Ambiguity of time dilation/twin paradox

[1977ub]

Also, u can include a 3rd person C, who moves such that A and B are equidistant from him, So he'll measure the speeds and accelerations of A and B to be exactly the same wrt himself (but in opposite directions), throughout the whole trip.

I take this to mean that he is moving - not only initially, but throughout the entire exercise - to keep himself midway between A & B. Neglecting issues of simultaneity, this means that he is not inertial all the way through, and not having a single IRF. If we treated him to only *initially* i.e. on the outgoing leg to move midway between A & B, then I think your diagram describes that.

 

[ghwellsjr]

Oh so I am OP? is it an abbreviation? Yeah i know that the second diagram of george is not according to what i mentioned. I meant it this way...
*________________*________________*
A(on earth) C(in spaceship) B(in spaceship)
B and C are moving in --------> direction. A and B are moving away in opposite directions as observed by C. So C measures A's speed as say 'v' (same as B's speed in opposite direction)

OP can mean either Original Poster (the person who created the thread) or Original Post. That's you.

I think I understood correctly what you said you wanted for C. You said C measures A's speed as 'v' to be the same as B's speed in the opposite direction. This can only be true for the first part of B's trip. Isn't that exactly what I drew in the second diagram? The fact that it cannot continue to be true after B turns around so that A and B end up at the same age is a mistake on your part. That's one of the things I'm trying to show in the second diagram. But I also wanted to provide a second IRF diagram where you can do radar measurements to show that it doesn't matter what IRF diagram you use. You can even calculate B's measurements of A position as a function of B's time using the second IRF diagram. It doesn't matter which one you use. In fact, I encourage you to use both just to convince yourself that it doesn't matter.

 

[ghwellsjr]

I take this to mean that he is moving - not only initially, but throughout the entire exercise - to keep himself midway between A & B. Neglecting issues of simultaneity, this means that he is not inertial all the way through, and not having a single IRF. If we treated him to only *initially* i.e. on the outgoing leg to move midway between A & B, then I think your diagram describes that.

OK, so when does C change speed so as to remain midway between A & B?

 

[ShreyasR]

Ok... But when B turns Back, even C turns back, so to C, it must appear that A(earth) starts moving towards C's Spaceship. In this case, Even A's trajectory should change from C's RF. Shouldn't it? How is it not possible for C to measure A and B to be equidistant after the turnaround?

 

[1977ub]

People (including myself) often scratch their heads initially when they hear that A & B can each find the other's clock to be slowed. This seems to indicate that since "motion is relative" that anybody with a coordinate system is just as good as anybody else, in terms of deciding whose clock is running slower. The element of inertia - not present in exercises of coordinate system gymnastics by empirically measurable - is what determines who actually ages less. It can take a few tries to really get this. Since the "outgoing" leg part - or the incoming leg part - both indeed feature situations wherein neither A nor B has any privileged view of the rate time is passing, inertia has to be brought into explain why B is different from A overall.

 

[ghwellsjr]

People (including myself) often scratch their heads initially when they hear that A & B can each find the other's clock to be slowed. This seems to indicate that since "motion is relative" that anybody with a coordinate system is just as good as anybody else, in terms of deciding whose clock is running slower. The element of inertia - not present in exercises of coordinate system gymnastics by empirically measurable - is what determines who actually ages less. It can take a few tries to really get this. Since the "outgoing" leg part - or the incoming leg part - both indeed feature situations wherein neither A nor B has any privileged view of the rate time is passing, inertia has to be brought into explain why B is different from A overall.

What do you do in a scenario where no observer is inertial?

 

[ShreyasR]

One main thing that I want to ask after learning about how to plot the space-time curve... I have now understood how to Measure A's position as a function of B's time. But is it possible to draw a spacetime curve wrt B's non inertial frame of reference.? If yes, should i just tabulate the position of A wrt B's time and plot it directly? I am asking this since i am not sure. If I just try it and this puts me off track, its again difficult to come back...

 

[ShreyasR]

What do you do in a scenario where no observer is inertial?

In this case too, Earth is non inertial, but we are neglecting that aren't we?

 

[ghwellsjr]

Ok... But when B turns Back, even C turns back, so to C, it must appear that A(earth) starts moving towards C's Spaceship. In this case, Even A's trajectory should change from C's RF. Shouldn't it? How is it not possible for C to measure A and B to be equidistant after the turnaround?

Why don't we postpone the issue of how C can move so in his rest frame A and B remain equidistant from him? After you thoroughly understand how each observer can construct his own rest frame based on radar measurements, we'll be better fit to address this issue.

 

[ghwellsjr]

One main thing that I want to ask after learning about how to plot the space-time curve... I have now understood how to Measure A's position as a function of B's time. But is it possible to draw a spacetime curve wrt B's non inertial frame of reference.? If yes, should i just tabulate the position of A wrt B's time and plot it directly? I am asking this since i am not sure. If I just try it and this puts me off track, its again difficult to come back...

Yes, that is what you should do to be able to draws B's rest frame showing how A moves.

 

[ShreyasR]

OK, so when does C change speed so as to remain midway between A & B?

C doesn't change speed, C turns back(towards earth) when C observes that B turns back...

 

[ghwellsjr]

In this case too, Earth is non inertial, but we are neglecting that aren't we?

At the speeds we are considering, the Earth is negligibly non-inertial, so yes, we are not considering the Earth's motions. Pretend like it is stationary and the only body in the universe.

 

[ShreyasR]

Yes, that is what you should do to be able to draws B's rest frame showing how A moves.

Thanks George! I shall do it tomorrow. Thanks a lot!

 

[ShreyasR]

I mean the A,B,C problem is like me chasing a dog running away twice as fast as i am, starting from home, and after sometime the dog turns around and starts chasing me, and i turn back and start running home... So at any point of time i am equidistant from my home and the dog...

 

[ghwellsjr]

C doesn't change speed, C turns back(towards earth) when C observes that B turns back...

If you look at the second diagram, you will see that C won't observe B turning around until C's time is at about 3.7 years, just before he actually reaches B, so he won't be midway between A and B. Do you want to reconsider?

 

[ghwellsjr]

I mean the A,B,C problem is like me chasing a dog running away twice as fast as i am, starting from home, and after sometime the dog turns around and starts chasing me, and i turn back and start running home... So at any point of time i am equidistant from my home and the dog...

Neither you nor your dog are running anywhere near the speed of light so you must avoid jumping to conclusions when extrapolating to high-speed scenarios. You have to actually work out the details and you must do it correctly.

 

[ShreyasR]

If you look at the second diagram, you will see that C won't observe B turning around until C's time is at about 3.7 years, just before he actually reaches B, so he won't be midway between A and B. Do you want to reconsider?

Yes i am aware that C won't be midway between A and B, but when the time is about 3.7 years just before B actually reaches C, C observes B turning back. Cant C now turn back at that time such as to maintain equidistance from A and B (as measured by C) and hence to avoid the meeting between the two? (as shown in the diagram)

 

[ShreyasR]

Oh! so for this to happen, speed greater than that of light is required which is not possible? Is this the explanation?

 

[ghwellsjr]

Yes i am aware that C won't be midway between A and B, but when the time is about 3.7 years just before B actually reaches C, C observes B turning back. Cant C now turn back at that time such as to maintain equidistance from A and B (as measured by C) and hence to avoid the meeting between the two? (as shown in the diagram)

Yes, C can turn back then to avoid meeting B but he won't be equidistant from A and B which I thought was important to you. He's going to have to accelerate long before he sees B turn around in order to maintain his equidistant posture between A and B. In other words, the whole scenario has to be carefully choreographed and agreed to by all participants before they even start out.

 

[ghwellsjr]

Oh! so for this to happen, speed greater than that of light is required which is not possible? Is this the explanation?

No, we can't have any speeds greater than light, they just have to plan out the scenario before hand as I mentioned in the previous post.

 

[1977ub]

OK, so when does C change speed so as to remain midway between A & B?

When he decides that B has turned around, presumably.

[and]

He could catch wind of B's itinerary beforehand and then plan to do everything that B does, only traveling at half the velocity. I guess that makes sense from A's frame.

 

[ghwellsjr]

OK, so when does C change speed so as to remain midway between A & B?

When he decides that B has turned around, presumably.

Why do you think that will fulfill the requirement of remaining midway between A & B? You need to actually work out the details. And you should be very much aware that the definition of being midway between A & B is frame dependent, so you need to address that issue, too.

He could catch wind of B's itinerary beforehand and then plan to do everything that B does, only traveling at half the velocity. I guess that makes sense from A's frame.

If you want to define everything from A's frame, then we know that at half the velocity which would be 0.48c, C will neither observe A and B to be traveling away from him at the same speed in opposite directions nor will he, in his rest frame, determine that he is midway between A & B.

Please work out the details for whatever you have in mind before you post your ideas, otherwise, it's a complete waste of time.

 

[1977ub]

Aaanyhoww, I take the OP's intention of introducing C was to find the *coordinates* of A & B to be identical but in reversed directions, and that this should somehow imply parity of their clocks slowing from his *coordinate* point of view. To make the long story short, It takes more than geometric spatial coordinates to create parity of aging, since A remains inertial while B does not.

 

[ghwellsjr]

Aaanyhoww, I take the OP's intention of introducing C was to find the *coordinates* of A & B to be identical but in reversed directions, and that this should somehow imply parity of their clocks slowing from his *coordinate* point of view. To make the long story short, It takes more than geometric spatial coordinates to create parity of aging, since A remains inertial while B does not.

Specifically, what more does it take?

 

[1977ub]

Specifically, what more does it take?

Inertia. It is often asked by beginners why the traveling twin cannot simply declare the rest twin to be going away from him and then coming back, and from B's perspective A's clock should be moving slowly the whole way. B changes inertial frames, and experiences acceleration, which cannot be seen simply from setting up dueling systems of space coordinates.

 

[ghwellsjr]

You just quoted the OP saying that C measures the speed of A and B to be the same but in opposite directions, so how can you now say that A remains at rest?

In the original scenario, A remains inertial - and still does once we add C to the mix. Does A remain inertial in your diagram?

Yes, in all the diagrams that I have drawn so far. Isn't that obvious? Why did you have to ask?

OP states "So he'll measure the speeds and accelerations of A and B to be exactly the same wrt himself (but in opposite directions)" - this can be done if we use the original scenario of A inertial, B goes out, turns around, and comes back. If C is simply creating coordinate system of distance and time, neglecting inertia, he can indeed merrily declare the movements of A (who remains inertial) and B (who does not) to have identical coordinate movements and accelerations, though only B *experiences* acceleration throughout the trip.

I originally thought that B is the only one that experiences acceleration and that's how I made my second diagram in post #42, but now I am told by the OP and here by you that C is supposed to remain midway between A and B but I don't know how this is possible unless C also experiences acceleration. Can you please work out the details of what you mean by your last sentence?

 

[ghwellsjr]

Specifically, what more does it take?

Inertia. It is often asked by beginners why the traveling twin cannot simply declare the rest twin to be going away from him and then coming back, and from B's perspective A's clock should be moving slowly the whole way. B changes inertial frames, and experiences acceleration, which cannot be seen simply from setting up dueling systems of space coordinates.

Inertia? Do you mean changing the scenario? If not, I need more details.

 

[1977ub]

I originally thought that B is the only one that experiences acceleration and that's how I made my second diagram in post #42, but now I am told by the OP and here by you that C is supposed to remain midway between A and B but I don't know how this is possible unless C also experiences acceleration. Can you please work out the details of what you mean by your last sentence?

C does indeed experience acceleration. I don't think OP realized that makes C's perspective unsuited to make determinations of how much time has passed for A & B.

 

[Nugatory]

Inertia. It is often asked by beginners why the traveling twin cannot simply declare the rest twin to be going away from him and then coming back, and from B's perspective A's clock should be moving slowly the whole way. B changes inertial frames, and experiences acceleration, which cannot be seen simply from setting up dueling systems of space coordinates.

Be aware that there are variants of the twin paradox in which the traveling twin experiences no acceleration and both twins are in free fall throughout: use a tight hyperbolic orbit around a massive object to turn the traveler around. (This is not necessarily a GR thought experiment, as we can arrange the conditions so that the turnaround is an arbitrarily small portion of the journey and the "change of inertial frame" time-gap analysis using SR works just fine).

Thus, we really should not think of the acceleration as anything more than the easiest way of introducing the necessary asymmetry (something has to be different for the twins to age differently) into the thought experiment. The explanation of the differential aging is found in observable phenomena (the relativistic doppler analysis) and in the mathematical formalism (calculate the proper time along each worldline).

 

[1977ub]

Be aware that there are variants of the twin paradox in which the traveling twin experiences no acceleration and both twins are in free fall throughout: use a tight hyperbolic orbit around a massive object to turn the traveler around. (This is not necessarily a GR thought experiment, as we can arrange the conditions so that the turnaround is an arbitrarily small portion of the journey and the "change of inertial frame" time-gap analysis using SR works just fine).
.

Ok. I'll look for that. The twins are brought back together?

 

[1977ub]

I'm seeing the older threads about this. We can say that there is no "acceleration" involved but there is unflat spacetime involved, it seems to *require* gravity even though nobody experiences acceleration. For flat spacetime, acceleration is required.

 

[Nugatory]

Ok. I'll look for that. The twins are brought back together?

Yes, because traveler does a tight hyperbola around the massive object and ends up heading back in the direction from which he came. (If they aren't brought back together then there's no "paradox", just routine relativity-of-simultaneity stuff).

To avoid acceleration at the beginning and end of the journey, you assume that traveler was already moving at the beginning, and that the initial clock synchronization and comparison of ages happened as traveler zoomed past stay-at-home. When traveler returns he zooms past stay-at-home in the other direction and they do another clock comparison and count of grey hairs to see who has aged more.

One oddity: If both twins are isolated in sealed rooms with no external observations except the two comparisons with each other when they pass... The only way that either twin could know whether he was traveler or stay-at-home would be through the differential aging comparison. They meet, they meet again later, they agree about which worldline experienced more proper time.

 

[ghwellsjr]

Let's see--the OP said that C remains midway between A & B. The OP says B & C both reverse direction. If massive objects are involved, there must be two of them. How does B avoid reversal when he (twice) passes by the massive object that reverses C?

 

[ShreyasR]

C does indeed experience acceleration. I don't think OP realized that makes C's perspective unsuited to make determinations of how much time has passed for A & B.

Oh! Yes you are right. I did not realize that. Hmmm so the trip must be pre-planned such that from C's non-IRF, after the completion of the trip, the distance values measured by C should be same for A and B in opposite directions at every instant. I can't predict whether it is possible or not. I guess this will take a lot of mathematical calculations and head scratching. I will do it when I am free. Thanks everyone

 

[1977ub]

Oh! Yes you are right. I did not realize that. Hmmm so the trip must be pre-planned such that from C's non-IRF, after the completion of the trip, the distance values measured by C should be same for A and B in opposite directions at every instant. I can't predict whether it is possible or not. I guess this will take a lot of mathematical calculations and head scratching. I will do it when I am free. Thanks everyone

Ok... I thought maybe you don't need to now that you know that A and B will age differently when all come together, even admitted by C. A will age the most, then C, then B who will age the least.

 

[ghwellsjr]

One main thing that I want to ask after learning about how to plot the space-time curve... I have now understood how to Measure A's position as a function of B's time. But is it possible to draw a spacetime curve wrt B's non inertial frame of reference.? If yes, should i just tabulate the position of A wrt B's time and plot it directly? I am asking this since i am not sure. If I just try it and this puts me off track, its again difficult to come back...

Yes, that is what you should do to be able to draws B's rest frame showing how A moves.

Thanks George! I shall do it tomorrow. Thanks a lot!

Can we see the diagram that you said you were going to make yesterday?

 

[ShreyasR]

Sorry George. I couldn't make myself free for that. :( Anyway I shall definitely do it and post it here ASAP.

 

[1977ub]

Let's see--the OP said that C remains midway between A & B. The OP says B & C both reverse direction. If massive objects are involved, there must be two of them. How does B avoid reversal when he (twice) passes by the massive object that reverses C?

B was described as accelerating, then decelerating in each leg. Rockets.
C was described as carefully remaining midway between A & B. Rockets again.
I did not see gravity as involved in this exercise.

 

[ghwellsjr]

B was described as accelerating, then decelerating in each leg. Rockets.
C was described as carefully remaining midway between A & B. Rockets again.
I did not see gravity as involved in this exercise.

I did not see gravity as involved in this exercise either. But the only discussion from you after I asked you what you meant about "inertia" involved gravity (posts #79 through #82).

You still have not answered my questions:

Inertia? Do you mean changing the scenario? If not, I need more details.

 

[1977ub]

I did not see gravity as involved in this exercise either. But the only discussion from you after I asked you what you meant about "inertia" involved gravity (posts #79 through #82).

You still have not answered my questions:

There were some misunderstandings. In particular I believe you had misinterpreted purpose of adding observer C. If you would like to start from scratch - not referring specifically to any particular past exchange - I would be happy to reply. Otherwise I don't see the point.

 

[ghwellsjr]

There were some misunderstandings. In particular I believe you had misinterpreted purpose of adding observer C. If you would like to start from scratch - not referring specifically to any particular past exchange - I would be happy to reply. Otherwise I don't see the point.

You're right, I didn't realize that the OP knew that C would have to accelerate and so my second diagram does not represent his understanding. But I did realize that he wanted C to remain midway between A and B although I don't believe C can do that at a constant speed like the OP stated. But I have no idea how C must accelerate to maintain that goal, do you?

 

[ShreyasR]

You're right, I didn't realize that the OP knew that C would have to accelerate and so my second diagram does not represent his understanding. But I did realize that he wanted C to remain midway between A and B although I don't believe C can do that at a constant speed like the OP stated. But I have no idea how C must accelerate to maintain that goal, do you?

I did not state that C must maintain constant speed. If i have said so, i am sorry. C is free to accelerate such that the condition is met.

Just thinking of it in the probabilistic way, C is somewhere in between A and B. So there is a non-zero probability that C can remain exactly equidistant between A and B. But the problem is how must C accelerate in order to achieve this? I am not so good at calculations. And I am not free this week. I have internal assessment in college. I will try solving the problem mathematically and graphically ASAP and if i get stuck, i'll post it here.

 

[1977ub]

Initially B had a period of acceleration, which I wished to shrink to zero. likewise the deceleration and the accel/decl on the return leg. In A's frame, to stay midway between A & B, C can just halve the velocity of B. Anyhow I'm pretty sure the point was to ask about C's frame as if it was as adequate to measure time dilation as A's and we've discussed how it isn't. I don't think anything remains to be done.

 

[ghwellsjr]

You're right, I didn't realize that the OP knew that C would have to accelerate and so my second diagram does not represent his understanding. But I did realize that he wanted C to remain midway between A and B although I don't believe C can do that at a constant speed like the OP stated. But I have no idea how C must accelerate to maintain that goal, do you?

I did not state that C must maintain constant speed. If i have said so, i am sorry. C is free to accelerate such that the condition is met.

You did in post #61:

OK, so when does C change speed so as to remain midway between A & B?

C doesn't change speed, C turns back(towards earth) when C observes that B turns back...

I assumed you meant that just like B doesn't change speed (he just changes direction), that C also doesn't change speed, he just changes direction (both according to A's rest frame).

Just thinking of it in the probabilistic way, C is somewhere in between A and B. So there is a non-zero probability that C can remain exactly equidistant between A and B. But the problem is how must C accelerate in order to achieve this? I am not so good at calculations. And I am not free this week. I have internal assessment in college. I will try solving the problem mathematically and graphically ASAP and if i get stuck, i'll post it here.

You are correct, as long as you establish what you mean by "equidistant between A and B". As 1977ub pointed out in post #71:

He could catch wind of B's itinerary beforehand and then plan to do everything that B does, only traveling at half the velocity. I guess that makes sense from A's frame.

However, you originally stated that you not only wanted C to remain equidistant from A and B (trivial to define in any frame) but you wanted C to make the measurements:

Also, u can include a 3rd person C, who moves such that A and B are equidistant from him, So he'll measure the speeds and accelerations of A and B to be exactly the same wrt himself (but in opposite directions), throughout the whole trip. This should mean that The calculations of C should result in A and B being the same age after the trip isn't it?

I took this to mean that you also wanted C to measure that A and B were equidistant from him in his own rest frame. But since we all agree that his own frame is non-inertial we have the added problem that there is no standard way to establish a non-inertial frame. However, prior to my posting on this thread, the discussion centered around using radar methods to establish a frame and that's what I was focusing on when I drew the three diagrams in post #42.

With that in mind, I was using the second diagram to show how C's non-inertial frame must start out. In that frame, C will be inertial and "he'll measure the speeds and accelerations of A and B to be exactly the same wrt himself (but in opposite directions)" and he will measure the distances to A and B to be the same.

We also know that approaching the end of the scenario, an inverted situation applies where C is inertial and makes all the same measurements of A and B (but in opposite directions). However, somewhere in between, C has to accelerate in order to get from the first inertial position to the last inertial position and I believe he can do this in such a way that he will always measure A and B to be equidistant from himself but his measurements and observations of the speeds and accelerations and times of A and B cannot remain identical.

But I'm encouraging you to put this problem on the back burner until you learn how an observer uses radar to establish a reference frame. You need to feel real comfortable with that relatively easy task before tackling the very difficult task of establishing C's trajectory to maintain equal distance between A and B.

 

[ghwellsjr]

Initially B had a period of acceleration, which I wished to shrink to zero. likewise the deceleration and the accel/decl on the return leg.

Yes, we have all agreed to make B accelerate instantly at the start, midpoint and end of the scenario. To do otherwise would only add complication and not add understanding.

In A's frame, to stay midway between A & B, C can just halve the velocity of B.

Yes, since distance is frame dependent, we can use A's frame to establish a trajectory for C in which he remains midway between A & B. However, this is not the only thing that the OP wanted. He also wanted C to measure the speeds of A and B to be identical but in opposite directions. This won't ever be true for this trajectory defined by A's frame.

Anyhow I'm pretty sure the point was to ask about C's frame as if it was as adequate to measure time dilation as A's and we've discussed how it isn't.

True, because the standard definition of Time Dilation requires an Inertial Reference Frame (IRF) and A is the only observer who remains inertial in this scenario, but that doesn't mean that C's initial IRF (the second diagram in post #42) is not just as adequate to show Time Dilation, just like B's initial (and final) IRF is. Furthermore, we can use C's non-inertial reference frame to show how he can observe the Proper Time on the clocks of A and B during his travel, just like we can use B's non-inertial reference frame to show how he observes the Proper Time on the clocks of A and C, and how we can use A's inertial reference frame to show how he observes the Proper Time on the clocks of B and C.

I don't think anything remains to be done.

Well we could show a trajectory for C (in A's IRF) such that C measures himself to be equidistant from A & B at the beginning and end of the scenario and show how he's not equidistant in between. We could also show C's observations of A's and B's clocks and how they agree with the final outcome. We could show other trajectories for C, maybe even find one in which he measures himself to remain equidistant from A & B during the entire scenario and then see how he observes the others clocks and still comes to the correct conclusion at the end. There're lots of fun things left to do. I hate to leave anyone with the notion that only one frame (A's) is adequate to analyze or demonstrate what is happening in a scenario.