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Why do tidal forces exist in black holes?

Physics Asked on June 12, 2021

If a person (free)falls into a black hole at some point or the other, he will be stretches by a ‘force differential’ between the two ends of his body. This is the process of spaghettification.

But, general relativity also tells us that any observer who falls into the black hole will be inertial, as he is just moving along a geodesic that leads to the interior of the black hole.

The Equivalence Principle surely cannot be wrong. But the observer also feels pulled and stretched by a force (which means he is accelerating for sure). So how do these tidal forces emerge (which clearly turn an inertial observer into an accelerating one, while he is in free fall)?


Not probably relevant to the question, but I thought up this explanation:

If we consider the observer to be like two spheres (one a the feet, other at the head) connected by a string (the rest of the body), and if we consider this object to be falling into a black hole, then du to the curvature of spacetime, the sphere nearer to the singularity will experience higher time dilation, and hence might take a different geodesic than the other sphere (which experiences lower time dilation).

In that case, as the two spheres travel in two separate geodesics toward the same singularity, maybe one of them accelerates faster (through time if not through space) than the other. The discrepancy in the accelerations causes the string to go taut, and eventually (as the discrepancy increases) to break, leading to the same effect of spaghettification.

Is this explanation correct, or at least am I thinking in the right direction?

2 Answers

You're on the right track, but not quite there - in particular you don't need GR to see where tidal forces arise from.

But, general relativity also tells us that any observer who falls into the black hole will be inertial, as he is just moving along a geodesic that leads to the interior of the black hole.

This only happens for a point particle. A point particle will indeed experience no tidal forces. You can get this result from Newtonian mechanics, where the force experienced by the point particle is simply $F = GMm/r^2$.

Tidal forces arise because real objects are not point particles. If instead of the hypothetical point particle falling into the black hole, you had a 1-meter long object, and if the force experienced by the bottom of the object is $F = GMm/r^2$, then the force experienced by the top of the object is $F = GMm/(r+1)^2$. The difference between the two is the tidal force, which increases as $r$ decreases. It is why spaghettification occurs. When the tidal force is large enough, then for this object, the frame is no longer inertial. Still, if you decrease the size of the object, you can make the frame inertial (until you hit the singularity).

Answered by Allure on June 12, 2021

The equivalence principle is a local principle. It does not rule out detecting a gravitational field by making non-local observations. Tidal forces can only be detected by comparing the trajectories of objects that start from different locations in spacetime. So tidal forces are a non-local phenomena, and using tidal forces to detect a gravitational field does not contradict the equivalence principle.

Answered by gandalf61 on June 12, 2021

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