1. Jun 23, 2013

### Alasdair Beal

Dear Friends,
Discussion of the Einstein's Special Relativity clock paradox is often complicated by the role of accelerations and objects following curved or polygonal paths. In the following example considers the clock paradox in a situation which only involves linear motion.
Imagine a rod with a clock and observer at each end. Following the Einstein/PoincarĂ© method of clock synchronisation by light flashes, the observer at clock A sends a light flash to clock B, observes the reflected return flash and establishes that clocks A and B are synchronised. He also establishes that the length of the rod is L = t/c, where t is the time taken for a light flash to go from A to B.
Imagine that there is a second rod with clocks (and observers) C and D at its ends, which is moving at velocity v towards the first rod, travelling on a line parallel to rod A-B which passes immediately alongside A-B. The observer at clock C sends a light flash to D, observes the reflected return flash and establishes that (in his frame of reference) clocks C and D are synchronised. He establishes that the length of the rod is also L (i.e. the same as the length that observer A has measured between A and B).
When the rods are positioned with clock C alongside clock B, the observer at B compares their readings and finds that both clocks show t=0.
Questions:
(i) when the rod moves on and clock C is alongside A, what times does observer A see on clocks A and C?
(ii) when clock D is alongside B, what times does observer B see on clocks B and D?
Also
(iii) when clock C is alongside A, what times does observer C see on clocks A and C?
(iv) when clock D is alongside B, what times does observer D see on clocks B and D?

2. Jun 23, 2013

### Staff: Mentor

Well, what do you think? How would you figure that out?

3. Jun 24, 2013

### ghwellsjr

I will answer your questions for a specific example where L = 5 feet and v = 0.6c and where c = 1 foot per nsec. First, a spacetime diagram in which A and B are at rest:

I think you can see that this meets all your specifications.

Now a spacetime diagram in which C and D are at rest:

This second diagram was not drawn from scratch, rather it was created simply by using the Lorentz Transformation process to convert the coordinates of significant events in the first diagram into one moving at -0.6c with respect to the first one. Since the worldlines in this diagram are mirror images of the ones from the first diagram and since the LT was used to create this diagram, this indicates that the first one was correctly drawn in terms of the placement of C and D.

Now we're ready to answer your questions. If you count the number of dots representing clock ticks for each clock, you can see that the time on A and D at their conjunction with B or C is 8.333 nsec and the time on B and C at the same conjunction is 6.667 nsec. From this you should be able to determine formulas as a function of v and L.

By the way, this is not what is commonly referred to as the clock paradox. That requires the clocks that were set to the same time at the beginning to rejoin for a subsequent comparison.

And finally, just for the fun of it, here is another spacetime diagram in which all observers are traveling at the same speed:

This spacetime diagram was created by transforming the first diagram into one moving at -0.333c with respect to it.

Note that in all three of these spacetime diagrams, the same answers to your questions can be determined. Also note that we needed the first one (or something equivalent) in order to establish the clock synchronization for clocks A and B and we needed the second one to establish clock synchronization for clocks C and D.

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4. Jun 24, 2013

### Alasdair Beal

Thanks for this.

5. Jun 25, 2013

### Alasdair Beal

PS I see now that this is how the mathematics must work out. However I still struggle to picture the physical reality of a situation where, even though the clocks, rods and observers pass right alongside one another, the observers on the two rods cannot agree on anything - the observers at A and B conclude that rod CD is shorter than AB and clocks C and D are running slower than A and B but, after witnessing the same events, the observers at C and D conclude that rod CD is longer than AB and clocks C and D are running faster than A and B. I'll need to read and think more about all of this. Thanks for explaining the mathematical aspects anyway.

6. Jun 26, 2013

### ghwellsjr

I don't know why you say that. When B and C coincide, they immediately agree that both their clocks read zero and the other two observers will also agree some time later when they see that event. When C reaches A, they immediately agree that C's clock read 6.667 and A's clock read 8.333 and so will the other two observers eventually. Similarly, when B reaches D, they immediately agree that B's clock read 6.667 and D's clock read 8.333 and so will the other two observers eventually. Finally, when A and D coincide, they immediately agree that both their clocks read 15 and so will the other two observers when they see that event. Every observer agrees on the observations that all observers make, they just can't see them at the same time.

The observers don't actually see or observe the other rod to be shorter than their own nor do they observe Time Dilation, they establish these features using radar methods by making an assumption about the propagation of light and then by making some measurements and doing some calculations. Let me show you by going back to the third diagram on my previous post except that I have extended some of the worldlines:

In this drawing, we will focus on the red observer A and how he uses radar to establish his own rest frame. Although he would typically be sending out radar signals continually, I only show the two most important of them. The first one he sends out at his Proper Time of -5 nsecs. It travels upwards to the right along a 45 degree diagonal and bounces off both B and C at their Proper Time of 0 nsecs. The reflection arrives back at A at his proper time of 5 nsecs and so the radar signal was gone for 10 nsec. Now here's where he makes his assumption. He assumes that the radar signal took the same amount of time to get to B and C as it did to get back. From this, he concludes that B and C were 5 light-nsecs away at the midpoint of his measurement (half way between sending and receiving the radar signal) so that means that when his Proper Time was 0 nsecs, B and C were 5 light-nsecs or 5 feet away.

Now he watches C traveling towards him and observes that he gets to him at his Proper Time of 8.333 nsecs. From this he concludes that C is traveling at 5 feet per 8.333 nsecs or 0.6 feet per nsec.

He continues to watch the rod and at his Proper Time of 15 nsecs, the other end at D passes him. Since it took 15-8.333 = 6.667 nsecs for this to happen and since the rod is traveling at 0.6c, he calculates that its length is velocity multiplied by time or 0.6*6.667 = 4 light nsecs or 4 feet. Since he measured his own rod to be 5 feet, this is in accord with the Lorentz contraction of 0.8 at 0.6c.

Now what about Time Dilation? He bases this on the fact that his measurement established that C's clock read 0 nsecs at the same time his read 0 nsecs and then 8.333 nsecs later according to his clock, C's clock read 6.667. This is a ratio of 1.25, the Time Dilation factor at 0.6c.

Meanwhile, he has sent out another radar signal at his Proper Time of 1.667 nsecs. It hits B and D and then returns at his Proper Time of 11.667. Like before, the radar signal was gone for 10 nsecs so he once again declares that B and C were 5 feet away at his Proper Time of 6.667 nsecs. He can also see that when he receives the radar signal, B's clock read 6.667 nsecs and D's clock read 8.333 nsecs.

He can put all this information together and eventually create a diagram of his rest frame based on his measurements and assumptions about the propagation of light. It looks like this:

Note that I have included Proper Time labels only for the moving observers since A and B are not moving, their Proper Times are always the same as the Coordinate Times so all you have to do is look over at the scale on the left-hand side to see any Proper Time for A or B.

Now we can repeat the whole process to show how B can use radar measurements and here's another diagram to show that:

I'll leave it to you to verify the details.

Finally, we could repeat the whole process showing the radar measurements for C and D. I think you can see that all the numbers would be the same, just the labels would be different and it would be a kind of mirror image. Can you see how it is possible for each observer to establish their own frame of reference based on their own assumption that light propagates at c with respect to themself and that this leads to them both determining that the other ones moving rod is Length Contracted and the other ones moving clocks are Time Dilated?

You're welcome. I hope this further explanation helps.

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7. Jul 4, 2013

### Alasdair Beal

Dear George,
Thanks for taking the time to respond.
I understand how observers A and B may calculate from light or radar signals that when rod CD is approaching them at speed, CD appears shorter than AB, with an apparent length of 4 feet, rather than its physical length of 5 feet.
However if A sends out a light flash when C is right alongside him, it will be reflected back to him instantaneously. However the physical lengths of both rods are the same, then presumably if B emits a light signal he will find that it is instantaneously reflected back to him from D. Alternatively if A and B simply reached out when rod CD was alongside they would find that they touched C and D at the same time.
Any thoughts?
Regards,
Alasdair

8. Jul 4, 2013

### Staff: Mentor

It doesn't just "appear" shorter. In any meaningful sense, the moving rod as measured by observers A and B is shorter. The moving length is no less physical than the proper length.

You seem to think that at the moment that C aligns with A that both frames will observe that D also aligns with B. Not so. In fact, neither frame will observe the ends of the rods aligned at the same time. Remember, according to A-B, the rod C-D is shorter. So when (according to A) A aligns with C, the end of the moving rod D will have already passed point B.

9. Jul 4, 2013

### ghwellsjr

Yes, this is true in all frames.
You should say, if the Proper Lengths of both rods are the same..., then, again, this is true in all frames.
This is not true in all frames. "At the same time" is only true in certain frames. It is true in the last frame of post #3 but not in the first two frames. You cannot draw invariant conclusions regarding remote times being simultaneous.
You should think precisely about what you mean by "A and B simply reached out".