# Spacetime curvature due to acceleration causing gravity?

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In summary, the conversation discusses Einstein's prediction about light bending in gravity and in an accelerating elevator in space. It is noted that Einstein's original prediction was off by a factor of 2 due to his lack of understanding of curved spacetime. The Equivalence Principle is mentioned, which states that the degree of light bending in a gravitational field is the same as that in a moving elevator. It is then questioned whether an accelerating elevator in space also causes a warping of spacetime. The conversation then delves into the concept of tidal forces and their role in spacetime curvature. It is also mentioned that there is no solution in general relativity for an infinite flat mass with a uniform gravitational field and no tidal forces.
The point I was trying to make was that without spacetime curvature at the Earth, you would not feel a proper acceleration while standing on the Earth because you would not be forced to deviate from an inertial path through spacetime.
Again, no. It is the unbalanced real contact force with the ground that causes the proper acceleration.

Any scenario in which there is an unbalanced real force will lead to the sensation of proper acceleration (with or without curvature). Any scenario in which there is no unbalanced real force will lead to the absence of proper acceleration (with or without curvature). The presence or absence of curvature is irrelevant.

Remember that spacetime curvature is tidal gravity, the sensation of proper acceleration is not. So what does spacetime curvature do in this scenario? It allows you to have proper acceleration in the opposite direction of a person on the other side of the world without the distance between you increasing.

That being said, doesn't it follow that if there is no curved spacetime at the surface of the Earth, one would not feel proper acceleration while standing on the surface of the Earth?
This is getting annoying. No matter how many times you repeat it the answer will still be no.

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PeterDonis said:
You're making an implicit comparison here which isn't really meaningful. When you say "without spacetime curvature at the Earth", what are you comparing the actual situation with? And how are you going to pick out "the same place" in whatever you are comparing the actual situation with?

Am I not comparing it to empty space where there is no curvature? I visualize spacetime curvature as something that shapes the geodesics through spacetime and as a result, when you stand on the Earth, you and the Earth are being forced to deviate from an inertial path through spacetime. This causes the "push" between you and the Earth and the result is the sensation of proper acceleration. Is this not a good way to look at this?

Am I not comparing it to empty space where there is no curvature?

How are you going to do the comparison? Empty space has no distinguishable places; how are you going to pick out the place where the "Earth" is and compare the proper accelerations at the two places?

Is this not a good way to look at this?

No. As has already been said a number of times in this thread. And it has been repeatedly explained why.

Sorry @PeterDonis, I wanted to post one last explanation of what curvature actually does
I visualize spacetime curvature as something that shapes the geodesics through spacetime
This is true, but spacetime curvature is not just any shape. The specific shape that spacetime curvature gives to geodesics is that spatially separated geodesics diverge. Around Earth the shape of the curvature means that geodesics at your head and foot are pulling apart while geodesics on your left and right are squeezing together. Hopefully you recognize this as tidal gravity.

In a given region with a fixed spacetime curvature you can get rid of the proper acceleration simply by removing the floor. The proper acceleration disappears but the curvature doesn’t. However, the tidal force is present whether you are in free fall or not, it does not disappear under any condition as long as there is curvature.

Dale said:
The specific shape that spacetime curvature gives to geodesics is that spatially separated geodesics diverge.

Or converge.

Dale

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