# Twin paradox - a further variant

I am still troubled by the "twin" paradox (thanks to those who have answered my previous question and suggested some reading which I will follow up). I have however thought of a variation on the experiment though - what if there are five clocks?

• Clock 1 is of an observer.

• Clocks 2 and 3 are travelling away from clock 1 at let's say 0.5 c in opposite directions.

• After one year (say from clock 1's perspective) they each encounter a clock (clock 4 and clock 5) travelling towards clock 1 at 0.5 c (from clock one's perspective) and clock 4 and clock 5 synchronise their times to clock 2 and 3 respectively.

One year on the clocks 4 and 5 are passing clock 1, and each other. What do each of the clocks show as the time ? In a sense clock 1 is not needed for this thought experiment but is still interesting to have it in the mix. But the basic issue is, will clocks 4 and 5 "see", each of them, the other clock as having run more slowly?

• Do you understand about the relativity of simultaneity? It's really the key to all these kinds of questions. In the frame of clock 5, the event of clocks 2 and 4 passing one another and synchronizing happens much earlier than the event of clocks 3 and 5 passing and synchronizing, whereas in the frame of clock 4, the 3/5 synchronization happens much earlier than the 2/4 synchronization. Thus each one can believe the other is running slow, but still make sense of the fact that 4 and 5 read the same time when they meet. – Hypnosifl Jan 29 '15 at 22:59
• @Hypnosifl that should probably be an answer – David Z Jan 29 '15 at 23:21

## 3 Answers

If the incoming clocks pass the outgoing clocks simultaneously according to clock 1, then this is an easy problem.

According to clock 1, the event that clock 2 and clock 4 pass each other occurs when $$t_2 = \frac{1}{\gamma_{0.5}}\,\mathrm{yr} = 0.866\, \mathrm{yr}$$

Similarly for $t_3$.

Thus, at this event,

$$t_4 = 0.866\, \mathrm{yr}$$

Similarly for $t_5$.

One year later, the incoming clocks pass clock 1. At this event,

$$t_4 = 1.73 \, \mathrm{yr}$$

Similarly for $t_5$.

But the basic issue is, will clocks 4 and 5 will each "see" the other clock as having run more slowly?

Yes. However, their times will agree at the event of passing clock 1. This is not a paradox. According to either incoming clock, the meetings of the incoming and outgoing clocks on either side were not simultaneous.

Yes, they will see each other as both having the same time, which is less than the time of clock one.

In the coordinate system corresponding to clock 1, clocks 2 and 3 passed off times simultaneously.

In a coordinate system corresponding to a stationary clock 4, first clock 5 received some time from clock 3. In this frame clocks 3 and 4 are not dilated at all. Later, clock 4 received precisely the same time from clock 2, which had been heavily dilated. Clock 5 continues ticking at a dilated pace while clock 4 ticks at a normal pace, and by the time they've intersect everything is all caught up and they have the same times.

Draw a Minkowski diagram!

There is full symmetry between what happens with the clocks in the pair 2 and 4, and what happens with the clocks in the pair 3 and 5. Thus, for simplicity of thinking, we can do the calculus of the 4D events of the pair 3, 4, from the point of view of the system of the clock 4. For the other pair, and in the system of the clock 5 the calculus is identical. Thus, the readings of 4 and 5 when they pass by the Earth will be the same.

With respect to the clock on the Earth, the clocks 4 and 5 will show a shorter time because they copied the local time of 2 and 3 (the time in the system in which the clock is at rest), and then, until reaching the Earth, the clocks 4 and 5 measured also local time.