# Why can clocks not be compared unless they are meeting?

In the answer here to a special relativity question about clock synchronization: https://physics.stackexchange.com/a/485517/141472 it says (bolding mine):

As long as the two space ships are not meeting, their clocks cannot be compared, and each of the space ship considers that itself is at rest. When they are meeting, both spaceships traveled from A to B, but it is their path which is decisive for their respective aging, and the synchronization is done by the means of the path integral of their respective velocity. The spaceship with the higher path integral of velocity has aged less than the other spaceship.

I don't understand why it's not possible to compare clocks unless they are meeting. Consider a scenario where we have clock 1 on a spaceship travelling at $$0.8c$$ from A to B (1 light year apart) and clock 2 grounded on A. From clock 2's point of view it take clock 1 1.25 years to get from A to B. Due to length contraction the spaceship sees the distance between A and B as 0.6 Ly and so clock 1 shows 0.75 years when it arrives at B.

All well and good so far. Now as soon as the spaceship arrives at B and stops it can send a light signal back to A with the message that it showed 0.75 years had elapsed upon arrival at B. This message would take 1 year to get to A, and clock 2 at A knows this (since it knows the distance between A and B) and as soon as this message arrives, clock 2 would take its current time (2.25 years elapsed), subtract 1 and get that at the time clock 1 arrived at B, clock 2 was showing 1.25 years while clock 1 was showing only 0.75 years.

This to me seems like a way of comparing the two clocks even though we don't need any meeting between the actual clocks. What am I missing here?

• Sounds like you might be describing something called "radar time", which is used, but I can't find a good reference ATM, so I'll leave it at that for now. Jul 3, 2023 at 19:46
• OK here is one, "On the radar method in general-relativistic spacetimes", does that help? arxiv.org/abs/0708.0170 Jul 3, 2023 at 19:50
• Thanks!, I'll take a look. Jul 3, 2023 at 20:13

It is rather misleading to say that two clocks moving relative to each other cannot be compared at a distance- of course they can, in exactly the way you describe. However, the point is that there are two ways to perform the comparison, both equally valid, and they yield different results. One way is to assume clock 1 is stationary; the other is to assume clock 2 is. Given that, it would be better to say there is no single authoritative way to compare two moving clocks at a distance.

• Hmm, why can't we resolve this two different methods yielding different results problem in the same way we resolve the twin paradox by saying that clock 1 is the one that accelerates and decelerates, so it's going to experience less time as having passed? Jul 3, 2023 at 19:52
• @Hadi Sure, in that case the ship's worldiine has 2 (or more) kinks in it, so it will accrue less proper time than the clocks in the AB frame (which have straight worldlines). However, we can't do that if the ship travels at constant velocity for the whole experiment, synchronising its clock with the clock at A as it passes, and then comparing its clock reading at B as it passes. Jul 3, 2023 at 20:13
• @Hadi (cont) It may help to consider two (co-linear) sets of synchronised clocks, one set in the AB frame, and one set moving at 0.8 c. The clocks in each set are in synch with each other, but they are out of synch with the clocks in the other set. Jul 3, 2023 at 20:14
• Yes you can do that, but you haven't 'resolved' the two methods- you have just arbitrarily picked one of them and ignored the other. You have just picked the clock 1 frame as the frame in which to perform the experiment. Jul 3, 2023 at 21:06
• @HadiKhan Acceleration is absolute, not relative, so both parties can agree on which one has accelerated and which one hasn’t. But motion is relative, so the two parties can’t agree on which one is moving and which is stationary. Jul 5, 2023 at 6:55

It can be done the way you describe. But it is more direct and conceptually clear in my opinion to suppose that there is a clock $$2'$$ in B that was previously synchronized with clock $$2$$ in A. It is possible because they are in the same frame of reference.

When the spaceship reaches B, clock $$1$$ will show $$0.75$$ years) and clock $$2'$$ will show $$1.25$$ years. That way the proper time of the ship is being compared with the coordinate time of the frame AB. The ship sees the time dilation directly without any calculation.

Note that it is not necessary for the ship to stop in B to that conclusion. But only if it stops, the comparison between clocks $$1$$ and $$2$$ can be done, because they will now be in the same frame.

The problem is when the same calculation is made from the spaceship's frame of reference, ie when the spaceship is at rest, B moves towards the ship (over the full 0.8 ly), and A moves away from the ship.

From that frame of reference, more time will elapse on the (nonmoving) spaceship clock than on the (moving) clock on A.

• How does that prohibit the procedure I described above? You only send the signal with the elapsed time value on clock 1 from B to A once the spaceship has arrived at B and stopped. We are only considering the inertial frame of reference where both A,B are stationary here. Jul 3, 2023 at 18:55
• From the frame of the spaceship, the spaceship is at rest and B is moving towards it at 0.8c from a distance of 1 ly. So in that frame, the spaceship will reach B (or rather, B will reach the spaceship) when the spaceship clock shows 1.25 years, not 0.75. Jul 3, 2023 at 19:02
• Hmm, this just confuses me even more now. I thought that since A and B are 1 Ly apart in their rest frame, when B starts moving towards the spaceship (and A starts moving away) the distance between A and B will be contracted down to 0.6 Ly and B is still approaching the spaceship at 0.8c, and so B will reach the spaceship in just 0.6/0.8=0.75 years, no? Jul 3, 2023 at 19:10
• It's easier to see the problem when considering two spaceships departing from a point equidistant to their destinations. The first one arrives, and announced clock 1 reads 0.75 years. The second ship uses your method to determine that this equals 1.25 years on clock 2. Then the second ship announced that it landed in after 0.75 years, which likewise equals 1.25 years on clock 1. How can 0.75 Y (C1) = 1.25 Y (C2) AND 0.75 Y (C2) = 1.25 (C1)? Jul 3, 2023 at 19:32