# Thread: Clock traveling perimeter of a regular polygon

1. ## Clock traveling perimeter of a regular polygon

Initial Setup
Let clock K be stationary at the center of a regular polygon. Let clock K' travel once around the perimeter of the regular polygon, at a constant speed. For simplicity we can use a square, but any regular polygon will do.

According to the reference frame of clock K
This is a simple matter of time dilation. Clock K' will always tick slower than clock K, so it is clear that when clock K' completes the polygon, it will have accumulated fewer ticks than clock K.

According to the reference frame of clock K'
This is not as simple. My calculations using the Lorentz transforms tell me that clock K will tick slower than clock K' along the length-contracted sides of the polygonal path, but that clock K will jump ahead in time whenever a corner (vertex) is encountered.

Summary
In both cases, after clock K' completes the polygon, it will have accumulated fewer ticks than clock K. So there is no contradiction at this stage.

Adding a beacon to clock K
Let clock K send out its current time on an expanding sphere of light. This beacon will require some calculable amount of time to reach clock K', but given the calculable distance between the clocks, it seems like it should be possible for clock K' to calculate the actual "now" time on clock K.

Question regarding the beacon received at any given vertex
In the instants before and after any given vertex is encountered, there is essentially only one value for the time carried on the beacon. Also, clock K' calculates the same distance between the clocks in the instants before and after any given vertex is encountered. So how exactly would clock K' use the beacon information to calculate two different "now" times on clock K, before and after any given vertex is encountered?

2. Originally Posted by Neddy Bate

Question regarding the beacon received at any given vertex
In the instants before and after any given vertex is encountered, there is essentially only one value for the time carried on the beacon.
From the perspective of the frame of reference attached to K, yes.
From the perspective of the frames attached to K', no. K' "jumps" frames at the corner since it changes direction.

Also, clock K' calculates the same distance between the clocks in the instants before and after any given vertex is encountered.
Clocks don't calculate distances, observers do. Anyway, the above is not relevant to the problem.

So how exactly would clock K' use the beacon information to calculate two different "now" times on clock K, before and after any given vertex is encountered?
Hint: the Doppler effect changes with every change in direction at each vertex. What you put together is , in essence, a form of the twin paradox where the traveling twin follows a polygon. The calculations are easy from the perspective of the "stay at home" twin. To see how to do the calculations from the perspective of the traveling twin, you need to read this.

3. Originally Posted by Tach
Hint: the Doppler effect changes with every change in direction at each vertex.

I knew clock K' would find the beacon from K to be redshifted just before encountering each vertex, and blueshifted just after. However, there is essentially only one time encoded in the beacon at each vertex, regardless of the Doppler shift. So, I was not sure how K' could calculate two different times for clock K.

After thinking about this further, I think I have it figured out. According to clock K', the beacon must have originated at a slightly earlier time, because light travels at a speed of $c$, not an infinite speed.

Well, at a slightly earlier time before encountering each vertex, clock K would have been closer to clock K', and at a slightly earlier time after encountering each vertex, clock K would have been farther from clock K', (if clock K' had always been moving in that direction). So, in calculating the "now" time, there are two different distances involved.

4. Originally Posted by Neddy Bate
You are welcome.

I think I have the answer now.

I knew clock K' would find the beacon from K to be redshifted just before encountering each vertex,
I don't think this is correct. If you consider all 4 vertices traversed by the traveling twin, the frequency will be redshifted entering vertices 1 and 2 and blueshifted when entering 3 and 0 (where the order of vertices is 0->1->2->3->0).

and blueshifted just after.
I don't think this is correct either. For example, the frequency is redshifted when both entering and exiting vertex 1. The frequency is blueshifted when both entering and exiting vertex 3.

Either way, if you look at all 4 legs of the trip, they are all different and the effects do not cancel because the setup is not symmetric (the "stay at home" twin is stuck in vertex 0). Also, none of the above helps you calculate the time elapsed on K as a function of the time elapsed on K', the stuff that you wanted to calculate in first place.

After thinking about this further, I think I have it figured out. According to clock K', the beacon must have originated at a slightly earlier time, because light travels at a speed of $c$, not an infinite speed.

Well, at a slightly earlier time before encountering each vertex, clock K would have been closer to clock K', and at a slightly earlier time after encountering each vertex, clock K would have been farther from clock K', (if clock K' had always been moving in that direction). So, in calculating the "now" time, there are two different distances involved.
I do not know if the above is correct or not. How would you use it to calculate the elapsed time for K? can you write down the math that goes along with the above statements?

5. Originally Posted by Tach
If you consider all 4 vertices traversed by the traveling twin, the frequency will be redshifted entering vertices 1 and 2 and blueshifted when entering 3 and 0 (where the order of vertices is 0->1->2->3->0).
It sounds like you have clock K located at vertex 0. That would explain why you have redshift on 1 and 2, and blueshift on 3 and 0.

However, I had defined clock K as being located at the center of the polygon, not at any of the vertices. That means clock K' will find redshift before entering each vertex, and blueshift after.

I have quite a few equations, but they are messy. I will try to simplify them and then post them here. Thanks again.

6. Originally Posted by Neddy Bate
It sounds like you have clock K located at vertex 0. That would explain why you have redshift on 1 and 2, and blueshift on 3 and 0.

However, I had defined clock K as being located at the center of the polygon, not at any of the vertices. That means clock K' will find redshift before entering each vertex, and blueshift after.
Yes, I see now that you have K in the center, this makes it difficult for K and K' to compare elapsed times at the end of K' trip. If that is the case, the situation is symmetrical. Yet, the fact that you have redshift entering a vertex and blueshift exiting it does not help in calculating the total elapsed time on K as measured by K'.

I have quite a few equations, but they are messy. I will try to simplify them and then post them here. Thanks again.

By placing K in the center you only need to do the calculation for K' traversing only one edge, from vertex 0 to vertex 1, for example. Then, you can multiply the answer by 4. For example, if you assume that K' accelerates with acceleration $+a$ for half the trip and $-a$ for the other half (and the scenario repeats for every edge) then:
$\Delta t=\frac{c}{a}sinh{\frac{a \Delta t'}{c}}$
$\Delta t > \Delta t'$

7. Originally Posted by Tach
Yes, I see now that you have K in the center, this makes it difficult for K and K' to compare elapsed times at the end of K' trip.
Yes, I noticed that. At first I thought I had made a mathematical error because the two clocks did not agree on the time displayed on clock K after one complete trip around the polygon. It turns out that they won't agree at that point, because it is a vertex, and they never agree at any of the vertices, (as long as K' keeps moving).

Of course if K' stops moving anywhere, even at a vertex, then they must both agree on the time displayed on clock K.

Originally Posted by Tach
If that is the case, the situation is symmetrical. Yet, the fact that you have redshift entering a vertex and blueshift exiting it does not help in calculating the total elapsed time on K as measured by K'.
I can already calculate the total elapsed time on K as measured by K' by using the Lorentz transforms for time. When I do it that way I notice that, according to clock K', the time displayed on clock K must jump ahead in time as every vertex is encountered. What I am trying to solve for now is how to calculate that jump using only the beacon of light sent from K to K'.

Originally Posted by Tach
By placing K in the center you only need to do the calculation for K' traversing only one edge, from vertex 0 to vertex 1, for example. Then, you can multiply the answer by 4. For example, if you assume that K' accelerates with acceleration $+a$ for half the trip and $-a$ for the other half (and the scenario repeats for every edge) then:
$\Delta t=\frac{c}{a}sinh{\frac{a \Delta t'}{c}}$
$\Delta t > \Delta t'$
That is an interesting way to do it, but I don't think $\Delta t > \Delta t'$ is always true, according to K'. Before K' rounds the first vertex, he would say $\Delta t < \Delta t'$. Then after he rounds the first vertex, he would say $\Delta t > \Delta t'$.

8. Originally Posted by Neddy Bate
I can already calculate the total elapsed time on K as measured by K' by using the Lorentz transforms for time. When I do it that way I notice that, according to clock K', the time displayed on clock K must jump ahead in time as every vertex is encountered.
I don't think so. Can you show your calculations?

That is an interesting way to do it, but I don't think $\Delta t > \Delta t'$ is always true, according to K'.
Actually, if you read the references I gave earlier in the thread you would have found out that :

$\Delta t > \Delta t'$ is always true, where K is associated with the inertial frame and K' with the accelerated one(s). (there are multiple accelerated frames)

Before K' rounds the first vertex, he would say $\Delta t < \Delta t'$. Then after he rounds the first vertex, he would say $\Delta t > \Delta t'$.
$\Delta t$ represents the total elapsed time, you seem to be thinking in terms of incremental times. Did you read the references I linked in earlier in this thread?

9. Originally Posted by Tach
I don't think so. Can you show your calculations?
Yes, I'm working on writing them up.

Originally Posted by Tach
Actually, if you read the references I gave earlier in the thread you would have found out that :

$\Delta t > \Delta t'$ is always true, where K is associated with the inertial frame and K' with the accelerated one(s). (there are multiple accelerated frames)
You seem to be saying that, in the Twin Paradox, one twin is always younger than the other. That is true at the end of the trip, but not during the outbound leg of the journey. During the outbound leg, the stay-at-home twin would say that the traveling twin is younger than himself, but the traveling twin would ALSO say his stay-at-home brother is younger than himself. That is the whole "paradox" part of the "Twin Paradox".

Originally Posted by Tach
$\Delta t$ represents the total elapsed time, you seem to be thinking in terms of incremental times. Did you read the references I linked in earlier in this thread?
Yes, I am thinking in terms of incremental times. I have clock K' moving at a constant speed, even when he encounters a vertex. I am interested in the times he finds on clock K before and after he encounters a particular vertex. I just realized that you had clock K' decelerating to be at rest with each vertex. If you try it with constant speed, you should see what I am talking about.

I am not interested in the total elapsed time, because I already know that clock K' must elapse fewer ticks than clock K.

10. Originally Posted by Neddy Bate
That is true at the end of the trip, but not during the outbound leg of the journey.
This is false, the twin that remains in the inertial reference frame ages more than the accelerated twin.

During the outbound leg, the stay-at-home twin would say that the traveling twin is younger than himself,
True.

but the traveling twin would ALSO say his stay-at-home brother is younger than himself.
False. You really need to read the references I provided you.

11. Tach, have you really never heard of the reciprocal nature of relative velocity time dilation?

http://en.wikipedia.org/wiki/Time_di..._time_dilation

"When two observers are in relative uniform motion and uninfluenced by any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock."

"In the situations of velocity time dilation, both observes saw the other as moving slower (a reciprocal effect)."

"Common sense would dictate that if time passage has slowed for a moving object, the moving object would observe the external world to be correspondingly "sped up". Counterintuitively, special relativity predicts the opposite."

12. Originally Posted by Neddy Bate
Tach, have you really never heard of the reciprocal nature of relative velocity time dilation?
Yes, I did. The twin paradox is not an exercise in time dilation.This is where you keep repeating your mistake.

13. Originally Posted by Tach
Yes, I did. The twin paradox is not an exercise in time dilation.This is where you keep repeating your mistake.
From the same article called "Time Dilation":

"A question arises: If Ship A and Ship B both think each other's time is moving slower, who will have aged more if they decided to meet up? With a more sophisticated understanding of relative velocity time dilation, this seeming twin paradox turns out not to be a paradox at all (the resolution of the paradox involves a jump in time, as a result of the accelerated observer turning around)."

Notice how they say there will be a "jump in time" at the turning point? The turning point is the first vertex of the square, and the "jump in time" is what causes clock K to jump from being behind clock K' to being ahead of clock K', according to K'.

14. Originally Posted by Neddy Bate
From the same article called "Time Dilation":

"A question arises: If Ship A and Ship B both think each other's time is moving slower, who will have aged more if they decided to meet up? With a more sophisticated understanding of relative velocity time dilation, this seeming twin paradox turns out not to be a paradox at all (the resolution of the paradox involves a jump in time, as a result of the accelerated observer turning around)."
Actually, contrary to your beliefs, the traveling twin is already younger that the "stay at home twin" before he hits the turnaround point. Nothing to do with any "jump in time".

Notice how they say there will be a "jump in time" at the turning point? The turning point is the first vertex of the square, and the "jump in time" is what causes clock K to jump from being behind clock K' to being ahead of clock K', according to K'.
There are a lot of bad articles in wiki. So, you don't want to read the references?

15. Originally Posted by Tach
So, you don't want to read the references?
I don't think I understand what you are claiming. Are you still denying that while clock K' is traveling, clock K' will find clock K to tick at a slower rate than clock K'? If so, I don't think you'll be able to find a reference. If not, then what are you claiming?

16. Originally Posted by Tach
The traveling twin is already younger that the "stay at home twin" before he hits the turnaround point.
That is only true according to the stay at home twin. According to the traveling twin, before the turnaround point, it is the stay at home twin which is the younger of the two. I thought everyone knew this. That is why it seems like a "paradox". Your crank comments are going to get this thread locked. Please desist.

17. Originally Posted by Neddy Bate
I don't think I understand what you are claiming. Are you still denying that while clock K' is traveling, clock K' will find clock K to tick at a slower rate than clock K'? If so, I don't think you'll be able to find a reference. If not, then what are you claiming?
That you do not understand what you are dealing with despite being provided with references. Here is another one, for the case of the traveling twin going in a circle while the "stay at home" twin remains on one point of the circumference. There is no "turning point" and there is no "jump in time", the traveling twin is younger merely by virtue of the fact that his path through space-time is longer read exercise 9.25 here). THIS is the reason he's younger.

18. Originally Posted by Neddy Bate
That is only true according to the stay at home twin. According to the traveling twin, before the turnaround point, it is the stay at home twin which is the younger of the two. I thought everyone knew this. That is why it seems like a "paradox". Your crank comments are going to get this thread locked. Please desist.
Let's try a simple exercise:

The distance between two points, A and B is L.
Twin 1 stays "home" in A while Twin 2 travels with (variable) speed $v(t)$ from A to B. The twins clocks are reset to 0 when Twin 2 starts on his trip from A. What does Twin 1 clock show when Twin 2 arrives in B? What does Twin 2 clock show when Twin 2 arrives in B? Formulas, please, no more mouth flapping.

19. Originally Posted by Tach
That you do not understand what you are dealing with despite being provided with references. Here is another one, for the case of the traveling twin going in a circle while the "stay at home" twin remains on one point of the circumference. There is no "turning point" and there is no "jump in time", the traveling twin is younger merely by virtue of the fact that his path through space-time is longer read exercise 9.25 here). THIS is the reason he's younger.
For a circular path, every point is the "turning point". That has nothing to do with the arrangement described in my opening post. My OP clearly states that K' is moving at constant speed along the perimeter of a regular polygon. Thus, along each each side of the polygon, there is nothing but constant relative velocity between K and K'. During such times, SR says there is reciprocal time dilation, contrary to your crank claims.

Then, each vertex represents a "turning point". That is when clock K' says that clock K jumps ahead in time. Please get on topic.

20. Originally Posted by Neddy Bate
For a circular path, every point is the "turning point". That has nothing to do with the arrangement described in my opening post. My OP clearly states that K' is moving at constant speed along the perimeter of a regular polygon. Thus, along each each side of the polygon, there is nothing but constant relative velocity between K and K'. During such times, SR says there is reciprocal time dilation, contrary to your crank claims.

Then, each vertex represents a "turning point". That is when clock K' says that clock K jumps ahead in time. Please get on topic.
It's not a crank claim. Tach's setting K frame at rest. The right choice considering the geometry of your example.

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