Does the Twin Paradox Break Symmetry in the Zig-Zag Scenario?

In summary: A.When B arrives at T, he emits a light signal containing his current clock reading and the current clock reading on T. Let's suppose, for definiteness, that B travels at 87% of the speed of light relative to...A.A sees B's signal arrive after a time interval which is 100 years less than the time it would have taken for the signal to have travelled to T from Earth. Because T and Earth are moving with respect to each other, A's clock will be slightly slower than B's clock, by a small amount (say 0.87 seconds per year).
  • #71
ash64449 said:
the first statement is false

Why? How does TT account for that portion of HT's worldline in his calculation?
 
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  • #72
PeterDonis said:
How does TT account for that portion of HT's worldline in his calculation

At the end of the trip, Elapsed proper time for HT as measured by TT is still 20 years and TT's elapsed proper time is 12 years. So why should we account for it?
 
  • #73
ash64449 said:
why should we account for it?

You should account for it because you are claiming to show how things look "from TT's viewpoint". If "TT's viewpoint" is correct, TT should be able to use his "viewpoint" to predict that HT's clock will read 20 years when they meet up again. But your analysis "from TT's viewpoint" does not show how TT makes that prediction.
 
  • #74
PeterDonis said:
But your analysis "from TT's viewpoint" does not show how TT makes that prediction.

Didn't i do that? the last part of the post #66 explains that.
 
  • #75
That is at the turn around, the elapsed proper time jumped from 3.6 years to 16.4 years for the clock at the earth(from TT's point of view) but for planet the clock still reads 10 years. Therefore simultaneity convention changes.
 
  • #76
i used the word 'left out'. That was wrong. Never was elapsed proper time of HT left out from my explanation of TT's point of view.
 
  • #77
ash64449 said:
at the turn around, the elapsed proper time jumped from 3.6 years to 16.4 years for the clock at the earth(from TT's point of view)

Ok, so what if TT were moving towards HT, but then turned around to move away again? (For example, suppose we are analyzing a modified scenario in which TT goes out, comes part way back, turns around and goes out again, and then comes all the way back.) That would make HT's "elapsed proper time now" go backwards. Do you consider that a viable "point of view" for TT to take?
 
  • #78
PeterDonis said:
Ok, so what if TT were moving towards HT, but then turned around to move away again? (For example, suppose we are analyzing a modified scenario in which TT goes out, comes part way back, turns around and goes out again, and then comes all the way back.) That would make HT's "elapsed proper time now" go backwards. Do you consider that a viable "point of view" for TT to take?
Especially given that what you visually see has no such anomaly, nor is there any such anomaly if you are continually exchanging messages with HT.
 
  • #79
PeterDonis said:
For example, suppose we are analyzing a modified scenario in which TT goes out, comes part way back, turns around and goes out again, and then comes all the way back

Do you mean the scenario in which TT leaves HT, in the middle of the journey,TT rotates,reach planet and TT reach HT again?
 
  • #80
ash64449 said:
Do you mean the scenario in which TT leaves HT, in the middle of the journey,TT rotates,reach planet and TT reach HT again?

I don't think so, but I'm not sure what you are describing here.

The scenario I am talking about, which I don't think had been discussed in this thread before I brought it up, would be described using the HT's rest frame as follows: TT leaves and travels outbound to a distant location (planet, space station, whatever); then TT turns around and heads back towards HT; halfway back, TT turns around again and heads back to the planet/space station/whatever; then TT turns around and heads back all the way to HT and they meet up again.

In this scenario, when TT turns around the second time (when he's gone halfway back to HT, and turns around to head outward again), by your definition of the TT's "point of view", the HT's elapsed proper time will go backwards.
 
  • #81
PeterDonis said:
The scenario I am talking about, which I don't think had been discussed in this thread before I brought it up, would be described using the HT's rest frame as follows: TT leaves and travels outbound to a distant location (planet, space station, whatever); then TT turns around and heads back towards HT; halfway back, TT turns around again and heads back to the planet/space station/whatever; then TT turns around and heads back all the way to HT and they meet up again.

In this scenario, when TT turns around the second time (when he's gone halfway back to HT, and turns around to head outward again), by your definition of the TT's "point of view", the HT's elapsed proper time will go backwards.

I mentioned the difficulties of zigzag paths for Fermi-Normal coordinates - that you have to restrict them to a world tube around the 'traveler'. For sudden zigzags, this word tube touches the traveler on one side or the other.
 
  • #82
ash64449 said:
No.it won't

EDIT: As i exactly calculated, it seems to me that it will.. How can this anomaly be addressed?
By admitting that the momentarily comoving inertial frame is not a reasonable model for a zig zag traveler's measurements because their past is different and any operational approach to setting up coordinates will be completely different.

To try to shorten my long essay some posts back, the basis of inertial coordinates is procedures that can be used to define them. If a zig zag traveler uses some reasonable operational approach to setting up coordinates, the problem does not arise. That is, nothing in their observations have this jump backwards in time, so if they use observationally based coordinates, there is no problem.
 
  • #83
A way to describe a criterion for reasonable observation based coordinates is that anything you see is considered simultaneous to some event in your past later than anything else you've already seen (and assigned). Basically, you are modeling light delay. Any model meeting this criterion will provide consistent coordinates for anything you can see.
 
  • #84
PeterDonis said:
Do you consider that a viable "point of view" for TT to take

No.. This is something that TT cannot accept. To address this issue, A complicated situation arises in which i have to consider turn back at the half-way is not instantaneous.

Even that is not viable. Then another question will come as to why this one is not instantaneous while turn around at the planet is instantaneous?

Then even if i consider the turn around is not instantaneous at the half-way, i see after the process, HT should read the same time as that of the time when TT reached half-way and didn't turn.

So everything gets complicated and hence not viable.

Can you give the answer to your scenario? What will be the reading of the clocks when TT and HT meet up again according to your scenario according to my analysis?
 
Last edited:
  • #85
PAllen said:
A way to describe a criterion for reasonable observation based coordinates is that anything you see is considered simultaneous to some event in your past later than anything else you've already seen (and assigned). Basically, you are modeling light delay. Any model meeting this criterion will provide consistent coordinates for anything you can see.

Can you show me what your telling by answering to the question as to what HT and TT will see their clocks reading when both meet again according to PeterDonis's scenario?

I think that is the best way to understand what you are telling.
 
  • #86
I think If i use HT's rest frame, TT's travel along the space-time is a zig-zag path and by using the relationship between HT's proper time and TT's proper time, i can see what their clocks will read at the end of the scenario.
 
  • #87
By calculation, I see HT should read 30 years while TT should read 18 years.
 
  • #88
Yeah. I think i found the solution. The solution is "go for it". Let us consider that TT's measure of HT's proper time went backwards.

When TT reaches mid-way and turns,

Clock at HT reads 11.8 while clock at the planet reads 18.2.

When TT reaches the planet, TT now predicts Clock at HT should read 13.6 years while planet should read 20 years.

Now when he makes another turn, Clock at the HT reads 26.4 years and clock at the planet reads 20 years.
When he reaches the earth, HT clock now reads 30 years. And his clock should read 18 years.This is all predicted by TT's point of view and i can see that Point of view method is not viable since at certain instants we need to think that proper time of HT should run backwards- which is not possible.

Anyway as i said earlier, Point of view is something that doesn't have real existence.

I think PeterDonis was trying to say this.
 
  • #89
ash64449 said:
Can you show me what your telling by answering to the question as to what HT and TT will see their clocks reading when both meet again according to PeterDonis's scenario?

I think that is the best way to understand what you are telling.
What they see when they meet is invariant - the same for any valid coordinates.

The most straightforward coordinates meeting the criteria I describe are called radar coordinates. TT simply uses Einstein simultaneity convention - send a signal out to HT, receive a signal back, assign the event of HT receipt to half way between TT send and TT receive. These are called rader coordinates.
 
  • #90
PAllen said:
What they see when they meet is invariant - the same for any valid coordinates.

I can see. I used a different method "point of view" but still answer came same.
 
  • #91
This raises me a question. Does TT actually 'see' proper time of HT go backwards as i addressed?
 
  • #92
ash64449 said:
This raises me a question. Does TT actually 'see' proper time of HT go backwards as i addressed?
I already answered this a couple of times. TT sees HT clock going forward at all times in a zig zag trajectory. That is why TT should find it hard to accept a model that has HT clock going backwards, contrary to what they see.

They see HT clock going slow until first turnaround, then fast until second, then slow again, then fast again. Never backwards. The rate seen is given by the Doppler factor, which is:

√((1+β)/(1-β))

where β = v/c, and v is taken positive if towards (e.g. HT is seen approaching), negative if away.

If you apply this over the different legs, you will find that it correctly shows the final resulting clock comparison.
 
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  • #93
PAllen said:
If you apply this over the different legs, you will find that it correctly shows the final resulting clock comparison.
Yes. For this example Doppler factor is 1/3 for the case when TT moves away from HT and 3 when TT moves towards HT. By using this I can explain Twin Paradox.

When TT reaches planet, His clock will read 6 years and TT will see 2 years as reading in the clock of the HT(with the help of a large telescope). And finally, when TT reaches the earth,6 years have passed, But now TT during these 6 years, TT will see HT's clock moving faster by the factor 3 and hence he will consider 18 years as passed in HT and hence total age of HT is 20 years.

Now for the Peter's Scenario:

When TT reaches planet,6 years passed and TT will see 2 years passed for HT, When TT reaches mid-way,additional 3 years passed and clock runs faster for HT as seen by TT and hence considers 9 years as passed for HT, and again when TT reaches planet,additional 3 year passes and HT clock beats slower as seen by TT and he considers only 1 year as being passed. And finally when TT reaches the earth,TT considers HT's clock to beat additional 18 years.

So total time elapsed for HT as seen by TT is 2+9+1+18=30 years.

Total elapsed proper time for TT is 6+3+3+6=18 years.

So i understand that what is actually seen by TT is accounted by Doppler factor and not frame switch method i was talking about. So my method doesn't actually describe point of view of observer. What is actually seen is correctly explained by Doppler factor-which means time slowing down(Time Dilation) cannot be seen by any observer.

But still Last Answer of my method was correct.
 
  • #94
PAllen said:
Especially given that what you visually see has no such anomaly, nor is there any such anomaly if you are continually exchanging messages with HT.

Yes. When you were talking like this you were talking about the Doppler Factor. I didn't know that.
 
  • #95
ash64449 said:
[..] What is actually seen is correctly explained by Doppler factor-which means time slowing down(Time Dilation) cannot be seen by any observer. [..] .
What is actually seen includes time dilation; I suppose that you used "relativistic Doppler" which includes time dilation, and it's necessary for symmetry of observations between inertial reference systems that are in relative motion. Classical Doppler doesn't include it.
- https://en.wikipedia.org/wiki/Doppler_effect
 
  • #96
harrylin said:
What is actually seen includes time dilation

Yes. It includes Time Dilation .But we don't see Time Dilation.

I can explain Doppler factor PAllen was talking about .I involve only calculation for a single outgoing scenario.

Consider HT send a signal when it's 1 year for him.

Then the signal will reach TT after 4 years which is 5 years of travel for TT in HT's rest frame.

Now using Time Dilation formula, we find only 3 years elapsed for TT.

So TT think 1 year has passed for HT when 3 years is passed for TT. This is what TT 'see'.

If i consider to work in TT's rest frame,only 1.8 years has been passed for HT and not 1 year when 3 years is passed.

Doppler factor includes Time Dilation but we don't see the Effect of Time Dilation. If so, Then TT should have thought 1.8 years passed for HT when 3 years has been passed for TT, But TT see 1 year pass for HT. So we don't see Time Dilation.
 
  • #97
ash64449 said:
What will be the reading of the clocks when TT and HT meet up again according to your scenario according to my analysis?

I don't know because it looks like your analysis is not self-consistent. But we don't need to do the analysis your way to get the answer. There are plenty of other ways to do it.

ash64449 said:
I think If i use HT's rest frame, TT's travel along the space-time is a zig-zag path and by using the relationship between HT's proper time and TT's proper time, i can see what their clocks will read at the end of the scenario.

ash64449 said:
By calculation, I see HT should read 30 years while TT should read 18 years.

Yes, this is one valid way to do it.
 
  • #98
ash64449 said:
Yes. It includes Time Dilation .But we don't see Time Dilation.

I can explain Doppler factor PAllen was talking about .I involve only calculation for a single outgoing scenario.

Consider HT send a signal when it's 1 year for him.

Then the signal will reach TT after 4 years which is 5 years of travel for TT in HT's rest frame.

Now using Time Dilation formula, we find only 3 years elapsed for TT.

So TT think 1 year has passed for HT when 3 years is passed for TT. This is what TT 'see'.

If i consider to work in TT's rest frame,only 1.8 years has been passed for HT and not 1 year when 3 years is passed.

Doppler factor includes Time Dilation but we don't see the Effect of Time Dilation. If so, Then TT should have thought 1.8 years passed for HT when 3 years has been passed for TT, But TT see 1 year pass for HT. So we don't see Time Dilation.
I'm afraid that I can't follow your reasoning, which looks unnecessarily complex to me. A feature of the so-called "twin paradox" is that in the end the clocks are compared side by side, and we can see that the times they display differ considerably. Can you think of a more striking way to "see" time dilation?
 
  • #99
harrylin said:
"see" time dilation

I think Time Dilation means this: When a observer moves relative to us, Time slows down for him relative to us.(not depend on which direction observer moves)

The question that needs to be addressed here is whether we see Time Dilation with our eyes.

The Answer is No. Because We see Relativistic Doppler Effect which include Time Dilation. We 'see' time beats at slower rate for an observer which is moving away from you and we 'see' time run at a faster rate when the observer move towards you. This effect is the one we 'see' which is not Time Dilation. What we see is Relativistic Doppler Effect which includes Time Dilation.

This is the one what i said at the previous post.
 
  • #100
ash64449 said:
I think Time Dilation means this: When a observer moves relative to us, Time slows down for him relative to us.(not depend on which direction observer moves)

The question that needs to be addressed here is whether we see Time Dilation with our eyes.

The Answer is No. Because We see Relativistic Doppler Effect which include Time Dilation. We 'see' time beats at slower rate for an observer which is moving away from you and we 'see' time run at a faster rate when the observer move towards you. This effect is the one we 'see' which is not Time Dilation. What we see is Relativistic Doppler Effect which includes Time Dilation.

This is the one what i said at the previous post.
"Seeing" always includes a certain amount of interpretation of what it is that you think you are seeing. :wink:
 
  • #101
harrylin said:
"Seeing" always includes a certain amount of interpretation of what it is that you think you are seeing. :wink:
What is that interpretation?
 
  • #102
ash64449 said:
What is that interpretation?
Ehm no. Just try to come up with an example of such a thing that you are looking for, and in which you do not interpret! (and compare for example post #7 of this thread).
 
  • #103
ash64449 said:
What we see is Relativistic Doppler Effect which includes Time Dilation.
You may wish to read about "transverse Doppler".
 
  • #104
I think some spacetime diagrams might help to illustrate the two scenarios specified in this thread. The first one comes from post #55:
ash64449 said:
Let the distance between the Earth and planet be 8 light years.(rest length between them)
Let speed of the traveling twin(TT) be 0.8c in both part's of the scenario.
...
In the frame HT uses, TT is moving and hence time runs slower for TT. So HT predicts TT would measure 20*0.6=12 years to reach earth. And hence concludes that in between the events 'TT leaves the earth' and 'TT reaches the earth', proper time elapsed for TT is 12 years and proper time elapsed for HT is 20 years and hence TT is 8 years younger than HT.

HT is shown in blue and elapses 20 years while TT is shown in red and elapses 12 years (the dots mark off 1-year increments of Proper Time for each twin):

TP&Simulataneity1.PNG

The issue of relativistic Doppler was brought up as a way for each twin to observe and keep track of the other ones aging. Here is a diagram showing how HT views TT's aging:

TP&Simulataneity2.PNG

Please note that HT cannot see the progress of TT in real time, he can only see the red signals coming to him from TT at the moment they hit him. But since he knows that he has been inertial the whole time, he can calculate the speed of TT and plot his trajectory as shown in the above diagram based just on what he sees and knows of Special Relativity. But this only works for him because TT starts out coincident with him so he can start integrating TT's speed to determine his distance from him. I did not show the planet in the diagrams but HT would not be able to tell how far away it was simply by looking at the relativistic Doppler coming from it. He could however determine the distance to the planet after he sees TT arrive there.

Similarly, TT can use relativistic Doppler to track the aging of HT and since he presumes that HT remained inertial while he did all the accelerating, he can construct a diagram of the scenario just like HT did:

TP&Simulataneity3.PNG

Another method that was discussed is radar coordinates. In this scheme, each twin continually sends out coded signals earmarked with the time they were sent and later when the echo is received, the received time is averaged with the sent time to establish the time of the measurement and the difference of the received time and the sent time divided by two establishes the distance (assuming units where c=1). Here is a diagram showing just a few of the radar signals sent (in blue) and received (in red) by HT:

TP&Simulataneity4.PNG

You should be able to confirm in the three radar examples that the coordinates that HT calculates match the coordinates in the diagram.

Now let's see what TT does:

TP&Simulataneity5.PNG

I've only shown two radar examples which are enough, plus the start and stop data for TT to construct a diagram but since he is not inertial, it will look different than the one that HT makes or the one he made using relativistic Doppler:

TP&SimulataneityNL6.PNG

This may look like a surprising result but what is significant about this is that it maintains all the same radar and Doppler signals that the previous diagrams enacted. In fact, after TT makes this diagram, he can do the same exercise over again from the point of view of HT and get the inertial reference frame that HT got.

One question that might come up is why would TT bother to use radar coordinates when the relativistic Doppler got him the inertial reference frame more directly. The most important answer is that this method works for distant objects such as the planet. Another answer is that it provides a method to show "the point of view" of a non-inertial observer. However, this particular POV (or any other) does not provide any more information to an observer than an inertial reference frame does but it does provide a mechanism for a non-inertial observer to first construct a non-inertial frame and then convert it to an inertial frame (provided that there is an inertial object that he can take data off of.

In my next post I will show Peter's zig-zag scenario.
 
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  • #105
Finally, we get to Peter's zig-zag scenario:
PeterDonis said:
The scenario I am talking about, which I don't think had been discussed in this thread before I brought it up, would be described using the HT's rest frame as follows: TT leaves and travels outbound to a distant location (planet, space station, whatever); then TT turns around and heads back towards HT; halfway back, TT turns around again and heads back to the planet/space station/whatever; then TT turns around and heads back all the way to HT and they meet up again.
TT can use radar coordinates just like before to construct this diagram but here I show some of the significant radar signals that TT employs:

TP&Simulataneity71.PNG

And from that data he will construct this non-inertial frame:

TP&SimulataneityNL72.PNG
 
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