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.

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    $\begingroup$ There is no contradiction! $\endgroup$
    – MBN
    May 5, 2011 at 9:47
  • $\begingroup$ Yes there is! Not being able to test the EP is experiencing something out of the ordinary. $\endgroup$
    – finbot
    May 5, 2011 at 10:27
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    $\begingroup$ 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). $\endgroup$ May 5, 2011 at 11:20
  • $\begingroup$ 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. $\endgroup$ May 5, 2011 at 11:32
  • $\begingroup$ 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. $\endgroup$
    – finbot
    May 5, 2011 at 12:35

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – finbot
    May 5, 2011 at 13:52
  • $\begingroup$ 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. $\endgroup$
    – finbot
    May 5, 2011 at 14:05
  • $\begingroup$ 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. $\endgroup$ May 5, 2011 at 18:16
  • $\begingroup$ 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. $\endgroup$ May 5, 2011 at 18:22
  • $\begingroup$ 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. $\endgroup$ May 5, 2011 at 18:30

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".

  • $\begingroup$ 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. $\endgroup$ May 11, 2011 at 19:03
  • $\begingroup$ 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. $\endgroup$
    – finbot
    May 12, 2011 at 5:51
  • $\begingroup$ 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. $\endgroup$ May 12, 2011 at 12:42
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    $\begingroup$ Since black holes arise in metric theories of gravity, there can be nothing about them that can contradict the equivalence principle. $\endgroup$ May 13, 2011 at 12:07
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    $\begingroup$ @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. $\endgroup$ May 15, 2011 at 23:06

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