Twin paradox, virtual clock on ship with Earth time, discontinuity

  • #106
PeterDonis said:
The blue line in your chart, the one titled "Universal Time Simultaneity", is not a 45 degree line, which is what ##t = t'## describes.

When the ship stops at turnaround point the world line is vertical. That means rate is the same for ship's clock and earth's clock, so a 45 degree line in this graph. It is exaggerated on purpose to reflect this fact, but instead of a stoppage of a few seconds it seems like a stoppage of several months.

Regarding the rest of the questions, could you read this document? I think it's from a physics professor at Princeton, and maybe he explains it better than I can.

Kirk T. McDonald said:
Moments When vB = 0
Thus, twin B can compute the “age” of twin A (presuming that twin A has remained at rest) at those moments when he (twin B) is at rest with respect to twin A, whether or not the two twins are in the same place.

Twin B’s Trip Includes Frequent Stops
The discussion in sec. 2.2.1 above offers a kind of solution to the issue of distant simultaneity(“age” of a distant clock) for an accelerated observer such as twin B, provided the distant clock A is somehow known to remain in a single inertial frame at all times. Namely, the accelerated observer B brings himself to rest with respect to the distant clock in inertial frame A whenever he wants to know its “age”, which is then the value, eq. (7), of the time on the clock frame of A that is next to the accelerated observer B.In principle, such “stops” could be very brief for the accelerated observer B, such that the result of the computation (7) is little different from that made just before a “stop”. Hence,the accelerated observer B could reasonably omit the (disruptive) “stops”, and simply assign the “age” of the distant clock (i.e., of twin A in the present example) as the value tB(τ B) he computes from his accelerometer data.
http://kirkmcd.princeton.edu/examples/clock.pdf
 
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  • #107
LikenTs said:
When the ship stops at turnaround point
I'm talking about the graph you posted in post #102. It doesn't look like that graph has a single turnaround point.

LikenTs said:
could you read this document? I think it's from a physics professor at Princeton
That doesn't make it a valid reference. It looks like a personal document that is not published anywhere in the literature.

That said, the document does say something very important in the last paragraph of section 3.4, which you should consider carefully:

It remains a difficulty for many that special relativity, a classical theory, does not have
a unique answer to the apparently simple question as to what twin B thinks is the “age” of
twin A during the trip. Instead, an interpretation is required, which affects the answer.
 
  • #108
LikenTs said:
this document
Btw, the document also mentions Mike Fontenot, who is well known here at PF, although it's been quite a while since the last discussion of his claims. The basic issue with him was that he refused to accept the statement I quoted at the end of post #107 just now. Again, you might want to consider that carefully.
 
  • #109
LikenTs said:
maybe he explains it better than I can
The fact that you are trying to construct a way for twin B to calculate "what time does twin A's clock read right now?" does not need to be explained. Everyone else in this now quite overlong thread already understands that thoroughly. (We also understand the limitations of taking this point of view in the first place, which I'm not sure you do.)

What we don't know, because you haven't shown any calculations, just a bunch of graphs with no math behind them, is how, specifically, you are having twin B make that calculation. If the paper you referenced contains formulas you are using, which ones? We are not going to try to reverse engineer your graphs to figure out what formulas you are using. You need to tell us.
 
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  • #110
LikenTs said:
The key factor is the two ways of measuring other frame's time. One can follow one foreign clock and in that case both inertial frames measure the slowdown of the other. Or one can follow the synchronized clocks of another frame while passing through them and in this case both frames observe that the other frame's time goes faster.
More precisely:
You can imagine an inertial reference frame that has a grid of Einstein-synchronized clocks at rest in that frame, which represent the coordinate time of that frame at their respective location.

If a single clock passes through this grid of clocks, then the proper time of the single clock can be compared always to that clock of the grid, that happens to be next to it. These comparisons show, that the coordinate time of the reference frame goes faster than the proper time of the single clock, which passes through the grid.

This shows, that the reciprocity of time dilation is not paradoxal.

LikenTs said:
The Lorentz transformation is more focused on the first type of time measurement.
No. The Lorentz transformation contains coordinate-times (represented by clock grids) of both frames.

For example, the inverse Lorentz transformation for time is

##\Delta t = \gamma (\Delta t' + \frac{v}{c^2}\Delta x')##

But...
If a clock is i.e. at rest at ##x'=0## in the primed frame, then two consecutive tick-events of the clock have ##\Delta x' =0## as coordinate-difference. Plugging this into the above transformation formula:

##\Delta t = \gamma (\Delta t' + 0)##.

At the location of the clock in it's (primed) rest frame, it's proper time between the tick-events is equal to the elapsed coordinate-time of that (primed) frame. $$\Delta \tau = \Delta t'=\frac{1}{\gamma }\Delta t$$
 
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  • #111
LikenTs said:
TL;DR Summary: Twin paradox, special relativity, general relativity, computerized clock in ship showing time on earth.

Conventional Twin Paradox. A ship with speed v = 0.8c makes a round trip from Earth and back. It lasts 6 years and on earth 10 years have passed.

The ship carries its own clock and also a computerized clock that always shows the time on earth at that moment.

It is about knowing the T - T' graph that the computer clock in the ship shows during the trip.
You don't need a computerized clock, because you can have a real clock that shows the time on Earth (in fact the time for the stay-at-home twin). How? Instead of considering a hypothetical case/experiment, something that was never done (trip to another star system and back), you can extend the Hafele-Keating experiment. The stay-at-home twin would stay on the highest point of the equator and the traveling-twin would fly around the Earth, over the equator, at the same altitude as stay-at-home twin. You can build a tower for stay-at-home twin, in order to be at the exact same altitude with the traveling-twin. Also, both would have large displays for their respective atomic clocks in order to compare their time. In this way you can get the best graph on how their clocks/age are really evolving relative to each other.

Of course, the speed of the traveling twin, even it is the orbital speed for the respective altitude, would be much lower than 0.8 c. Still, with atomic clocks, it would be enough, especially if the trip/experiment is taking years, as in the "classical" twin paradox experiment.

With this real life twin paradox experiment we can learn much more than we could from the classical/hypothetical one. One thing is that acceleration is not the cause for ageing less, because the traveling/accelerating twin would age more if flying westwards with the same speed as Earth rotates eastward.

Another thing is that the change of frame may be also irrelevant. Why? Consider the Earth as not spinning and flat, with mirrors placed all along the equator. The twins would see each other in mirrors, exactly as described in the classical experiment. The speed (0.8 c) and distance (4 ly) can also be the same, if the Earth is big and dense enough. Now, consider that at the "arrival point", on the opposite side (in relation with the tower) of the big, non spinning, Earth, we have another plane traveling with the same speed, relative to surface, in the opposite direction and we synchronize the clock on the second plane with the one in the first/original plane. What difference would be between the clocks on the 2 planes when they meet again at the tower? Is the turning back (in this case with relay), and the respective change of frames, really relevant/important, or the first plane's clock would display the same time as the second one on arrival at home/tower?
 
  • #112
DanMP said:
How?
A question your setup does not answer.
DanMP said:
In this way you can get the best graph on how their clocks/age are really evolving relative to each other.
How are you planning on doing this? If you are thinking of each twin watching the other clock through a telescope and mirror system this shows the time in the past, not the time "now", exactly the same as doing this in the normal setup.
DanMP said:
Still, with atomic clocks, it would be enough,
Walking speed is enough. You could do the twin paradox experiment with a van and a couple of lab-grade atomic clocks if anyone wanted to do it.
DanMP said:
With this real life twin paradox experiment
In what sense is this a "real life" version of the experiment? Has it been done? With what precision does your plane have to hold altitude, speed and course that makes this different from Hafele-Keating?
DanMP said:
One thing is that acceleration is not the cause for ageing less, because the traveling/accelerating twin would age more if flying westwards with the same speed as Earth rotates eastward.
In that case, both twins are accelerating.
DanMP said:
Another thing is that the change of frame may be also irrelevant.
Of course the frame change is irrelevant. The frame change is only there because people instinctively adopt the instantaneous rest frame of an object. You don't need to, and the twin paradox pretty much exists solely to illustrate the dangers of doing it.
 
  • #113
Ibix said:
How are you planning on doing this? If you are thinking of each twin watching the other clock through a telescope and mirror system this shows the time in the past, not the time "now", exactly the same as doing this in the normal setup.
Thank you for considering that "each twin watching the other clock through a telescope and mirror system" is "exactly the same as doing this in the normal setup". Probably you are referring at the case with a huge planet, with 8 l.y. circumference. The thing is that with a much smaller planet, similar to Earth in size (but not spinning), the traveling tween gets to be at the same place (inches apart) with the stay-at-home (in the tower) twin many many times in its 4 l.y. journey, while flying near the tower. I wrote that the twins
DanMP said:
would have large displays for their respective atomic clocks in order to compare their time
Basically they can take pictures with both displays in the same frame, so they both can monitor the time differences between them all along the trip/experiment. That's why a "computerized" clock is no needed.

Ibix said:
In what sense is this a "real life" version of the experiment?
By "real life twin paradox experiment" I meant not only that we really did similar experiments (Hafele-Keating, GPS satellites), but also that we don't neglect gravity. The conventional setup is in an imaginary world, without gravity.

Ibix said:
In that case, both twins are accelerating.
No, the twin in the tower does not accelerate, at leat not to increase its speed, as the traveling twin does. The increase/decrease in speed is not the cause/reason/enough for ageing less.

Ibix said:
Of course the frame change is irrelevant.
I'm happy that you agree to that, but I'm not sure that you get what I meant.

I realized/remembered that there is no need for a second traveling-twin to travel in the opposite way. It is enough for the traveling twin to orient its telescope forwards at the "turning point". Doing such they can see, through mirrors, that they are travelling "back", towards the other twin, the return leg of the 4 l.y. trip. It is like a relay version with no actual return. But the traveling twin can also look backwards, seeing that he/she still travels away from the tower ...
 
  • #114
DanMP said:
The thing is that with a much smaller planet, similar to Earth in size (but not spinning), the traveling tween gets to be at the same place (inches apart) with the stay-at-home (in the tower) twin many many times in its 4 l.y. journey, while flying near the tower.
So they cannot monitor their ages at all times. They still need to use a simultaneity criterion to judge their relative ages when they are not co-located.

You have simply designed an unnecessarily complex kind of repeated twin paradox, where the twins travel out-and-back repeatedly instead of just once and check their ages each time they meet. You have not removed (you cannot remove, in fact) the need for a simultaneity convention (aka "computerised clock") when the twins are not co-located.
DanMP said:
By "real life twin paradox experiment" I meant not only that we really did similar experiments (Hafele-Keating, GPS satellites), but also that we don't neglect gravity. The conventional setup is in an imaginary world, without gravity.
So it would be more accurate to describe it as "an imaginary experiment where DanMP thinks the idealisations (such as a six trillion kilometer diameter planet) are irrelevant" than a "real world" experiment.

There is no problem with neglecting gravity in the twin paradox. Relativistic gravitational effects are negligible in the usual setup.
DanMP said:
No, the twin in the tower does not accelerate
Really? How are they kept weightless then?
DanMP said:
I realized/remembered that there is no need for a second traveling-twin to travel in the opposite way. It is enough for the traveling twin to orient its telescope forwards at the "turning point". Doing such they can see, through mirrors, that they are travelling "back", towards the other twin, the return leg of the 4 l.y. trip. It is like a relay version with no actual return. But the traveling twin can also look backwards, seeing that he/she still travels away from the tower ...
None of that has anything to do with frame changes.
 
  • #115
Ibix said:
So they cannot monitor their ages at all times. They still need to use a simultaneity criterion to judge their relative ages when they are not co-located.
If the traveling-twin flies (over the Earth's equator, at the same altitude as the tower-twin) with orbital speed, as I proposed in my first post here, the twins would be able to compare/monitor their ages/clocks every 1.5 hours. It is more than enough for years long trip. And if the friction with the Earth's atmosphere bothers you, you can build/imagine a higher tower or a similar setup on the Moon.

Ibix said:
You have simply designed an unnecessarily complex kind of repeated twin paradox, where the twins travel out-and-back repeatedly instead of just once and check their ages each time they meet.
No, in fact it is a simplified version, because there is no need for the traveling twin to stop and turn back ... Also there is no need of a second traveling-tween for a relay. You just turn the telescope at the "turnaround" point.

And, yes, it is not really necessary to complete more than a trip around the Earth, if you are only interested to see/debunk the influence of acceleration and frame-change, but if you want to see/monitor, as the OP requested, the stay-at-home tween time over years long trip, this is a better way to do it than with a "computerized" clock on a starship going to another star.

Ibix said:
There is no problem with neglecting gravity in the twin paradox.
Yes, there is, because without gravity (or some kind of stunt) you cannot reunite with the stay-at-home tween by just going in one direction. You won't find the Earth 4 light years past Alpha Centauri :) Only by using gravity you can learn that there is no need to stop and turn the ship around in order to age less than stay-at-home twin.

Ibix said:
Really? How are they kept weightless then?
The stay-at-home tween, the one in the tower, is (or must be) kept weightless?!?
 
  • #116
DanMP said:
Yes, there is, because without gravity (or some kind of stunt) you cannot reunite with the stay-at-home tween by just going in one direction. You won't find the Earth 4 light years past Alpha Centauri :) Only by using gravity you can learn that there is no need to stop and turn the ship around in order to age less than stay-at-home twin
Sure, but that is a very different scenario from what most people mean when they talk about the twins paradox. You should be clear rather than expecting people to read your mind.

At this point this thread is more than done. I am locking it. @DanMP if you would like to continue to discuss you are welcome to do so. This is not a closure of the topic, just this thread. If you open a post to discuss a gravitational-based scenario then do so clearly and explicitly so that people don't waste their time talking past each other as just happened here.
 

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