B Do Twins Age Differently in Space?

[Chris Miller]

Twins leave Earth in opposite directions, accelerate at the same rate until their relative v = ~c, hold this v for 20 years (or however long), then decelerate at the same rate to v=0, and accelerate at the same rate back towards Earth until their relative v again =~c, hold for same amount of time, then decelerate at the same rate until they reunite where they started. For the entire journey, wouldn't each have seen the other as aging much more slowly? Are they the same age now? How much time has passed on earth?

 

[Ibix]

They'd be the same age. A triplet at home would be older. Look up Orodruin's Insight on the geometric resolution of the twin paradox.

Edit: one of the learning points of the twin paradox is that time dilation isn't anywhere near the whole story.

 

[Chris Miller]

Thanks, that's what I thought. So even though each twin "saw" the other as aging more slowly for the entire trip, they are the same age when they reunite. It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

 

[Orodruin]

To boil it down: The time dilation formula is based on the assumption of the simultaneity in a particular frame. This notion of simultaneity will differ from frame to frame and will therefore change as an observer accelerates.

Note that time dilation does not describe what an observer actually sees, but what it can deduce happened. In order to find out how it would actually look you would need to take the finite travel time of light into account.

 

[Ibix]

Thanks, that's what I thought. So even though each twin "saw" the other as aging more slowly for the entire trip, they are the same age when they reunite. It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

No. Their observations are always consistent with the notion that the travellers will be the same age when they return. The Earth frame does nothing. It's just a choice of coordinates.

 

[Orodruin]

Look up Orodruin's Insight on the geometric resolution of the twin paradox.

Oh, and the easiest way to find this text is by following the "My Insights" link in my signature.

 

[rede96]

even though each twin "saw" the other as aging more slowly for the entire trip

If it were possible for each twin to observe the other for the entire trip then they couldn't see each other as ageing slower for the entire duration. That would be like Twin A watching Twin B's clock running slower for the whole trip and then suddenly at the end of the journey the clock jumps forward to the same time as Twin A

I'm not too sure what twin A would observe if he could watch Twin B's clock for the entire journey but what ever he sees it must symmetrical (e.g. if he sees twin B's clock running slower then at some point he must see it running equally faster.) as the two clocks have to read the same time again at the end of their journeys.

 

[Bandersnatch]

Have a look here:
http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html
It includes diagrams with 'messages' being sent between twins. The short story is, during the outbound trip the messages arrive in longer intervals than they are sent (equivalent to 'seeing time slow down'), while during the return trip, the intervals are shorter.

 

[Chris Miller]

To boil it down: The time dilation formula is based on the assumption of the simultaneity in a particular frame. This notion of simultaneity will differ from frame to frame and will therefore change as an observer accelerates.

Note that time dilation does not describe what an observer actually sees, but what it can deduce happened. In order to find out how it would actually look you would need to take the finite travel time of light into account.

Thanks, Orodruin. I get that acceleration impacts t dilation, which is why I kept it symmetrical. I also get that neither can really observe in "real" time the other's clocks (biological or mechanical). But, for the purposes of this thought experiment, can I not assume some sort of instantaneous (say entanglement based) method of observation? In any case, what both deduced was occurring, via SR, clearly was not, or was retroactively undone?

 

[PeroK]

Thanks, that's what I thought. So even though each twin "saw" the other as aging more slowly for the entire trip, they are the same age when they reunite. It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

If you are serious about learning SR, you need to knuckle down and study the relativity of simultaneity and the so-called leading clocks lag rule, which lies at the heart of these problems.

Coming up with a pseudo explanation like this is not the real understanding.

 

[Orodruin]

If it were possible for each twin to observe the other for the entire trip then they couldn't see each other as ageing slower for the entire duration. That would be like Twin A watching Twin B's clock running slower for the whole trip and then suddenly at the end of the journey the clock jumps forward to the same time as Twin A

I'm not too sure what twin A would observe if he could watch Twin B's clock for the entire journey but what ever he sees it must symmetrical (e.g. if he sees twin B's clock running slower then at some point he must see it running equally faster.) as the two clocks have to read the same time again at the end of their journeys.

This is exactly what you will find if you compute what the observers will actually see, as mentioned in my previous post. This is not what time dilation is about. Time dilation is about what time coordinates are assigned to different events in a particular frame and its comparison to the proper time.

This is based on the notion of simultaneity in the particular frame and changes when you change reference frame by accelerating.

 

[Chris Miller]

If it were possible for each twin to observe the other for the entire trip then they couldn't see each other as ageing slower for the entire duration. That would be like Twin A watching Twin B's clock running slower for the whole trip and then suddenly at the end of the journey the clock jumps forward to the same time as Twin A

I'm not too sure what twin A would observe if he could watch Twin B's clock for the entire journey but what ever he sees it must symmetrical (e.g. if he sees twin B's clock running slower then at some point he must see it running equally faster.) as the two clocks have to read the same time again at the end of their journeys.

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe, which is why it's called a paradox, I guess.

 

[PeroK]

Thanks, that's what I thought. So even though each twin "saw" the other as aging more slowly for the entire trip, they are the same age when they reunite. It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

If you are serious about learning SR, you need to knuckle down and study the relativity of simultaneity and the so-called leading clocks lag rule, which lies at the heart of these problems.

Coming up with a pseudo explanation like this is not the real understanding.

 

[Orodruin]

can I not assume some sort of instantaneous

No

Even if you could it would depend on which frame you want it to be instantaneous in. If it is instantaneous in one frame it will not be in another (it will be an FTL communication that could a priori go back in time). This is at the very core of relativity and is called the relativity of simultaneity: what is "at the same time" in one frame is not "at the same time" in another.

Failing to understand the relativity of simultaneity is likely responsible for more than 90 % of the misconceptions beginners have of relativity. Understanding it is crucial if you want to have any chance of understanding SR at a deeper level than popular science.

 

[PeroK]

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe, which is why it's called a paradox, I guess.

It's what you think SR predicts, because you only know about time dilation. SR is more than just time dilation.

 

[jbriggs444]

I get that acceleration impacts t dilation

Acceleration does not affect time dilation. Acceleration results in a change of synchronization convention.

 

[Chris Miller]

Acceleration does not affect time dilation. Acceleration results in a change of synchronization convention.

so acceleration is not the same as a gravity (feels the same though), been wondering

 

[Chris Miller]

No

Even if you could it would depend on which frame you want it to be instantaneous in. If it is instantaneous in one frame it will not be in another (it will be an FTL communication that could a priori go back in time). This is at the very core of relativity and is called the relativity of simultaneity: what is "at the same time" in one frame is not "at the same time" in another.

Failing to understand the relativity of simultaneity is likely responsible for more than 90 % of the misconceptions beginners have of relativity. Understanding it is crucial if you want to have any chance of understanding SR at a deeper level than popular science.

I'm not sure I have the grey matter or mathematical chops to understand SR well. It'd be nice not to totally embarrass myself writing SF, though. And, so far, this site's been helpful in that regard.

PS

Aren't entanglement's changes simultaneous across any distance? And so FTL?

 

[jbriggs444]

so acceleration is not the same as a gravity (feels the same though), been wondering

No. You continue to misunderstand.

Gravity is the same as an acceleration. That's the equivalence principle. But gravitational acceleration is NOT associated with time dilation. Gravitational potential is. The distinction is that potential is a combination of an acceleration and a distance.

Accelerations in special relativity can be interpreted as an incremental change of reference frame and an associated incremental change of clock synchronization. The amount of clock change scales with distance. A mere acceleration is not enough. You need an acceleration and a remote clock some distance away.

 

[Herman Trivilino]

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe, which is why it's called a paradox, I guess.

It's not a real paradox, it's an apparent paradox. This is why you often see it referred to as the Twin Trip or some such, as it really doesn't deserve the name Twin Paradox. A less complex apparent paradox is the symmetry of time dilation, which really ought to be resolved before tackling the twin paradox as glossing over it means you'll forever find the twin paradox baffling.

But to address the issue you raise, imagine each twin with a flashing strobe light. Let's call these strobe lights clocks, as they send out the flashes at regular intervals, say once per minute. When each is at rest relative to the other, the rate at which the flashes are seen by one twin is equal to the rate at which they are sent by the other. But if there is relative motion between the twins, that will not be the case. When moving apart each will see the other's flashes as slow, but when they approach each other, each will see the other's as fast. Now this speeding up and slowing down of clock rates is due only partially to time dilation, the other part being due to light travel time. So each will see the other's clock as running fast or slow depending on whether they are approaching or receding, but if you subtract off the part of the effect due to light travel time, then each will observe that the other's clock is running slow, regardless of whether they are approaching or receding.

So what you see is different from what you observe.

Anyway, if you go through the details you will find that after the twins reunite to compare notes, the number of flashes that each sent will equal the number of flashes that the other received. But they may not agree on the time that elapsed between the flashes, and when the total time between flashes is added up, they may get different sums.

Now, if the journeys are symmetrical in the way you propose, the sums will match. But a triplet that stayed home will get a larger sum.

There's a hokey old cartoon that can you find by searching YouTube for "Paul Hewitt Twin Trip". The physics is solid.

 

[Dale]

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe

No, this is not at all what SR predicts. Not only is it not what they visually observe, there is also no reference frame where that is what they calculate to be true.

What they would observe is the other twins clock being redshifted during the first part, normal in the middle, then blueshifted during the last part. They end with the same elapsed time.

In the Earth's frame they are always equal. They end with the same elapsed time.

In an inertial frame where one twin is initially at rest then the other will start slow, but the first will end slow. They end with the same elapsed time.

In a non inertial frame the math is complicated. They end with the same elapsed time.

No matter what frame you pick, if you actually do the math... They end with the same time.

It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

No, it is nothing like that at all.

 

[Chris Miller]

It's not a real paradox, it's an apparent paradox.

That's why I quoted it in my question.

...but if you subtract off the part of the effect due to light travel time, then each will observe that the other's clock is running slow, regardless of whether they are approaching or receding.

This, and your entire explanation was extremely helpful/interesting (even the cartoon!). Although I still don't understand how, if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict) how their (discounted) sums match at the end.

 

[jbriggs444]

This, and your entire explanation was extremely helpful/interesting (even the cartoon!). Although I still don't understand how, if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict) how their (discounted) sums match at the end.

Read up on the twin's paradox. The explanation that I gravitate to is that the accounting from the traveling twin's point of view skips part of the aging process of the stay-at-home twin. At turnaround his notion of what age his twin is "right now" changes. He matches up the first bit of his twin's aging to his outbound trip. He matches up the last bit of his twin's aging to his return trip and never accounts for the part in between.

 

[robphy]

If it were possible for each twin to observe the other for the entire trip then they couldn't see each other as ageing slower for the entire duration. That would be like Twin A watching Twin B's clock running slower for the whole trip and then suddenly at the end of the journey the clock jumps forward to the same time as Twin A

I'm not too sure what twin A would observe if he could watch Twin B's clock for the entire journey but what ever he sees it must symmetrical (e.g. if he sees twin B's clock running slower then at some point he must see it running equally faster.) as the two clocks have to read the same time again at the end of their journeys.

Here's a diagram based on my Relativity on Rotated Graph Paper Insight.
(You can use https://www.geogebra.org/m/HYD7hB9v to draw parts of the diagram below.)

In the frame of the stay-at-home twin (on which the gridlines of the rotated graph paper are based),
we draw the ticks for each "traveling" twin (each with speed (3/5)c).
Then, I show the light signals sent by the "positive-x" twin... Think of it as the "negative-x" twin watching an 8 hour tv show broadcast by the "positive-x" twin.
Due to their relative motions, the viewer sees that show with irregular timing.
(You can imagine that each has the same 8-hr movie started at separation...and that one twin is comparing the received broadcast with her own local playback of the movie.)

The ticks of the stay-at-home twin are not shaded in... but you can shade them into realize that
at reunion, 10 ticks elapse for the stay-at-home twin while 8 ticks elapsed for each of these symmetrically-traveling traveling-twins.

You could also analyze what each (traveling or stay-at-home) twin sees from the other two twins.
You could also try asymetrical twins, as well as trips with asymmetrical there and back legs.

(One could try to interpret the situation using light-travel time corrections and change of lines of simultaneity, etc... but I feel that this presentation is direct and to the point.)

 

[Nugatory]

Although I still don't understand how, if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict) how their (discounted) sums match at the end.

The quick answer is that special relativity does not predict that each one SEES the other's clock running slow . We've already discussed what the twins SEE at some length, and there's also a good explanation of what they do SEE in in the "Doppler analysis" section of the twin paradox FAQ.

If you haven't already read that FAQ in its entirety, do so now. It covers the asymmetric case in which one twin stays on Earth while the other flies out and back, but until you clearly understand that case you won't be ready to take on the symmetric case.

 

[Chris Miller]

The quick answer is that special relativity does not predict that each one SEES the other's clock running slow . We've already discussed what the twins SEE at some length, and there's also a good explanation of what they do SEE in in the "Doppler analysis" section of the twin paradox FAQ.

If you haven't already read that FAQ in its entirety, do so now. It covers the asymmetric case in which one twin stays on Earth while the other flies out and back, but until you clearly understand that case you won't be ready to take on the symmetric case.

I've read the Twin paradox and get EM Doppler effect and that acceleration (GR) muddies the waters. "See" is clearly a bad word to use. SM doesn't predict what they would literally "see." But does it predict the rate at which each would expect the other's clock to be running?

 

[Ibix]

Special relativity, as normally presented, provides a way to relate the tick rate of someone else's clock after lightspeed delay is corrected for, yes. However, there is some flexibility in what "correcting for the lightspeed delay" means, and the "natural" way of doing it leads to different answers for different people, and different ideas about what "at the same time" means. Accounting for the latter is one way to resolve the twin paradox, as @jbriggs444 says above.

 

[Dale]

if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict)

SR does not predict this. This is not predicted as a visual observation, and it is not predicted (after discounting light travel time) in any inertial frame nor in any valid non-inertial frame.

Here is an example of a valid non-inertial frame, with the analysis for the standard twins scenario. It also explains why the naïve approach is not a valid non inertial frame.

https://arxiv.org/abs/gr-qc/0104077

 

[Dale]

But does it predict the rate at which each would expect the other's clock to be running?

SR predicts the rate of change of proper time with respect to coordinate time for any clock and for any valid coordinate system. You must specify both the clock and also the coordinate system in order to get such a rate.

 

[PeterDonis]

does it predict the rate at which each would expect the other's clock to be running?

It predicts this given the assumption that each twin is at rest in a fixed inertial frame. But if the twins separate and then meet up again, this assumption must be false for at least one of them. In the standard twin paradox, it's false for the traveling twin. The "time gap" page of the Usenet Physics FAQ article that Nugatory linked to discusses this.

 

[DrGreg]

It's not a real paradox, it's an apparent paradox.

To be pedantic, it is a paradox, because the word "paradox" has more than one meaning:

Paradoxically, the word "paradox" has (amongst others) two meanings that are almost opposites:

1. an argument that comes to a false conclusion (e.g. contradicts itself), via steps that appear, at first, to be valid
   
2. an argument whose conclusion may appear, at first, to be false, but is actually true

The "twins paradox", like some other paradoxes in maths and physics, is a paradox of the second type.

In this post I used "paradoxically" with a third meaning, "having apparently contradictory characteristics".

 

[Ibix]

I've argued that it is a paradox by definition 1 if one misunderstands relativity in a particular way common among beginners. It's not a paradox under a correct understanding of relativity. Hence the resolution is to realize that "time runs slowly at high speed" is not really true. Or at least, is rather incomplete.

 

[Herman Trivilino]

This, and your entire explanation was extremely helpful/interesting (even the cartoon!). Although I still don't understand how, if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict) how their (discounted) sums match at the end.

Thanks, but let's sort something out here. The cartoon uses the Doppler effect to analyze the usual twin paradox, where one twin stays home and the other travels. In this scenario it is not appropriate to say each sees the other's clock running slow, because sometimes they see them running fast. When you do the sums at the end there is no need to worry about any delays due to light travel time because each twin is simply taking a sum of the times elapsed between flashes at his own location.

If you want instead to analyze things a different way and focus on that oft-repeated phrase that each observes the other's clocks to be running slow during both the outbound and inbound phases, then you have to account for the fact that when the traveling twin turns around his entire notion of what's happening "simultaneously" back home gets shifted. It's an understanding of that shift in the notion of what's simultaneous that's needed to understand how the symmetry of time dilation can account for the twins' different ages.

 

[Jeronimus]

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe, which is why it's called a paradox, I guess.

They don't see but measure/calculate each others' clocks to go slower. More precise, they calculate the clock count of the instances of each others' clocks which cross the simultaneity axis to be lagging behind the clock count of a stationary clock local to them.

It is also not true that they measure each others' clocks to be running slower for the ENTIRE trip. The stay at home twin will measure/calculate the traveling twin's clock instance that lies on the simultaneity axis( where the worldline of the clock crosses the simultaneity axis) to be running slower by the same factor as the traveling twin measures the stay at home twin's clock to be going slower, with the exception of the acceleration phase.

At the acceleration phase/turn around phase, the stay at home twin will measure the travellings twin's clock to be running slower still while the traveling twin "breaks" to a halt. At the halt point, logically their clocks would be running at the same pace again for a very brief moment of time, followed by further acceleration towards the stay at home twin where the stay at home twin measures the traveling twin's clock to start ticking slower again until it ticks slower by the same factor as before, once max relative velocity is reached.The traveling twin measures something different in the acceleration phase. Whole accelerating back, the traveling twin will measure/calculate the stay at home twin's clock to be ticking much faster than his. Faster the higher the magnitude of the acceleration. If the acceleration was to be near instantaneous, he would see the clock count of the stay at home twin's clock to "jump" up a considerable amount. After the acceleration, back to the same relative velocity as before, he would again measure the stay at home twin's clock to be ticking slower by the same factor as before.

You can see what happens exactly in the video here, where the lorentz transformation formulas have been used to map events that happen at x,t measured by the stay at home twin (left diagram) to x', t' in the right diagram, which represents the traveling twin's perspective.



On the left diagram, the cyan coloured clock represents the stay at home twin's clock measured from the stay at home twin's perspective, while the white clock is is the what we _interpret_ at the traveling twin's clock.

In the right diagram, the white clock represents the traveling twin's clock from the traveling twin's perspective, while the blue clock is what we _interpret_ as the stay at home twin's clock.

What are the numbers and what are the moving circles/dishes in the diagrams? Generally speaking, those are all events which in the left diagram happen at x,t and are then mapped to the right diagram where they are measured at x',t', using the Lorentz transformation formulas.

More detailed, the numbers represent clock counts. In the left diagram, all but the traveling twin's clock are at rest seen from the stay at home twin's perspective, which is why you see the instances of the clocks, displaying different clock counts, all being on an (imagined) straight, vertical worldine.

A clock that is measured to be at rest from the perspective of the stay at home twin, would be measured to be moving from the perspective of the traveling twin, and therefore the instances of a certain clock which is at rest relative to the stay at home twin and are on a (imagined) vertical line on the left diagram, will be on a angled line in the right diagram(between 0 and 45° for this type of diagrams).
The reason why i am pointing out that those instances of the clocks on the simultaneity axis are merely what we interpret as moving clocks which are simultaneous to us, is because this is just a physical definition we agreed upon.

If you look at the video, particularly at stage 3 and 4, you see how the blue clock in the right diagram (the filled blue circle), ticks much faster in phase 3 and 4 when the traveling twin accelerates. "Its count goes up much faster" relative to the traveling twin's clock local to him.

This means, the traveling twin will measure a future instance of the blue clock to be on the simultaneity axis post acceleration. However, he could decide to accelerate in the opposite direction again, and then the instance of the blue clock which lies on the simultaneity axis as measured by the traveling twin would be one of the past compared to the one before accelerating into the opposite direction.

The traveling twin could keep doing this "acceleration dance" and the instance of the blue clock which is measured by the traveling twin to be on the simultaneity axis, would seemingly be "moving forwards and backwards in time" on repeat.

So one has to be careful in what the simultaneity axis really is. It is an axis which has been defined in physics accurately, but it NOT the axis where events happen "at the same time" to us when by "at the same time" we are asking "what is my friend doing/feeling right now". They happen at the same time only in the sense of having the same t or t' coordinate.

Unfortunately, consciousness or subjective experience is beyond the scope of physics but it is nevertheless important to understand that when we talk about events happening at the "same time" in relation to the twin paradox, as physicists, we merely mean they have the same t or t' coordinates and therefore lie on a specific axis we defined as the simultaneity axis precisely.

Physical time has been defined precisely, but it completely ignores the "present" or "now (subjective)experience". Hence, physicists do not care how consciousness travels through the block universe on a worldline and or when two worldlines cross, the two consciousnesses traveling those worldlines would actually meet at the cross-points or not at all.
Yet, unless we are ok with meeting zombies at the worldline crossing points, we would have to come up with a "meta-physics" which takes this into account as well. (I will be starred at for typing this, i know... but someone had to say it)

It would be absurd to assume that events happening at the same physical time as in having the same t or t', actually happen at the same time as in what two people experience at the same time, shown by the example with the "acceleration dance" above.

 

[Dale]

then the instance of the blue clock which lies on the simultaneity axis as measured by the traveling twin would be one of the past compared to the one before accelerating into the opposite direction.

The traveling twin could keep doing this "acceleration dance" and the instance of the blue clock which is measured by the traveling twin to be on the simultaneity axis, would seemingly be "moving forwards and backwards in time"

Note that what you describe here is not a valid coordinate system. A coordinate chart on spacetime is a one to one mapping between events and coordinates.

 

[Jeronimus]

Note that what you describe here is not a valid coordinate system. A coordinate chart on spacetime is a one to one mapping between events and coordinates.

What i was describing there is the clock count of what we _interpret_ as the stay at home twin's clock, measured by the traveling twin during his acceleration period. In this particular case, the traveling twin would accelerate towards and away of the stay at home twin on repeat (while at a distance), which would result in the clock count of the stay at home twin's clock to be measured going up and down on repeat. Hence the traveling twin would be measuring future and past instances of the same clock to be simultaneous to him on repeat, according to how we define simultaneity in SR.

I don't consider this describing a coordinate system but maybe i am missing something here. Maybe you could elaborate.

 

[Dale]

Hence he would be measuring future and past instances of the same clock to be simultaneous to him on repeat, according to how we define simultaneity in SR

Actually, this is not quite true. That is how we define simultaneity for an inertial observer, but the observer in question is not inertial. What you are describing is not simultaneity according to the non inertial observer, but rather according to a series of momentarily comoving inertial observers.

If you apply the Einstein synchronization convention procedure to a non inertial observer then you get a different result that yields a single valid coordinate system with a one to one mapping. See here:

https://arxiv.org/abs/gr-qc/0104077

 

[Jeronimus]

Actually, this is not quite true. That is how we define simultaneity for an inertial observer, but the observer in question is not inertial. What you are describing is not simultaneity according to the non inertial observer, but rather according to a series of momentarily comoving inertial observers.

If you apply the Einstein synchronization convention procedure to a non inertial observer then you get a different result that yields a single valid coordinate system with a one to one mapping. See here:

https://arxiv.org/abs/gr-qc/0104077

Without claiming i understand the linked document,

I cannot see how any convention could change the fact that after each acceleration step (assuming a near instantaneous acceleration), the clock we are referring to, which is the blue clock in my video animation, and which the traveling twin uses as a reference to compare to his local clock (white clock in the right diagram in the video animation), would not end up with a higher, lower, higher, lower count on repeat should the traveling twin perform near instantaneous accelerations in different directions on repeat.

So, ignoring the acceleration phase itself, and looking only post acceleration, the traveling twin would be faced with an instance of the blue clock which is simultaneous to him, having the same t' position. Hence simultaneous according to how we define simultaneous in SR for inertial frames.

And each time after the near instantaneous acceleration phases, being in an inertial frame, he would measure the blue clock to be showing a lower then higher then lower then higher counter etc etc... and if i am not mistaken, you would also agree that this blue clock being on the simultaneity axis _post acceleration_ is simultaneous to the local clock of the traveling twin according to Einstein's synchronisation convention.

Hence it would be absurd to compare "physical" simultaneity to what "one experiences at the same time" who is not local to yourself.

My attempts to somehow think of a resolution for this, would not work, unless the coordinate systems as i drew them in the animation are not really 100% accurate but spacetime as i mapped it out according to SR is merely an approximation which works only at short distances.

It is this part of the document you linked,

"Although this period of acceleration can indeed fix the gap between G and H, it cannot resolve the more serious problem (mentioned also in Marder[7] and in Misner et al.[8]) which occurs to Barbara’s left. Here her hypersurfaces of simultaneity are overlapping, and she assigns three times to every event! Also, if Barbara’s hypersurfaces of simultaneity at a certain time depend so sensitively on her instantaneous velocity as these diagrams suggest, then she would be forced to conclude that the distant planets swept backwards and forwards in time whenever she went dancing!"

which suggests the authors understood the very problem i am puzzled about and believe to have resolved it. But unfortunately i cannot see how.

 

[Ibix]

Einstein's coordinates work by filling space with a 3d grid of rods equipped with clocks (synchronised in a particular way) at each junction. That's fine, and you can have more than one grid in relative motion (at least conceptually).

The problem for the instantaneous turn around is that the plan is to have one grid, then vapourise at the same time as the turn around and replace it with another grid. "At the same time" is the problem - it means different things to different grids, so you end up with part of spacetime with no grid and part with two grids. That leads to some events having no coordinates and some having two, which is analogous to thinking that those street atlases with a small overlap between pages actually mean that the streets in the overlap exist in two places.

Dolby and Gull resolve this with a more practical arrangement than rods and clocks. They have a radar set with a clock. They assign distance and time to reflection events by the usual procedure for a radar set. Then they observe that, for an inertial radar set, the result is the same as Einstein's coordinates with the radar set as the spatial origin. But - the process can naturally assign unique coordinates through the acceleration phase because emission, reflection and reception only happen once (at most).

The rest is maths showing the paths assigned to the stay at home twin under various scenarios.

 

[Jeronimus]

Einstein's coordinates work by filling space with a 3d grid of rods equipped with clocks (synchronised in a particular way) at each junction. That's fine, and you can have more than one grid in relative motion (at least conceptually).

The problem for the instantaneous turn around is that the plan is to have one grid, then vapourise at the same time as the turn around and replace it with another grid. "At the same time" is the problem - it means different things to different grids, so you end up with part of spacetime with no grid and part with two grids. That leads to some events having no coordinates and some having two, which is analogous to thinking that those street atlases with a small overlap between pages actually mean that the streets in the overlap exist in two places.

Dolby and Gull resolve this with a more practical arrangement than rods and clocks. They have a radar set with a clock. They assign distance and time to reflection events by the usual procedure for a radar set. Then they observe that, for an inertial radar set, the result is the same as Einstein's coordinates with the radar set as the spatial origin. But - the process can naturally assign unique coordinates through the acceleration phase because emission, reflection and reception only happen once (at most).

The rest is maths showing the paths assigned to the stay at home twin under various scenarios.

But the problem is not really the instantaneous or near instantaneous acceleration as far as i can tell.

You could draw those diagrams using non-instantaneous accelerations. Nevertheless, if the traveling twin at a distance would do non-instantaneous accelerations in opposite directions on repeat, he would measure/calculate the stay at home twin's clock to be moving forward and backwards in time on repeat, as in he would measure the instance of the stay at home twin's clock (where the clock's worldline crosses the simultaneity axis) to be an instance with a lower, then higher then lower clock count on repeat, post acceleration, using his IFR to calculate the x' and t' position of the stay at home twin's clock.

Given a high enough acceleration at a far enough distance, post acceleration, the stay at home twin instance on the simultaneity axis could be an instance of a dead twin. Accelerating in the opposite direction the instance of the stay at home twin on the simultaneity axis would be that of an alive twin and so on.
And this is true for non instantaneous accelerations as well.

 

[Orodruin]

But the problem is not really the instantaneous or near instantaneous acceleration as far as i can tell.

You could draw those diagrams using non-instantaneous accelerations. Nevertheless, if the traveling twin at a distance would do non-instantaneous accelerations in opposite directions on repeat, he would measure/calculate the stay at home twin's clock to be moving forward and backwards in time on repeat, as in he would measure the instance of the stay at home twin's clock (where the clock's worldline crosses the simultaneity axis) to be an instance with a lower, then higher then lower clock count on repeat, post acceleration, using his IFR to calculate the x' and t' position of the stay at home twin's clock.

Given a high enough acceleration at a far enough distance, post acceleration, the stay at home twin instance on the simultaneity axis could be an instance of a dead twin. Accelerating in the opposite direction the instance of the stay at home twin on the simultaneity axis would be that of an alive twin and so on.
And this is true for non instantaneous accelerations as well.

I do not understand why you consider this a problem. The events for the stay-home twin that can be simultaneous with some given event for the traveling twin in some inertial frame are all space-like separated from that event for the traveling twin. It is just an arbitrary matter of what you call "simultaneous".

 

[Jeronimus]

I do not understand why you consider this a problem. The events for the stay-home twin that can be simultaneous with some given event for the traveling twin in some inertial frame are all space-like separated from that event for the traveling twin. It is just an arbitrary matter of what you call "simultaneous".

I consider this a problem because i want to believe that in spacetime a world exists with things happening while i am not local to those events. And i want to believe that there is an event at an exact x and t position relative to me where the stay at home twin would exist in the "now" even when i am not close.

Unfortunately, while i can mark simultaneous events on my diagrams, i am not capable of marking that exact "now" event for the stay at home twin and vice versa if i were to describe the whole scenario from the stay at home twin's perspective.

edit: Related to my example above, where one would accelerate in opposing directions on repeat when at the distance to the stay at home twin.
Supposed that someone was asked by another who happened to be flying along with him, "what is the stay at home twin doing right now?" after each acceleration phase, the traveling twin would have to answer with "he is dead" after the acceleration phase in one direction was over, and "he is alive" after the acceleration phase in the other direction, given a high enough acceleration at a great enough distance and given he would consider "now" equivalent to the definition of simultaneity we use in SR.

This of course is absurd. But then where exactly is the "now" of the stay at home twin? How can i calculate it and render it in my video with the two diagrams?

 

[Ibix]

That's what radar coordinates do. All Einstein's coordinate system is is a process for assigning coordinates to events in spacetime. But it doesn't work for an axis that is curved because it asserts that "now" is the plane perpendicular to your chosen axis, and if the axis is curved those planes overlap.

All Dolby and Gull do is provide a system for defining planes that curve when the axis is not straight, and revert to inertial planes when the axis is straight. So there is a unique now under this system. There are other ways of doing it. But attempting to glue together inertial frames will not work.

 

[Jeronimus]

That's what radar coordinates do. All Einstein's coordinate system is is a process for assigning coordinates to events in spacetime. But it doesn't work for an axis that is curved because it asserts that "now" is the plane perpendicular to your chosen axis, and if the axis is curved those planes overlap.

All Dolby and Gull do is provide a system for defining planes that curve when the axis is not straight, and revert to inertial planes when the axis is straight. So there is a unique now under this system. There are other ways of doing it. But attempting to glue together inertial frames will not work.

So let me ask this in a way i would understand.

I am the traveling twin. Currently not accelerating. I am in my rocket and draw a standard x-t diagram. On this diagram i place myself at x=0 and t=0. I take a look at my clock and the clock count shows 10 seconds. So i draw a clock at x=0 and t=0 which has a clock count of 10 seconds. That is an event if you want.

Now my question is: Can i draw this "unique now" of the stay at home twin on my x-t diagram? If yes, where would i place it? Which formula would i use to calculate the coordinates?

 

[Ibix]

Your now is a horizontal line on your map, by definition. Dolby and Gull show how your now would appear on a Minkowski diagram (i.e. in the stay-at-home's inertial frame). They are the pale grey lines in figures 5 and 6. Figure 5 is for an instantaneous turnaround, figure 6 for a smooth turnaround.

 

[Jeronimus]

Your now is a horizontal line on your map, by definition. Dolby and Gull show how your now would appear on a Minkowski diagram (i.e. in the stay-at-home's inertial frame). They are the pale grey lines in figures 5 and 6. Figure 5 is for an instantaneous turnaround, figure 6 for a smooth turnaround.

But i did not ask for worldlines. Worldlines are not the "unique now" i was asking for. I was asking how to plot the unique now of the stay at home twin into my simple x-t diagram as described above. That "unique now" would have specific x,t coordinates, not be a worldline. And i need the formula which would allow me to calculate those coordinates.

 

[m4r35n357]

I consider this a problem because i want to believe that in spacetime a world exists with things happening while i am not local to those events. And i want to believe that there is an event at an exact x and t position relative to me where the stay at home twin would exist in the "now" even when i am not close.

That belief is the root cause of your problem.

 

[Ibix]

But i did not ask for worldlines. Worldlines are not the "unique now" i was asking for. I was asking how to plot the unique now of the stay at home twin into my simple x-t diagram as described above. That "unique now" would have specific x,t coordinates, not be a worldline. And i need the formula which would allow me to calculate those coordinates.

Those aren't worldlines. They're lines of equal t' (where t' is your time coordinate). So figure 5 is a classic twin paradox viewed in the stay at home twin's frame. If your clock reads t'=10 at the point that one particular grey line crosses your worldline then all events on that line were at t'=10.

 

[Orodruin]

I consider this a problem because i want to believe that in spacetime a world exists with things happening while i am not local to those events. And i want to believe that there is an event at an exact x and t position relative to me where the stay at home twin would exist in the "now" even when i am not close.

Well then I am sorry, this is not how relativity works. Nature does not care for your wants.

Also, there is no need for you to be local for events to occur. It is just that you cannot have any information on events that have space-like separation to you, just as you cannot have any information on events in your future light-cone. The difference is that events that have space-like separation to you cannot have any information about your current state either.

The entire point is that "now" is not something that is uniquely defined. You seem to not have grasped the concept of space-time being a single entity and wish to separate space and time. This is doing yourself a huge disfavour in willingly electing to not accept one of the most astonishing physical insights in history. "Now" is a convention and there are several different conventions that work and there is no real reason to favour one over the other. Technically, any split of Minkowski space into space-like surfaces works to define simultaneities and thus different possible definitions of "now".

 

[Jeronimus]

Well then I am sorry, this is not how relativity works. Nature does not care for your wants.

Also, there is no need for you to be local for events to occur. It is just that you cannot have any information on events that have space-like separation to you, just as you cannot have any information on events in your future light-cone. The difference is that events that have space-like separation to you cannot have any information about your current state either.

The entire point is that "now" is not something that is uniquely defined. You seem to not have grasped the concept of space-time being a single entity and wish to separate space and time. This is doing yourself a huge disfavour in willingly electing to not accept one of the most astonishing physical insights in history. "Now" is a convention and there are several different conventions that work and there is no real reason to favour one over the other. Technically, any split of Minkowski space into space-like surfaces works to define simultaneities and thus different possible definitions of "now".

Then you misunderstood me. If anything, i am not trying to separate space and time but would go even further to rather talk about 4-space than spacetime.

However, i realize now that this is not a problem i should have brought up in a pure physics forum. Only a philosopher which is also a physicists could possibly solve it.
Pure physicists cannot deal with consciousness and subjective experience. "The entire point is that "now" is not something that is uniquely defined" is quite an indication for that.