I A version of the twin paradox without accelerations

[George Plousos]

I will study your notes, Ibix. In the meantime I have prepared this thought experiment:

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

Fig 1: The distance that separates Bob from Alice in the yellow reference frame is the same as the distance that separates Alex from Helen in the pink reference frame. The clocks of the observers in each separate reference frame are synchronized with the Einstein method. Tom is considered immobile, while Bob and Alex move at the same speed in Tom's view. At the meeting point of the three boys, their three clocks are synchronized.

Fig 2: At some point along the way, Alex will meet Alice and they will be photographed with each other. At some point the same thing will happen between Bob and Helen. We will return to figure 2 later.

Fig 3: Tom meets Alice and Helen. Because of the symmetry, Alice and Helen's clocks remain synchronized in Tom's view. If you want, the two girls are photographed with each other. Immediately after this observation, Tom sees Alice and Eleni slowing down at the same speed and stopping. Obviously, in the short time between Tom's last observation and the immobilization of the girls, there is no dramatic change in their clocks, but this is true anyway because of the symmetry of the system with respect to Tom.

Now all observers are motionless with each other - they are in the same frame of reference. Alice is in sync with Bob and Helen, who is in sync with Alex. So Bob, Alice, Alex and Helen are all in sync with each other. But then all four observers must have been in sync at all stages of the journey between Figures 1 and 3. Consequently, in the events described above for Fig 2 these four observers would also be in sync. We can now forget the involvement of Helen and Tom in the experiment, to find that the same conclusions apply to the original question - we now refer to the facts of figure 2 without Helen and Tom.

The answer to the original question was that the photos between Alex and Alice would show that Alex's watch was going backwards relative to Alice's watch before Alex stopped, but based on the new arguments the same photos must show that their clocks have remained in sync.

I can not see any error in the above arguments, unless I am blind because of my prejudice, or I have not learned everything I should.

 

[Nugatory]

or I have not learned everything I should.

You haven’t.
Go back and read post #37 by @Ibix in this thread. Then follow the recipe he outlined there, writing down the coordinates assigned to each relevant event using one frame and then applying the Lorentz transformations to find the coordinates using the other frame.

 

[PeterDonis]

To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants.

It has already been pointed out that, if you don't have the basic scenario with just three people (Bob, Alice, and Alex) in it straight on its own, adding more people won't help, it will just confuse you.

I strongly suggest that you take a step back and really consider this advice.

 

[Ibix]

In the meantime I have prepared this thought experiment:

You are playing the "yes, but..." game I mentioned in #37. As I said in that post, it confuses you, frustrates us, and you don't learn anything. Focus on applying the maths instead of verbal reasoning about ever more complex scenarios.

The answer to the original question was that the photos between Alex and Alice would show that Alex's watch was going backwards relative to Alice's watch before Alex stopped, but based on the new arguments the same photos must show that their clocks have remained in sync.

There cannot be contradictions in this type of scenario because the Lorentz transforms are analogous to rotations in the same way interval is analogous to length. Claiming a contradiction here is like drawing a triangle on a piece of paper, turning it round, and claiming that the edges no longer connect - clearly you've done something wrong if you predict such a result.

SR may or may not be an accurate model of reality, but it is not self-contradictory.

 

[Janus]

I will study your notes, Ibix. In the meantime I have prepared this thought experiment:

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

View attachment 267159

Fig 1: The distance that separates Bob from Alice in the yellow reference frame is the same as the distance that separates Alex from Helen in the pink reference frame. The clocks of the observers in each separate reference frame are synchronized with the Einstein method.

According to which frame? From what you say later, I assume you mean according to the frame in which Tom is considered at rest.

Tom is considered immobile, while Bob and Alex move at the same speed in Tom's view. At the meeting point of the three boys, their three clocks are synchronized.

Fig 2: At some point along the way, Alex will meet Alice and they will be photographed with each other. At some point the same thing will happen between Bob and Helen. We will return to figure 2 later.

Fig 3: Tom meets Alice and Helen. Because of the symmetry, Alice and Helen's clocks remain synchronized in Tom's view. If you want, the two girls are photographed with each other. Immediately after this observation, Tom sees Alice and Eleni slowing down at the same speed and stopping. Obviously, in the short time between Tom's last observation and the immobilization of the girls, there is no dramatic change in their clocks, but this is true anyway because of the symmetry of the system with respect to Tom.

Now all observers are motionless with each other - they are in the same frame of reference. Alice is in sync with Bob and Helen, who is in sync with Alex. So Bob, Alice, Alex and Helen are all in sync with each other. But then all four observers must have been in sync at all stages of the journey between Figures 1 and 3.

Prior Alex, Bob, Helen and Alice coming to a rest with respect to Tom, Alice and Bob would not say that their clocks were in sync. Neither would Alex and Helen say that their clocks were in sync.
So while Tom might say that all four of these clocks started in sync, started their deceleration at the same time, decelerated at the same rate, and thus remain in snyc the whole time, Non of the other observers would say the same. Alice, for example, would say that Bob's clock reads ahead of hers at the start. He will begin his "deceleration" before she does, and thus after that his clock will run slow compared to hers.
Then when she finally reaches Tom, and begins her "deceleration", the effects of being in an accelerated frame causes here to say that Bob's clock starts running even slower. By the time she comes to a rest with respect to Bob, she will say that both their clocks do read the same. But that does not mean that She and Bob agree that their clocks read the same during the whole exercise.
If we look at figure 2, it shows Alex and Alice lining up at the same time that Bob and Helen do. We also assume that all four clocks read the same time as this happens. But all this is according to Tom's rest frame.
In Alice and Bob's frame, The distance between Alex and Helen is length contracted, and thus is shorter than the distance between Alice and Bob. Thus Bob has to line up with Helen before Alice reaches Alex. The one thing all frame will agree on is that The times on Bob's, Alice's, Helen's and Alex's clock all read the same value during these passings ( Alices clock reads the same as Alex's when they meet and so do Bob's and Helen's, and in addition if Alice's and Alex's clock read 2:00 when they pass, Bob's and Helen's Clocks read 2:00 when they pass. However, since Bob's and Helens's clock meet before Alice and Alex do, according to Alice and Bob, Bob's and Helen's clocks has to read 2:00 before Alice's and Alex's do.

Here's an animation which is similar to what you are trying to do here:
We have two rows of clocks, in motion with respect to each other. We have arranged things, so that from the rest frame of the lower row of clocks, the clocks in the upper row are spaced the same distance apart as those in the bottom row, run at the same rate as those in the bottom row and are snychronized with the bottom rows clocks, and themselves. Thus, as each clock passes the one below it, they read the same time.

However, if we switch to being at rest with respect to the upper row, we see this:

The clock's in the lower row run slower than the upper row clocks, the distances, are different, and none of the clocks are in sync in either row.
However, every time a upper and lower row clock pass each other, they still read the same time.
Thus the two frames remain in agreement about what happens as any two clocks pass each other, they just don't agree on how this comes about.

 

[kent davidge]

View attachment 267167

However, if we switch to being at rest with respect to the upper row, we see this:

View attachment 267166

why is the second clock in the first row brighter than all others? does this have something to do with the physics?

 

[Janus]

why is the second clock in the first row brighter than all others? does this have something to do with the physics?

That's a lighting artifact. This animation was done using ray-tracing software, So it simulates the appearance of a real object being lit by a real light. Basically, you are just seeing a specular highlight due to the relative postions of the light, object, and "camera" in the scene.

 

[Ibix]

why is the second clock in the first row brighter than all others?

@Janus is pretty good at 3d modelling, if his avatar didn't give that away.

 

[George Plousos]

Indeed, Janus's moving clocks show what is happening, and a few simplified calculations can actually show that the diagrams for Figures 1 - 3 work properly, however, the impression is created that the mathematical consequence is due to the inherent impossibility of synchronizing clocks in moving reference frames - ie the theory slips like an eel to the point where a clearer explanation is needed. On the other hand, Einstein's two axioms seem to make it necessary to modify the concepts of space and time. I have no objection to that, but perhaps nature hides surprises as is often the case from the point of view of quantum physics, and a (reliable?) experiment that combines the two theories will test the "twin paradox" in a few years, as long as it does not have the luck of similar experiments that have been abandoned in the past:

https://scitechdaily.com/physicists-put-einstein-to-the-test-with-a-quantum-mechanical-twin-paradox/

 

[mitochan]

I think I can show that the answer to the original question may be different. To achieve this I will use two additional people, Helen and Tom, but in the end I will remove these two people as if they never existed, so only Bob, Alice and Alex will remain on stage, as the original question wants. The image helps a lot.

Say Alice and Bob have synchronized clocks in their IFR and Alex and Helen have synchronized clodks in their IFR. In the simultaneous snapshots in the IFR of Tom, the reading of the clocks according to their owners are ;

Figure 1
Alice = $T(1-\frac{1}{\Gamma})$
Alex =0
Bob =0
Helen = $T(1-\frac{1}{\Gamma})$
Tom =0

Figure 2
Alice = T
Alex =$\frac{T}{\Gamma}$
Bob =$\frac{T}{\Gamma}$
Helen = T
Tom = $\frac{t}{2}$

Figure 3
Alice = $T(1+\frac{1}{\Gamma})$
Alex =2$\frac{T}{\Gamma}$
Bob =2$\frac{T}{\Gamma}$
Helen = $T(1+\frac{1}{\Gamma})$
Tom = t

where
T=\frac{L_0}{V}
t=\frac{L_0}{\gamma v}
V=\frac{2v}{1+\frac{v^2}{c^2}}
\Gamma=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}
\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
$L_0$ : proper length of the bars
$\pm v$ : speed of the bars

 

[PeroK]

I have no objection to that, but perhaps nature hides surprises as is often the case from the point of view of quantum physics, and a (reliable?) experiment that combines the two theories will test the "twin paradox" in a few years, as long as it does not have the luck of similar experiments that have been abandoned in the past:

That is just a complete and utter intellectual cop-out! Testing the twin paradox is not something that has any relevance to physics in 2020. The whole of modern physics is built on SR, including the models of spacetime and energy-momentum that are tested every day in particle collisions. If you don't understand SR then that's no problem - it's not the easiest thing to learn. But don't pull the "SR might be wrong card" out of your pocket - that doesn't achieve anything.

 

[Ibix]

But don't pull the "SR might be wrong card" out of your pocket - that doesn't wash!

Indeed - in particular, the experiment cited seems to be a test of GR and quantum physics. If I understood right they're building a single-atom interferometry experiment and allowing the two paths to have different flight times due to gravity. This certainly isn't an "intro to SR" twin paradox scenario - it's something much subtler. It can't make the twin paradox anything other than what it is (we've already done that experiment), but may reveal something interesting about the interaction between quantum particles and gravity.

 

[George Plousos]

Agree. At least I've learned a lot here and the range of my doubts about Relativity is now more limited.

 

[vanhees71]

That is just a complete and utter intellectual cop-out! Testing the twin paradox is not something that has any relevance to physics in 2020. The whole of modern physics is built on SR, including the models of spacetime and energy-momentum that are tested every day in particle collisions. If you don't understand SR then that's no problem - it's not the easiest thing to learn. But don't pull the "SR might be wrong card" out of your pocket - that doesn't achieve anything.

Of course you are right, the kinematic effects of SR (and also of GR by the way) including the twin paradox have been tested by experiment to a very high precision. That does not imply that new tests using other contexts have no relevance to physics. The quoted test of time dilation in the context of quantum theory is at least interesting. Of course, I don't expect any surprises here, but still it's interesting and no such test of well-established theory is irrelevant.

Of course, the arguments of the typical "Einstein cannot be right, because it's contradicting my common sense", are unlikely to bring forward anything substantial ;-(.

 

[Isaac0427]

Agree. At least I've learned a lot here and the range of my doubts about Relativity is now more limited.

If you need some resources to help you understand relativity better, I'm sure some people here can help you out with that. I can also let you know some pointers I found helpful when I first learned it (I did do a fairly intensive study of relativity, I just had to get a refresher in this thread as it had been a year and a half since I did a relativity problem). If explained the right way, the "paradoxes" start making sense, to the point where they stop seeming like paradoxes.On a different note:

Of course, the arguments of the typical "Einstein cannot be right, because it's contradicting my common sense", are unlikely to bring forward anything substantial ;-(.

I just laugh at the irony of that argument, as it was precisely Einstein's argument in the EPR paper.

 

[vanhees71]

Well, it's well known that Einstein was not very satisfied with this infamous EPR paper, but that's another story.

 

[Tomas Vencl]

I will study your notes, Ibix. In the meantime I have prepared this thought experiment:
I think I can show that the answer to the original question may be different. ……..

Here is Minkowski diagram of your new experiment. Generally it is nothing more then Ibixs picture, maybe with more explanations. It is without scale.

You can arrange the synchonization of Alice and Bobs (Alex and Helens )clocks before, and set it up so, that Bob,Tom and Alex has the zero time at the same event.
You can see, that from Alice, Bob and Tom point of view Helens clocks started earlier (and symmetricaly Alices).
When Bob and Helen meets, Bobs clocks shows 1.x and Helens 2.x , but from Bobs point of view Helens clocks started much earlier, so even if from Bobs view Helens clocks are slower, they shows more.
At the end of experiment, when Alice Tom and Helen meets, Toms clocks show 4, Alice and Helen 4.x, but again their clocks ticks slower (from Toms view), but they started earlier. And also From Toms view Bobs and Alexs clocks show less (3.x) .

 

[PeroK]

Actually, if we want to have more than two observers we should have Carol and Ted in addition to Bob and Alice:

https://www.imdb.com/title/tt0064100/