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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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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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