I've got a problem setting up a general relativity (GR) thought experiment. Thanks for your help!

I'm the observer, in a lab that is falling through the event horizon of a black hole, observing a cloud of freely falling test particles that initially straddles the horizon. The particles above the horizon are escaping to infinity in formation, as GR allows. My lab is small enough, or the black hole massive enough, that the tidal force is negligible throughout this experiment.

Here's my problem: I want all the particles to move in formation as I measure. GR's equivalence principle (EP) tells me that the laws of physics are the same in my lab as in any other inertial frame. In another inertial frame I could let all the particles move in formation just by giving them all the same velocity as I measure. But GR also tells me that this cloud must be splitting apart! GR demands that the particles below the horizon move inexorably inward, toward the black hole's singularity, whereas the particles above the horizon move ever outward, away from the black hole. Then how can I let all the particles move in formation, so that the EP is true?

If I simply let all the particles have the same velocity as I measure (the velocity of the escaping particles), how can they move toward multiple destinations? If instead my lab was falling toward the ground on Earth, with me observing a cloud of particles of which half are escaping to infinity and the other half are falling toward the ground, they won't all have the same velocity as I measure; they won't be moving in formation. This seems to be a real paradox. GR won't let all the particles move in formation, but its own EP demands that they can. What am I missing?

(I'll say in advance that I won't be satisfied with an answer that invokes the tidal force, since I've declared that to be negligible as GR allows. Humans and their labs can in principle survive a fall through an event horizon, without being worse for wear. You can assume that whether the cloud is splitting apart can be detected by me in an arbitrarily short time on my clock, before the particles have moved so much as a nanometer as I measure. Giving credit where credit is due, this question was inspired by an article you can find by searching for "no black holes finbot". It's not mine.)

  • $\begingroup$ In order for some of the cloud to escape, the particles that are still outside have to accelerate very rapidly (this acceleration tends to infinity near the surface). If you go to some escaping particle's frame, an additional event horizon will appear between the it and the particles that haven't yet passed, as the particle is so rapidly Lorentz transforming it's space-time, the light from particles inside the black hole won't reach us (See Rindler coordinates for more details - en.wikipedia.org/wiki/Rindler_coordinates). $\endgroup$
    – kristjan
    Feb 3, 2015 at 17:42
  • $\begingroup$ From the lab's frame, basically all the particles that want to escape must accelerate very fast to ultrarelativistic velocities. Those particles that almost fall into the BH, will "struggle" to escape for some time very near the horizon. This "struggle" can take arbitrarily long, much longer than the time for the lab (in it's own frame) takes to fall to the singularity. Thus from the lab's frame, tidal forces will become significant, as the lab and particles will separate before the "struggle" to get out ends. $\endgroup$
    – kristjan
    Feb 3, 2015 at 17:57
  • $\begingroup$ Do these test particles start as a sphere stationary with respect to you? If so how can any of them escape to infinity? Alternatively are you propelling the test particles away from you - sort of a test particle explosion with you at the centre? But then why would it be surprising that the ones you propel in a direction away from the event horizon escape. Or are you maybe waiting until you're right on the horizon the propelling them all in a radial direction away from the black hole? So half would escape and half wouldn't? $\endgroup$ Feb 3, 2015 at 19:13
  • $\begingroup$ @kristjan - the escaping particles needn't be accelerating. They are defined to be freely falling. To escape from any location where the escape velocity is less than c, the speed of light, a freely falling object need only have sufficient constant upward velocity, not acceleration. See "escape velocity" in Wikipedia for further info. $\endgroup$ Feb 3, 2015 at 19:58
  • $\begingroup$ @John Rennie - The cloud of particles as I described it could just happen to be there when I pass by it, and then I start observing it. But I suppose a device in my lab could've propelled them all in a radial direction away from the black hole as you say, such that the initial conditions I described above still apply. The mystery would remain as to how I get all the particles to move in formation as I measure, since GR both allows and disallows that. $\endgroup$ Feb 3, 2015 at 20:14

2 Answers 2


The event horizon is a null surface, which means that if you pick a very small region of spacetime that straddles the event horizon, small enough that the equivalence principle should apply, then any local inertial frame in this region should measure the event horizon to be moving outwards at the speed of light. Slower-than-light objects in this region can avoid crossing the event horizon by accelerating away from it, but in terms of distance measurements in the local inertial frame, they should be getting closer and closer to the horizon, similar to Rindler observers in flat spacetime who accelerate in such a way that their worldlines asymptotically approach a particular null surface, the Rindler horizon. Here's a diagram showing a Rindler horizon as a dotted line, and the worldline of accelerating Rindler observers as black hyperbolas, drawn from the perspective of an inertial frame (the gray lines represent surfaces of constant time in a non-inertial coordinate system in flat SR spacetime called Rindler coordinates):

enter image description here

Note that the diagram above is drawn from the perspective of an inertial frame (where the x and t axes are inertial space and time coordinates), not Rindler coordinates--in Rindler coordinates the Rindler horizon would just be a vertical line of constant position, as would the worldlines of all Rindler observers. And since the Rindler horizon is moving at the speed of light as measured in an inertial frame, naturally slower-than-light objects on the other side of the horizon must be getting farther away from the horizon as time passes no matter how they accelerate, as measured in an SR inertial frame.

edit: I originally said this means that the distance between slower-than-light objects that remain on opposite sides of the horizon must increase over time as measured in an inertial frame, but I now realize this isn't true--if you imagine copying and pasting one of the curves in the diagram above to the opposite side of the dotted line, the coordinate separation between the two curves along the x-axis would remain the same at each value of t. So in that sense, from the perspective of a local inertial frame in a small neighborhood of spacetime straddling the event horizon, I think objects on either side of the horizon can fly in formation in this small neighborhood, even though the one outside must be getting closer to the horizon in the coordinates of this inertial frame while the one inside must be getting farther from it. Also note that just because a particle is getting closer to the horizon in the coordinates of a local inertial frame, that doesn't rule out the possibility that the same particle is moving farther from the horizon in terms of its Schwarzschild radius, since curves of constant Schwarzschild radius would presumably themselves be getting closer to the horizon in a local inertial frame. This is what happens when you plot curves of constant Schwarzschild radius in Kruskal-Szekeres coordinates, which have the nice property that just like in an inertial frame in SR, all light rays traveling in a radial direction have worldlines that look like straight lines at 45 degrees from the vertical, as does the event horizon. For example, see the Kruskal-Szekeres diagram below from this page, where the white region represents the spacetime outside the event horizon, the dark diagonal is the event horizon, and the light gray region beyond it is the interior of the black hole. You can see the event horizon and the sides of light cones are both diagonals at 45 degrees from vertical, but a line of constant Schwarzschild radius r=2.75M looks like a curve in these coordinates, and one which gets closer and closer to the horizon with time in terms of the Kruskal-Szekeres radial and time coordinates, even though in Schwarzschild coordinates r=2.75M would remain at a fixed distance from the horizon at all times.

enter image description here

So, you could easily draw a curve outside the horizon that was getting closer to it with time in terms of Kruskal-Szekeres coordinates, but nevertheless crossed the r=2.75M curve in an outward direction (along with other curves of constant Schwarzschild radius), getting farther from the horizon with time in terms of Schwarzschild coordinates. I don't have the math to verify this, but I would imagine the relation between the coordinates of a local inertial frame and Schwarzschild coordinates works in a similar way.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Feb 27, 2015 at 7:07

Let's call the formation "spear". And we call an observer inside the free falling lab "Labguy". And an observer at infinity we call "Farguy".

The spear is thrown out of the lab.

Labguy says: "The spear keeps moving at constant speed, and its length stays constant"

Farguy says: "The spear's velocity compared to the local speed of light near the spear is decreasing, and the spear's length is increasing, it's the decreasing Lorentz contraction that causes the lengthening of the spear. Labguy does not observe any lengthening of the spear, because his acceleration away from the spear cancels the spear's acceleration towards him".

And this is what the Labguy thinks about the spear reaching the Farguy: "The Farguy is accelerating, the head of the formation reaches the Farguy, the end of the formation only reaches the place where the Farguy is now".

Basic special relativity:

A rod may Lorentz contract in one frame, but experience a decrease of Lorentz contraction in another frame.

Adding just a little bit of general relativity we can say:

A rod may have a constant Lorentz contraction in one frame, but experience a decrease of Lorentz contraction in another frame.

  • $\begingroup$ I don't disagree with this, but also don't see how it answers my question. I'm not concerned with what a faraway observer measures. I want to know how I can let all the particles of the cloud move in formation as I measure, and as the EP allows, when GR also demands that the cloud be splitting apart as I measure. Nothing is accelerating in my lab. Everything is freely falling in my lab, which is itself freely falling. So there's no decreasing or increasing Lorentz contraction. $\endgroup$ Feb 3, 2015 at 23:30

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