Twin Paradox (thorough explanation needed)

[bobc2]

Short cuts are never a good idea.

A short cut thru spacetime is a great idea if you would like to make a trip and later reunite with your twin in a way that will make you younger than your twin.

By the way, JesseM is one of the sharpest physicists on the forum and you can safely take his comments as true.

 

[Eli Botkin]

Of course the initial difference accounts for their final difference in B's frame, otherwise this would imply one frame was "right" and the other was "wrong"! For example, suppose they are approaching each other at 0.6c, and when they are 15 light years apart in A's frame, both clocks are synchronized to read 0 years in A's frame. Then 15/0.6 = 25 years later in A's frame they will meet, and A's clock will show an elapsed time of 25 years while B's clock shows an elapsed time of only 20 years, so A concludes B's clock was ticking at 0.8 times the rate of A's clock.

Now, consider things in B's frame. In B's frame, when B's clock reads 0 years, A's clock already reads 9 years, due to the relativity of simultaneity. Then 20 years later in this frame they meet, with B's clock reading 20 years and A's reading 25. So, B concludes that A's clock only ticked forward by 25-9=16 years during that 20-year period, meaning in B's frame it was A's clock that was ticking at 0.8 times the rate of B's clock. Their views of each other's clock rates are completely symmetrical thanks to differences in perceptions of simultaneity.

JesseM:
I agree with your calculations and the 0.8 ratio that that both compute. But the real issue is the trip time. If A does the synchronization then A says that he waited a period of 25 'til B arrived while B traveled a period of 20. A claims B's time is dilated.

From B's point of view A's clock was not at 0 at B's start but rather at 9. So indeed A waited only 16 during B's 20. B claims A's time is dilated.

Yes, their relative rates are the same (0.8), as it should be (and is, even in the classic twin paradox case). But it's the time differences, not the ratios, that make the point of time dilation.

The additional interesting point is that all the numbers are exchanged between A and B if B does the synchronization.

 

[GrayGhost]

JesseM:
I agree with your calculations and the 0.8 ratio that that both compute. But the real issue is the trip time. If A does the synchronization then A says that he waited a period of 25 'til B arrived while B traveled a period of 20. A claims B's time is dilated.

From B's point of view A's clock was not at 0 at B's start but rather at 9. So indeed A waited only 16 during B's 20. B claims A's time is dilated.

Yes, their relative rates are the same (0.8), as it should be (and is, even in the classic twin paradox case). But it's the time differences, not the ratios, that make the point of time dilation.

The additional interesting point is that all the numbers are exchanged between A and B if B does the synchronization.

This is one of those cases where what everyone says (you and Jesse) is correct, but each comes at it from a different angle. Neither disagrees with each other. It's one of those things where one doesn't want to write a book on the matter, and so just cuts to the chase. I could argue that the dilation is the result of the geometry of the situ, a geometry based upon Lorentz symmetry. It would be nice if we could look down on 4-space and see it all transpire from global perspective, collectively. However, we cannot.

The fellow who experiences dilation depends on how the interval is defined. In your example, you've defined the interval in 2 different ways, one way such that A experiences the dilation, and then the other way such that B experiences the dilation. It all comes down to this ...

The interval is defined by 2 events. The events are marked by 2 separated points at rest in one system, and you might consider those points as boeys. They have a proper separation, which must be measured contracted by anyone in relative motion. Since both observers A and B record the very same velocity of each other, he who records the boey separation the smallest, experiences the interval the quickest. Now here's the thing ...

In the one case, A defines 2 markers at rest in his own system for the interval, and so he must record the longest time interval. In the other case, B defines 2 markers at rest in his own system for the interval, and so he must record the longest time interval. So he who experiences the time dilation depends entirely on how the interval is defined, ie which frame possesses the interval's 2 event markers as stationary. The observer which holds both event markers at rest, is not the observer that experiences dilation, because of how the interval was defined. There are other ways to look at it as well.

GrayGhost

 

[JesseM]

JesseM:
I agree with your calculations and the 0.8 ratio that that both compute. But the real issue is the trip time. If A does the synchronization then A says that he waited a period of 25 'til B arrived while B traveled a period of 20. A claims B's time is dilated.

From B's point of view A's clock was not at 0 at B's start but rather at 9. So indeed A waited only 16 during B's 20. B claims A's time is dilated.

Yes, their relative rates are the same (0.8), as it should be (and is, even in the classic twin paradox case).

No it's not, not in the "classic twin paradox case"! In that case they both agree on who aged less, and on the ratio of their average rates of aging. In my example they disagree on who aged less, and one's ratio is the inverse of the other's (i.e. if you're considering the ratio of B's elapsed time to A's, A says it's 0.8 while B says it's 1/0.8=1.25).

But it's the time differences, not the ratios, that make the point of time dilation.

What do you mean "make the point of time dilation"? Time dilation is normally defined in terms of a ratio of clock time to coordinate time, and in each one's rest frame the other one's clock only elapses 0.8 the amount of coordinate time in that frame. Besides, the "start time" is totally arbitrary, you could imagine both their clocks have been ticking forever from -infinity until the time they meet, and they both read 0 at the moment they meet--in that case how would you define the "time differences" for each one?

 

[Gulli]

OnlyMe:

While in motion the spaceship is length contracted, but the distance between the planets is not.

If you park a whole lot of spaceships between the planets (the first ship touching one planet, the last ship touching the other planet) and park them stationary to the planets, they will all seem contracted once you start speeding past them, but the chain of ships still has to touch the planets on both ends (remember the parked ships are stationary to the planets and will stay that way for all eternity, so relativity of simultaneity can't make the ships not touch the planets, no matter what frame of reference you pick), so the distance is contracted. That's one way to visualize it, alternatively you could just remember that distance contraction has to supplement time dilation, otherwise the speed of light would not be the same in every frame (hell, you could even travel faster than light).

 

[Dale]

The ship's clock will not agree with the Esther stationary clock.

What is an "Esther stationary clock".

Clocks, just like rods, agree with other clocks when they are measuring the same interval. They may show a different accumulated time, but that is a comparison between a clock and an odometer, not a clock and a rod. Unfortunately, although we have two different words for those two different types of measuring devices for distances we use the same word for both types of measuring devices for time.

 

[ghwellsjr]

Clocks, just like rods, agree with other clocks when they are measuring the same interval. They may show a different accumulated time, but that is a comparison between a clock and an odometer, not a clock and a rod. Unfortunately, although we have two different words for those two different types of measuring devices for distances we use the same word for both types of measuring devices for time.

Actually, we do have two different words for time measuring devices but nobody bothers to use them both. If we want to be precise, we should use metronome for measuring time intervals and rods for measuring distance intervals and then use clock for measuring accumulated time and odometer for measuring accumulated distance.

 

[Eli Botkin]

GrayGhost:
You are correct, there are many ways to skin this pussycat. I got involved because I wanted SR novices to know that the twin paradox's acceleration phase not needed in showing the existence of time dilation

 

[Dale]

If we want to be precise, we should use metronome for measuring time intervals and rods for measuring distance intervals and then use clock for measuring accumulated time and odometer for measuring accumulated distance.

Hmm, interesting. You are of course correct, but metronomes have such a strong association with music and the acoustic output that I have never heard it used in a physics context.

 

[bobc2]

GrayGhost:
You are correct, there are many ways to skin this pussycat. I got involved because I wanted SR novices to know that the twin paradox's acceleration phase not needed in showing the existence of time dilation

I'm so glad you pointed that out, Eli Botkin. If you really are focused on the aging of a given observer, the path (spacetime proper distance) seems most relevant.

It's interesting to contemplate the sequence of LTs (letting the proper time increment between new origins of sequential traveling twin frames be as small as you wish--approach zero in the limit if you wish). So I would not want to take away the fascination with that. You can contemplate the stay-at-home changes in position along his world line as presented in the sequence of changing frames of the accelerating traveler (you can of course make those stay-at-home changes appear whatever factor greater than speed of light as you wish, i.e. stay-at-home world line ds/dT, where ds is stay-at-home incremental proper distance divided by traveling twin's incremental proper time, dT).

And note that the stay-at-home's speed along his own world line is still always the speed of light, c, totally indpendent of any other frames. But one must be careful about the meaning attached to that kind of accelerated motion analysis, notwithstanding that the accelerated analysis (when analyzed properly) results in the same age difference between twins upon reuniting as you would obtain by a straight forward comparison of the proper path lengths traveled in spacetime.

You can marvel at the stay-at-home twin seemingly moving along his world line at faster than c (even moving faster than light along the moving twin's X1 axis), but I assign no more relevance to that than I would for the following experiment: Shine a spot of light on the moon at night with a laser beam that is mounted on a platform with rapid rotation capability. Then sweep the spot of light across the moon at a speed (spot on moon) greater than the speed of light. When the traveling twin accelerates, he is just rotating in spacetime, just like the laser beam, yielding apparent motion in the analysis of the sequence of cross-section views of 4-D space (moving twin's frames).

I realize you can claim that all frames of reference are equally valid. However, you are talking about every individual inertial frame being as valid as ever other inertial frame. Each one of the traveling twin's frames are as valid as all of the others in the accelerating sequence. So, does that necessarily mean that all of those frames collectively (yielding the bazzar faster-than-light result) are as valid as any individual inertial frame? I don't think so, but I'll need to go back and review Rindler and a couple of others to see if they comment on that aspect. Maybe bcrowell, JesseM, or DaleSpam already have the answer.

I know one thing--if you are not careful with your analysis and conclusions, you will find yourself believing that the stay-at-home guy moves along the moving twin's rest frame X1 axis at speeds greater than c, and we know that is not correct. And please do not confuse acceleration with velocity.

 

[Eli Botkin]

No it's not, not in the "classic twin paradox case"! In that case they both agree on who aged less, and on the ratio of their average rates of aging. In my example they disagree on who aged less, and one's ratio is the inverse of the other's (i.e. if you're considering the ratio of B's elapsed time to A's, A says it's 0.8 while B says it's 1/0.8=1.25).

What do you mean "make the point of time dilation"? Time dilation is normally defined in terms of a ratio of clock time to coordinate time, and in each one's rest frame the other one's clock only elapses 0.8 the amount of coordinate time in that frame. Besides, the "start time" is totally arbitrary, you could imagine both their clocks have been ticking forever from -infinity until the time they meet, and they both read 0 at the moment they meet--in that case how would you define the "time differences" for each one?

JesseM:
Let's look at a classic twin paradox case. A and B start at T=0 and meet when A's clock reads 25. Their relative velocity is 0.6 (where c=1). That means that B turns around when A says that B is at X=7.5. B's clock at turn-around reads 10 (and reads 20 when they meet).For A the dilation ratio will be 20/25 = 0.8.

What is it for B? Well B is aware of A's clock readings from 0 t0 8 and from 17 to 25. As far as B is concerned A instantaneously reset his clock from 8 to 17. So from B's point of view A's clock ticked through 25-9 = 16. Therefore for B the dilation ratio is 16/20=0.8.

And this dilation symmetry makes sense because we expect that each one will see the other clock dilated in the same way. The symmetric two-way time dilation always exists with or without acceleration, but the non-symmetric outcome when they finally meet is a result of only B having that blind-spot on A's worldline.

 

[JesseM]

JesseM:
Let's look at a classic twin paradox case. A and B start at T=0 and meet when A's clock reads 25. Their relative velocity is 0.6 (where c=1). That means that B turns around when A says that B is at X=7.5. B's clock at turn-around reads 10 (and reads 20 when they meet).For A the dilation ratio will be 20/25 = 0.8.

What is it for B? Well B is aware of A's clock readings from 0 t0 8 and from 17 to 25. As far as B is concerned A instantaneously reset his clock from 8 to 17. So from B's point of view A's clock ticked through 25-9 = 16. Therefore for B the dilation ratio is 16/20=0.8.

B can clearly see that A has aged 25 years, not 16 when he returns--proper time is an objective frame-invariant quantity in relativity, there can be no ambiguity about the fact that 25 years of proper time elapse on A's worldline between the event of B departing and the event of B returning. You can't just mix and match results from different inertial frames to define B's "point of view" that way, the phrase "point of view" doesn't even have any well-defined meaning for an observer who doesn't move in a 100% inertial way, there would be many different ways to construct a non-inertial coordinate system where B is at rest and all would be equally valid as far as relativity is concerned. Think of it this way, what if B does not turn around perfectly instantaneously but accelerates in a continuous way to turn around for 1 hour or 1 second or 1 nanosecond, what would you say happens to A's clock from B's "point of view" then?

 

[OnlyMe]

There is a planet, the Earth and another planet, the colony. The two, planets are 20. LY apart and at rest relative to one another. A spaceship takes off from the Earth traveling at 0.5 c, accelerating instantly. And travels directly to the colony planet.

Nothing else exists in this thought experiment.

What an observer on the Earth sees is the colony 20 LY away, holding steady at that distance and the spaceship moving away from the Earth toward the colony, at 0.5 c.

What an observer on the colony sees is the Earth 20 LY away, holding steady at that distance and a spaceship approaching the colony at 0.5 c from the direction of the earth.

What an observer on the spaceship sees is the Earth moving away from the spaceship and away from the direction of the colony at 0.5 c and the colony approaching the spaceship in the direction of the Earth at 0.5 c.

All three see their own clocks as working properly and keeping good time.

An observer in the spaceship cannot observe the distance between the Earth and the colony directly, as they are in front of and behind the spaceship. The observer in the spaceship can only measure the distances of each relative to itself.

In idealized thought-experiments it's assumed that each observer has a grid of rulers and clocks at rest relative to themselves and extending out arbitrarily far, which they use to assign coordinates in their own rest frame--this sort of network is how Einstein originally defined the notion of reference frame.

While this is true the thought experiment as restated above was set up as defining observer perspectives.

Either way, in the spaceship's rest frame the position coordinates of the Earth and station at a single moment of coordinate time will be 17.32 light years apart.

I do not disagree. I should have continued the example to include the time dilated observation. The intent was to demonstrate that with the spaceship essentially moving at a known velocity within the at rest frame of reference of the planets, calculating the proper distance/length between each of the planets and the spaceship was easily demonstrated. Not that this example is unique, just that it lent itself to the purpose.

This applies whether the observer is moving relative to the object or the object is moving relative to the observer.

Do you understand that this is a completely meaningless distinction? In relativity there is no difference between "A is moving relative to B" and "B is moving relative to A"

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

If there is an observer in both the moving frame of reference and the stationary frame

Do you understand that "moving" and "stationary" can only be defined relative to a choice of reference frame, and that all reference frames are equally valid in SR?

I am pretty sure the thought experiment included sufficient information to know which frame of reference was in motion. One at rest frame of reference, one moving frame of reference and a total of three observer dependent "perspectives".

Yes, all reference frames are equally valid. SR also includes the means to reconcile observed differences between two frames of reference in uniform rectilinear motion relative to one another.

While in motion the spaceship is length contracted, but the distance between the planets is not.

Relative to what frame? In the spaceship's frame the spaceship is not in motion, it's stationary, and therefore its length is not contracted while the distance between planets is contracted since they are the ones in motion in this frame. And this frame's perspective is every bit as valid as the perspective of the frame where the planets are stationary and the spaceship is in motion. Do you disagree?

No and yes. This gets to the heart of my intent. The difference between proper distance/ length and length contracted observation of distance/length. Proper distance/length is the same for all observers. Where a moving frame of reference is involved the proper distance/length appear length contracted. As long as the relative velocity is known, the Lorentz transformations provide the means to calculate the proper distance/length from the observed contracted distance/length.

This is not a new or unique discussion. There are theoreticians on both sides of the issue. My position is that the proper distance/length is the real distance/length.

In my original post I excluded velocity dependent time dilation and length contraction.

If you assume two planets, moving uniformly with respects to one other through space

Moving with respect to one another, or with respect to some observer's frame of reference? Obviously if they are moving with respect to one another the distance between them is changing!

This was misstated. It should have been, "at rest relative to one another and moving uniformly in space...".

the planets can be length contracted while in motion, the distance between them remains constant. It is not length contracted.

Distance is neither an object.., matter nor energy. Though it may appear length contracted under some circumstances, distance itself does not move and so cannot be length contracted, in a way similar to rods and spaceships.

Both of these can be viewed as similar to, Bell's spaceship paradox. I don't think you will find it in a textbook, but a description of the thought experiment should be available on the net.

Very briefly, it involves two spaceships that begin at rest with a string stretched tight connecting them. They accelerate uniformly such that the distance between them remains the same when observed from the "rest" frame of reference from which they began. Does the string break as it is length contracted? There were and are some very bright theorists on both sides. The CERN theory group decided that the string would not break. I am not yet sure but I like their answer.

I did not raise this example earlier, because it deals with the length contraction of the string, an object and that carries the conversation further than was my intent.

I do understand SR and the math involved. I did not and do not believe that math is necessary to present the perspective.

 

[pervect]

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

Well, relativity says that one cannot determine which frame of reference is actually moving.
Thus, as a consequence, the whole twin paradox is "meaningless" - i.e. it's a false paradox. It's only a paradox if one has the belief (which is inconsistent with relativity) that one can determine which frame of reference is "actually" moving.

One would hope we could wrap this long thread up at this point - my intuition says that that's not going to happen. But it really is that simple!

 

[Eli Botkin]

B can clearly see that A has aged 25 years, not 16 when he returns--proper time is an objective frame-invariant quantity in relativity, there can be no ambiguity about the fact that 25 years of proper time elapse on A's worldline between the event of B departing and the event of B returning. You can't just mix and match results from different inertial frames to define B's "point of view" that way, the phrase "point of view" doesn't even have any well-defined meaning for an observer who doesn't move in a 100% inertial way, there would be many different ways to construct a non-inertial coordinate system where B is at rest and all would be equally valid as far as relativity is concerned. Think of it this way, what if B does not turn around perfectly instantaneously but accelerates in a continuous way to turn around for 1 hour or 1 second or 1 nanosecond, what would you say happens to A's clock from B's "point of view" then?

JesseM:
I don't disagree that "B can clearly see that A has aged 25 years." I was addressing the point of how to compute the dilation ratio (DR). For even in the classical twin paradox problem it should turn out that every observer "sees" a moving clock as running more slowly than his own.

If the acceleration is continuous, then instead of being a constant the DR would be a function having value 1 when the relative velocity passes through zero. Though I haven't yet made the computation, I expect that both DR functions would have the same average value.

Jesse, I appreciate your insightful replies. They've made me think more. But I don't have more to offer so I'll sign off on this thread. Thanks.

 

[ghwellsjr]

If we want to be precise, we should use metronome for measuring time intervals and rods for measuring distance intervals and then use clock for measuring accumulated time and odometer for measuring accumulated distance.

Hmm, interesting. You are of course correct, but metronomes have such a strong association with music and the acoustic output that I have never heard it used in a physics context.

Well then, you never read this post:

The point I was trying to make is that time and distance do not behave the same way in SR. The fact that the time difference persists while the length difference does not (quite aside from how it happens "physically") underscores that difference.

But time and distance both persist, you only think they don't because you are making an invalid comparison of a clock to a ruler. Time dilation does not directly affect the time on a clock, it directly affects the tick rate of a clock and then the clock integrates (or counts) the ticks to keep track of elapsed time.

To get similar behavior, we should use a metronome (which does not count ticks) and a ruler. Take them both on a high speed trip, during which the metonome slows down and the ruler contracts, and when we come back to the starting point, the ruler is the same length as one that did not take the trip and the metronome ticks at the same rate as one that did not take the trip.

Now if you want to get similar behavior to a clock, you need an odometer. This will integrate distance traveled just like the clock integrates time. And you could have a speedometer which calculates divides the (contracted) distance traveled by the (dilated) time.

For example, let's suppose that we take a vehicle with a clock, an odometer and a speedometer. We accelerate the vehicle to 0.6c and take it on a round trip for 50 years according to the starting frame. It's speedometer will read 0.6c and from the point of view of an observer that was stationary with the vehicle before it left, the vehicle's speed is also 0.6c. The gamma factor at this speed is 1.25 which means the clock will be running slow by 1/1.25 according to the rest frame. Its lengths along the direction of motion will also be contracted to 1/1.25 of what the observer in the rest frame sees.

So in our example, the vehicle will take 50/1.25 or 40 years to make the complete trip and this is what will be indicated on its clock. Similarly, the distance traveled according to the rest frame is 0.6*50 or 30 light-years. But according to the on-board odometer, it has traveled 24 light-years. And the speedometer will have read 24/40 = 0.6c during the trip.

 

[bobc2]

Well then, you never read this post:

Very nice way to explain that, ghwellsjr.

 

[bobc2]

Well, relativity says that one cannot determine which frame of reference is actually moving.
Thus, as a consequence, the whole twin paradox is "meaningless" - i.e. it's a false paradox. It's only a paradox if one has the belief (which is inconsistent with relativity) that one can determine which frame of reference is "actually" moving.

pervect, the twin paradox is not about which one moves with constant velocity. It's about which twin accelerates. Relativity definitely does not say that you cannot determine which twin is accelerating. More to the point, it's about which twin took the shorter world line path through spacetime from the event of the departure to the event of the reunion. That can certainly be determined--and any observer can figure that one out (no matter what reference frame you wish to use).

By the way the reference frame does not move. Just pick a reference frame and then analyze the motion of some object (or observer) with respect to that reference frame.

 

[IsometricPion]

Both of these can be viewed as similar to, Bell's spaceship paradox. I don't think you will find it in a textbook, but a description of the thought experiment should be available on the net.

I found it and its solution (that the string breaks) on wikipedia http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox" .

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

It is not much more difficult to analyse the twin paradox using an inertial frame in which both planets are in motion. The phrase "actually moving" does not have meaning in the context of SR (since it implies there is some coordinate independent notion of (spatial) velocity, which there isn't).

The difference between proper distance/ length and length contracted observation of distance/length. Proper distance/length is the same for all observers. Where a moving frame of reference is involved the proper distance/length appear length contracted. As long as the relative velocity is known, the Lorentz transformations provide the means to calculate the proper distance/length from the observed contracted distance/length.

This is not a new or unique discussion. There are theoreticians on both sides of the issue. My position is that the proper distance/length is the real distance/length.

This was recently discussed on the linked page of https://www.physicsforums.com/showthread.php?t=484405&page=7" thread. As was said there, distance has a well known definition which is not frame invariant. The proper length between two points in space-time is frame invariant and corresponds to the distance between the two points in only one frame. In the frame in which the spaceship is at rest in, e.g., its outbound journey, the proper length defined by the path connecting points simultaneous in frame in which Earth is at rest does have the same value as the distance measured in Earth's rest frame, but it is not a distance (since it has a temporal component).

 

[JesseM]

pervect, the twin paradox is not about which one moves with constant velocity. It's about which twin accelerates.

How are the two different? If one moves at constant velocity and one doesn't, that's exactly equivalent to saying one accelerates and one doesn't.

By the way, pervect doesn't post that much any more but I remember from when he did he was one of the most knowledgeable posters here, if you think he's getting some basic aspect of relativity wrong chances are you're misunderstanding what he's trying to say.

 

[GrayGhost]

JesseM:
I don't disagree that "B can clearly see that A has aged 25 years." I was addressing the point of how to compute the dilation ratio (DR). For even in the classical twin paradox problem it should turn out that every observer "sees" a moving clock as running more slowly than his own.

Eli,

My understanding is that twin B can see the twin A clock ticking faster than his own "while B himself is non-inertial". He has to, for otherwise he could not agree with twin A on their relative aging when they are again co-located. The faster ticking is not the result or proper time speeding up, but only a change in the twin B POV due to frame transitioning, which causes the A-clock to advance faster along its worldline (per B).

GrayGhost

 

[Dale]

Well then, you never read this post:

Yes, I don't know how I missed that one, since I was an active participant on that thread. Good post!

 

[bobc2]

How are the two different? If one moves at constant velocity and one doesn't, that's exactly equivalent to saying one accelerates and one doesn't.

You can tell the difference when the twins reunite and compare proper distances traveled.

By the way, pervect doesn't post that much any more but I remember from when he did he was one of the most knowledgeable posters here, if you think he's getting some basic aspect of relativity wrong chances are you're misunderstanding what he's trying to say.

Either he was not thinking clearly at the moment or I was not. I do that a lot, so I'm probably the culprit (I get a little distracted when my boss starts giving me those quizzical looks--I try not to post very often from my desk). But thanks for the heads-up.

 

[Dale]

I realize you can claim that all frames of reference are equally valid. However, you are talking about every individual inertial frame being as valid as ever other inertial frame. Each one of the traveling twin's frames are as valid as all of the others in the accelerating sequence. So, does that necessarily mean that all of those frames collectively (yielding the bazzar faster-than-light result) are as valid as any individual inertial frame? I don't think so, but I'll need to go back and review Rindler and a couple of others to see if they comment on that aspect. Maybe bcrowell, JesseM, or DaleSpam already have the answer.

The collective frames do have some problems, most notably that there are events in the spacetime where this approach assigns multiple sets of coordinates. This violates one of the defining properties of a coordinate system. However, you can fix this simply by saying that the regions where there is ambiguity are excluded from the coordinate system. This makes your system a bit "choppy" and you cannot analyze physics in areas that are not covered by the chart, but similar things happen in the Rindler and most other non-inertial coordinate systems.

If you do it right, it is OK, but most people using this approach don't recognize that there are hidden pitfalls, don't know that there are equally valid alternatives, and don't realize that there is no standard convention.

 

[SeventhSigma]

So what does it mean if, say, the universe is a hypersphere and going in one direction means you come out the other? e.g. zooming off Earth and then arriving back at Earth without turning around?

 

[GrayGhost]

So what does it mean if, say, the universe is a hypersphere and going in one direction means you come out the other? e.g. zooming off Earth and then arriving back at Earth without turning around?

It does seem to be the case, assuming the universe is a closed unbounded system. Now as to whether the Earth still exists, has yet to begin existing, or what the date-time happens to be upon your arrival, is the big question.

GrayGhost

 

[Gulli]

One handy formula to know when thinking about the relativity of simultaneity is this: if two clocks are synchronized and a distance D apart in their own rest frame, then that means that in the frame of an observer who sees the clocks moving at speed v (parallel to the axis between them), then at a single moment in the observer's frame the clock at the rear will show a time that's ahead of the clock at the front by an amount vD/c^2. So in this example, since the Earth clock and the colony clock are synchronized and a distance of 20 light-years apart in their own frame, then for the spaceman who sees them moving at 0.5c, at any given moment in his frame the two clocks must be out-of-sync by (0.5c)(20 light-years)/c^2 = 10 years, with the rear colony clock ahead of the front Earth clock by that amount. So, at the same moment the Earth clock reads 600,000 AD, the colony clock must read 600,010 AD, according to the definition of simultaneity in the spaceman's rest frame.[\QUOTE]

I'm sorry to bother you again but I'm still confused by that 10 year head start: I mean, I sort of understand why that would happen (it's similar to the Andromeda paradox, which doesn't seem like much of a paradox to me, after all, both observers on Earth still see the same event through a telescope and that event happened in the past anyway), but I don't understand why it doesn't happen the other way around. Why doesn't the colony see the spaceship with a (0.5c)*(17.32LY)/c^2 = 8,66 year head start? And why does the D stand for the distance in the rest frame of a point, does "a point" even have a rest frame?

 

[JesseM]

You can tell the difference when the twins reunite and compare proper distances traveled.

The difference between what? You think there's a difference between "A moved at constant velocity, B did not" and "B accelerated, A did not?" The two statements are exactly equivalent.

 

[JesseM]

One handy formula to know when thinking about the relativity of simultaneity is this: if two clocks are synchronized and a distance D apart in their own rest frame, then that means that in the frame of an observer who sees the clocks moving at speed v (parallel to the axis between them), then at a single moment in the observer's frame the clock at the rear will show a time that's ahead of the clock at the front by an amount vD/c^2. So in this example, since the Earth clock and the colony clock are synchronized and a distance of 20 light-years apart in their own frame, then for the spaceman who sees them moving at 0.5c, at any given moment in his frame the two clocks must be out-of-sync by (0.5c)(20 light-years)/c^2 = 10 years, with the rear colony clock ahead of the front Earth clock by that amount. So, at the same moment the Earth clock reads 600,000 AD, the colony clock must read 600,010 AD, according to the definition of simultaneity in the spaceman's rest frame.

I'm sorry to bother you again but I'm still confused by that 10 year head start: I mean, I sort of understand why that would happen (it's similar to the Andromeda paradox, which doesn't seem like much of a paradox to me, after all, both observers on Earth still see the same event through a telescope and that event happened in the past anyway), but I don't understand why it doesn't happen the other way around.

When I talk about the 10 year head start I'm not talking about what the ship sees visually, but rather what's true in the ship's rest frame (remember that I said the ship doesn't actually see the colony clock reading 600,010 AD until the ship's own clock reads 600,017.32 AD). The colony and the Earth share the same rest frame, so there's no difference between how the Earth defines simultaneity and how the colony defines it; if the Earth thinks the colony clock read 600,000 AD simultaneously with the ship clock reading 600,000 AD, the colony agrees. And naturally all frames must agree that the ship clock read 600,000 AD at the same time the Earth clock read 600,000 AD, since they were both at the same position in space and all frames must agree about localized facts, the relativity of simultaneity only comes in when you consider events at different locations in space.

Why doesn't the colony see the spaceship with a (0.5c)*(17.32LY)/c^2 = 8,66 year head start?

The formula only deals with two clocks at rest relative to each other. If there was a second ship at rest relative to the first one, and the ships were 17.32 LY apart in their mutual rest frame (which would make the distance between them smaller in the Earth/ship frame), then if their clocks were synchronized it would be true in the Earth/ship frame that they were 8.66 years out of sync.

And why does the D stand for the distance in the rest frame of a point, does "a point" even have a rest frame?

? I never said D stands for the distance in the rest frame of "a point", I said the formula applied in a situation where "two clocks are synchronized and a distance D apart in their own rest frame"

 

[Gulli]

When I talk about the 10 year head start I'm not talking about what the ship sees visually, but rather what's true in the ship's rest frame (remember that I said the ship doesn't actually see the colony clock reading 600,010 AD until the ship's own clock reads 600,017.32 AD). The colony and the Earth share the same rest frame, so there's no difference between how the Earth defines simultaneity and how the colony defines it; if the Earth thinks the colony clock read 600,000 AD simultaneously with the ship clock reading 600,000 AD, the colony agrees. And naturally all frames must agree that the ship clock read 600,000 AD at the same time the Earth clock read 600,000 AD, since they were both at the same position in space and all frames must agree about localized facts, the relativity of simultaneity only comes in when you consider events at different locations in space.

I know it's not what he actually sees when he leaves Earth, otherwise he could look into the future. He can only infer a head start through reasoning, like you said, and that's what I meant.

The formula only deals with two clocks at rest relative to each other. If there was a second ship at rest relative to the first one, and the ships were 17.32 LY apart in their mutual rest frame (which would make the distance between them smaller in the Earth/ship frame), then if their clocks were synchronized it would be true in the Earth/ship frame that they were 8.66 years out of sync.

I'm still trying to figure out for myself (intuitively) when I should factor in relativity of simultaneity and when not, your example with a second spaceship will probably help me out.

? I never said D stands for the distance in the rest frame of "a point", I said the formula applied in a situation where "two clocks are synchronized and a distance D apart in their own rest frame"

Sorry, my mistake.