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Imagine a second earth one lightyear away from our earth. Now imagine a bunch of scientists with a plane and some atomic clocks who are living on this second earth and who perform the Hafele-Keating Experiment, which is the experiment that was done in 1971 on our earth to prove the relativistic effects of kinematic and gravitational time dilation:

Joseph C. Hafele, a physicist, and Richard E. Keating, an astronomer, took four cesium-beam atomic clocks aboard commercial airliners. They flew twice around the world, first eastward, then westward, and compared the clocks against others that remained at the United States Naval Observatory. When reunited, the three sets of clocks were found to disagree with one another, and their differences were consistent with the predictions of special and general relativity. (Wikipedia)

Now lets suppose both earths are not moving or at least that they are not moving in relation to each other. Ignoring the gravitational time dilation, we can tell already from the experiment of 1971 that the atomic clocks on the fast moving plane will run a bit slower than the static atomic clocks on earth-two due to the kinematic time dilation of special relativity.

Because earth-one is not moving in relation to earth-two we can expect that if there was another atomic clock on earth-one it would run just as fast as the clocks on earth-two, hence also faster than the ones on the plane.

Now my question: Does this asynchronity between the time on earth-one and the time on the highspeed-plane flying around earth-two start the moment the plane starts (ignoring the short time of acceleration the plane needs) or does it start much later, when the light (information) of the flying plane arrives at earth-one (which would be one year delayed because of the distance of one lightyear between the two earths)?

In other words: Do the effects of special relativity occur instantaneously, independed of the distance between the observing and the observed frame of reference?

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When we use the Lorentz transformations we are transforming between coordinate systems. Both Earth One and Earth Two share the same inertial frame and the same coordinate system (ignoring gravitational effects as you suggest).

The aeroplane occupies a different inertial frame and coordinate system, and the Lorentz transformations tell us how spacetime points in the plane's coordinates map to spacetime points in the two Earths' coordinates, and vice versa. So if take some point $(t, x)$ in the plane's coordinates and apply the Lorentz transformations we'll get the point $(t', x')$ in the two Earths' coordinate system.

The key point is that the point $(t', x')$ is the same whether you're on Earth One or Earth Two because both Earths are using the same coordinates. So if $t \ne t'$, as would be the case due to time dilation, both Earth's will agree how much the planes time is dilated. There is no sense in which the time dilation has to travel from Earth Two to Earth One.

I'd guess the confusion is because in SR the act of observing means the assigning of events to spacetime points. Observing doesn't mean you have to wait some non-zero time for the light from an event to reach you - it doesnt mean the same as seeing. Suppose the plane takes off from Earth Two at some time $t_1$, the time it takes off is also $t_1$ on Earth One. It isn't $t_1$ plus the time light takes to reach Earth One from Earth Two.

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Imagine a triangle in ordinary Euclidean geometry, with vertices A, B, C. The sum of the lengths of the sides AB and BC is larger than the length of the side AC (this is the triangle inequality). Where is this extra length? Clearly the question is meaningless; one path from A to C is simply longer than the other. There is no part of the longer path that corresponds naturally to the whole of the shorter path, leaving the rest as the extra length.

Spacetime is very similar to four-dimensional Euclidean space, except that the distance formula is different. The answer there is the same. If you synchronize two clocks (at A) and find that they have different readings when they meet against later (at C), it's meaningless to ask where that difference happened. The spacetime equivalent of Euclidean distance is proper time, and that is what clocks measure, so this is a close analogy.

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