The clock hand off version of the twin paradox is when clock info is transferred at the point an outgoing ship meets an incoming ship. For example an outgoing ship at .6c, clock sync'd to earth on takeoff, will sync an incoming ship's clock 3 ly from earth so that when he meets up with the earth clock at .6c, the time on the ship's clock is 2 yrs less than the earth clock at earth in agreement with the results from a normal twin paradox scenario.

Why do I ask if this is fake?

The ships and earth are all engaged in constant relative velocity so they are all ageing at the same proper time rate relative to each other as if they were all stationary. The only thing subject to a frame jump, which is what causes age difference in the twin paradox, is the clock info. The incoming ship itself has not experienced a frame jump so the clock in the ship will not reflect the ageing the rest of the ship does. The captain will not end up ageing a different total time from a valid start than the earth clock as he would after a frame jump in a true twin paradox scenario. (A valid start is determined by syncing the two clocks at co-location and working backward to see when they would have both been zero.)

The info is not subject to the Rindler metric that a real clock would experience in a frame jump. The info instantaneously changes the clock whereas the Rindler metric's effect on time itself takes time to affect a physical clock having frame jumped.

  • $\begingroup$ Well, imagine that, when the two ships sync their clocks, a person jumps from the first ship to the second one. Then the "clock info" represents the proper time of this hypothetical person, even if he/she doesn't exist at all. So yes, in the end the ship's clock is less than the Earth's clock. $\endgroup$ – HicHaecHoc Nov 1 '19 at 17:46
  • $\begingroup$ Yup if the two ships change captains it becomes a proper twin paradox example as opposed to a clock info transfer example which I say is not really the same thing. The captains will experience a proper Rindler metric induced age difference which I guess will eventually agree with the fake clock info transfer induced age difference. $\endgroup$ – ralfcis Nov 1 '19 at 17:54
  • $\begingroup$ You say that "the info instantaneously changes the clock". I don't know if I exactly understand what you mean by this. Anyway, the second ship's clock doesn't age in the same way as the hypothetical traveling twin. It simply registers the proper time of said twin, and this is simply because at the time the two ships meet, the clock was set up to do so. The fact that in the end this clock is 2 yrs less than Earth's one is not so surprising, then. $\endgroup$ – HicHaecHoc Nov 1 '19 at 18:07
  • $\begingroup$ Maybe I 'm improperly trivializing but this situation is somewhat analogous to when I change the hour on my clock for Daylight saving time. Obviously, when I do so I'm not experiencing a time travel :) $\endgroup$ – HicHaecHoc Nov 1 '19 at 18:11
  • $\begingroup$ What I want to say is: there is nothing mysterious in changing the hour of a clock to make it match the value that I want. $\endgroup$ – HicHaecHoc Nov 1 '19 at 18:20

The twin paradox doesn't work like that. I can't blame you for being confused since special relativity tends to be taught in a horribly misleading way.

Here are the real rules:

  • Clocks measure the length of their own worldlines (i.e. their paths through spacetime).

  • The length is like Euclidean distance, except that there's a minus sign in it: $\sqrt{Δt^2-Δx^2}$ instead of $\sqrt{Δx^2+Δy^2}$.

That's all you need to know to answer any question about any number of clocks moving in any way in Minkowski space.

The Euclidean version of your example is this: You have three lines in general position that describe a triangle. Call the intersection points A, B, C. Three "clocks" (rulers) measure the distances AC, AB, BC. There's a "handoff" at B which means that you simply add the lengths of AB and BC and don't add or subtract any additional length at the point B. Then you'll always find (because of the triangle inequality) that AB+BC > AC.

In Minkowski space it's the same except that because of the minus sign you always find AB+BC < AC.

If there's a brief acceleration at the point B, then instead of a triangle you have a triangle with a rounded corner at B. But if the rounded part is small enough it scarcely affects the result. And no matter how wiggly you make the second path from A to C, it will always be longer (resp. shorter) in the Euclidean (resp. Minkowskian) case because a straight line segment is the shortest (resp. longest) distance between two points.

Rindler coordinates are Minkowskian polar coordinates. In the Euclidean case, if the rounded bit at B is an arc of a circle then you can use polar coordinates to analyze the path there. But there's no good reason to do that; it only makes the problem more complicated. You can just use the formula for the length of a circular arc, or approximate the length as though the radius of curvature was 0. All of this is also true in the Minkowskian case, for the same reason.

It's not true that to analyze something in special relativity you have to use coordinates in which it's at rest to first order or to second order. All coordinate systems are equivalent, so you can use whatever coordinate system makes the problem easiest to solve.

  • $\begingroup$ Thanks. So the world line of the incoming ship can co-exist independently from the world line the clock traced out over the outgoing and incoming ships. The incoming ship does not necessarily age 2 yrs less than the earth clock but the ship's clock does. $\endgroup$ – ralfcis Nov 1 '19 at 18:36
  • $\begingroup$ I slept on this and now recognize this is the most important answer I've ever finally understood. Everything else is just noise. If relativity was taught like this, physics forums would be mostly empty. Run your clock across each path, compare your totals, distance is a negative. $\endgroup$ – ralfcis Nov 2 '19 at 13:28


No, it is not 'fake'. All you need for the twin effect is a change to another reference frame moving relative to the first. Strictly speaking you don't even need any clocks-the result is just the consequence of transforming coordinates between moving reference frames.

  • $\begingroup$ So a real clock has traced a real world line and ends up ageing less because its world line includes distance and the earth's does not. This real world line is independent of the world lines the ships follow. Ok, I think I finally got it. $\endgroup$ – ralfcis Nov 1 '19 at 18:16

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