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Imagine a simple, non-rotating black hole and a massive rocket that is free-falling past the event horizon of the black hole, linearly towards the singularity. The rocket is massive enough that it carries an appreciable momentum, and the linear velocity of the black hole will be measurably changed by addition of the rocket's momentum to that of the black hole.

An observer outside of the black hole is carefully, continuously monitoring the position and velocity of the black hole event horizon, and is able to detect the change in its motion dynamics caused by the free-falling rocket combining with the black hole.

However the rocket also has very powerful engines, and the captain of the rocket is able to make a decision after passing the event horizon as to whether to engage these engines to accelerate (briefly) at full power towards the singularity. The captain is instructed to only engage the engines in the case that some specific condition inside the event horizon is observed, and the external observer is also aware of this plan.

So my question is, will the dynamics of two cases differ in any way, from the perspective of the external observer? In the first case (no engine), there is a single collision between the rocket and the singularity at a predictable point in time. In the second case the initial collision will take place at a different (earlier) time from both the rocket and the external observer's perspectives, and with higher initial momentum - noting this is a more complex dynamic scenario as the rocket exhaust gases will also interact with the singularity via both gravitational attraction and eventually their own secondary collisions.

If the position of the black hole event horizon in the two scenarios varies at all for any instance, then the observer can accurately detect the signal that the captain of the ship issued and information can be transmitted out of the black hole via momentum signals. Even if the final momentum of the system may be the same in both scenarios, the intermediate dynamics may vary and this can be detectable.

The scenario could be simplified from a rocket to two masses held together with an explosive charge between them, without loss of generality. In this case the decision is whether the explosive charge is detonated or not.

It seems to me that the main transfer of momentum from the falling object to the black hole must happen when it collides with the singularity. So if the timing of the singularity collision(s) can be changed, then the timing of the momentum change(s) of the black hole will also vary and this should be visible to an external observer.

What do I miss in this scenario, that prevents information being transmitted externally?

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    $\begingroup$ You cannot transmit information backwards in time. From my answer physics.stackexchange.com/a/696441/123208 "once it crosses the event horizon it can no longer change the spacetime curvature outside the horizon". $\endgroup$
    – PM 2Ring
    Commented Sep 15 at 9:06
  • $\begingroup$ @PM2Ring so how & when does the transfer of linear momentum occur between the infalling object and the black hole? If I understand correctly you are saying this happens before the object crosses the event horizon, is that right? So there must be a mechanism causing a black hole to be 'pushed' by an object approaching it in free-fall, before the object crosses the event horizon? $\endgroup$
    – roblev
    Commented Sep 15 at 9:42
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    $\begingroup$ The object and black hole fall towards each other, just like in Newtonian gravity, although the equation of motion is slightly different, especially when the distance becomes small. And a distant observer never sees the object actually cross the event horizon, as described here: physics.stackexchange.com/q/21319/123208 $\endgroup$
    – PM 2Ring
    Commented Sep 15 at 10:09
  • $\begingroup$ Also see physics.stackexchange.com/q/815189/123208 $\endgroup$
    – PM 2Ring
    Commented Sep 15 at 10:14
  • $\begingroup$ I thought that "distant observer never sees the object cross the event horizon" applied only in the Schwarzchild case where the infalling object is massless. In my case the infalling object is massive enough to change the black hole dynamics and so the Scharzchild approximations are not valid. Given that we know two black holes can combine in finite time (from our observer perspective via LIGO), then why would the massive rocket still take infinite time to pass the EH of the BH? $\endgroup$
    – roblev
    Commented Sep 15 at 10:45

2 Answers 2

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The point that you are missing is that the geometry of spacetime behind the event horizon is such that the light cones of any observer that cross it are "tilted" in such a way that there are no causal worldlines originating at the observer that point away from the singularity, cf. this question. Any detectable signal (In this case, gravitational radiation) must travel along a causal worldline and can therefore never cross the horizon from the inside. That being said, if the rocket is massive enough to impart measurable momentum on the black hole, this momentum transfer must happen when the rocket crosses the event horizon. This causes the horizon to slightly distort, and as it returns to a spherical shape, the black hole "rings", emitting gravitational radiation. Furthermore, conservation of momentum holds within black holes as well. If the rocket were to fire its thrusters after crossing the horizon, any momentum gained by the rocket is compensated by gas being ejected into the opposite direction, so that the total energy-momentum density inside the horizon stays constant (This is obvious from the fact that nothing can escape), thus yielding no measurable change in the shape or momentum of the black hole after the rocket has crossed the horizon.

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  • $\begingroup$ Thanks for this answer, do you know what is the mechanism for the momentum transfer when the rocket crosses the event horizon (beyond just "General Relativity"!)? $\endgroup$
    – roblev
    Commented Sep 15 at 11:01
  • $\begingroup$ @roblev The shape of the horizon is determined from the energy-momentum distribution in the spacetime region containing the rocket and the black hole, and changes accordingly. Momentum is not necessarily transferred in a mechanical way, but rather as a consequence of how spacetime geometry changes in response to a changing energy-momentum distribution. But this is "just General Relativity", so I can't really give you a much deeper answer here. However, there are already questions addressing this topic on phys.SE, and if they don't help, you can ask a new one. $\endgroup$
    – paulina
    Commented Sep 15 at 13:22
  • $\begingroup$ @safesphere yes, my point was that nothing you do inside the black hole region can be detected from outside. "After the rocket has crossed the horizon" just means that it is inside the black hole region. Perhaps that was poor choice of wording. $\endgroup$
    – paulina
    Commented Sep 17 at 19:23
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    $\begingroup$ Fair enough. I’ve deleted my comment. The key here is that the direction from the horizon to the inside is not spatial, but temporal. The same as the direction from midnight to tomorrow. After the moment of midnight is crossed, nothing can affect yesterday, because yesterday no longer exists. The same way an external observer cannot even try to detect anything from the inside, because nothing crosses to there for eternity of the external universe. $\endgroup$
    – safesphere
    Commented Sep 17 at 20:10
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Event Horizon is a global notion. Technically speaking, a BH is a region from which no causal signal can travel out into the spacetime , to the infinite future. EH which is the boundary of this region is therefore defined in a way taking into account all events which are going to happen in the future. This includes the choice of captain to accelerate and his intention to communicate the observer outside. In other words, if captain can communicate to the observer far away, then he is not inside an BH by definition.

Further, physically an isolated rocket can only accelerate in one direction if it ejects some mass in the opposite direction. However, at the very instant the velocity of center of mass doesn't change following conservation of momentum. Similar argument for angular momentum holds true. The observer outside can only measure the total momentum, angular momentum of the BH + rocket system, both of which will remain conserved in this process.

However, there is still a way to infer that some event has indeed taken place by exploiting how quantum fields can evolve under the effect of curvature or relativistic motion leading to a change in entropy. For eg you can devise some strategy along the lines discussed in these articles here and here. The observer and captain can share some entangled state and any attempt by the captain to accelerate will result in interference and particle creations thereby changing the entanglement entropy. However, this doesn't imply that you are sending new information out to the observer. The entropy is just a measure of correlation : you simply know since the start of the experiment how and when a state is going to evolve if there is such an acceleration.

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