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Can the equivalence principle be tested to high precision in a human-sized lab falling through the horizon of a black hole, in principle? By "human-sized lab" I mean a lab the size of the International Space Station, say, with typically-sized lab equipment. I also mean the absolute kind of (event) horizon, not the apparent kind.

The #1 answer at this question says:

For large enough black holes, space is still weakly curved at the event horizon, so of course we should expect that normal physics still exists there. An infalling observer wouldn't experience anything out of the ordinary when crossing an event horizon.

This suggests that the answer to my question here is "yes". However, when I asked a question involving such a test, the highest-rated answer with 23 votes clearly says "no":

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.

I've seen this contradiction many times elsewhere. Certainly the EP has actually been tested to high precision in larger-than-infinitesimal labs. Yet once one talks about testing the EP in a lab falling through the horizon of a black hole it seem respondents in that case think the lab must be infinitesimal in size, so that the EP can't be tested in any practical sense, even though "An infalling observer wouldn't experience anything out of the ordinary when crossing an event horizon" if the black hole was sufficiently large.

I understand that the EP is strictly true for only a point in spacetime. But that hasn't prevented it from being validly tested to high precision in a larger lab, where the tidal force in the lab was weak enough that it didn't change the result at that precision. I'm not asking about testing the EP to infinite precision, nor am I asking about a test involving a black hole of any size, like one where the lab would be spaghettified by the tidal force before reaching the horizon. That's why I qualified the question with "in principle".

Also note that I'm talking about a test being done as the lab equipment (temporarily) straddles the horizon, not a test limited to one side of the horizon or the other within the lab. A test using equipment straddling the horizon wouldn't be allowed according to the other answer with 23 votes, which says:

The infinitesimal flat patch in which you're allowed to play with the EP does not include ... anything beyond the horizon ...

I don't see how a test of the EP would be invalidated due to the equipment straddling the horizon.

Community
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finbot
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    There is no contradiction! – MBN May 05 '11 at 09:47
  • Yes there is! Not being able to test the EP is experiencing something out of the ordinary. – finbot May 05 '11 at 10:27
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    This horizon is not that horizon. As the event horizon is teleologically determined, you can have the event horizon experiencing no curvature whatsoever. (Imagine the gravitational collapse of a spherical shell of matter, when the shell becomes sufficiently compact, there can be regions inside the shell that is completely flat, yet inside the event horizon.) The apparent horizon is (almost; statement is true in spherical symmetry) locally determined, and its presence necessarily signals non-flat geometry (again, spherical symmetry blah blah blah by considering the Hawking mass). – Willie Wong May 05 '11 at 11:20
  • The comments about large versus small blackholes, however, is something altogether different yet again. At the apparent horizon, the area radius $r$ is roughly the same as the (Hawking) mass $M$, while the curvature is roughly $M/r^3$ on a (slowly evolving) blackhole spacetime. So at the apparent horizon you expect a curvature of about $1/M^2$, hence that for large black holes, even at the apparent horizon, the curvature effects may be weak. – Willie Wong May 05 '11 at 11:32
  • Wikipedia says "In the context of black holes, the term event horizon refers almost exclusively to the notion of the absolute horizon." I clearly specified the absolute horizon and said I'm not talking about an apparent horizon. – finbot May 05 '11 at 12:35
  • @finbot: and I claim the question is meaningless with regards to the absolute horizon in general, since the absolute horizon is not locally definable. For stationary black holes, the two horizons coincide, and so there is no difference. But for dynamical black holes the issue of searching in local measurements for gravitational artifacts only makes sense if you are speaking about apparent horizons. – Willie Wong May 05 '11 at 13:05
  • @finbot: if it's the event horizon, it can be completely non-locally testable. There are constructs (most notably, certain Vaidya models) where an event horizon is present in perfectly flat spacetime--it is, in principle, impossible to detect a horizon in this case. – Zo the Relativist May 05 '11 at 13:19
  • Also, any earth-based lab is going to be infinitesimal relative to the radius of curvature of the earth, which is very long, as the radius of the local spacetime curvature is very large, roughly of order $g,c$. – Zo the Relativist May 05 '11 at 13:21
  • @Willie: I'd like to stick to an absolute horizon. Go ahead and consider this a spherically symmetric, uncharged, nonrotating black hole, where the absolute horizon is always at r = 2M in geometric units. – finbot May 05 '11 at 14:07
  • Oops, error above--the radius of curvature should be roughly $\frac{c^{2}}{g}$, but that's even worse, of course. – Zo the Relativist May 11 '11 at 19:30

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The answer to the question posed in your title is no, but not for the reason which you may think is the case.

The problem is that the Einstein equivalence principle is not a precise principle, so the statement "testing the equivalence principle to high precision" is in itself meaningless. The principle is usually stated informally as

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

with the caveat in the definition of "local" and "independent". Indepdence is generally taken to mean that indistinguishable to the precision allowed by your experiment. The definition of local, besides the obvious criteria of non-interaction with objects outside the laboratory except through gravitational coupling, contains also the requirement that throughout the space-time region occupied by the laboratory during the extent of the experiment, the gravitational field changes negligibly (where negligible is defined, again, as below experimental precision). So in particular you can not have large tidal forces. To achieve this you need the total curvature contained in the space-time region occupied by the laboratory during the experiment (obtained by something like integrating the Kretschmann scalar over the space-time domain occupied by the laboratory) to be small.

What am I getting at? The equivalence principle only requires that at any space-time location, ignoring quantum limits, you can build a sufficiently small laboratory that runs for a sufficiently short (proper) time, so that any (non-gravitational) experiments performed within it is indistinguishable from one performed in ideal Minkowski space. So the interesting question to ask is: given that your laboratory is roughly the size of the international space station (roughly $10^3 m^3$), and you run your experiment for a fixed lenghth of time (say 1 day in proper time), and your laboratory is straddling the apparent horizon of a black hole, what is the minimum mass of the black hole such that you won't be able to tell the difference, to say, one part in $10^{10}$.

A rough estimate for this can be done with some dimensional analysis. As is well known, the Kretschmann scalar is rougly on the order of $10r_S^{-4}$ for a black hole at the apparent horizon, where $r_S$ is the Schwarzschild radius, given by the black hole mass $M$ via the formula $2GM/c^2 = r_S$. Integrating the Kretschmann scalar over a space-time region will then be a dimensionless quantity, which we would to be small $\sim 10^{-10}$. The space-time volume occupied by our laboratory is $cT V$ where $V$ is the spatial volume of the lab and $T$ is the amount of time in which to run our experiment. In our assumptions above, we have that $cTV \sim 3 \times 10^{8} \frac{m}{s} \times 9\times 10^4 s \times 10^3 m^3 \sim 2.5 \times 10^{15} m^4$. So we need

$$ r_S^4 \sim 2.5\times 10^{15} m^4 \times 10^{10} \times 10 = 2.5 \times 10^{26} m^4 \implies r_S \sim 4 \times 10^6 m$$

This gives a mass of

$$ M = \frac{c^2 r_S}{2G} \sim 9 \times 10^{16} \frac{m^2}{s^2} \times 2\times 10^6 m \times \frac{1}{7} \times 10^{11} \frac{kg \cdot s^2}{m^3} \sim 2\times 10^{33} kg $$

which is about 1000 solar masses.

To summarise, if you drive the international space station to the vicinity of a 500 solar mass black hole, and you have some experimental apparatus with sensitivity 1 part in $10^10$, and you accidentally crossed the apparent horizon without looking out the window, then you won't notice until about a day later, which is, well, rather late.

Let me make a note here that it is rather impossible for a free falling laboratory to straddle the apparent/event horizons for any length of time. The event horizon is null, and for Schwarzschild black holes (as well as black holes in certain other nice matter models) the apparent horizon is achronal. So for your laboratory to remain straddling the horizon, an upperbound of the time you can do so is roughly $T = L/c$, where $L$ is the linear size of your laboratory. For something of the size of the ISS, that makes $T$ on the order of $10^{-6}$ seconds in proper time.

So if you were to perform an experiment only when you are straddling the horizon, you gain a factor of $10^{10}$ compared to the previous computation, and the minimal mass of the black hole such that you won't be able to tell the difference is on the order of the mass of the Earth.

Willie Wong
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  • If the statement "testing the equivalence principle to high precision" is in itself meaningless, then why can I find tests of the EP to high precision? Do you think those tests are invalid? – finbot May 05 '11 at 13:52
  • If those tests are valid then, looking at the rest of your post, the answer should be "yes", not "no". You haven't given any reason it wouldn't be possible in principle to test the EP at the same high precision in a human-sized lab falling across a horizon. Have you? I realize the crossing of the horizon would take only a fraction of the test's duration. I'm not suggesting that the crossing of the horizon would be happening during the entire test. – finbot May 05 '11 at 14:05
  • Without seeing the fine prints of those test that you found, I cannot say whether they are valid or not. You seem to be missing the point of my post, so let me say it again. There are two spots which will cause a disagreement between your measurement and the dynamics as predicted in flat space. First is due to unavoidable experimental error (and I assume when you say "high precision" you mean to try to minimize this as much as you can). The second is due to the inherent fuzziness in the statement of the equivalence principle. This fuzziness comes from local curvature effects. – Willie Wong May 05 '11 at 18:16
  • My post tried to explain why it is pointless to try to stick a "human sized lab" everywhere you can and measure the deviation of dynamics from flat space dynamics by reducing measurement error. For any fixed scale of experiment and any precision your experimental devices can discern, there is a mass limit below which, at the apparent horizon, the fuzziness built-in to the statement of EP is already bigger than your experimental error, so your experiment can't confirm EP more than a less precise experiment. – Willie Wong May 05 '11 at 18:22
  • To say it another way, at regions with large curvature, EP is only expected to hold (to a certain fixed precision) for experiments at a sufficiently small scale. To try to probe the theory with larger scale apparati is meaningless in the sense that your designed experiment already falls outside the regime for which the theory is applicable. – Willie Wong May 05 '11 at 18:30
  • In my question I say "in principle". Yes there is "inherent fuzziness" to use your words. But that doesn't mean the EP can't be tested to high precision in a human-sized lab in principle; those tests can be easily found with a simple web search. I've read your post and comments a few times; what I see is that you're just saying the black hole can't be too small or else it'll invalidate the result at some precision. Yes that's true. But it doesn't prevent a high-precision test of the EP in principle, like for a larger black hole. If you think I've misinterpreted you please tell me how. – finbot May 06 '11 at 00:01
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The answer is yes. To Willie Wong's "no" answer I commented:

Yes there is "inherent fuzziness" to use your words. But that doesn't mean the EP can't be tested to high precision in a human-sized lab in principle; those tests can be easily found with a simple web search. I've read your post and comments a few times; what I see is that you're just saying the black hole can't be too small or else it'll invalidate the result at some precision. Yes that's true. But it doesn't prevent a high-precision test of the EP in principle, like for a larger black hole.

The EP has indeed been tested to high precision in labs on Earth. In principle could one such lab fall through the horizon of a black hole during the test, without the experimenters (or anyone else on Earth) noticing? Yes; the tidal force in the lab needn't noticably change in that event, when the black hole is sufficiently massive. There needn't been anything out of the ordinary noticed at all--many relativity texts note that. "Nothing special need happen there", they say. The tidal force is the only reason there is any "inherent fuzziness" when testing the EP. In the question I included "in principle" meaning "in some cases according to the theory".

finbot
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  • Locally, the generating surface of the horizon will be a pair of outgoing null rays. To a sufficiently small local observer, for a sufficiently small time, these will just appear as any ordinary local light rays. There will be two perpendicular spacelike directions, and another null direction perpendicular to both these spacelike directions. All of this is identical to what an observer in flat space would see. the only difference you get in the neighborhood of a black hole is that these directions will appear tilted relative to those of a distant observer. There is no contradiction. – Zo the Relativist May 11 '11 at 19:03
  • There is indeed a contradiction. One cannot logically say that nothing special need happen at the horizon, yet one cannot test the EP there to high precision (like one can in a lab on Earth). Those are mutually exclusive things. Not being able to test the EP to high precision is a special thing happening. Your "no contradiction" is talking about something different that I never said was contradictory. – finbot May 12 '11 at 05:51
  • Yes, you can test the equivalence principle anywhere. The lab need only be smaller (in all four dimensions) than the local radius of curvature of spacetime in the region. All of your horizon effects come from confusing local reference frames near and far from the horizon. – Zo the Relativist May 12 '11 at 12:42
  • Thanks; it seems that you agree the answer to the question above is "yes". I never mentioned something far from the horizon. Even in the other question referenced above, everything was local to the horizon (completely contained within an arbitrarily small, freely falling frame, that is falling through the horizon). – finbot May 12 '11 at 20:29
  • Ok, fair enough. But you do need to realize that the equivalence principle is one and the same as a metric theory of gravity. they mean the same thing. – Zo the Relativist May 12 '11 at 22:39
  • That statement seems to have no relevance to my questions here or elsewhere on physics.stackexchange. – finbot May 13 '11 at 06:18
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    Since black holes arise in metric theories of gravity, there can be nothing about them that can contradict the equivalence principle. – Zo the Relativist May 13 '11 at 12:07
  • Not only is that a false conclusion (the conclusion doesn't follow from the premise), it's disproven here. – finbot May 13 '11 at 16:47
  • @finbot: you're not setting up your experiment correctly.The rod in your experiment has a timelike interval in its restframe.You're not accounting for the fact that the null cones in the neighborhood of the horizon are turned over relative to infinity.You're mapping the coordinates of a distant observer onto the local coordinates of the horizon, and getting nonsense.The statement about metric theories is true-for any arbitrary $\delta$ you can find an $\epsilon$ such that geodesics curve less than $\epsilon$ for $\delta$ time.That is all the EP says, and is a theorem of differential geometry. – Zo the Relativist May 13 '11 at 17:02
  • It's not that complicated! The EP says a sufficiently small freely falling frame is equivalent to an inertial frame (yes that is a correct formulation of it), so drop all mention of "timelike interval" and "null cones"--unnecessarily complicated. All one needs to know is that an object (e.g. piece of toast) that straddles the horizon can't have the same velocity (or higher speed in the outward direction) in such frame as an escaping particle does, even if part of the toast is right beside the particle. That's a clear violation of the EP. There is no such restriction in a true inertial frame. – finbot May 14 '11 at 15:27
  • And no, that does not "map the coordinates of a distant observer" in any way whatsoever. Nothing in principle prevents a local experiment on an escaping particle. It's escaping, not escaped. – finbot May 14 '11 at 15:29
  • One more comment then we're done, since you have no interest in actually discussing anything, or learning anything from anyone. For the rod to represent a rod, it has to be a spacelike interval. This means that it can't contain two timelike or null seperated points. Doing this thought experiment using Schwarzschild coordinates gives you a junk answer. Rigid bodies are a subtle, subtle thing to treat with the correct mathematics in general relativity, and you're just dismissing the insight of several people who actually know what the hell they're talking about. – Zo the Relativist May 14 '11 at 21:34
  • Seriously, it's not that complicated! We're talking about a freely falling frame here, given to be small enough that the tidal force in it is negligible. According to the EP there are no issues with rigidity for free-floating objects in such frame, just like there isn't in an inertial frame. The EP says it's simple SR in such frame, no "timelike or null seperated points" need be mentioned. I'm not dismissing your insight, I'm showing how it's incorrect, or at least obfuscated, and that can be a good discussion. You're the one who is ignoring my comments. I'm not ignoring yours. – finbot May 15 '11 at 09:48
  • YES, it is that complicated (and what I"m saying isn't all that complicated). The freely falling frames near the horizon are not the freely falling frames far from the horizon. And once you are straddling the horizon, there is no Lorentz transformation that can make them equivalent. If you think of this problem in terms of null cones, then it is extremely apparent what the issue is--you are placing the bar in a physically unreasonable configuration--space and time are mixed in such a way to make your setup not actually work. – Zo the Relativist May 15 '11 at 17:18
  • The EP does apply to a sufficiently small freely falling frame falling through the horizon of a black hole--the whole frame, both sides of the horizon at any given moment in that frame. Yes a rod can straddle the horizon. What do you think happens when it's there, a gnome eats it? It disappears? Either would violate the EP, and you can't refute my thought experiment by violating the EP. You're contradicting a score of relativity texts including that of Kip Thorne, Taylor, and Wheeler. – finbot May 15 '11 at 22:17
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    @finbot: thinking of the horizon as a "location in space" is wrong-headed. it is a null surface, not a spacelike surface. If you try to set up some t=constant restframe for the rod, and then stretch out the rod in an r-coordinate, as your intuition seems to be having you do, you are destined to represent something that a freely falling observer would not call a rod. – Zo the Relativist May 15 '11 at 23:06
  • You can read Free-Float Frame, pg. 2-5 to see what Taylor and Wheeler predict for an (inertial-equivalent) frame falling through a horizon. A rod (or the key, coin, or coffee cup that T&W use) straddling the horizon must be able to pass an escaping particle in the outward direction, or else the EP is violated. You can't refute that with "null" or "spacelike"--that doesn't change the fact that GR violates its EP. In a true inertial frame a rod, key, coin, or coffee cup can pass any other object in the frame in any direction. GR must allow that. – finbot May 16 '11 at 03:47
  • It seems to me you're trying to explain (with "null" and "spacelike") why GR doesn't allow the rod straddling the horizon to be passing the particle in the outward direction. What you're not realizing is that you're trying to justify GR's violation of its own EP. No logical justification is possible. GR must allow the rod to do that, or else it violates the EP. Nothing, absolutely no explanation whatsoever, can prevent the rod from doing that and save GR; those are mutually exclusive. – finbot May 16 '11 at 03:59