Ambiguity of time dilation/twin paradox

[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.