywhere in an inertial frame and let them be at rest relative to each other. But GR does not allow the box and the particle to be at rest relative to each other. When GR says that SR applies in X but then overrides SR in X, it is self-inconsistent.

[Qfwfq]

You're just plain arguing pointlessly and not making much sense Zanket.

 

It contradicts SR because SR does not require that the box and the particle move relative to each other. In SR you can put any two objects anywhere in an inertial frame and let them be at rest relative to each other. But GR does not allow the box and the particle to be at rest relative to each other.

This doesn't contradict SR. Who says that GR does not allow the box and the particle to be at rest relative to each other?

[Guest Zanket]

Who says that GR does not allow the box and the particle to be at rest relative to each other?

The supporting info in the OP does. The box cannot pass outward through the horizon, because nothing can do that, according to the definition of a horizon. The particle is given to be escaping to r=infinity, which GR allows. The particle’s radial position is increasing; the box’s cannot. Then they cannot be at rest relative to each other.

[andrewgray]

Just one thing to Qfwfq:

 

I just wanted to remind you that I showed that the curvature tensor for the Schwarzchild metric goes as

 

So near r=2M, the curvature tensor goes as 1/M². For M not too small, the curvature tensor can be arbitrarily small, which means that spacetime near r=2M can be arbitrarily flat relative to the size of a human.

 

And Zanket, this means that your spanning box can escape.

 

But I know we disagree. So I will make one last attemp to make you see the light.

 

First of all, we know that in relativity, velocities are relative. If one has 2 inertial frames, one moving with velocity V with respect to the other, all physics is defined and all transformations are defined between them.

 

Next if one wants to know things from the point of view of the "moving" frame, one must only substitute -V for V and the same equations from the point of view of the other frame work fine, and again all physics and transformations hold.

 

For accelerations between frames, this is simply not true if one of them is inertial. If one frame is inertial and another frame has acceleration a, then indeed all physics is known as are all transformations. However, the laws of physics apply in the inertial frame, and they do not apply in the accelerated frame. Example: in our reference frame here on Earth, one must apply a fictitious Coriolis Force to projectile motion.

 

One cannot simply change a to -a and get the same result as one does with velocities. In other words, accelerations are not relative. Inertial frames have no acceleration with accelerometers measuring nothing. Accelerated frames have accelerometers that measure something.

 

This is General Relativity at its core. In GR, the freefall geodesic equation has a=0 by definition:

 

[math]\frac{D^2 x^\alpha}{d \tau^2} = 0 [/math]

 

or

 

[math]\frac{d^2x^\alpha}{d \tau^2} + \Gamma ^\alpha _{\mu \gamma} \frac{dx ^\mu}{d \tau} \frac{dx^\gamma}{d \tau} = 0 [/math]

 

where the Γ's are the Christoffel symbols found in curvilinear calculus.

 

These are the frames, defined in GR, where accelerometers measure no acceleration, and the laws of special relativity physics apply. No fictitious Coriolis forces. Just simple flat spacetime physics in the small flat neighborhood around these frames. For example, in a spaceship that is floating in space, one can throw a protein bar to your buddy, and it will go in a straight line to him. However, if the spaceship is accelerating, the protein bar will go in a parabolic path to your buddy.

 

Finally, according to GR, one cannot tell the difference if the spaceship is accelerating at g out in the middle of free space, or if the spaceship is accelerating simply by standing at Cape Canaveral ready to lift off. In both cases, if one throws a protein bar to your buddy, the bar makes a parabolic path over to him. This confirms that both cases are accelerating, according to GR, and that they are not inertial. In inertial frames, thrown protein bars go in a straight line. This is the equivalence principle at its core.

 

Now back to Zanket's logic. Zanket, if one takes a cube that is 1 cm on each side, and puts it out at large r, then obviously, the (equatorial) Schwarzchild coordinates measure:

 

[math]r d\phi = 1 cm[/math]

[math]r d\theta = 1 cm[/math]

[math]\frac{dr}{(1-2M/r)} \approx dr = 1 cm[/math]

 

However, when one approaches r=2M, do you know what the cube measures on each side?

 

You guessed it. Each side measures 1 cm! However, the Schwarzchild coordinates give:

 

[math]r d\phi = 1 cm[/math]

[math]r d\theta = 1 cm[/math]

[math]\frac{dr}{(1-2M/r)}= \infty[/math]

 

We see that r is no longer a valid radial coordinate at r=2M. For r<2M, these coordinates give negative distances for positive dr's. So it is clear that we have a coordinate problem, since spacetime can be arbitrarily flat here. The cube is indeed just 1 cm on each side.

 

Coordinate problems are not the fault of a theory. Take Norwegian Roald Amundsen trekking across the South Pole. His path crossed a coordinate singularity right at the pole. If one wanted to describe his motion or his velocity derivative going across the pole, one would have to change coordinates, as the normal r,θ,φ coordinates are unusable. So the theory of walking across the pole is not inconsistent if the coordinates used are singular there. The prescription is to just change coordinates and describe the motion with them. If one puts the South Pole on the new coordinate equator, then all is well, and Roald's trek across the South Pole is not unusual.

 

The same is true for the Schwarzchild coordinates at r=2M. If one insists on using these coordinates for crossing the horizon, t→±∞. Here are the plots of these crossings:

 

 

We see that they are exactly time reversal symmetric, but the t coordinate is ill-behaved at r=2M. Lets compare these crossing with the crossing of R=0 of the hyperbolic Rindler Horizon in flat spacetime. If one insists on extending the hyperbolic coordinates across the horizon (as we have done with Schwarzchild), then one gets something that looks like this:

 

 

These coordinates satisfy the flat spacetime metric:

 

[math]-d\tau^2 = -R^2dT^2 + dR^2 +dy^2+dz^2[/math]

 

Let's take a trip across the R=0 horizon:

 

 

Notice that below the horizon when R<0, the R coordinates are spacelike, and it is impossible to stay at R=constant. The acceleration required to "stay" at constant R would be infinite.

 

Next, the analogous trips across the horizon are like this:

 

 

Look familiar? These are the brute force extensions of the flat spacetime geodesic x=1 across the flat spacetime horizon of the singular hyperbolic coordinates. Notice that T→±∞ as we cross the horizon. And even though it impossible to "stay" at constant R when R<0, it is surely possible to cross these R coordinates in both directions! The acceleration required to "stay" at constant R is infinite, yet we can cross these R coordinates in both directions! And this is flat spacetime with no inconsistencies here!

 

Zanket, let's stop here. This is alot. Are you in agreement that the above x=1 geodesic looks like this across the horizon in flat spacetime hyperbolic coordinates?

 

If so, then onward.

 

If not, then I give up trying to make you understand, and what is your full name so I can give you credit if I decide to submit a paper using some of your ideas?

 

Andrew A. Gray

 

P.S. Sorry if I take a long time to respond, but you must realize that these posts take up a lot of my time and thought.