# SHM caused by a black hole's gravitational force

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## Main Question or Discussion Point

I hope I posted in the correct forum...

So, to put it simply. Let's say we have a point mass "m" at rest on the event horizon of a black hole of mass "M" and we throw it directly toward the location of the black hole's singularity. The particles only does linear motion and does not orbit the black hole at any point. IT falls directly into the spacetime curvature caused by the black hole. This means it will follow the shortest path and only travel the distance represented by the blackhole's Schwarzschild radius until it reaches the singularity.

Let's say that initially the black hole and the point mass are stationary relative to each other. There's no spinning of either the black hole or the point mass.

When this happens, it accelerates. My question is, will it come off from the other side of the singularity and execute simple harmonic motion?

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Ibix
Once you've crossed the event horizon, your path ends in the singularity, I'm afraid. All paths do. Once inside the horizon you can't move further from the singularity any more than you can move back in time outside it. So no, SHM is not on the cards. The singularity is the future once an object has crossed the horizon.

A minor aside: a massive object cannot be at rest at the event horizon.

pervect
Staff Emeritus
I hope I posted in the correct forum...

So, to put it simply. Let's say we have a point mass "m" at rest on the event horizon of a black hole of mass "M"
This is not possible :(. To explain why we need some special relativity. An object of mass m will always move at less than the speed of light, in the laungauge of SR it would "follow a timelike worldline". The event horizon of a black hole is, in some sense, moving at the speed of light. In the language of SR, it is a null worldline, sometimes called a "lightlike worldline". Lightlike worldlines don't have a "point of view". No massive particle can move at the speed of light, so the sorts of worldlines that massless particles follow (which are lightlike worldlines) are different in kind from the sort of wordlines that massive particles follow (which are called timelike wordlines).

What's probably unfamiliar about this idea is the idea that the event horizon is moving. If we consider things from the point of view of a massive particle, falling into the black hole hings become simpler.

From the point of view of the massive particle, the event horizon is always moving outwards at the speed of light.

Thus the relative velocity of any massive particle and the event horizon will be equal to "c" when the particle is at the event horizon.

... and we throw it directly toward the location of the black hole's singularity. The particles only does linear motion and does not orbit the black hole at any point. IT falls directly into the spacetime curvature caused by the black hole. This means it will follow the shortest path and only travel the distance represented by the blackhole's Schwarzschild radius until it reaches the singularity.
We can imagine a particle falling into a black hole, a particle that was initially accelerating outwards from the black hole to maintain a constant Schwarzschild R coordinate that was outside the black hole, that suddenly stops accelerating and falls into the black hole.

This falling particle will follow a geodesic path, but I would not characherize this path as "the shortest path". I would characterize the path that the fallign particle follows as a space-time geodesic. From the particles point of view, we can say that it measures not a distance (it's not moving in its own frame), but a time. There well be a finite time interval (by the particles clock) between when itt cuts its acceleration and the event horizon rushes up to meet it, and another finite time interval after that at which the particle is destroyed by the central singularity.

Let's say that initially the black hole and the point mass are stationary relative to each other. There's no spinning of either the black hole or the point mass.

When this happens, it accelerates. My question is, will it come off from the other side of the singularity and execute simple harmonic motion?
No, in standard GR, it will be destroyed by the central singularity. There are some theories in which this does not hapen that are not standard GR, such as those published by Nikoderm Poplawski. But it would be misleading to think of what happens in those theories as "simple harmonic motion". I'm not sure how much more to explain, if there is some interest in what these theories predict, perhaps it'd be best to ask them in a different thread since they aren't GR. I'll give a popular reference to the class of theories though - https://www.insidescience.org/content/every-black-hole-contains-new-universe/566, and suggest that for clarity we make a discussion of what happens in those theories a different thread from what happens in standard GR. What happens in standard GR is that physics becomes singular at the central singularity. I suppose the most accurate thing to say might b that GR makes no prediction because the theory becomes singular there, but it's also common to say that the particle is destroyed (and ceases to exist as a particle) when it encounters the singularity.

PeterDonis
Mentor
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Let's say we have a point mass "m" at rest on the event horizon of a black hole of mass "M" and we throw it directly toward the location of the black hole's singularity.
As pervect points out, no object can be at rest at the event horizon. But you could have one at rest a little bit outside the event horizon, and then release it into radial free fall.

This means it will follow the shortest path and only travel the distance represented by the blackhole's Schwarzschild radius until it reaches the singularity.
The Schwarzschild "radius" of the hole is not a radial distance between the horizon and the singularity. It is $\sqrt{A/4 \pi}$, where $A$ is the area of the hole's horizon. The concept of "radial distance to the singularity" actually does not make sense, because the singularity is not a certain distance inside the horizon; it is a certain time to the future of the horizon. (Pervect pointed this out as well.)

When this happens, it accelerates.
Only in the sense of coordinate acceleration--but as the name implies, this kind of acceleration depends on the coordinates you choose. The object is in free fall, so it feels no acceleration, and felt acceleration (more precisely, "proper acceleration") is the only kind of acceleration that is invariant, independent of coordinates.

will it come off from the other side of the singularity and execute simple harmonic motion?
No, because, as above, the singularity is not a certain distance inside the horizon, it is in the future. So the object will reach the singularity and (at least according to classical GR) be destroyed, as pervect said.