If I'm an object in free fall near earth, then I'm an inertial frame of reference and I see the earth accelerating towards me with no force acting upon it. What causes that acceleration? The spacetime curvature I cause should be negligible. Is the spacetime curvature caused by earth itself responsible for its acceleration in my frame?
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$\begingroup$ If you are accelerating, the frame in which you are stationary is not inertial, so it is not valid to apply the same physics. (It is like asking what causes the centrifugal force) $\endgroup$– user195162Commented May 18, 2018 at 0:50
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3$\begingroup$ @Quantumness In general relativity (which this post is tagged), free-fall is an inertial frame. $\endgroup$– Chris ♦Commented May 18, 2018 at 1:43
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$\begingroup$ @Chris do you know of a source to read more about that? As I know it inertial frames never accelerate with respect to each other. $\endgroup$– user195162Commented May 18, 2018 at 2:16
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3$\begingroup$ @Quantumness Any GR textbook. In flat spacetime, your statement is true, but the coordinate acceleration between two inertial observers doesn't need to be zero in curved spacetime. $\endgroup$– Chris ♦Commented May 18, 2018 at 2:36
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3 Answers
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In classical (Newtonian) mechanics, you are not in an inertial frame, so your observations are not valid: the acceleration is fictitious.
In general relativity, the solution is more subtle. You are in an inertial frame, and so is the earth. But in general relativity, inertial frames are not global. The correct way to think about the earth's motion is to determine that it is following a geodesic, and so it has no proper acceleration. The apparent acceleration of earth is purely an artifact of the frame you've chosen, just as it is in the Newtonian case.
You are both in inertial frames, that happen to have a relative acceleration between them. This can't happen with flat spacetime, but there is no contradiction once you introduce curved spacetime.
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$\begingroup$ What if earth's mass were concentrated in a small volume and I, the falling object, were very close to it? Would my local assumptions approximately hold, then? $\endgroup$ Commented May 18, 2018 at 1:26
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2$\begingroup$ No, as you make Earth smaller and smaller the spacetime around it gets more and more curved, so if anything that's worse. At the extreme limit, you get a black hole. $\endgroup$– Chris ♦Commented May 18, 2018 at 1:35
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$\begingroup$ @TuringMachine What Chris said. As spacetime gets more curved, you have to cut it into smaller "chunks", where each chunk is flat enough to use Special Relativity in it, and SR notions like locality are applicable. $\endgroup$– PM 2RingCommented May 18, 2018 at 2:59
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It's important to distinguish between proper acceleration and coordinate acceleration.
An observer in the rest frame of the Earth would see you accelerating with an acceleration of 9.8 m/s$^2$. At the same time, you would see the earth accelerating at the same rate. Both of these are examples of coordinate acceleration - each observer labels the position of the other with some numbers, and they see those numbers changing at an increasing rate as time goes on.
However, neither observer is experiencing proper acceleration. In your reference frame, the fact that the coordinates of the earth are changing at an increasing rate is an artifact of the fact that you have chosen a coordinate system which is moving through curved spacetime.
answered May 18, 2018 at 1:36
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1$\begingroup$ This is wrong. The earth-fixed observer is experiencing a proper acceleration, and that is not coordinate-dependent. $\endgroup$– user4552Commented May 18, 2018 at 3:11
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1$\begingroup$ @BenCrowell Are you referring to the fact that somebody standing on the surface of the earth is not in free fall? That's not what I meant - I was referring to a reference frame attached to the center of the earth, in free fall towards the person. $\endgroup$ Commented May 18, 2018 at 3:20
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The force acting on the ground is the normal force of the layer of the earth right beneath the ground. The force acting on the layer right below the ground is the normal force from right below that.
Ultimately, however, you can not really understand gravity just by thinking about inertial frame. The whole inertial frame business only works locally: globally, you have to know how spacetime is curved everywhere, and solve the full Einstein equations, etc, to figure out how the earth will behave in the presence of the gravitational field. As I have said, the "force" which acts on each particle of the earth, that makes it move in non geodesic motion, is the pressure (normal force) from its fellow particles.
answered May 18, 2018 at 1:09
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$\begingroup$ So the flaw in my thinking is that the earth occupies a large enough volume that we can't assume the problem is approximately local, if I understand correctly. What if earth's mass were concentrated in a small volume? Since I, the falling object, am near the earth (just about to touch it), my local assumptions should approximately hold, right? $\endgroup$ Commented May 18, 2018 at 1:16
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$\begingroup$ By "local" I mean that the gravitational force is approximately constant on the length scale being considered. If the wart was smaller, the region of space you could consider and still reasonably call "local" would shrink accordingly. $\endgroup$ Commented May 18, 2018 at 2:50