Acceleration is not important in the twin paradox

[xyzt]

Let's not. Let's put an end to this nonsense. The scenario from the exercise is perfectly equivalent with a scenario where B turns around instantaneously , with infinite acceleration. In other words, the "skipping" midway from observer B to observer C is equivalent to a turnaround with not any kind of acceleration but with infinite acceleration. So, the claim in the title of your thread is patently false.

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To show that transfer of information is not important, we can further modify the experiment by putting markers (flags, or planets) at the location where A and B meet, and where B and C meet.

[md65536]

Incorrect, there is an infinity of observers that will actually disagree that B and C show the same time. Only one observer agrees, that observer is the one that measures B and C to move with equal and opposite velocities wrt him.

Yes, I have not followed your argument. I agree that relativity of simultaneity is not an issue, but I don't understand the above argument. It seems to argue that relativity of simultaneity is an issue.

 

Observers B and C "pass through" the event of their passing. Everyone agrees on the proper time of that event measured by B, and everyone agrees on the proper time of that events as measured by C. They each have separate clocks, which don't need to be synchronized, so they can show a different time, but everyone agrees on what each of their clocks shows at that single event.

[xyzt]

Did you just miss the post before yours?

[md65536]

Let's not. Let's put an end to this nonsense. The scenario from the exercise is perfectly equivalent with a scenario where B turns around instantaneously , with infinite acceleration. In other words, the "skipping" midway from observer B to observer C is equivalent to a turnaround with not any kind of acceleration but with infinite acceleration. So, the claim in the title of your thread is patently false.

Okay, I'll agree with that interpretation.

 

However I still assert that the infinite acceleration we're talking about here is that of an abstract observer. We're making the switch between observers by choice, and no one experiences any physical effect of us choosing a different observer.

 

(Edit: And besides, if we're just adding up proper times, there is no need even for an abstract acceleration, the effect is identical.)

Edited May 5, 2013 by md65536

[xyzt]

Okay, I'll agree with that interpretation.

 

 

 

Then stop arguing, your OP is wrong.

Edited May 5, 2013 by xyzt

[md65536]

Then stop arguing, your OP is wrong.

I'll stop arguing too, OP is correct.

[xyzt]

Okay, I'll agree with that interpretation.

 

However I still assert that the infinite acceleration we're talking about here is that of an abstract observer. We're making the switch between observers by choice, and no one experiences any physical effect of us choosing a different observer.

 

Then, your scenario has a fatal flaw: C is perfectly symmetrical with A , so, when they meet, their clocks show ....exactly the same proper elapsed time.

Edited May 5, 2013 by xyzt

[md65536]

Then, your scenario has a fatal flaw: C is perfectly symmetrical with A , so, when they meet, their clocks show ....exactly the same proper elapsed time.

Each of A and C are measuring proper time since passing B.

If A and C are observed symmetrically (or any other way), A passes B earlier than C passes B, and measures a greater proper time.

 

The velocity of B relative to A is 0.866c after passing (gamma=2). The velocity of B relative to C is about 0.990c (gamma=7).

[xyzt]

Each of A and C are measuring proper time since passing B.

If A and C are observed symmetrically (or any other way), A passes B earlier than C passes B, and measures a greater proper time.

 

The velocity of B relative to A is 0.866c after passing (gamma=2). The velocity of B relative to C is about 0.990c (gamma=7).

No, it doesn't, A and C are perfectly symmetrical. You have rendered B totally irrelevant.

Edited May 5, 2013 by xyzt

[md65536]

No, it doesn't, A and C are perfectly symmetrical. You have rendered B totally irrelevant.

I don't understand that. I don't see the mistake I've made.

[xyzt]

I don't understand that. I don't see the mistake I've made.

Simple: proper time is frame invariant. Pick the frame in which A and C apprach each other at equal speeds. Both will measure the same exact proper time. By removing acceleration, you remove the asymmetry. By removing the assymetry, the proper times are equal. The end.

[Iggy]

No, it doesn't, A and C are perfectly symmetrical. You have rendered B totally irrelevant.

 

B recedes from A and approaches C during the experiment. In no frame are the events where A and C collocate with B simultaneous. Clearly those things are asymmetrical. The only issue, then, is whether the results of the experiment depend on those facts involving B.

 

If the results do rely on B then I don't fully understand Md's assertion that information transferred from B is unimportant, or that A, B, and C don't have to know about each other.

 

If the results don't rely on B, if B is totally irrelevant, then I don't fully understand your assertion that this is a well known scenario in which acceleration is hidden.

 

 

Simple: proper time is frame invariant.

 

Proper time is invariant for asymmetrical observers, accelerating observers, or any other kind. It makes me very curious about something you said earlier:

 

...you claimed that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." When I showed you that the claim is false...

 

That claim is not false. It is 100% true. Different observers can disagree about the rate at which a clock ticks, but everyone must agree about what the clock reads at a particular event. We can't very well have one person say that an astronaut lands at his destination (or passes a waypoint) at 20 years old and someone else say the astronaut was 40 when he landed or passed.

 

"Everyone will agree on what time the two passing clocks read at the moment they pass each other." couldn't be said more clearly or be more correct, yet you seem to know what you're talking about and disagree -- so, I'm curious what you meant.

 

 

 

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Yes, but the turn around doesn't cause the time dilation, it causes the frame switch, which doesn't actually require that a physical object switch frames.

 

The force of the argument is maybe between people who define acceleration such that switching frames applies, and people who don't.

 

I would say that it is the nature of spacetime which makes the inertial path between events more extremal than any curved or deviated path. It doesn't matter if one, or 100, clocks contribute toward measuring the paths, or indeed if nothing measures them at all. The length of the path in spacetime is the most direct cause, not necessarily the proper acceleration of the device measuring it.

 

You might like this article: Clock Postulate.

 

It would allow you to argue that acceleration is not important for the twin paradox even when there are only two observers and one of them accelerates. It isn't a universally supported postulate.

[hypervalent_iodine]

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Moderator Note

In following with the other two notes, xyzt, this is a reminder that you need to readjust your posting attitude. It is okay to criticize ideas, but it is not okay to insult the people making them. It is not productive for discussion and is a generally rude way to engage with people.

[xyzt]

It is very simple , really: by removing any acceleration from the scenario, there is no asymmetry between A and C. Since A and C describe the same path through spacetime (in opposite directions), there can be a frame where their velocities are equal and opposite. In that frame, the elapsed proper time \tau_A=\tau_C, meaning that in any frame \tau_A=\tau_C. This is what I explained above.

Edited May 5, 2013 by xyzt

[Delta1212]

The asymmetry is clearly introduced by B, which A and C will observe to be ticking at a different rate because it is moving at different velocities with respect to each clock. Because they agree on the number of times that B ticked between AB and BC, but observe it to be ticking at a different rate, they must observe the elapsed time between the events AB and BC as being different.

 

 

 

This whole experiment is measuring the time between events AB and AC. A measures the time between events fairly simply because it is at both of them. Because neither B nor C are at both events, but at some point collocate between those events, it is possible to use B's measurement of the time between the first event and that middle event with C's measurement of the time between the middle event and the final event to come up with a total time which, because it was measured partially in one frame and partially in another, will be shorter than the total elapsed time as measured by A, exactly as in the traditional twin experiment.

 

The reason that md insists that the clocks don't need to know about each other or directly transfer information, is because, since every observer will agree on the time shown by the passing clocks at each event, it is possible for anyone to calculate the duration as measured by B between AB and BC, and to calculate the duration as measured by C between BC and AC, and compare it to the duration as measured by A.

 

It's just as easy to say that C measured 100 hours while passing B and 200 hours while passing A, and to then add 100 hours of elapsed time to B's total, as it is for B to send its total to C when it passes and have C start counting from that point.

 

 

The force of the argument is maybe between people who define acceleration such that switching frames applies, and people who don't.

 

I think this is basically the source of the problem. The experiment requires frame switching of the measurement, but at no point does anything need to physically switch frames. I think md was trying to illustrate that the act of acceleration doesn't physically cause time dilation, while xyzt is trying to point out that you have to switch frames in order to get an asymmetrical reading, and everyone is getting lost in differences over what acceleration means.

[xyzt]

The asymmetry is clearly introduced by B, which A and C will observe to be ticking at a different rate because it is moving at different velocities with respect to each clock. Because they agree on the number of times that B ticked between AB and BC, but observe it to be ticking at a different rate, they must observe the elapsed time between the events AB and BC as being different.

No, in the absence of acceleration there is no asymmetry:

 

1. A measures \tau_A=Integral{\sqrt{1-(v/c)^2}dt}

2. C measures \tau_C=Integral{\sqrt{1-(v/c)^2}dt}

 

The interval of integration for above are the same, the speeds are the same so, \tau_A=\tau_C.

 

You are confusing total elapsed time with ticking rate.

Edited May 5, 2013 by xyzt

[Delta1212]

There are two measurements, one by A, and one by a combination of B and C.

 

Showing symmetry between A and C doesn't describe the complete experiment.

[xyzt]

There are two measurements, one by A, and one by a combination of B and C.

 

Showing symmetry between A and C doesn't describe the complete experiment.

Removing the acceleration from the problem has rendered B irrelevant.

[Delta1212]

How is B irrelevant if part of the measurement of elapsed time between events AB and AC is being made using the reading of clock B?

[xyzt]

How is B irrelevant if part of the measurement of elapsed time between events AB and AC is being made using the reading of clock B?

B isn't used. You need to pay attention to the scenario, at the point of meeting, clock C shows the same elapsed time as clock B.

Edited May 5, 2013 by xyzt

[Delta1212]

 

It isn't. You need to pay attention to the scenario.

Yes it is. It's rather integral to the whole scenario. What exactly is it that you think we're talking about measuring?

 

 

Edit: To respond to your edit: First, in the original scenario, C will show the same elapsed time as B at the point of meeting, but A will not be in agreement with B at this point, so even with A and C being symmetrical, the total elapsed time will not be in agreement when C reaches A.

 

Second, in addressing your objection that photons traveling from B to C count as acceleration, md created an alternative formulation whereby no information need be directly exchanged between clocks but measured by an outside observer with the same results. In this formulation of the experiment, B and C do not need to match at the point of passing.

Edited May 5, 2013 by Delta1212

[xyzt]

Yes it is. It's rather integral to the whole scenario. What exactly is it that you think we're talking about measuring?

 

 

Edit: To respond to your edit: First, in the original scenario, C will show the same elapsed time as B at the point of meeting, but A will not be in agreement with B at this point, so even with A and C being symmetrical, the total elapsed time will not be in agreement when C reaches A.

 

Second, in addressing your objection that photons traveling from B to C count as acceleration, md created an alternative formulation whereby no information need be directly exchanged between clocks but measured by an outside observer with the same results. In this formulation of the experiment, B and C do not need to match at the point of passing.

 

 

We are well past this point. It is very simple , really, there is a frame F where A travels from x_A to x_C and C travels from x_C to x_A, at equal and opposite velocities. The result is that, when A and C meet half-way (remember, motion is relative, you cannot think A as being at "rest" in the frame F) their clocks show the same exact elapsed proper time.

The fact that C exchanged (identical) clock readings with B along its path is irrelevant. I do not know how much simpler to make this.

[Delta1212]

 

 

We are well past this point. It is very simple , really, there is a frame F where A travels from x_A to x_C and C travels from x_C to x_A, at equal and opposite velocities. The result is that, when A and C meet half-way (remember, motion is relative, you cannot think A as being at "rest" in the frame F) their clocks show the same exact elapsed proper time.

The fact that C exchanged (identical) clock readings with B along its path is irrelevant. I do not know how much simpler to make this.

Because the experiment isn't to compare the total elapsed time of A with the total elapsed time of C. It's comparing the total elapsed time between events AB and AC as measured by A on the one hand, and the combined elapsed time between AB and BC as measured by B, and between BC and AC as measured by C on the other.

 

If you think that isn't what is being measured, then you are misunderstanding the entire conversation.

Edited May 5, 2013 by Delta1212

[Iggy]

It is very simple , really: by removing any acceleration from the scenario, there is no asymmetry between A and C. Since A and C describe the same path through spacetime (in opposite directions), there can be a frame where their velocities are equal and opposite. In that frame, the elapsed proper time \tau_A=\tau_C, meaning that in any frame \tau_A=\tau_C. This is what I explained above.

 

What you keep repeating is indeed absurdly simple, and it also does nothing to address the objection that three people have now raised to you. You did say:

 

You have rendered B totally irrelevant.

 

Clearly A and C's clocks run the same rate from the perspective of the other, but B interacts with each differently. That is an asymmetry. Either it is a relevant asymmetry, or an irrelevant one, to the thought experiment proposed. Either you can describe the thought experiment with the correct results without reference to B or the events caused by B, or you can't. If you can't then "B is irrelevant" is irrelevant.

 

 

...you claimed that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." When I showed you that the claim is false...

 

Do you still maintain that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." is false?

 

Please don't ignore the question again, because the claim is not false. It is true, and if you were simply mistaken about that (as it appears) then that would be rather telling.

[xyzt]

Because the experiment isn't to compare the total elapsed time of A with the total elapsed time of C. It's comparing the total elapsed time between events AB and AC as measured by A on the one hand, and the combined elapsed time between AB and BC as measured by B, and between BC and AC as measured by C on the other.

 

If you think that isn't what is being measured, then you are misunderstanding the entire conversation.

That is false, you can only compare co-located clocks, i.e. A and C. When A and C are next to each other you can compare them. and they will show the same exact time since the scenario has been dumbed down by taking away the acceleration.

 

Clearly A and C's clocks run the same rate from the perspective of the other, but B interacts with each differently. That is an asymmetry. Either it is a relevant asymmetry, or an irrelevant one, to the thought experiment proposed. Either you can describe the thought experiment with the correct results without reference to B or the events caused by B, or you can't. If you can't then "B is irrelevant" is irrelevant.

B simply transfers to C its reading. At the time they meet, B and C show the same reading. This makes B totally irrelevant to the exercise, since C simply accumulates proper time as if B didn't exist in the exercise. So, as shown above, when A and C meet, \tau_A=\tau_C.

Edited May 5, 2013 by xyzt