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There are many variations of questions and answers concerning the twin paradox. However, what is interesting is multiple travelers starting out at about the same point all accelerating to different velocities near light speed, traveling vast distances, and then returning to the same point and checking their clocks . . . what would they find?

As I recall, the twin paradox (or non-paradox) came out of special relativity - no gravity and acceleration was not discussed. Do we need both SR & GR to address this?

I'm not interested in what is observed along the way; just what is observed when they all return to the same point. Further, I understand the point of view that we are all traveling through space-time at the speed of light as well as other formulations - so just the simple answer .. what is observed and why?

Dr. Jerald Feinstein
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What is observed when they all return to the same point is that each traveler will have aged less than the non-traveler.

Here's an analogy: in ordinary geometry, even on curved surfaces, between any two points you can always find some path connecting the two points which is shortest. Furthermore, you can always find arbitrarily long paths by making the path "squiggle" around.

For the (psuedo-)geometry of spacetime the situation is similar but with an important twist: there will be some path connecting the two points in spacetime which is longest (i.e. the traveler aged the most), and you can always find arbitrarily short paths (i.e. the traveler ages as close as you want to "0" time).

The path in spacetime which is the longest corresponds to the traveler who stays at rest (they age the most). Roughly speaking, for the other travelers, whichever traveler "accelerates the most" or will be the traveler with the shortest path (aged the least).

Jbag1212
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  • "What is observed when they all return to the same point is that each traveler will have aged a different amount." Not necessarily. – Math Keeps Me Busy May 05 '21 at 00:55
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    @MathKeepsMeBusy Yes, true. What is actually observed depends on the path taken. – Jbag1212 May 05 '21 at 01:00
  • I would suggest adding words to the effect of ". . . and all of them age less than the non-traveler" to the end of your first sentence. – m4r35n357 May 05 '21 at 11:05
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Clocks measure the length of the path taken through spacetime. So multiple observers who leave point A in spacetime and return to the same place in space (but at a different time, so this is a new point B in spacetime) will see different things on their clocks, depending on the exact path they've taken through spacetime. This depends on their acceleration and the time they spent at various velocities. The longest possible time would be measured by an observer who remained inertial the whole time, i.e. who never changed velocity and hence never "left" the original point in space.

The analogy to a car's odometer is close: if multiple observers start at point A and drive to point B along different paths, when they all meet up again their odometers will read different amounts of distances traveled. In this case the shortest distance would be measured by an observer who does not change speed or direction (who goes directly from A to B at a constant speed). So the analogy with clocks is not quite exact, because spacetime is hyperbolic rather than Euclidean; basically the sign of the time contribution to distances in spacetime is opposite to the space contribution.

Eric Smith
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Here is a spacetime diagram on rotated graph paper that demonstrates what the other answers say... between the separation event and reunion event, the elapsed proper time will generally differ along each worldline between those events. As mentioned in the other answers, the ticking wristwatch measuring proper time along a worldline in spacetime is analogous to the odometer measuring distance along a path in space.

(The diamonds are actually traced out by the light-rays in a light-clock along a piecewise-inertial worldline. Lorentz invariance requires that the areas of all such light-clock diamonds are equal. So, count the diamonds along each worldline to get the proper time along that worldline. This is the "clock effect".)

robphy-RRGP-multiple-paths

(The image was taken from my answer at Equivalence of two definitions of proper time in special relativity .)

robphy
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  • I agree . . . but I think it danced around the question . . . what is it they notice about the time each has on their own clocks at the termination of the trip. Same with random paths of high velocity particles in a constrained space. – Dr. Jerald Feinstein May 05 '21 at 01:44
  • @Dr.JeraldFeinstein Each traveler will present his or her elapsed wristwatch's elapsed time... and they will generally be different among them. However, all will agree on how much time elapsed on traveler-A's wristwatch, on traveler-B's wristwatch, etc... By analogy, each odometer will display differing amounts of wear depending on the length of the distance traveled along their path. – robphy May 05 '21 at 01:54
  • @Dr.JeraldFeinstein: if you want a more precise answer, you'll have to ask a more precise question. The time elapsed on any individual traveler's wristwatch can be any value between (but not including) 0 and the time on the stay-at-home observer's wristwatch, depending on exactly how fast and how long the traveler went. If you impose some constraint like "each traveler $i$ always traveled at constant speed $v_i$ both outbound and inbound" then you could deduce that the elapsed time is proportional to $T \sqrt{1-\frac{{v_i}^2}{c^2}}$ where $T$ is the stay-at-home time. – Eric Smith May 05 '21 at 12:29
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As I recall, the twin paradox (or non-paradox) came out of special relativity - no gravity and acceleration was not discussed. Do we need both SR & GR to address this?

SR handles acceleration just fine. As long as we assume a relatively empty, non-expanding universe, SR is sufficient, and GR is not needed.

However, what is interesting is multiple travelers starting out at about the same point all accelerating to different velocities near light speed, traveling vast distances, and then returning to the same point and checking their clocks . . . what would they find?

That all depends on the conditions of their travel. All the travellers could come back with the their clocks in sync, or they could come back with their clocks out of sync. You need to specify the details of their trips in order to determine what the results will be.

Math Keeps Me Busy
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