Please consider the following thought experiment:
Case 1:
Twins are side by side to begin with. They speed off into opposite directions with equal accelerations (and velocities) in magnitude, approach and travel at c/2 for sometime (say 1 year), then slow down with symmetric velocity and acceleration vectors, stop and start coming back in the same manner, till they meet again and stop.
Case 2:
Same (accelerations & velocities) with case 1, but 10 years of (bothways) travel at c/2.

How will relativity handle this and what would each twin think about the other's age ?

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Since you have defined a central, stationary, inertial frame in which comparisons are made and about which both cases are symmetric they will both agree that they are the same age when they analyse what has happened in this frame (i.e. compare ages).

The twins, meanwhile, are in non-inertial frames. The precise details of what they each measure is more complex but if they regard themselves as at rest whenever they are not accelerating then they will both think that the other twin is ageing at a different rate. They will also both agree that it is the acceleration (which of course they will each work out to be different as they are at rest in their frame) is what will bring their ages back into synch in the final measurement frame.

3. Originally Posted by Tempeststrawberry
[...] and what would each twin think about the other's age ?
How can we determine the state of mind of people -- hypothetical people no less?

Do the twins know their Physics?

4. Originally Posted by Shaula
Since you have defined a central, stationary, inertial frame in which comparisons are made and about which both cases are symmetric they will both agree that they are the same age when they analyse what has happened in this frame (i.e. compare ages).

The twins, meanwhile, are in non-inertial frames. The precise details of what they each measure is more complex but if they regard themselves as at rest whenever they are not accelerating then they will both think that the other twin is ageing at a different rate. They will also both agree that it is the acceleration (which of course they will each work out to be different as they are at rest in their frame) is what will bring their ages back into synch in the final measurement frame.
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
These cases should not produce same result, mathematically.
What do you think?

5. Originally Posted by Tempeststrawberry
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Time dilation is only due to relative velocity. (In special relativity.)

Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
All that means is that the period where it is easy to see that A sees B's clock running slower and B sees A's clock running slower is different. The period of acceleration where they bring their clocks back to a common frame of reference will still ensure that their clocks are synchronised when they meet.

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Originally Posted by Tempeststrawberry
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
These cases should not produce same result, mathematically.
What do you think?
I think you should show the maths.

Also what I have written in no way suggests what you claim it does. You are ignoring the fact that the two twin frames are, when considered over the course of the round trip, non-inertial. Which means that relying on simple SR equations and only considering the steady speed component will give you an incomplete picture. You need to analyse the turnaround component of the trip - and when you do this while specifying the frames you are making the measurements in carefully you find there is no paradox.

I'd need to do the math to be sure but I am pretty sure your issue is that you are mixing frames of reference. I think that if you take measurements from the twins frame (which means that the time in uniform motion is different for each of them) then the measured accelerations are not equal. And if you measure in the central frame they later meet up in then the travel times are identical, as are the accelerations. So the case you consider doesn't actually exist as it is a hybrid frame with results from each mixed into it.

7. Originally Posted by Tempeststrawberry
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
These cases should not produce same result, mathematically.
What do you think?
Note that the standard explanations of this analyse the case where only one twin accelerates (because that is the only case where the "paradox" arises). If you apply the same logic to both twins, then you get the expected result.

8. Originally Posted by Shaula
I think you should show the maths.

Also what I have written in no way suggests what you claim it does. You are ignoring the fact that the two twin frames are, when considered over the course of the round trip, non-inertial. Which means that relying on simple SR equations and only considering the steady speed component will give you an incomplete picture. You need to analyse the turnaround component of the trip - and when you do this while specifying the frames you are making the measurements in carefully you find there is no paradox.

I'd need to do the math to be sure but I am pretty sure your issue is that you are mixing frames of reference. I think that if you take measurements from the twins frame (which means that the time in uniform motion is different for each of them) then the measured accelerations are not equal. And if you measure in the central frame they later meet up in then the travel times are identical, as are the accelerations. So the case you consider doesn't actually exist as it is a hybrid frame with results from each mixed into it.
Shaula,
Sure you can take your time and do the math but what I am asking is a simple question of logic.
These cases are identical, except for the duration of uniform speed at c/2.
All accelerations, brakings, turnarounds etc.. everything is exactly the same.
So in your equation for total time dilation, only the component due to relative velocity will be different, tenfold.
And I have a hard time understanding how this equation will yield exactly the same result, when you change a variable significantly, others staying the same.
Last edited by Tempeststrawberry; 2017-Mar-10 at 08:24 AM.

9. Originally Posted by Tempeststrawberry
Shaula,
Sure you can take your time and do the math but what I am asking is a simple question of logic.
These cases are identical, except for the duration of uniform speed at c/2.
All accelerations, brakings, turnarounds etc.. everything is exactly the same.
So in your equation for total time dilation, only the component due to relative velocity will be different, tenfold.
I'm not sure what you expect to be different in the two cases. Obviously the total elapsed time and distance travelled is different. The amount why which each sees the other's clock get "out of synch" with theirs differs.

But, in the end, when the clocks are brought back together, they will record the same time (because they took identical journeys through space-time).

10. Originally Posted by Strange
I'm not sure what you expect to be different in the two cases. Obviously the total elapsed time and distance travelled is different. The amount why which each sees the other's clock get "out of synch" with theirs differs.

But, in the end, when the clocks are brought back together, they will record the same time (because they took identical journeys through space-time).
Time dilation formula has 2 components, one relates to acceleration (also referred as gravitational td) and the other deals with relative motion.
This second component says that an observer's clock will tick at a different rate from the clock of a speeding traveller, as seen from the observer. And of course same applies to the traveller. As this is a continuous process, the longer you sustain relative velocities, the longer time difference will get. Interestingly enough, when twins are coming back toward each other this does not correct itself (as opposed to Doppler redshift).

So twins that are separated for 2 years (in case 1) should have a hugely different idea about each other's age, than the others in case 2 (which are separated for 20 years).

11. Originally Posted by Tempeststrawberry
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
These cases should not produce same result, mathematically.
But, they do. In the sense that they agree, because they're symmetrical in one of the frames. Such is the genius of Einstein.
What do you think?
Try the links at the bottom of this old (2008) post. (They've been "fixed", except I dunno why the last one was spelled "pardox")
Originally Posted by hhEb09'1
Originally Posted by surdrawrod
It seems to me that the twin paradox cannot be real unless there is a preferred frame of reference, that of the Earth which stays behind.
This is a hard question to answer.

There is no paradox of course, but I'm sure that's not what you meant by it cannot be real.

A "preferred frame of reference" means a frame of reference where the physics works, but it doesn't work in others. For intance, in the rotating earth frame, in Newton's mechanics, we must take into account so-called fictitious forces to account for coriolis effects, for instance. Newton's laws do not work in that rotating frame, without adding additional terms.

Not so, with relativity. It just gets more complicated

Twin Pardox, etc

Welcome to BAUT!

12. Originally Posted by grapes
But, they do. In the sense that they agree, because they're symmetrical in one of the frames. Such is the genius of Einstein.

Try the links at the bottom of this old (2008) post. (They've been "fixed", except I dunno why the last one was spelled "pardox")
Please consider that Case 1 and Case 2 will yield quite different results due to relativistic time dilation, and this has nothing to do with symmetry or reference frames or acceleration, because these are exactly the same in both cases. This is a paradox to me.

I'll check the links you have provided, thx..

13. Originally Posted by Tempeststrawberry
Please consider that Case 1 and Case 2 will yield quite different results due to relativistic time dilation, and this has nothing to do with symmetry or reference frames or acceleration, because these are exactly the same in both cases. This is a paradox to me.
I understand it is a paradox to you. But that's just relativity.

You think that the two cases should end up with different results, but they don't. If they did, there would be a (simple) problem with relativity, but there's not. When two observers are side by side, looking at a clock, it doesn't matter which reference frame is being used, they will both report the same value on the face of the clock. That's why we can use the "stay at home" reference frame to determine that both cases will experience symmetrical effects--and agree once they're re-united.
I'll check the links you have provided, thx..
yw
Last edited by grapes; 2017-Mar-10 at 09:38 AM. Reason: Two observers...

14. in laymans terms, (subject to minor variations through human error) the two twins are always in indentical frames of reference,so when they meet up it will all add up and there will have been no net relativistic effects. however during transit they would each observe identical things in the other as the other was observed to approach light speed and relativistic effects would be taking place they would each observe time to be passing slower for the other. the key thing is that in each case this will be identical to both observers at each point and they are always travelling at the Sam velocity and thus it all cancels out in the final case.

ETA:however if you put in a 3rd neutral observer who wasn't travelling then effects would be seen. he would see time slow down for both as they approached C and they would observe the same for him. but because of the difference in reference frames, it wouldn't cancel out and less time would have passed for them than for him despite the fact that in each case they would have both observed time dilation in the other. also of course the two twins would see the other approach C long before either would think so relative to the neutral observer. each is accelerating at 1G but will oberve a 2G acceleration relative to the other.

it's all about frame of reference rather than observation at the end of the day. when these differ there are net effects when they don't there are not. it's part of what I have been saying elsewhere, it's all about relative velocity and you can only state such relative to something else and observation is based upon this, but frame of reference isn't something that depends upon the observer it's a physical thing. I hope you can see the distinction here.
Last edited by malaidas; 2017-Mar-10 at 10:26 AM.

15. Originally Posted by malaidas
in laymans terms, (subject to minor variations through human error) the two twins are always in indentical frames of reference,so when they meet up it will all add up and there will have been no net relativistic effects.
There will be net relativistic effects

The effects will just be the same for both travelers. That's why the OP set them up symmetrically.

The OP is just trying to understand how the mathematics of relativity works--how the calculations in the moving/accelerating frames result in the same conclusion.

16. getting your head around the difference between observation and the observed is critical here. space-time doesn't depend upon the observer but yields some oddities to observation, however in some absolute sense your experience of it does depend upon your velocity. independent of any observation you can make. there is therefore apparant time dilation and actual time dilation or space contraction. the two are Separate things but so long as you are in the same far of reference as someone else everything will add up even if this causes contradiction to your observation.

17. Originally Posted by grapes
There will be net relativistic effects

The effects will just be the same for both travelers. That's why the OP set them up symmetrically.

The OP is just trying to understand how the mathematics of relativity works--how the calculations in the moving/accelerating frames result in the same conclusion.
picky picky, ok yes I meant no apparent net effects unless they then compared with a neutral party.

18. another way to separate this is to drop back to basics. let's take two cars approaching one is doing 70 the other 60. however each will observe the other doing 130. but of course neither of these are correct each is actually doing their measure speed relatuve to the earth that is rotating and travelling through space etc. your observation is always relative to something else, you are stationary.

however despite our everyday observations, space-time isn't so simple. your experience of such ultinately depends upon an absolute velocity even if you cannot observe it. the faster in this sense you go, the more it 'warps' for you. this effect however becomes apparent to observation relatively. the observer is part of the measurement, he is not part of what I actually going on. The net effect of this is that your twins are always experiencing the same things because they are always travelling in the same frame of reference. but they will both observe something different to this. Each will observe time slowing down for the other, but in reality their clocks are going to be identical and equally different to the neutral observer, in a global sense.
Last edited by malaidas; 2017-Mar-10 at 12:01 PM.

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Originally Posted by Tempeststrawberry
What you have written suggests that "time dilation due to relative motion" is irrelevant and possibly does not exist.
Because if you consider case 1 & case 2 side-by-side, the accelerations are identical but time spent in uniform motion is quite different (10 times).
These cases should not produce same result, mathematically.
What do you think?
What prevents you from doing the actual math?

Seriously. You said that there were "solid" reasons to support the so-called "twin paradox". All I've seen from you is a commitment to not actually work through the scenario. Whenever someone works through the scenario, the "paradox" disappears.

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Originally Posted by Tempeststrawberry
Shaula,
Sure you can take your time and do the math but what I am asking is a simple question of logic.
Again and again in physics, I find people critiquing the reasoning on the basis of "logic" without being willing to engage in the logic.

What is the reasoning here? The math. Where is the math in the OP? Nowhere. So the "logic" of the scenario is missing.
These cases are identical, except for the duration of uniform speed at c/2.
All accelerations, brakings, turnarounds etc.. everything is exactly the same.
So in your equation for total time dilation, only the component due to relative velocity will be different, tenfold.
And I have a hard time understanding how this equation will yield exactly the same result, when you change a variable significantly, others staying the same.
Nobody says that the twins will have the same age as if they had not moved, which is the only way I can imagine you think that there is some discrepancy here. Because you have done no math, your presentation and analysis is too vague to be useful.

21. yes precisely, their age will be different relative to a neutral observer than if they hadn't made the journey, but their relative ages will still be identical.

22. Originally Posted by Kwalish Kid
Again and again in physics, I find people critiquing the reasoning on the basis of "logic" without being willing to engage in the logic.

What is the reasoning here? The math. Where is the math in the OP? Nowhere. So the "logic" of the scenario is missing.

Nobody says that the twins will have the same age as if they had not moved, which is the only way I can imagine you think that there is some discrepancy here. Because you have done no math, your presentation and analysis is too vague to be useful.
I am not talking about their absolute ages. Pls read what I said carefully.
Relative to me, you are zooming off and coming back 20 years later, having travelled at relativistic speeds.
Thus in my viewpoint you are younger than me, because of relativistic time dilation.
From your viewpoint same aplies, because I have travelled in the same manner and aged less, relative to you..
The formula says so, not me.
Again none of us is stationary, we move symmetrically and have similar acceleration experiences.
Our absolute ages are irrelevant.
I don't always have to plug in numbers to see where the equation is really going.
If you need, just do it.
Unless the Universe has a directional preference, there is a problem.
One of us is wrong.

You don't need math to know that an apple is bigger than a cherry.

23. Originally Posted by Tempeststrawberry
Please consider the following thought experiment:
Case 1:
Twins are side by side to begin with. They speed off into opposite directions with equal accelerations (and velocities) in magnitude, approach and travel at c/2 for sometime (say 1 year), then slow down with symmetric velocity and acceleration vectors, stop and start coming back in the same manner, till they meet again and stop.
Case 2:
Same (accelerations & velocities) with case 1, but 10 years of (bothways) travel at c/2.

How will relativity handle this and what would each twin think about the other's age ?
So, this is a slightly different case than the standard twin paradox. Let's imagine that our two travellers are Alice and Bob, and we'll introduce a third observer (Charlie), who remains inertial throughout the experiment. From Charlie's perspective, he always remains midway between Alice and Bob. Also, it's much easier to work out the math if, instead of having them accelerate constantly on the journey, they accelerate briefly to some substantial fraction of the speed of light at the beginning and at the turnaround points, but then coast the rest of the way. The net result comes out the same, but if we end up having to work out the details, it will be easier with constant speeds for the journey.

Now, if we were to just consider Alice and Charlie for a moment, these two constitute what is essentially the classic twin paradox, with one twin leaving and returning, and the other remaining inertial. From relativity, both Alice and Charlie will agree that when they reunite, Alice will have experienced less time than Charlie by a certain amount. The exact amount of time elapsed will be different in your cases 1 and 2, but both Alice and Charlie will be unsurprised, at least if they both understand relativity.

Now if we look instead at just Bob and Charlie, the situation will be essentially the same, since direction doesn't matter. So Bob will have aged less than Charlie during the trip, and again, both will agree on just how much less if they understand relativity. And since the journeys were of the same duration and velocity, the difference in time will be the same. Since Alice and Bob will both have experienced less time pass than Charlie, and everyone will agree how much less time has passed for them, it shouldn't be surprising that Alice and Bob will thus find themselves having aged the same amount as each other during the trip. Which is also what Charlie expects, since from his perspective, they should have experienced equal time dilation. You're correct that in case 1 and case 2, the various times will be different, but in each case, Alice, Bob, and Charlie will all agree on the expected results.

Am I right about where your confusion with the situation lies, and does my explanation help clear it up at all?

24. Thanks Grey.
I am thinking about what you have written.
Hope I will put it in a new perspective this time.

25. Hey, here we go. Someone has been kind enough to draw out the spacetime diagram for this problem (they used Alice and Beth as the travellers) in a sample problem. Does the explanation there help? Note that, just as with the classic twin paradox, it's the sudden change in simultaneity at the turnaround point that resolves the apparent discrepancy.

26. And here's another look at the problem. Now, for this site, I question some of the author's reasoning, so I'm not sure I would take all of the text too seriously (he seems to ascribe some special status to the observer in the center being at rest with respect to the cosmic microwave background), especially the second half. However, he does a pretty nice job of drawing a second version of the spacetime diagram that applies, but this time from the frame of reference of one of the travellers, along with the light signals that would be going back and forth. It might help in thinking about the problem.

27. Originally Posted by Grey
Hey, here we go. Someone has been kind enough to draw out the spacetime diagram for this problem (they used Alice and Beth as the travellers) in a sample problem. Does the explanation there help? Note that, just as with the classic twin paradox, it's the sudden change in simultaneity at the turnaround point that resolves the apparent discrepancy.
"Both Alice and Beth observe the time in the other spaceship to be flowing slower in the outbound and inbound parts of their trips, but because of the "jump forward" effect when they turn around, they will be of the same age when they meet at C."

Yes, this does the job in a spacetime diagram.
Now I am trying to visualize how this 'jump forward' effect will materialize in physical terms, without violating continuity..
Thx..

28. cheers for that link Grey, got it saved

29. The key Tempeststrawberry as I said, is to separate your view of the observer and the observed. what Alice actually sees isn;t the same thing as what has happened in this case. So yes she will see time seem to rush forwards, it doesn;t mean it has, its just the way it appears as she returns to the same reference frame as Charlie, before setting off again. The symetry is in the observation

30. Originally Posted by malaidas
The key Tempeststrawberry as I said, is to separate your view of the observer and the observed. what Alice actually sees isn;t the same thing as what has happened in this case. So yes she will see time seem to rush forwards, it doesn;t mean it has, its just the way it appears as she returns to the same reference frame as Charlie, before setting off again. The symetry is in the observation
Yes this was the problem with my approach.
Couldn't be summarized better.

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