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It seems to me that a body under the event horizon of the black hole should not experience any tidal forces, because the singularity is not in any spatial direction from the body but in the future.

As such, the space should look flat under the horizon, should not it? There should be perceived no spatial curvature or acceleration, again because there is no spatial direction to the singularity.

Am I wrong?

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  • $\begingroup$ @safesphere since nothing can penetrate the event horizon, all the matter is sticked at the BH surface, so its interrior is hollow. $\endgroup$
    – Anixx
    Jun 26 at 4:24
  • $\begingroup$ @safesphere may question was simple: we have an infalling object with 3 axes, x,y,z. The axis z is oriented towards the BH. After crossing the horizon, which axis will be oriented along t? I assume the object remains intact. Where the rocket's nose will point to? $\endgroup$
    – Anixx
    Jun 26 at 8:55
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The tidal forces experienced by a freely falling observer is due to the curvature of spacetime, not where any "sources" are located. For weak fields, the acceleration along direction $\mathbf{x}$ are $a^i = -R_{0j0}^i x^j$ where $R$ is the Riemann tensor: it is entirely a local property of the nearby spacetime.

The basic Schwarzschild solution is curved everywhere, including under the horizon. Indeed, trying to patch in a piece of flat spacetime on the inside would have trouble making the derivatives match up: it doesn't work.

The fundamental confusion many have about black holes is thinking that they are discrete "things" surrounded by horizons and other phenomena. But they are actually extended spacetime curvature structures (that imply the various phenomena). The singularity is not doing anything and is not responsible for the gravitational field, it is a consequence of the field.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jun 26 at 10:34

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