How does the following brief thought experiment fail to show that general relativity (GR) has a major problem in regards to black holes?

The full thought experiment is in my blog post. The post claims that GR violates its own equivalence principle at the horizon of a black hole. The principle says that the laws of physics in any sufficiently small, freely falling frame are the same as they are in an inertial frame in an idealized, gravity-free universe. Here's a condensed version of the thought experiment:

In an arbitrarily small, freely falling frame X that is falling through the horizon of a black hole, let there be a particle above the horizon that is escaping to infinity. A free-floating rod positioned alongside the particle and straddling the horizon couldn't be escaping to infinity as well, or else it'd be passing outward through the horizon. However, if instead the rod didn't extend as far down as the horizon, then in principle it could be escaping, possibly faster than the particle beside it. In an inertial frame, unlike in X, a body's freedom of movement (in principle and if only relative to other free objects in the frame) doesn't depend on the body's position or extent. Then a test of the laws of physics can distinguish X from an inertial frame. If X was equivalent to an inertial frame, I wouldn't be able to tell whether the rod could possibly be passing the particle in the outward direction, by knowing only whether the rod extends as far down as an imaginary boundary (the horizon) within the frame. If X was equivalent to an inertial frame, the rod could in principle be passing the particle in the outward direction regardless of its extent within X.

The thought experiment above takes place completely within X, which is arbitrarily small in spacetime (arbitrarily small both spatially and in duration). That is, the experiment is completely local. That the particle is escaping to infinity is a process occurring within X; it tells us that the particle won't cross the horizon during the lifetime of X. The particle needn't reach infinity before the experiment concludes.

It isn't necessary to be able to detect (by some experiment) that a horizon exists within X. It's a given (from the givens in the thought experiment) that a horizon is there. Likewise, I am free to specify the initial conditions of a particle or rod in relation to the horizon. For example, I am free to specify that the rod straddles the horizon, and draw conclusions from that. The laws of physics in X are affected by the presence and properties of the horizon regardless whether an observer in that frame detects the horizon.

It seems to me that the only way the equivalence principle is satisfiable in X is when in principle the rod can be escaping to infinity regardless of its initial position or extent in X, which would rule out black holes in a theory of gravity consistent with the principle. Otherwise, it seems the bolded sentence must be incorrect. If so, how? In other words, how can I not tell whether the rod can possibly be passing the particle in the outward direction, by knowing only whether it extends as far down as the horizon?

I'd appreciate hearing from Ted Bunn or other experts on black holes. A barrier to getting a satisfactory answer to this question is that many people believe the tidal force is so strong at the horizon that the equivalence principle can't be tested there except impossibly, within a single point in spacetime. An equation of GR (see my blog post) shows that a horizon isn't a special place in regards to the tidal force, in agreement with many texts including Ted Bunn's Black Hole FAQ. In fact the tidal force can in principle be arbitrarily weak in any size X. To weaken the tidal force in any given size X, just increase the mass of the black hole. (Or they might believe it's fine to test the principle in numerical approximation in a frame larger than a point, but not fine to test it logically in such frame anywhere. Kip Thorne disagrees, in a reference in my blog post.) Note also that the Chandra X-ray Observatory FAQ tells us that observations of black holes to date aren't confirmations of GR, rather they actually depend on the theory's validity, which is to say the existence of black holes in nature isn't proven.

Edit to add: I put a simple diagram, showing GR's violation of its own EP, at the blog post.

Edit to add: I'm awarding the bounty to dbrane, whose answer will likely retain the lead in votes, even though it's clearly incorrect as I see it. (In short, the correct answer cannot be that an infinitesimally small frame is required to test the EP. It is in fact tested in larger labs. The tidal force need only be small enough that it doesn't affect the outcome. Nor is the horizon a special place in regards to the tidal force, says GR.) I do appreciate the answers. Thanks!

Edit to add: this question hasn't been properly answered. The #1 answer below made a false assumption about the question. I've beefed up the question to address the objections in the answers below. I added my own answer to recap the objections and reach a conclusion. Please read the whole post before answering; I may have already covered your objection. Thanks!

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    $\begingroup$ The extent of your object is what spoils the freely falling frame. It is not equivalent to an inertial frame. $\endgroup$ Commented Feb 28, 2011 at 12:12
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    $\begingroup$ Two comments: tidal forces don't have to be large at the horizon; they can be negligible for very large black holes. Also, the evidence for black hole horizons has been discussed here physics.stackexchange.com/questions/3349/… $\endgroup$
    – dbrane
    Commented Feb 28, 2011 at 12:58
  • $\begingroup$ @Raskolnikov: In a reference in my blog post, Kip Thorne specifically says that the equivalence principle applies in a small, freely falling frame that is falling through the horizon of a black hole, and says that the laws of physics are testable in such frame, just like they are in any other small, freely falling frame anywhere else. So yes, according to the equivalence principle X should be equivalent to an inertial frame. @dbane is correct that the tidal forces can be mild at the horizon. In principle they can be arbitrarily weak in any given size X. $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 4:20
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    $\begingroup$ @finbot: Nobody is going to win the bounty, since many good replies have been given but you just don't understand them. As I said, freely falling frame $\neq$ intertial frame. They are only equal locally. Forget black holes and GR, what you are proposing would be equivalent to saying that because you hurl a stone away during free fall, the stone must absolutely escape because the free fall frame is equivalent to an inertial frame. This is obviously untrue. $\endgroup$ Commented Mar 5, 2011 at 12:29
  • $\begingroup$ @Raskolnikov: According to bounty rules, half must go to the highest-voted answer. I'll give the other half to the second-highest-voted answer. Yes, they are equivalent only locally, so it's a good thing X is defined to be local, by virtue of being defined to be arbitrarily small. What I'm proposing isn't like your "hurling stone" example at all. The thought experiment concerns whether it's possible for the rod to also be escaping, given how far down within X it extends. The experiment isn't trying to prove that the rod must be escaping. $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 2:13

10 Answers 10


I just read your blog post and it's clear to me where you've gone wrong.

The equivalence principle only allows you to transform to an inertial frame locally. This means that if your spacetime is curved, then the falling observer can only choose Minkowski coordinates for an infinitesimal region around her.

Think of a curved surface and having to choose a very small patch on it for it to appear flat. Clearly, you can't extend that flat patch indefinitely and call it an inertial frame of infinite extent (which you require in order to argue that the frame would allow you to send signals out to infinity).

The horizon is a global object that you realize exists when you patch together all the infinitesimal coordinate systems and examine its causal structure.

So, yes, the falling observer can do experiments to realize the horizon exists, but this does not violate the Equivalence Principle because such experiments are not done locally in an infinitesimal region. This applies to the rod that you seem to want to send away to infinity after crossing the horizon too. The infinitesimal flat patch in which you're allowed to play with the EP does not include infinity (or anything beyond the horizon), so you can't throw things outside of the horizon once you've crossed.

  • $\begingroup$ @dbrane, I don't think this answers the question, because in fact it deals only with the laws of physics locally, completely within X. The particle is escaping to infinity; it doesn't have to reach infinity for the question to apply. Nor is the question about an experiment to realize the horizon exists. If I can tell whether the rod can possibly be passing by the particle in the outward direction as measured entirely within X, by knowing only whether the rod touches the horizon in X, I don't see how the question is answered. (That's a local, hence valid test of the principle.) $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 3:19
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    $\begingroup$ If I understand correctly, the gist of it is that you can't fit a rod inside a region of infinitesimal extent. $\endgroup$
    – David Z
    Commented Mar 4, 2011 at 0:11
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    $\begingroup$ @finbot: but note that I said infinitesimal extent, not arbitrarily small (which would imply some finite size). $\endgroup$
    – David Z
    Commented Mar 4, 2011 at 3:21
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    $\begingroup$ @finbot: when you're at the event horizon of a black hole, sufficiently small is infinitesimally small. This comment thread is getting kind of long and I'm not inclined to say anything further on the matter. $\endgroup$
    – David Z
    Commented Mar 4, 2011 at 4:36
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    $\begingroup$ @dbrane--- this answer is not totally wrong, but its false. The scale at which the equivalence principle holds is determined by the curvature scale, not by the distance to the horizon. $\endgroup$
    – Ron Maimon
    Commented Aug 20, 2011 at 6:28

Dbrane's answer contains the essential points. However I should point out that General Relativity is more sophisticated than your models suggest.

  1. The Inertial Frame concept (as used in the Equivalence Principle) is really only valid infinitesimally (whence it matches Minkowski space and "idealised gravity-free universe"). Some authors have critized the EP for this, and you are too. Most authors accept this and just present the EP "locally" - with "local" meaning no large deviations via curvature. Near the Event Horizon of a Black Hole is not a good place to find such flatness - especially if the BH is rotating - so we would be dealing with small Frames at best. All this makes "Law K" in your post suspect. (EDIT ADD FOR CLARITY) Thus the Blog phrase "Then law K is false in X" needs to say "Then law K is false in General Relativity".

  2. A different problem here is the status of "Event Horizon" (presumed at R=2M in your post). Put simply Event Horizons are difficult to find for an active Black Hole (one that is still eating up matter): its position is actually mobile until the Black Hole finally settles down (at the end of the Universe). This is a very counterintuitive behaviour of Black Holes and of General Relativity and arises because the "M" in "R=2M" hasnt been determined until the Black Hole has stopped growing!

  3. Concerning this:

    In an inertial frame, unlike in X, a body's freedom of movement (in principle and if only relative to other free objects in the frame) doesn't depend on the body's position or extent.

The "freedom of movement" that I think you are referring to whether the object can be accelerated beyond the speed of light, which cannot be done in any Inertial Frame. As it cannot be so accelerated then no physical process in momentarily passing frame X can stop the momentarily straddling rod from entering the Event Horizon (remember that the Frame X is entering the Horizon too).

  • $\begingroup$ @Roy Simpson: On #1, X is defined to be arbitrarily small, so it meets the condition of local, hence nothing suspect about the laws of physics being tested. On #2, that's a difficulty in practice only, which is irrelevant here. In principle (in theory) we are free to test the laws of physics in X exactly as that frame is defined; Kip Thorne agrees too. On #3, that doesn't answer my question. For the EP to be satisfied in X, I mustn't be able to tell whether the rod can possibly be passing by the particle in the outward direction, by knowing only whether it touches the horizon. $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 4:48
  • $\begingroup$ Unfortunately there is a gap between the GR physics in Black Hole books and GR in general. So all we can do is give some pointers and advice: you will have to study up on these answers. My primary advice is to draw a diagram (or 3) of what you think is actually happening in this thought experiment. You will notice that all GR books are filled with diagrams, so must this experiment. When you do that you will have to "draw a border" around X, you will be constrained to put it inside/outside/straddling the EH. You will then add rods, particles, observers and anything else relevant. $\endgroup$ Commented Mar 2, 2011 at 17:02
  • $\begingroup$ Note that diagrams are usually drawn at a time t, but you might need to consider several such times to get a good picture of these events. In presentation of your arguments ensure that the logic and definitions are watertight. The phrase "freedom of movement" has come from nowhere in this description, so others cannot help here. Nevertheless the 3 points I have made are where there are surprising counterintuitive aspects of GR. For example when the books say "A must happen near a BH" they mean "A would not happen if only it could exceed the speed of light in some inertial frame." $\endgroup$ Commented Mar 2, 2011 at 17:12
  • $\begingroup$ I have added an EDIT on the Law K aspect, just to focus that point. $\endgroup$ Commented Mar 2, 2011 at 18:16
  • $\begingroup$ I think a diagram would be superfluous. I think the thought experiment amply shows that I'm able to tell whether the rod can possibly be passing by the particle in the outward direction, by knowing only whether it touches the horizon. Which violates the EP, it seems. $\endgroup$
    – finbot
    Commented Mar 2, 2011 at 22:27

The answers to this question all get the answer wrong. The answer is that an accelerating frame has exactly the same horizon as the black hole, so that the equivalence principle holds. It does not hold infinitesimally as you approach the horizon, it holds including the horizon, if you identify the black hole horizon with the Rindler horizon.

The length scale at which the EP fails is the inverse curvature, which is as large as you like compared to the distance to the horizon. So the motion of the ball and the rod is the same in a uniformly accelerated frame as it is next to a black hole.

This type of equivalence principle, with a short distance to the horizon, was never used by Einstein, but it's sort of folklore by now!

LATER EDIT: I see that this answer might be interpreted as lending support to the claimed violation of the equivalence principle in the OP's question. There is absolutely no violation of the equivalence principle, and this can be easily seen.

Given a rigid rod L in the horizontal direction, it is impossible to accelerate it horizontally while keeping it rigid with an acceleration any greater than

$a_{\mathrm{max}} = c^2/L$

because then the left-most point would be past the Rindler horizon of the right most point. If you try to do this to a rod, it gets properly longer, because the acceleration on the left point can't keep up (this is easily seen in a space-time diagram). The intuition that fails is that there is such a thing as "uniform acceleration of a rigid rod". So when the rod is longer than the distance to the horizon it will not be able to pass the particle in the inertial frame before the whole frame reaches the horizon and the question is moot.

More generally, it is impossible to find a contradiction between a black hole horizon and the EP, because the near horizon metric is Rindler, up to curvature corrections which are arbitrarily small, so it is equivalent to a flat space, and there is no thought experiment which can refute this in a black hole which doesn't work in flat space just the same.

  • $\begingroup$ For completeness' sake: this answer of course only talks about large black holes. Smalish ones have arbitrarily large curvature on the horizon. $\endgroup$
    – Marek
    Commented Aug 20, 2011 at 11:46
  • $\begingroup$ This answer doesn't answer the question. It doesn't refute the OP's noted difference between X and an inertial frame, a difference that contradicts the EP. An accelerating frame doesn't have the same horizon as a black hole, the OP shows. You can't assume such sameness in the answer, when the OP refutes that sameness. $\endgroup$
    – finbot
    Commented Aug 23, 2011 at 6:07
  • $\begingroup$ I just fixed that. The OP doesn't show any violation, but has just noticed the fact that rigid bodies in special relativity can't accelerate rigidly with arbitrarily great acceleration. There is a nice exposition of this by Bell in the special relativity paper in "Speakable and Unspeakable in Quantum Mechanics". $\endgroup$
    – Ron Maimon
    Commented Aug 23, 2011 at 18:18
  • $\begingroup$ Nothing is accelerating in the OP, so Bell's spaceship paradox doesn't apply here. The rod is defined to be free-floating, i.e. freely falling. (No gravitational acceleration either, since the tidal force is defined to be negligible. As measured in frame X, the rod's velocity is constant, as is the particle's.) As noted in the OP, the rod cannot be passing the particle in the outward direction, as it could in a true inertial frame. That's a violation of the EP. This violation doesn't require that the rod completely pass the particle. As to your last paragraph see "Rindler" in the blog. $\endgroup$
    – finbot
    Commented Aug 26, 2011 at 8:05
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    $\begingroup$ Um--- you set up a thought experiment where one particle is outside, and the other is inside the horizon. Then you look at it in a free-falling frame. In the free-falling frame, the outside-particle must be accelerating fast, and has a Rindler horizon, while the inside-particle passes this Rindler horizon. Whatever happens near a black hole in terms of communication loss happens exactly the same way for a particle accelerated in a freely falling frame. That's because the near-horizon geometry is flat, and you can apply the equivalence principle there, as you say. $\endgroup$
    – Ron Maimon
    Commented Aug 26, 2011 at 8:48

You're choosing a "freely falling" inertial frame. There's a natural set of coordinates for a non rotating black hole for this called "Gullstrand-Painleve" coordinates. They correspond to the natural coordinates for a particle falling into a black hole from infinity. See the wikipedia article.

In these coordinates, the speed of light is different for light trying to move away from the black than light moving towards it. As the little patch enters the black hole, the speed of light moving away from the black hole becomes negative, that is, even light moving away from the black hole still gets sucked into the singularity.

A well written, sort-of introductory and very intuitive paper you might find enlightening on these coordinates, and their generalization to a rotating and/or charged black hole, is:

Am.J.Phys.76:519-532,2008, Andrew J. S. Hamilton, Jason P. Lisle , The River Model of Black Holes

  • $\begingroup$ The other answers(+1) are correct; the problem is that the inertial frame has to be small. I've added this answer in the hope it will give a better intuitive understanding of what one of those small inertial frames looks like. $\endgroup$ Commented Mar 4, 2011 at 1:17
  • $\begingroup$ @Carl Brannen, frame X is defined to be arbitrarily small. So "it has to be small" isn't an answer to this question. I'm familiar with the river model (it's one of my favorites). In any model, however, the escape velocity above the horizon is less than the speed of light. The horizon is defined as the highest place where light cannot possibly escape. The particle can definitely be escaping by virtue of being above the horizon. The entire thought experiment takes place locally, within an abitrarily small (in spacetime) freely falling frame. No better arena for testing the EP exists. $\endgroup$
    – finbot
    Commented Mar 4, 2011 at 1:49
  • $\begingroup$ @Carl Brannen: The question doesn't say "escape to infinity". It says "escaping to infinity". That's different. Escaping to infinity is something that can be happening entirely locally, in an arbitrarily small region of spacetime. Likewise if a thought experiment specifies that a car is moving at 70 kph, it doesn't mean that the car moves 70 km during the experiment. Search above for "The thought experiment above" for a further explanation. $\endgroup$
    – finbot
    Commented Mar 4, 2011 at 3:37
  • $\begingroup$ @Carl Brannen: I think "escaping toward infinity" would be ambiguous. Only "escaping to infinity" indicates it's never coming back. I did a web search of .edu sites to see that I'm using the phrase properly. It's okay to talk about something escaping to infinity in terms of a local process. The particle reaches some great distance after the experiment is over. If I say "escaping toward infinity" there's no guarantee the particle won't fall back and cross the horizon during the lifetime of X. It indicates only the direction it's escaping, not what it's escaping. $\endgroup$
    – finbot
    Commented Mar 4, 2011 at 4:37
  • $\begingroup$ @Carl Brannen: "Just because a particle is escaping towards infinity doesn't mean it gets there." True. But escaping to infinity is different. In physics that term clearly means it's moving toward infinity and never falling back. And it doesn't mean that the situation can't be analyzed locally. I see 2K+ .edu links mentioning "escaping to infinity" and everywhere I look they're analyzing locally. After all, it would take forever to reach infinity. The mere existence of the word "infinity" in a thought experiment, or even "escaping to infinity", doesn't necessarily mean a global experiment. $\endgroup$
    – finbot
    Commented Mar 4, 2011 at 7:29

Saggitarius A* has been confirmed to be a black hole, and many others have been discovered; by observing the movement of stars around Saggitarius A* over many years , ballistic trajectories for stars that can only be explained by a DEEP gravitational well (millions of solar masses) exerting a major influence in what appears to be an empty spot. It is useful to remember that the existence of black holes has been confirmed by Astronomers, and that if your thought experiment somehow precludes their existence, the problem lies with your thinking.

I think the biggest issue you're having is with your treatment of this "rigid rod" as something that could actually physically exist. Any rod in this universe is made up of atoms, and its rigidity and elasticity are entirely the result of electromagnetic forces between the atoms in the rod. Therefore, saying that the "rod" is half-inside and half-outside the event horizon is only saying that half of the rod's constituent atoms are inside the event horizon and the other half are outside of it. The rod that you call upon in your thought experiment seems to have properties that are not of this universe.

The EH is not a physical boundary, and if the black hole is large enough, tidal forces will be negligible on infalling matter when crossing the EH. From the reference frame of the infalling matter, it would not experience being instantaneously teleported from one side of the EH to the other nor any of the other effects you've suggested. The matter will continue on a ballistic trajectory orbiting the center of mass; a ballistic trajectory that will never take it outside of the horizon (by definition), sure, but upon crossing the horizon - if our object was a guy in a space suit - he'd have no indication that he'd crossed the event horizon (except perhaps that his radio to home base has stopped working).

Importantly, the astronaut (or the rod) would not be stretched into infinity or torn apart at the EH; a black hole was recently discovered that has a radius of 4 light-days, and a density that is FAR below the density of Earth's atmosphere at sea level, for example. Our astronaut would pass the horizon of that black hole sedately, and asphyxiate and die and freeze before ever encountering any forces strong enough to cause even mild discomfort, even at Voyager-1 speeds headed directly at the singularity.


As this is a Bounty question (and because my other answer has a long set of comments) I have decided to add another answer. This answer is somewhat different from other answers provided, although it is consistent with them. There has been some challenge to understanding the other answers, and there may be some difficulty in understanding this answer too, but I shall record it here for those interested in this physical scenario.

In short: there is a contradiction in this physical scenario. The nature of this contradiction and its consequences I shall discuss at the end.

Let us consider the two physical assumptions which make up the scenario:

(A) A particle on a trajectory starting just above an Event Horizon, this trajectory being a timelike trajectory leading to timelike infinity (in an asymptotic model, say);

(B) A rigid rod (of length L say) straddling ie "half-in" and "half-out" of the Event Horizon at some time t say (in some appropriate coordinates).

The scenario continues by discussing Frames and so on, but this answer does not depend on anything else. Let us examine each of these assumptions a little more carefully.

(A) Is it possible for such a trajectory to exist? It depends on the Black Hole metric, but for the Schwarzchild metric if the particle has energy greater than some minimum E(R), then it may escape the region. If its energy and angular momentum is less than this value, it may orbit the Black Hole, or may directly plunge in. So let us assume that we are dealing with a Schwarzchild metric and that the particle can be given sufficient energy to follow the escaping trajectory.

Now let us consider (B) in more detail:

Is (B) possible? I claim that (B) is an inconsistent assumption. I shall outline a proof below.

First we need to return to the trapped surface property of an Event Horizon: a particle P is inside an Event Horizon if every trajectory leads to the Singularity. So now consider the rod. This rod is "rigid" in some sense, although we do not require any properties of "rigid" in this proof - so it could be (normal matter) elastic. However it is simpler to assume a regular model of a rigid rod of length L (much smaller than R=2M, say). Consider two points on the rod at time t: P is inside the Event Horizon and Q is outside. Since Q is outside there exists, by definition, some trajectory $\gamma$ such that $\gamma$ does not lead to the Singularity.

Now let the rod trajectory be such that point Q follows the $\gamma$ trajectory. The point P necessarily will follow a trajectory leading to the Singularity and so the proper distance PQ will extend to at least R = 2M in finite proper time. Thus the rod will break, and so the rod was not rigid as assumed but composed of at least two separate components (this is Hawking radiation!). So we have a contradiction as we assumed that the rod was rigid. So assumption (B) is inconsistent with General Relativity and cannot be used in any thought experiment.

A first objection might be via a "fluid model" intuition of the Event Horizon which allows a rod to be outside, then part straddling, then maybe entirely contained in the "fluid". But this intuition is not valid here: either the rod is or is not in the Event Horizon.

A second objection might be that this implies that the Event Horizon (fluid) has "moved" superluminarily, and this is not possible. The explanation is that the Event Horizon is not a local physical object and not constrained by the restrictions of Special Relativity. In fact it is a Global General Relativity object with counterintuitive properties: discontinuity and achronicity (see Hawking and Ellis 1973) and, as shown here, superluminarity.

We can now understand in outline the paradox that the original question identified with Inertial Frames. These are local objects in which a Global GR entity was analysed - but any attempt to account for the behaviour of a global object by purely local analysis will result in the sort of contradictions and paradoxes that the question has uncovered.


There is a further objection to the stark conclusion of this answer, that can be understood in terms of further thought experiments. I shall discuss these and how they relate to the original question.

Let us assume that the rod is actually a long spacecraft with a removable capsule at the top. If the lower part of the spacecraft is within the EH, then the capsule might be fired off, escaping to infinity. Thus in this case it is not true that the entire spacecraft either is or is not contained within the EH, and the EH fluid model somewhat applies. Of course then the rod is not actually rigid as we assumed, so our conclusion is still valid, but only just. We could generalise this scenario to a spacecraft with N modules. A further generalisation simply assumes that the matter of the rod is such that at any distance along its length an explosion can occur (perhaps caused by a striking antiparticle) which would cause the rod to split and the top part to escape (to infinity). In this case (and within the modelling approximation) it would be appropriate to consider a fluid-like model for the EH and thus talk about the rod "straddling" the Horizon.

However this "straddling" model assumes a wider physical scenario than did the original thought experiment, which merely considered the rod as an inert object, and certainly did not consider explosions, the quantum matter in the rod, and colliding particles which might happen (occasionally) in realistic physical situations. When these other factors are present one can discuss "straddling" rods: in the bare model as presented the "straddling" concept becomes inconsistent as discussed.

So in this wider sense the answer to the part of the thought experiment about how the upper straddling rod could in principle be detected in frame X apparently escaping to infinity faster than a nearby particle, is that experiments in X will have detected that the rod has been in some form of explosion near the EH (unlike any non-EH straddling rod which happened also to be escaping). However the original thought experiment was so limited (to a rigid rod idealisation) that "explosions" and the like cannot occur; as a consequence "straddling" cannot occur either.

  • $\begingroup$ It seems that your "the Event Horizon is ... not constrained by the restrictions of Special Relativity" is in contradiction with the EP. The EP does apply to a small, freely falling frame that falls through the horizon of a black hole (SR applies in such frame). You can't refute the thought experiment by noting that the rod can't be passing outward through the horizon. That is a premise of the thought experiment. You can refute the experiment (or any purportedly sound argument) only by showing that either one of its premises are false or that its conclusion doesn't follow from its premises. $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 5:01
  • $\begingroup$ Yes the EH is not (local) physical and so the EP indeed does not apply.(In this sense your overall intuition is correct.) Premise (B) is a premise of the thought experiment however, which although one might initially think is allowable, is shown here to be false ie inconsistent with GR. Thus your argument has led from a false (ie unusable) premise to a conclusion and is thus logically invalid. Read the answer again at least once before replying - I am saying some quite surprising things in this answer. $\endgroup$ Commented Mar 8, 2011 at 10:53
  • $\begingroup$ You can't refute a thought experiment that purports to show that GR violates the EP, by disagreeing with the EP. The EP does apply at the horizon. Check out the last paragraph of this page, by Kip Thorne: tinyurl.com/66ygkla. If you don't think you're disagreeing with the EP then I'm not getting you, maybe you can explain further how you could be agreeing with the EP when "the EP indeed does not apply". $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 15:48
  • $\begingroup$ @finbot : I am not disagreeing with the EP (which is not mentioned in this answer). Indeed the EP applies at the Horizon, but not to the Horizon itself. It is the (dynamic) Horizon itself which could be said to violate SR. In fact I think that your real argument here is that this thought experiment shows that the (dynamical) Horizon violates SR (which is itself surprising) rather than GR. The fact that you dont care about the smallness of the EP Lab also shows that you are really testing SR itself with the Horizon - and finding something wrong (as have I in this answer). $\endgroup$ Commented Mar 8, 2011 at 22:30
  • $\begingroup$ To take this further you need to understand the difference between SR and GR. Understanding the non-SR behaviour of the Horizon (from other examples which are known) would also help. As a matter of fact I have found a different thought experiment (before this) which also suggests that Black Holes violate SR! $\endgroup$ Commented Mar 8, 2011 at 22:33

I also would like to point out that the notion of "escaping to infinity" violates the locality of the equivalence principle, as it requires an infinite amount of time. That, too, is not a local probe of the gravitational field, as it depends on all the dynamics of the spacetime between the probe point and the time the particle reaches infinity.

  • $\begingroup$ I covered that above, and in my answer. The -ing suffix indicates action, which occurs locally too. I can be moving to France even when I haven't yet left Spain. Escaping to infinity is something the particle is doing wholly within X, which is arbitrarily small in spacetime. You'd be correct if the thought experiment instead said "escaped to infinity". $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 1:58
  • $\begingroup$ @finbot: how do you detect the "escaping"? You have to allow a sufficient amount of time to pass, and measure a sufficient amount of separation. The weaker you make the local field, the more time it takes to see a signal of size X. Either way, you violate locality. $\endgroup$ Commented Mar 8, 2011 at 2:52
  • $\begingroup$ The "escaping" doesn't have to be detected; it's a given. The particle is let to be escaping to infinity. Anything that is possible in principle can be let in a thought experiment. $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 4:51
  • $\begingroup$ @finbot: then, you're sensitive to any variation in the gravitational field that occurs in <b>any</b> finitely sized area, and the only local frame is a single point--you can find gravitational forces on any finite sized object if you're taking your thought experiment to that extreme. $\endgroup$ Commented Mar 8, 2011 at 14:42
  • $\begingroup$ I've covered that objection in comments here and in my answer. While the EP is strictly true only at a point in spacetime, it's testable in a larger frame. It is tested in labs larger than a point, and otherwise it'd be outside the realm of science, for all ideas must be testable in science. There is a tidal force in a frame larger than a point, but that's completely ignorable when it has no effect on the outcome of the experiment. See my comments to Jerry Schirmer in my answer below, to see why the tidal force has no effect on the outcome of the thought experiment. $\endgroup$
    – finbot
    Commented Mar 8, 2011 at 16:09

As far I understand the physics, in the local free fall frame, this is just an accelerating rod. In order to get away the from event horizon, the rod needs to accelerate heavily and continuously. If both ends of the rod accelerate with the same acceleration, then the rod will break due to length contraction. In order to keep the rod the same size, the back end has to accelerate more. That is just normal Relativity.

Assuming the front end of the rod accelerates just enough to stay ahead of the event horizon, it will eventually make it to infinity. If the back end of the rod accelerates enough to keep the rod intact, it will also make it out. If not, the rod will break, and the event horizon will catch up with the back end.


It is impossible to have part of the rigid rod being under the horizon and part above.

Mathematically on the horizon itself the speed of rod's movement becomes the speed of light. This means the linear contraction of length of the rod to zero for a stationary observer. Thus the rod crosses the horizon WHOLE AT ONCE and after this moment its speed becomes greater than speed of light.

Of course this also means that the rod cannot be made of substance or carry any information since information cannot be transferred faster than light (in actual world the BH will evaporte earlier than any object can approach the horizon, to cross the horizon one would have to move towards it faster than light with only light rays in theory can reach the horizon exactly at the last moment of the BH's existence).

  • $\begingroup$ Evaporating horizons recede with finite local speed. A simple penrose diagram for a evaporating horizon will show you that the horizon forms a timelike surface in the enveloping spacetime. $\endgroup$ Commented Apr 22, 2011 at 22:21
  • $\begingroup$ @Jerry Schirmer I said exactly the same: the horizon is a timelike surface, so you only can cross it at once with the whole your ship, like crossing a moment in time. You cannot have part of your ship under the horizon and part above. $\endgroup$
    – Anixx
    Commented Apr 22, 2011 at 22:28
  • $\begingroup$ Sure you can. Just not for an infinite time. The average doorway forms a timelike surface in space, after all. The null expansions or the global structure of the spacetime are the things that make the horizon special. Not the fact that it is a timelike (or null, in the case of non-expanding/contracting horizons) surface. $\endgroup$ Commented Apr 22, 2011 at 23:01
  • $\begingroup$ "Sure you can. Just not for an infinite time" - completely incorrect statement if the spaceship is rigid and forms a connected body. $\endgroup$
    – Anixx
    Commented Apr 22, 2011 at 23:44
  • $\begingroup$ @Anixx: solve the geodesic equation in non-singular coordinates yourself. The result is that the tidal force on a finite sized object at the horizon is proportional to $\frac{M}{r^{2}}$, which is finite, and can be made arbitrarily small for an arbitrarily large object. You aren't taking the light-cone tipping effect into consideration, and you really have to in this case. $\endgroup$ Commented Apr 23, 2011 at 0:22

The answer by dbrane nearly says what I am going to say but not as clearly and shortly as I will.

The OP states the Equivalence Principle as

« The principle says that the laws of physics in any sufficiently small, freely falling frame are the same as they are in an inertial frame in an idealized, gravity-free universe. »

This is wrong. This mistake has nothing to do with back holes, it's always wrong for any neighbourhood no matter how small it is. It's only true for a point, not for a neighbourhood. Or, to put it another way, it can be approximately true, up to first order, in a sufficiently small neighbourhood. But it can never be exactly true except at a point (unless the gravitational field is of a rather special kind, and even the Earth's field makes this impossible).

Mathematically, the values of the Christoffel symbols can be zeroed out at one point by an apropriate choice of coordinates, but they cannot be made zero for a neighbourhood no matter how small it is.

What the principle of equivalence says is that you cannot tell the difference between a gravitational field and a pseudo-force due to your choice of coordinates. It does not say you can find coordinates that makes the gravitational force zero. But you can find coordinates that make it zero at one point.

Now although I know nothing about black holes, I have to point out that if you fix what level of approximation you desire, and choose a small neighbourhood which is sufficiently small that within that tolerance there is a frame which is close to being an inertial frame, the required smallness of the neigbhourhood might change with time. With very violent dynamics, the needed smallness might shrink indefinitely, and if there were a singularity, it might be the case that no degree of smallness was sufficient for the wished for tolerance, and this would not violate the EP.

  • $\begingroup$ -1: I am sorry, but you are repeating the wrong answer by DBrane. The scale of violations of EP is the inverse curvature scale, which can be made arbitrarily large compared to the distance to the horizon. It is not a mistake to apply EP to a patch which includes the horizon, but you need to realize the horizon coordinates make a Rindler horizon in this patch. The "required smallness of the neighborhood" is bigger than the domain of the thought experiment, so the resolution does not come from this line of thinking. I have to downvote, because this is patiently explained in my answer + comments $\endgroup$
    – Ron Maimon
    Commented Jan 17, 2012 at 16:50
  • $\begingroup$ I carefully said « might », and my point is that it is incumbent on the OP to do an analysis of the validity of the approximation used in his thought experiment...to check whether this happens or not. It is not incumbent on me to do it for the OP! But it was incumbent on me to point out the mistake about what the EP is and what it is not. $\endgroup$ Commented Jan 17, 2012 at 17:13
  • $\begingroup$ @RonMaimon My answer is not a repetion of dbranes excellent answer since I isolate the typical student confusion about the EP and correct it more clearly, and since I qualify the other point with a « might.» I answer the OP since obviously the OP mistake about EP has to be fixed first, and then the analysis of the limits of the validity of the approximation have to be made...by the OP. $\endgroup$ Commented Jan 17, 2012 at 18:07
  • $\begingroup$ DBrane's answer is (atypically) not excellent, it is wrong. It is also absurdly upvoted, and grossly misleading. This is why I had to vote down both (sorry). Saying might doesn't help--- you need to check whether it is or is not to answer the question, or determine that it is unanswerable (it isn't). If you didn't do this homework, the answer is not satisfactory, it doesn't answer the question. Anyway the OP is a crackpot, and there is no point in communication. $\endgroup$
    – Ron Maimon
    Commented Jan 17, 2012 at 21:52
  • $\begingroup$ @dbrane I do not know enough about black holes to contradict Mr. Maimon, but I do know that the OP was making a serious mistake about the EP, and that the OP needs to double-check the validity of the usual approximation if they want to extend it to monkeywrenches and event horizons.... $\endgroup$ Commented Jan 17, 2012 at 21:55

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