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I'm working on a paper that also addresses the topic of general relativity (among other topics).

The most common answer I get to the question above (why do objects fall) is that the objects are not really stationary, they are moving through spacetime, and thus they are forced to follow spacetime geodesic lines. I have a hard time understanding this explanation... How is an apple moving through spacetime? When looking at the spacetime continuum from the outside, isn't the "moving" 3d apple actually a stationary 4d object?

Even more, in the common representation of the Earth bending the fabric of spacetime, I understand that the 2d grid is meant to be a representation of the 4d spacetime continuum. Then why is the Earth represented as a sphere? Shouldn't it be represented as a (long) cylinder instead, so it's existence in time is made clear?

Next, shouldn't the falling apple be represented as a much thinner cylinder, which at some point (when the stem breaks) starts getting closer to the larger cylinder (the Earth)?

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    $\begingroup$ How is it stationary? It is "moving" along the time axis. $\endgroup$ – Rob Jeffries Feb 16 '18 at 17:03
  • $\begingroup$ If there was a deep hole, all the way down to the center of the Earth, the apple would fall all the way down. If there were a billion holes that deep, all around the Earth, and a billion apples would be dropped in them, then all of them would make it to the center of the Earth. $\endgroup$ – user185092 Feb 16 '18 at 17:56
  • $\begingroup$ So even if we say, for lack of a better language, that the seemingly stationary 3d apples are actually "moving" along the time axis, when they are released, do they all follow geodesic lines to the center on the Earth? I find this to be very weird, because in the graphic representations, the t-axis geodesic lines never appear to converge to the center of the planet. Actually, they never appear to intersect each other. So, do I need to see a better, more accurate, graphical representation? Thanks a lot! $\endgroup$ – user185092 Feb 16 '18 at 17:56
  • $\begingroup$ " shouldn't the falling apple be represented as a much thinner cylinder, which at some point (when the stem breaks) starts getting closer to the larger cylinder (the Earth)?" - In a graphic of space-time, yes exactly, lookup "world volume". $\endgroup$ – Tom B. Feb 16 '18 at 21:49
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    $\begingroup$ " in the common representation of the Earth bending the fabric of space-time, I understand that the 2d grid is meant to be a representation of the 4d space-time continuum" -not really, those graphics can be confusing, you're supposed to imaging a marble or something rolling on that fabric and being able to curve around the earth, (but think about it, that would mean in the graphic, the force of gravity would be "down", like from the north pole to the south.) There is no time axis on those graphics. $\endgroup$ – Tom B. Feb 16 '18 at 22:00
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First, only a test particle "falls" along a geodesic. A test particle is an idealized object not only at rest, but which also does not itself contribute to the curvature of spacetime (no mass, no energy). An apple can be considered as a test particle in a system including earth, but it is a simplification, as the overall spacetime is determined by the dynamics of all objects that supposedly live "in" it; see this related answer of mine.

Now a massive object would also follow a geodesic, if we take this object into account in the spacetime itself by considering how it itself distorts spacetime. See this question - as John Rennie says there, it is a matter of terminology. The main point I want to make here is that spacetime is not a background. There is no "fabric" of spacetime.

Second, and as you correctly state in the question, a geodesic is not a purely spatial trajectory, it is a 4-dimensional curve, so it is actually misleading to think about "falling along" a geodesic, because the dynamical aspect of "falling" is already a part of the geodesic itself. A geodesic represents the world line of a test particle, and as such it is static: it tracks the object position in the most generalized sense, in all possible observer-relative decompositions in space and time of the unified spacetime it lives on. In other words it says where the particle is at any time in a way that is independent of any observer, an absolute way.

This means that "falling" along a geodesic is simply equivalent to "being somewhere" for a relativistic object. There is nothing forcing an object to follow a geodesic, a geodesic just tells when/where an object is, for its whole existence, an existence during which nothing ever messes with its position/momentum (nothing except gravity, but precisely because gravity is replaced by geometry in general reativity, it amounts to a non-interaction; see the answers to this question for a related discussion of the equivalence principle).

In short, a (timelike) geodesic is a geometric statement about inertia: where are you when you do nothing and nothing does anything to you? On a geodesic.

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In a curved manifold, as spacetime in GR (general relativity), a geodesic is a curve followed by a non interacting (free falling) particle, whether massive or massless (photon). It is the extension of the straight line concept of SR (special relativity) as in Minkowski flat spacetime.
To figure out why objects are described as falling, let us consider the cartesian coordinates in Minkowski spacetime. Even if a massive particle is at rest, its time coordinate does not stop, hence the path is a straight line, parallel to the time axis. If the particle is moving with a uniform speed the straight line will present an inclination. In a curved spacetime the worldline of a particle has a less simple shape, but it describes a path anyway.
As for the other points related to the pictorial representation of the fabric of spacetime, probably it is an artistic view.

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  • $\begingroup$ Thanks a lot, Michele I get the idea. But what happens if there is a large cylindrical hole all the way from the Noth to the South Pole of a planet with no atmosphere, and someone drops an apple at the North Pole? The apple will reach maximum speed at the center of the planet, then will slow down to zero right before getting to the South Pole, and then again, and again, etc... So what exactly does this mean, what exactly does the geodesic look like in 4d? According to the behavior of the apple, it should look like a sinus, right? Or am I missing something here? Thank you. $\endgroup$ – user185092 Feb 16 '18 at 19:39
  • $\begingroup$ @user185092 I think you're overthinking this. If you're locally experiencing 0-g forces, you're moving on a geodesic. $\endgroup$ – JEB Feb 16 '18 at 20:10
  • $\begingroup$ @user185092, yes the apple will oscillate, and the geodesic will be a wave of some sort, though I'm not sure it will be sinusoidal. Also remember that geodesics of falling objects don't have to intersect at the center of the Earth. In your example, if you tossed the apple with some horizontal velocity, it may miss the center, and orbit it instead. $\endgroup$ – Tom B. Feb 16 '18 at 22:13

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