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

[Sagittarius A-Star]

I'll study it.

The Dolby & Gull paper refers under Figure 8 to the spacetime-metric of a uniformly accelerated frame in radar coordinates (Lass coordinates). Instead, the 2nd linked paper in your posting #19 uses in equation (8) on page 4 the spacetime-metric of a uniformly accelerated frame, expressed in Kottler–Møller coordinates.

 

[LikenTs]

I think I'm beginning to have a clearer vision of the problem and SR in general.

In particular, I understand the thought experiment that I proposed in post #80, about the ship that makes stops during its trip.

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. The Lorentz transformation is more focused on the first type of time measurement.

If we suppose a universal inertial frame of convenience, for example the galactic frame, and we imagine that it contains a mesh or net of synchronized clocks, all relativistic ships could see their time offset with respect to galactic time when passing through these clocks. If they also stop close to one of these, they share the simultaneity criterion with the galactic frame, so they can assign that time to that of a clock on Earth.

Fortunately, that mesh is not necessary. A ship, with a local clock, an accelerometer (more possibly gyroscopes) and a computer, can reconstruct dynamically or in real time its speed and position with respect to the galactic frame, and with this can build a display that shows the galactic time. In fact, the ship's frame is also the most privileged for this computation, since all other frames see things with a signal delay and require greater infrastructures.

What this inertial or galactic time display would show on all relativistic ships is always a higher ratio than their local clocks and an increasing time offset. The greater the faster they go or the longer they travel (without direct dependence on acceleration). This inertial clock display would only go at the same rate as the local clock when the ships stop (And they would know at all times how old their relatives are).

The way to understand this inertial or galactic time display is that the simultaneity criterion of the Galaxy is adopted. While in most of the graphs that I put above the criterion of simultaneity of the ship is adopted. The computer can also reconstruct a second display to follow Time on Earth (or any Galactic place) with the ship's simultaneity criteria. This one does show dependence on accelerations and distances and can even go backwards in time.

Here is the graph of both virtual clocks on the ship, that of universal or galactic time, with galactic's frame simultaneity , in blue, and the time on Earth according to the simultaneity of the ship, in black. At the return point, green, the stopping time is somewhat exaggerated, a slope of 45 degrees. At that point in the ship is observed the same ratio of the 3 clocks , the local one and the two virtual ones. You can also see how the virtual ones coincide at the stops.

Some reference articles:

About the reconstruction of the world line from the ship.

Differential aging from acceleration, an explicit formula
https://arxiv.org/abs/physics/0411233
Journal reference: Am.J.Phys. 73 (2005) 876-880

On the two ways of measuring time and the different relevance for both twins
http://kirkmcd.princeton.edu/examples/clock.pdf

 

[PeterDonis]

inertial frames

What inertial frames are you talking about? You are talking about ships that accelerate and decelerate, i.e., they are not inertial. That means there is no inertial frame in which they are always at rest. You don't appear to be properly taking that very important fact into account. For example, you talk about "the" Lorentz transformation, but a Lorentz transformation is only valid between two inertial frames, and at best there is only one that applies to your entire scenario.

If we suppose a universal inertial frame of convenience, for example the galactic frame

If you are talking about our actual galaxy, there is no such thing. You can construct a frame in which the center of the galaxy is at rest, but it will not be an inertial frame. Gravity is present, and in the presence of gravity there are no global inertial frames.

If you are talking about an imaginary case where there is no gravity, but you have labeled some region of space as "the galaxy" and are choosing an inertial frame in which that region is at rest, then you can use such a frame in your imaginary case, but many conclusions that you reach will not generalize.

Here is the graph of both virtual clocks on the ship, that of universal or galactic time, with galactic's frame simultaneity , in blue, and the time on Earth according to the simultaneity of the ship, in black.

How are you calculating these lines?

 

[LikenTs]

What inertial frames are you talking about? You are talking about ships that accelerate and decelerate, i.e., they are not inertial. That means there is no inertial frame in which they are always at rest. You don't appear to be properly taking that very important fact into account. For example, you talk about "the" Lorentz transformation, but a Lorentz transformation is only valid between two inertial frames, and at best there is only one that applies to your entire scenario.

In this paragraph I speak in general of two inertial frames. Both see each other's time slowed down or faster, depending on what kind of time measurement they do. Even so, if what they measure is the time of the other's clock mesh, any accelerated frame always sees that the inertial frame's mesh goes at a higher rate.

If you are talking about our actual galaxy, there is no such thing. You can construct a frame in which the center of the galaxy is at rest, but it will not be an inertial frame. Gravity is present, and in the presence of gravity there are no global inertial frames.

I am ignoring in this case the gravitational fields, since that would be the scope of GR. But surely a galactic frame with standard time for low gravitational fields could be defined, and the computer on the ship could make adjustments according to GR. So everyone on the ship will know, seeing the inertial virtual clock, the time lag with Earth, even when the ship orbits a black hole. I think this type of virtual clock will be mandatory on relativistic ships, if they are ever developed.

How are you calculating these lines?

The blue line, Galactic or Universal Time virtual clock on ship, is easy, T = T' . gamma (1.66 for v=0.8c). Being T' the local clock on ship. It is what you would calculate by integrating the proper time of world line from the inertial frame. The black line, the time of the clock on Earth according to the criterion of simultaneity in the ship, is calculated in the accelerated frame (There are several documents cited above), but the key is that when the ship is inertial during both legs the ship sees the Earth clock going slower, while at the return point it sees a rapid jump in time on the earth clock. The trip is better understood anyway by adopting the galactic simultaneity criterion, and thinking that the own simultaneity criterion is distorted with respect of the Galaxy. The traveling twin sees on the inertial clock display that he is progressively falling behind in galactic time during the trip.

 

[PeterDonis]

In this paragraph I speak in general of two inertial frames.

Ok. But I don't see the point since there is only one global inertial frame in your scenario.

Both see each other's time slowed down or faster, depending on what kind of time measurement they do.

This is a very sloppy way of talking, which I suspect is causing many of your problems.

It is impossible for observer A, spatially separated from observer B, to directly measure "what time it is now for observer B". The only direct observation observer A can have from observer B is receiving light signals from B, and those can only show what B's clock read at the event where they were emitted. Observer A can of course calculate all manner of things, but those are calculations, not direct observations. Your use of the word "see" and "time measurement" to refer to such calculations obfuscates this important point.

if what they measure is the time of the other's clock mesh, any accelerated frame always sees that the inertial frame's mesh goes at a higher rate.

I'm not sure what this means.

I am ignoring in this case the gravitational fields, since that would be the scope of GR.

Ok.

But surely a galactic frame with standard time for low gravitational fields could be defined

You can define a frame in which the center of the galaxy is at rest, and you can define coordinate time in such a frame. But you cannot assume that this frame will work like a global inertial frame in SR. It won't.

the computer on the ship could make adjustments according to GR.

If the computer knows the spacetime geometry, and knows where it is in that spacetime geometry, it can compute appropriate adjustments, yes.

everyone on the ship will know, seeing the inertial virtual clock, the time lag with Earth

If they also know Earth's location in the defined coordinates, and are using that simultaneity convention, then yes, they can compute this according to that convention. But, as has been said many times now, this is just a convention. It has no physical meaning. It is certainly not a direct observation. But you appear to really, really want to treat it as such. That won't work.

The blue line, Galactic or Universal Time virtual clock on ship, is easy, T = T' .

The blue line in your chart, the one titled "Universal Time Simultaneity", is not a 45 degree line, which is what $t = t'$ describes.

The black line, the time of the clock on Earth according to the criterion of simultaneity in the ship, is calculated in the accelerated frame (There are several documents cited above)

Which specific paper's definition are you using for "the accelerated frame".

when the ship is inertial during both legs the ship sees the Earth clock going slower, while at the return point it sees a rapid jump in time on the earth clock.

Your diagram doesn't show this. This is a reasonable description of one possible "rest frame" for the traveling twin in the standard twin paradox, but your diagram doesn't look like a diagram of the standard twin paradox.

The trip is better understood anyway by adopting the galactic simultaneity criterion, and thinking that the own simultaneity criterion is distorted with respect of the Galaxy.

Perhaps this is true for you. (And even that implies that your understanding is correct, which, given how long this thread has gone on, is not obvious.) That doesn't mean it will be true for everyone.

 

[LikenTs]

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.

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

 

[PeterDonis]

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.

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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.

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. $$\Delta \tau = \Delta t'=\frac{1}{\gamma }\Delta t$$

 

[DanMP]

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?

 

[Ibix]

How?

A question your setup does not answer.

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.

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.

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?

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.

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.

 

[DanMP]

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

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.

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.

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.

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 ...

 

[Ibix]

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.

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.

No, the twin in the tower does not accelerate

Really? How are they kept weightless then?

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.

 

[DanMP]

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.

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.

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.

Really? How are they kept weightless then?

The stay-at-home tween, the one in the tower, is (or must be) kept weightless?!?

 

[Dale]

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.