The Twin Paradox: Triplets Edition

[Dale]

DaleSpam called it gravitational blueshift. But if Jane only accelerates for almost zero duration, why would grav blueshift last for a long period of time?
Unfortunately I'm not well versed in GR and I'm not going to pretend that I am. :shy:

If you go to the Dolby and Gull paper and look at figure 9 you can see the answer, although they did not describe it much in the text.

The instantaneous acceleration of the traveling twin (Barbara for Dolby and Gull) causes a "shock like scale discontinuity" that travels away from the twin in a future-directed light cone and also a past directed light cone.

Those "shocks" are what cause the turning of the stay at home twin's worldline (Alex for Dolby and Gull), one shock brings Alex to rest in the non-inertial frame and the other accelerates him towards Barbara. Those shocks also cause light to undergo gravitational red/blueshift. It redshifts as it goes from region P to region I and blueshifts as it goes from region I to region F.

 

[greswd]

If you go to the Dolby and Gull paper and look at figure 9 you can see the answer, although they did not describe it much in the text.

The instantaneous acceleration of the traveling twin (Barbara for Dolby and Gull) causes a "shock like scale discontinuity" that travels away from the twin in a future-directed light cone and also a past directed light cone.

Those "shocks" are what cause the turning of the stay at home twin's worldline (Alex for Dolby and Gull), one shock brings Alex to rest in the non-inertial frame and the other accelerates him towards Barbara. Those shocks also cause light to undergo gravitational red/blueshift. It redshifts as it goes from region P to region I and blueshifts as it goes from region I to region F.

Hmm, they wrote this:

It is often said of the twin paradox that[4] “a complete explanation of the problem can only be given
within the framework of general relativity”. However, as we have just shown, Barbara’s hypersurfaces
of simultaneity depend only on the kinematics involved, and can be fully understood without resorting
to general relativity

 

[Dale]

Hmm, they wrote this:

Yes, they did write that, and it is correct, there is no tidal gravity in this scenario so no need for GR.

 

[greswd]

Yes, they did write that, and it is correct, there is no tidal gravity in this scenario so no need for GR.

I see. Anyway I can't say anything otherwise for the time being; till I've learned GR. Thanks Dale.

 

[ghwellsjr]

Here's my original diagram.

http://img826.imageshack.us/img826/1289/vvvvvi.png

Now I'm just going to use the names Jane and Joe.

After extrapolating, Jane calculates that the 9th signal was sent in between Joe's 7th and 8th b-days. Extrapolating for the 7th and 8th signals, Jane wonders why one was sent before schedule and another sent behind schedule.

I know you don't like my long explanations so I made a bunch of diagrams for you to look at. First is the rest frame for Joe:

I drew this one upside-down so that I could transform it to the next two diagrams that show Jane's two inertial rest frames:

Please note that all three of the above diagrams correctly show Jane receiving the signals from Joe.

I need to start a new post since there is a limit of three images uploaded per post.

 

[ghwellsjr]

Now I've copied the last part of Jane's IRF and pasted it on to her first IRF to get a correct diagram of your image in the previous post:

Now you observe that the signals that Joe sends after his 6th one, get changed when they hit the interface between the two IRFs. For example, Joe's 7th signal suddenly changes to his 10th signal. But this merely shows that you cannot just connect the two IRFs along a vertical line. Instead, you could have at least drawn it like this:

But why not draw it like Dr Greg did in his last diagram in post #39 or like your referenced website did as shown in post #119?

Now between these last two diagrams, we can see that the problem is not a time gap but a distance gap. How is Jane to know if the 6th signal that she receives from Joe was sent when he was 4.5 light years away or 18 light years away? Same thing with all the rest of his signals up to his 24th signal.

One way she can know is to measure how far away he was when he sent each signal. I'll go into this in the next post.

 

[ghwellsjr]

I've already explained the radar method of determining how far away an object is but I'll go into more detail now. I've gone back to Jane's first IRF and added in several radar signals shown in different colors:

The first one is shown in blue and was sent when Jane's clock read 3 and the echo was received when her clock read 12. The return echo is aligned with Joe's 6th signal. She measures his distance by calculating how long it took for the radar signal to make its round trip and dividing that in half. Since it took 9 years, she divides that in half to get 4.5 years and so she concludes that Joe was 4.5 light years away at 4.5 years before she received the echo which puts it at 7.5 years. This agrees with the closer of the two distances making up the gap in the previous post.

She can repeat the process every year as shown by the different colored radar signals. For example, the next one was sent at year 4 and received at year 13 for a difference of 9 years. So again, she determines that Joe was 4.5 light years away. In fact, she will conclude this for all the signals sent up to the time when she changed directions at 12 years. But she can continue this process all the way to the end and when she gets done, she can make a diagram that will look like this:

Don't you agree that this is one possible way for Jane to solve the problem of how far away Joe is when he sent each signal?

 

[greswd]

I know you don't like my long explanations so I made a bunch of diagrams for you to look at.

Thanks for taking my needs into consideration.
You must've spent a lot of time and effort on these diagrams and I really appreciate it.

If the image is too wide I just use [MB3 + Drag]

Don't you agree that this is one possible way for Jane to solve the problem of how far away Joe is when he sent each signal?

I agree it's possible. I had a whole bunch of questions but Dale seems to have answered them using a so-called "non-tidal GR".

 

[ghwellsjr]

Thanks for taking my needs into consideration.
You must've spent a lot of time and effort on these diagrams and I really appreciate it.

If the image is too wide I just use [MB3 + Drag]

Fortunately, this new PF software adjusts each post to the proper width instead of the entire page--very nice.

Don't you agree that this is one possible way for Jane to solve the problem of how far away Joe is when he sent each signal?

I agree it's possible. I had a whole bunch of questions but Dale seems to have answered them using a so-called "non-tidal GR".

Now I'm hoping you will be convinced that you can take the above diagram of Jane's non-IRF rest frame and have Joe use the same radar method that she used and construct his own IRF rest frame:

So the same technique applies to both IRF and non-IRF frames.

One further thing I want to make sure you understand is that the Doppler method works correctly on both these rest frames. In other words, they both show what Jane sees of Joe's clock and provide the same Doppler ratios. If we wanted to, we could make two more diagrams showing the signals going from Jane to Joe and show how the Doppler method would also apply to Joe looking at Jane's signals. Does this all make sense to you or should I draw those diagrams?

 

[tade]

Where is the time gap?

 

[greswd]

Where is the time gap?

ghwellsjr was trying to show that there is none.


@ghwellsjr

Ok, I understand it now.
Except for Dale's explanations, but that's because I haven't learned GR.

 

[ghwellsjr]

I think we are ready to go back and pick up the triplet scenario. Back at post #19 I had asked you about what Adam and Bob would see of each others clocks when they each turn around. In your subsequent answer we got diverted on to this long discussion that focused on just the Twin Paradox relationship between Charles and Adam and we have pretty well covered that. I have re-read all the posts leading up to #19 and I would suggest that you do the same and then please answer my questions again:

Excellent.

I said in post #11 that Adam and Bob are going to age (about) a couple years so let's say they travel away for exactly one year according to their own clocks and then turn around and get back to Charles in exactly one more year. We'll first deal with what happens between Adam and Bob and when we get done with that we'll figure out what goes on between each of them and Charles.

Now according to the Doppler Analysis, Adam and Bob will each see the other ones clock running slower than their own by the factor of 5.00250125×10^-4 (which is 1/1999). So the first question we want to answer is what time will each of them see on the other ones clock when they reach the point of turnaround? The answer is simple--we multiply 1 year by 5.00250125×10^-4 (or divide it by 1999), which is just a little over four and a half hours.

The next question is what Doppler Factor will apply at the moment of turn around? How fast will they each see the other ones clock ticking immediately after they each turn around? What do you think?

 

[greswd]

I think we are ready to go back and pick up the triplet scenario. Back at post #19 I had asked you about what Adam and Bob would see of each others clocks when they each turn around. In your subsequent answer we got diverted on to this long discussion that focused on just the Twin Paradox relationship between Charles and Adam and we have pretty well covered that. I have re-read all the posts leading up to #19 and I would suggest that you do the same and then please answer my questions again:

I really don't know how to answer it. I suppose you could tell me everything you have on your mind, otherwise we'll take another hundred posts to resolve this.By the way, could you simplify Dale's GR explanations?

 

[ghwellsjr]

I really don't know how to answer it.

I suppose you could tell me everything you have on your mind, otherwise we'll take another hundred posts to resolve this.

Study this thread and see if you can get some hints.

By the way, could you simplify Dale's GR explanations?

No, I don't know GR so you'll have to ask him for more help.

 

[greswd]

Study this thread and see if you can get some hints.

Not really. It seems to be more confusing .

 

[ghwellsjr]

Study this thread and see if you can get some hints.

Not really. It seems to be more confusing .

Back on post #65 (page 5) I showed you several diagrams including this one:

and in the next post you said:

I understand the diagrams well.

Now that diagram is the same as this one from post #92 (page 6):

except this one is rotated 90 degrees and flipped so that it is like a conventional spacetime diagram. So I'm sure you're not having a problem with that.

Now here is the first diagram from the other thread:

Can you see that the right half of this diagram is identical to the diagram above it? And can you see that the left half of the diagram is a mirror image of the right half? Does this have anything to do with your confusion?

 

[greswd]

Will the solution involve another trapezoid?

 

[ghwellsjr]

Will the solution involve another trapezoid?

The non-IRF diagram using the radar method for each of the outside triplets calculating the path of the inside triplet will produce the same trapezoid as in the Twin Paradox "solution" but the shape they each calculate for the other outside triplet will be a more complicated.

But before we get to the complicated non-IRF diagram, I think you need to resolve your confusion over the simple IRF solutions. Can you provide specifics on what you are confused about?

 

[greswd]

The non-IRF diagram using the radar method for each of the outside triplets calculating the path of the inside triplet will produce the same trapezoid as in the Twin Paradox "solution" but the shape they each calculate for the other outside triplet will be a more complicated.

But before we get to the complicated non-IRF diagram, I think you need to resolve your confusion over the simple IRF solutions. Can you provide specifics on what you are confused about?

I understand your previous trapezoid diagram well.

Will the shape both calculate for the other outside triplet be a trapezoid as well?

 

[ghwellsjr]

I understand your previous trapezoid diagram well.

Then you should have no problem calculating the non-inertial rest frame for one of the outside triplets using the radar method. I have changed the scenario slightly to make it easier for you to do this. The only difference is that the traveling triplets go out for 16 months and 16 months coming back instead of just 12 each way, and the inertial triplet ages by 40 months:

Will the shape both calculate for the other outside triplet be a trapezoid as well?

No. Work it out and you'll see what it is.

 

[greswd]

Dayum, I got to do this manually.

Why provide the LabVIEW graphing software only after we're done?

 

[ghwellsjr]

Dayum, I got to do this manually.

It's not that hard. Start by making a table like I did in post #92 and #93 (page 6).

Why provide the LabVIEW graphing software only after we're done?

Yes, otherwise, you won't appreciate it. But my software doesn't do non-inertial frames. It allows you to define a triplet scenario (speed and length of time for the traveling triplets) and draws the diagram in the inertial triplet's rest frame and then it uses the Lorentz Transformation to redraw it at whatever speed you want the new IRF to be moving relative to the original IRF.

When I did the non-inertial rest frames, it was all done by cut-and-paste in Paint and I have only done this one on graph paper.

 

[greswd]

I will appreciate it

 

[ghwellsjr]

Here is a diagram repeating the one from post #140 but showing just a pair of radar signals emitted by Red as he measures the distance to Blue:

Now he can make the following table to summarize his findings:

Radar   Radar   Calculated   Calculated    Blue's
Sent    Rcvd      Time        Distance      Time
  0       0        0             0            0
  4      16       10             6            8
 16      28       22             6           32
 32      32       32             0           40

Remember that the Calculated Time is determined by adding the Radar Sent and Rcvd times together and dividing by two and the Calculated Distance is determined by subtracting the Radar Rcvd time from the Radar Sent time and dividing by two. If you want, you can include as many more samples from post #140 as you want. For completeness, I have included the readings at the beginning and at the ending (even though those radar signals are instantaneous).

Now here is another diagram repeating the one from post #140 but showing just a few of the radar signals emitted by Red as he measures the distance to Black:

And here is a table showing these findings (including the ones at the beginning and ending):

Radar   Radar   Calculated   Calculated   Black's
Sent    Rcvd      Time        Distance     Time
  0       0        0             0           0
  1      16        8.5           7.5         4
  4      28       16            12          16
 16      31       23.5           7.5        28
 32      32       32             0          32

Now we are ready to construct the non-inertial rest frame for red as he determines the distances to Blue and Black:

You can confirm that this diagram represents the findings from the two tables and any additional samples you want to include.

If you want, you can download the diagram and draw in radar and/or Doppler signals (along 45-degree diagonals) for all three observers and confirm that this diagram conforms to all my other diagrams in terms of what each observer sees of the other observers. All the diagrams that I have drawn are equivalent and contain the same exact information.

Any questions?