Ambiguity of time dilation/twin paradox

[1977ub]

Thank you for your patience. I was aware of the general idea but not all the ways that it works out. Since B's motion was composed of 2 discreet stretches of inertial motion, I couldn't think past treating each as an inertial case which needed to be treated with radar coordinates - i.e. ending up the same as Einstein coordinates. I understood radar to be interesting in the case of *periods* of acceleration. In this example, there is no *period* of acceleration, so I didn't get my head around the implications of using one "session" of radar coordinates for the whole trip. Also, It really hadn't occurred to me that two observers could end up on the same trajectory and yet have different radar coordinates with different information regarding, say, the current time on Earth.

 

[Dale]

It really hadn't occurred to me that two observers could end up on the same trajectory and yet have different radar coordinates with different information regarding, say, the current time on Earth.

Yes, that is a surprising result. It is essentially due to the fact that the point that is identified as "now" on a distant worldline in my radar coordinates depends not only on my motion now, but on my motion between when I sent the radar pulse and when I receive the echo.

So after B meets up with C, for a while, B continues to receive echoes from pulses sent out before the acceleration. These are the ones where B and C disagree. Once that first radar pulse after the turnaround reaches A, from then on B and C agree about the current time at A.

 

[ShreyasR]

Hey guys! Well, I'm from India. I am an engineering student who is studying relativity as a side hobby or whatever you want to call it. And from all the replies that followed my last reply, you people are discussing about the physics of A,B and C. Reading them slowly, I am confused whether the 3 people A,B and C the ones I mentioned? Or are they some other A,B, and C?
I mean I had mentioned the problem A on earth, B in a spaceship going for a linear trip (wrt earth) and coming back, and C traveling such that he measures A and B to be equidistant from him...?
Since DaleSpam mentioned something about B meeting up with C, which according to me will happen only after they return to earth... I am confused :-( :-(

 

[1977ub]

Hey guys! Well, I'm from India. I am an engineering student who is studying relativity as a side hobby or whatever you want to call it. And from all the replies that followed my last reply, you people are discussing about the physics of A,B and C. Reading them slowly, I am confused whether the 3 people A,B and C the ones I mentioned? Or are they some other A,B, and C?
I mean I had mentioned the problem A on earth, B in a spaceship going for a linear trip (wrt earth) and coming back, and C traveling such that he measures A and B to be equidistant from him...?
Since DaleSpam mentioned something about B meeting up with C, which according to me will happen only after they return to earth... I am confused :-( :-(

Different C. Sorry for any confusion.

 

[ShreyasR]

Different C. Sorry for any confusion.

Oh then its ok... Thanks.

 

[bobc2]

ShreyasR, If you follow in the traditions of the great mathematician Ramanujan and one of our physics heroes Natyendra Nath Bose (notwithstanding your engineering rather than mathematics and physics endeavors) you will be a marked man here. A special welcome to you coming all of the way from India via Bose photon paths (it is only fitting that you arrive on Bose photons).

 

[ghwellsjr]

ShreyasR, I'm going to draw some spacetime diagrams to illustrate a scenario similar to the one you proposed. In your scenario, you had A remaining stationary while B traveled away and back at 2.8x10^8 m/s which is about 0.933c. You stated that B would come back to find that A had aged by 10 years and B would age about 1 year. In fact, at that speed, B would age 3.59 years. So I want to change B's speed to 0.96c because that will make the drawing easier to make and understand since the Relativistic Doppler factor is exactly 7.

Here is the first diagram for the Inertial Reference Frame (IRF) in which A remains at rest:

Please note that B ages by exactly 1.4 years at the point of his turnaround and then ages another 1.4 years for a total age gain of 2.8 years compared to A's 10 years.

Now you also suggested a different IRF in which a 3rd person C remains at rest while A and B start off traveling away from him at the same speed in opposite directions. That speed would be 0.75c and here is a diagram showing that:

Now you will quickly see what was brought to your attention in earlier posts that C does not remain equidistant from A and B.

Finally, I want to show you how A can use radar to measure B's trajectory. I added in three radar signals in green, red and orange to the first diagram:

Remember, A cannot see B as we see him on the diagram. A needs to wait for the image of B at each different location to propagate to him. So when A sends out a radar signal, he has to wait the entire time before the return signal comes back to him.

His first radar signal shown in green is sent at 0.1 years into the mission and he receives the return signal at his time of 4.9 years. Note how the signal propagates along diagonal lines of exactly 45 degrees. The calculation that A does is simple. He first figures out how much time progressed between sending and receiving the radar signals. That's just the difference between the emit and return times or 4.8 years. He divides this time in half to get 2.4 years which tells him that A was 2.4 light-years away, but when?. For that, he takes the average of the two times, which is the midpoint of the interval, to get 2.5 years. This tells him that when his clock was at 2.5 years, B was 2.4 light-years away from him. You can confirm this on the diagram.

The red radar signal was sent at 0.2 years and received at 9.8 years so he concludes that B was located 9.6/2 or 4.8 light-years away at his time of 5 years.

The orange radar signal was sent at 5.1 years and received at 9.9 years so he concludes that B was located 4.8/2 or 2.4 light-years away at his time of 7.5 years.

Obviously, I hand-picked these three particular measurements because they were easy to make and illustrate on the diagram. Twin A would likely be sending out a new radar signal at a repetitive interval such as every 0.1 years but this would merely fill in more points along B's path.

You can also use the second spacetime diagram to calculate the same radar measurements. I would suggest that you copy the diagrams and paste them into Paint documents so you can add your own radar signal paths along 45-degree trajectories.

After you get proficient at determining the location of B as a function of A's time in either IRF diagram, you can use the same process to calculate A's location as a function of B's time. This will be a little more challenging but it will be worth it for you to do it. You asked us in post #9 to tabulate this kind of information but now that I have shown you how to do it, I'm asking you to do it yourself. If you need help or have any questions, just ask.

 

[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. This should mean that The calculations of C should result in A and B being the same age after the trip isn't it?

Now you also suggested a different IRF in which a 3rd person C remains at rest while A and B start off traveling away from him at the same speed in opposite directions.

I took OP's invention of C to be a figure whose distance from A & B is kept the same though they are the same original A & B, i.e. A remains at rest. I took it to be the usual Twin Paradox confusion over the primacy of coordinate systems without regard for inertial / non-inertial.

 

[ShreyasR]

Thanks a lot ghwellsjr for making time to give such a nice explanation! I have understood how to plot space-time graphs now... So if i consider any A on Earth and his twin bro B traveling in a spaceship away and back at a speed v, the angle of B's trajectory with space axis will be tan^-1(c/v), and i have to mark the dilated time of B accordingly along his trajectory.

But then i don't understand where to start when i try to find A's position as a function of B's time. Now i have to consider the length contraction factor too. And also i have to plot the graph showing the total time spent by B is 2.8 years. But B's frame of reference is non inertial. So how do i go about it?

If i consider the same spacetime plot you sent me, and B sending a radar signal simultaneously as the green radar signal is reflected back by the spaceship (0.6(B)years), it will travel along the same path as the green radar signal, reflected back and reaches the spaceship in about 2.06(B) years. though i have this data i am not able to determine anything! :(

 

[ghwellsjr]

I took OP's invention of C to be a figure whose distance from A & B is kept the same though they are the same original A & B, i.e. A remains at rest. I took it to be the usual Twin Paradox confusion over the primacy of coordinate systems without regard for inertial / non-inertial.

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? Do you realize that I transformed the first diagram into the second diagram and then added C at rest?

I don't understand your statement "the primacy of coordinate systems without regard for inertial / non-inertial". What are you talking about?

 

[ShreyasR]

I took OP's invention of C to be a figure whose distance from A & B is kept the same though they are the same original A & B, i.e. A remains at rest. I took it to be the usual Twin Paradox confusion over the primacy of coordinate systems without regard for inertial / non-inertial.

What do you mean by OP's invention?
Yes that's what i meant... A is on earth... B and C leave the Earth in different spaceships in same direction such that C measures A and B to be equidistant from him (after considering the time dilation and length contraction of C). When B turns back, C also turns back at the same time. Please help me with my problem in my previous reply so that i can try solving this one on my own.

 

[1977ub]

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? 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.

 

[1977ub]

What do you mean by OP's invention?

Only that C is *added* by you to your original scenario.

 

[ShreyasR]

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)

 

[ghwellsjr]

Thanks a lot ghwellsjr for making time to give such a nice explanation! I have understood how to plot space-time graphs now... So if i consider any A on Earth and his twin bro B traveling in a spaceship away and back at a speed v, the angle of B's trajectory with space axis will be tan^-1(c/v), and i have to mark the dilated time of B accordingly along his trajectory.

But then i don't understand where to start when i try to find A's position as a function of B's time. Now i have to consider the length contraction factor too. And also i have to plot the graph showing the total time spent by B is 2.8 years. But B's frame of reference is non inertial. So how do i go about it?

You don't have to do anything special to consider length contraction, that's what the radar method does for you automatically. I did A's measurements for B on a diagram which already showed lengths correctly for B. That's why I wanted you to repeat the same process on the second diagram where A is not stationary and where the distances to B will calculate differently than the diagram shows but they will come out the same as what were calculated using the first diagram. If you first tabulate the data for each measurement you will see that. After you get the tabular data, you can reconstruct the first spacetime diagram from measurements obtained from the second spacetime diagram.

If i consider the same spacetime plot you sent me, and B sending a radar signal simultaneously as the green radar signal is reflected back by the spaceship (0.6(B)years), it will travel along the same path as the green radar signal, reflected back and reaches the spaceship in about 2.06(B) years. though i have this data i am not able to determine anything! :(

The first green radar signal is reflected back at 0.7(B) years, not 0.6. So you take the difference between 2.06 and 0.7 which is 1.36 and you divide that by 2 to get 0.68 light-years as the distance. Then you take the average of 2.06 and 0.7 which is 1.38 years as A's time for when B is 0.68 light years away. Just repeat the process for several other radar signals. Do some earlier and some later in time. I know you can do it, it's easy. Post your results.

 

[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.