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If you had a singularity (that had mass but took up no space), what would happen to the acceleration of an object as it approached this singularity? I would assume that it would be infinite, since as $r$ approaches $0$, the force of gravity approaches infinity.

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The problem with questions like this is that you can't answer the question unless you specify what observer you mean. This may sound nit picking, but if you drop an object into a black hole you'll see it freeze at the event horizon i.e. the acceleration that you measure falls to zero as the object approaches the event horizon (NB this is the surface of the black hole not the singularity at the centre).

On the other hand, if you jump into a black hole and fall freely you will feel no acceleration. Not when you pass through the event horizon and not as you approach the singularity (GR doesn't tell us what happens at the singularity). You will feel tidal forces - more on this below.

Typically we measure acceleration relative to a freely falling observer. The answer to What is the weight equation through general relativity? shows how this is calculated, but the calculation is only valid outside the event horizon because inside the event horizon there can be no stationary observers. Interestingly, the acceleration calculated in this way becomes infinite as you approach the event horizon. By this I mean that if you hover a distance $\Delta r$ above the horizon and drop an object the initial acceleration goes to infinity as $\Delta r$ tends to zero.

But there is a sense in which your question can be answered. At the end of the second paragraph I mentioned tidal forces. These happen because if for example you jump in feet first your feet are nearer the black hole than your head, so they feel a greater force. The net effect is a force that stretches you. The tidal force on a freely falling object is not infinite at the event horizon, but as you approach the singularity this force becomes greater and at the singularity the tidal force does indeed become infinite.

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Good answer. +1 – Eduardo Guerras Valera Jan 23 '13 at 0:48

I'm assuming you're talking about Newtonian gravity where the force goes as $1/r^2$ in 3D. Then yes, from Newton's 2nd law the acceleration goes to infinity as $r \to 0$. Mathematically it is what it is, there is nothing wrong with it. physically, a point mass is an idealization, most massive objects in our world have a finite size. So before a test object can get to $r = 0$ of the massive object, it hits its surface.

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Its wrong answer. None of Newtonian laws are valid here. – Evil Angel Jan 16 '13 at 6:33

In Newtonian mechanics, if the object has any transverse velocity at all (i.e. motion not directly towards or away from the singularity), it will miss it, passing it at a minimum distance and then returning outward. This shows objects starting from the same maximum distance but with different transverse velocities.


In the limit, with no transverse velocity, the object "bounces" off the singularity. At the point of contact it has infinite acceleration.

In general relativity, this doesn't apply. Once it falls inside the event horizon it doesn't come out, even if it has transverse velocity. As said above, it becomes tricky to define acceleration inside the event horizon. But you can still define the tidal forces locally, and these do go to infinity (as they would in the Newtonian case as well).

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Newtonian Gravity is no longer valid to explain such situation. Gravity is nothing, but warpage of Spacetime (Yes, force of gravity doesn't exist according to Einstein's General Theory of Relativity). Singularity of your question is Singularity in time. Singularity would lie in the direct future of falling object. As such singularity isn't a point in space for falling object, you can't calculate acceleration at all.

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Downvoters should explain the reasons of downvoting.. I am confident with this answer. – Evil Angel Jan 23 '13 at 12:38

protected by Qmechanic Jan 23 '13 at 12:51

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