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We know the phrase "all objects move at the same rate when freefalling regardless of their mass" (interestingly this isn't even true because the objects are exerting their own gravity, but we usually disregard this minuscule effect), but my question is inside the black hole, where time and space behave "oddly", the objects are freefalling towards tomorrow, and the singularity is not a point in space, but rather a future event.

I guess the best we could do is ask how fast the infalling observer observes the singularity to be approaching, though note that this is theoretical since no light from the singularity could ever reach the observer's eye. And the answer is that the speed would indeed exceed c inside the horizon, though I must emphasize again that you shouldn't assign any physical significance to this.

Speed of object falling into a black hole, below event horizon

Now, are the objects moving (freefalling) towards this future event all at the same rate regardless of their mass (stress energy)? Please note I am not asking about their actual speed or acceleration or if we have a way to calculate this. All I am asking is whether this "rate", whatever it is, is the same for all objects regardless of their stress energy. So do all objects move towards this future event (singularity) at the same rate?

Now I believe this question has some merit, since there are a lot of questions on this site about the freefall inside the black hole, and there are even questions about how to calculate something in this manner, that is, the proper time and distance it takes for an object to reach the singularity:

What's the proper distance from the event horizon to the singularity?

But my question is not about a specific way to calculate the time or distance itself. I am asking whether all objects "move" at the same "rate" towards the singularity regardless of their stress energy.

Do the feather and the rock example work inside the black hole?

Question:

  1. Are all objects freefalling at the same rate inside a black hole?
Qmechanic
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Árpád Szendrei
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    Well, the thing about everything falling at the same rate is really about the equivalence principle. And yes, there is no reason why the equivalence principle wouldn't hold in a black hole. – Javier Oct 05 '22 at 02:45
  • I am not sure if you mean that freefalling toward tomorrow is an odd behavior, but it is not. Not only freefall, but every motion of every object is toward tomorrow. If you would manage to stop traveling into the future, you would become the greatest discoverer of all times. – Umaxo Oct 05 '22 at 08:41
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    Is it okay if I point out the futility of us trying to answer a question about the physics of a region that we can't be sure is even accurately described by our current understanding of physics? We can give you an answer that we can't be sure is correct and we can't test the accuracy of. – Jim Oct 05 '22 at 13:52
  • @Umaxo ”If you manage to stop traveling into the future, you would become the greatest discoverer of all times” - and disappear from reality. – safesphere Oct 06 '22 at 06:09
  • @Jim it is okay to point out, but I do not think it is futile. Studying extreme cases of our theories deepens our understanding of them and the problems they face. From strictly empiristic viewpoint it is futile, but from the viewpoint of understanding the theory and its limits and for guidelines to further research it is not. – Umaxo Oct 06 '22 at 06:31
  • @safesphere not necessarily. You can travel on timelike path (as everyone else), then switch to spacelike (or timelike to the past) and then back to proper timelike. – Umaxo Oct 06 '22 at 06:32
  • @safesphere we have this discussion quite regularly on this site. "The only way you can cross it is when the universe no longer exists" Sure, but as long as my worldline ends on the event horizon in finite amount of proper time, I have no reason (in GR alone) to cease to exist, even if universe did. Thus I want to guess what will happen to me and I extrapolate from known physics/models. This may not be in the spirit of strict positivism for observers not falling into BH, but its the best guess I have and it is sensible to do. – Umaxo Oct 06 '22 at 07:45
  • @Umaxo I wasn't aware we were defining guidelines for further research on this site. – Jim Oct 06 '22 at 14:25

2 Answers2

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Disregarding the same miniscule effect that you mention in the question, then yes. The metric interval equation does not feature the mass of the falling test object. Thus any derivatives you construct (with respect to $t$ or proper time) do not feature the mass of the falling object.

ProfRob
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General relativity says: The motion of a free falling body is described by the geodesic equation $$\frac{d^2x^\mu}{ds^2}=-\Gamma^\mu{}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}$$ where $x^\mu$ are the spacetime coordinates of the body, $s$ is a scalar parameter of motion (e.g. the proper time of the body), $\Gamma^\mu{}_{\alpha\beta}$ are the (usually spacetime-dependent) Christoffel symbols which are independent of the test body, and the summation over repeated indices ($\alpha$ and $\beta$) is implied.

This equation governs the motion in all gravitational fields, not only in the gravitational field of a black hole.

An important thing to note is: The mass $m$ of the falling body does not occur in this equation. So, all free falling bodies fall in the same way.

Thomas Fritsch
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