Twin Paradox (thorough explanation needed)

[OnlyMe]

Right, so hopefully you retract your earlier statement that lengths work differently from distances--they both contract in exactly the same way in a frame where the two objects (or two ends of the same object, in the case of length) are moving.

There distinction is that you cannot move a distance. It is defined by the location of its end points. So length contraction of a distance is a perceived condition that does not change the actual distance. Length contraction of a distance can only be observed from a moving observer.

Length contraction and time dilation as measured for an object in motion, given a stationary observer and an observer in motion, looks the same to both observers. Both think the other is length contracted and running slow. Once they meet up and compare clocks they find that one clock is slow and the other unaffected. The velocity of the object/observer in motion results in length contraction and time dilation. The time dilation is supported by experiment and observation. Length contraction has no such confirmation that I am aware of but its close connection to time dilation suggests that it too is real for the object in motion, while it is in motion.

 

[Gulli]

Which one do you think won't have their mid-flight belief vindicated? The spaceman believes that at the moment he left Earth the station clock read 600,010 AD (as evidenced by the fact that it was 17.32 light-years away at the moment he left Earth, and 17.32 years later he sees an image of the station clock reading 600,010 AD), and he believes the station clock is running slower, so isn't this belief vindicated when he reaches the station and the station has only added 30 years to the time he thinks it read when he left Earth, while his own clock has moved forward by 36.64 years since leaving Earth? He has no reason to think the station clock will actually be behind his when he arrives, since he thinks it had that "head start" of 10 years.

Right, yeah well, that head start still confuses me a bit. But I understand now why it has to happen (only accounting for the travel time of the ray of light would leave simultaneity intact, which can't be).

Also, you seem to be avoiding my question about what happens when we simply add a second spaceship behind the first one, without changing anything else. Do you agree the scenario becomes totally invertible with this addition? If so, it seems that any statement about one observer's view being "vindicated" must be invertible as well.

Yes, that would mean there are two references in either frame of reference, two ships in one and two planets in the other and provided the distance between the ships seems 20 lightyears to the ships and 17.32 lightyears to the planets this would render the whole thing completely symmetrical and it illustrates what it is exactly that can break the symmetry, make the problem not invertible: namely, removing one of the planets or one of the ships.

So if we have two ships in the same frame of reference who drifted apart at non-relativistic speeds and then have them fly past a planet at relativistic speed (but at rest in relation to each other) the people on the planet will notice the crew of the second ship (which are clones of the crew on the first ship) has aged faster than them.

The traditional twin paradox then is actually the same as my thought experiment run twice, because the turning point is some point defined by Earth, stationary to Earth, it's not a planet, but that doesn't matter.

 

[JesseM]

There distinction is that you cannot move a distance. It is defined by the location of its end points.

What does "move a distance" mean? Can you explain what it would mean to "move a length"? If you just mean that an object like a rod can be moving in your frame, well, if you're talking about a "distance" between two planets or something then the planets can also be moving in your frame.

So length contraction of a distance is a perceived condition that does not change the actual distance.

It doesn't change the distance in the rest frame of the endpoints, but neither does length contraction of a rod change the length in the rest frame of the rod. In both cases different frames disagree about the length/distance, and all frames are assumed to be equally valid in relativity so there's no reason to call the distance/length in the rest frame the "actual" distance/length.

Length contraction of a distance can only be observed from a moving observer.

You understand that "moving" is relative, that if an observer is moving relative to planets there's no objective truth about whether the planets are at rest and the observer is moving, or if the observer is at rest and the planets are moving, right? Anyway, contraction of the distance between the planets can only be observed by an observer moving relative to the planets, and likewise contraction of the length of a rod can only be observed by an observer moving relative to the rod. So again I don't see why you are differentiating between length and distance here, anything you say about one is true of the other as well.

Length contraction and time dilation as measured for an object in motion, given a stationary observer and an observer in motion, looks the same to both observers. Both think the other is length contracted and running slow.

Do you mean each observer has a ruler and clock at rest relative to himself, and each measures the length and tick rate of the other one's ruler and clock? If so yes, both measure the other one's ruler to be contracted and the other one's clock to be running slow. It's likewise true that if each observer has a pair of objects beside them, each one measures the distance between the other one's pair of objects to be contracted.

Once they meet up and compare clocks they find that one clock is slow and the other unaffected.

Only if they started out at the same position and reunited at the same position could they agree on whose clock elapsed less time, which would require one of them to accelerate, and whichever one accelerated would be the one whose clock elapsed less time. If they were simply approaching each other at constant velocity, then they wouldn't be able to agree on whose clock had ticked more slowly, because of the relativity of simultaneity. For example, if they were approaching each other from afar observer A might say in his frame that his clock read 30 years "at the same time" that B's clock read 34 years, so if by the time they met their clocks both read 50 years, observer A would say B's clock only ticked forward by 16 years in the time his clock ticked forward by 20 years, so B's clock was only ticking at 0.8 times the rate of his own. But then B would say that A's clock read 30 years "at the same time" B's own clock read 25 years, due to his different definition of simultaneity, meaning that if their clocks both read 50 years when they met, B would say A's clock only went forward by 20 years in the time his clock went forward by 25, so B would say A's clock was only ticking at 0.8 times the rate of his own. As long as both are moving at constant velocity there is no way to break the symmetry and say that one clock is "really" running slower than the other.

The velocity of the object/observer in motion results in length contraction and time dilation. The time dilation is supported by experiment and observation. Length contraction has no such confirmation that I am aware of but its close connection to time dilation suggests that it too is real for the object in motion, while it is in motion.

Again, hopefully you understand that "in motion" has no objective meaning, that different inertial frames disagree about who is "stationary" and who is "in motion" and therefore disagree about which objects are more length contracted and which clocks are more time dilated? And that all inertial frames are equally valid, they all make identical predictions about local events like what two clocks read at the moment they meet at a single point in space?

 

[Dale]

Once they meet up and compare clocks they find that one clock is slow and the other unaffected.

No, once they meet up and compare clocks they will find that they are both running at the same rate. Neither is slower than the other while they are together. Rods and clocks are the same in that sense.

 

[Eli Botkin]

Yes, but in this case twin B will say in his frame that their clocks were not initially synchronized, that in fact twin A's clock started at a time well ahead of twin B's clock, and that this explains why twin B's clock reads less when they meet despite the fact that twin A's clock was running slower than B's in this frame. That was my only point, that without the two clocks starting and ending at the same location, there is no frame-independent fact about which ran slower on average throughout the journey.

JesseM:
Yes, if twin-A does the synchronization, then Twin-B will believe twin-A's clock to be advanced, but not by any initial difference that could account for their final difference.

The point is that this demonstrates time dilation occurs for clocks in relative motion irrespective of any relative acceleration. The classical twin paradox problem, where the twins are initially together, tends to lead the SR novice to think that acceleration leads to the difference. It doesn't. Time dilation is built into the structure of spacetime for relative motion in general.

 

[Eli Botkin]

Yes, but in this case twin B will say in his frame that their clocks were not initially synchronized, that in fact twin A's clock started at a time well ahead of twin B's clock, and that this explains why twin B's clock reads less when they meet despite the fact that twin A's clock was running slower than B's in this frame. That was my only point, that without the two clocks starting and ending at the same location, there is no frame-independent fact about which ran slower on average throughout the journey.

JesseM:
Yes, if twin-A does the synchronization, then Twin-B will believe twin-A's clock to be advanced, but not by any initial difference that could account for their final difference.

The point is that this demonstrates time dilation occurs for clocks in relative motion irrespective of any relative acceleration. The classical twin paradox problem, where the twins are initially together, tends to lead the SR novice to think that acceleration leads to the difference. It doesn't. Time dilation is built into the structure of spacetime for relative motion in general.

 

[OnlyMe]

No, once they meet up and compare clocks they will find that they are both running at the same rate. Neither is slower than the other while they are together. Rods and clocks are the same in that sense.

Yes, they will both run at the same rate. The ship's clock will not agree with the Esther stationary clock.

 

[JesseM]

JesseM:
Yes, if twin-A does the synchronization, then Twin-B will believe twin-A's clock to be advanced, but not by any initial difference that could account for their final difference.

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.

 

[OnlyMe]

Short cuts are never a good idea.

The thought experiment has been changed over and over, which makes it difficult to arrive at any reasonable conclusion. So, what I will comment on is the following...

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 from the opposite direction 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.

Quote from: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
"The length of any object in a moving frame will appear foreshortened in the direction of motion, or contracted. ..."

This applies whether the observer is moving relative to the object or the object is moving relative to the observer. If there is an observer in both the moving frame of reference and the stationary frame, they will both see the other as moving relative to theirselves and length contracted (i.e. "foreshortened in the direction of motion, or contracted).

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

---------

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

 

[JesseM]

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. There are also other methods which would give equivalent ways of assigning coordinates to events, like one based on sending out radar signals and seeing how long it takes them to return. 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. If you wish to reject the idea that the observer can assign coordinates to events in his frame, then you are basically rejecting the whole structure of SR!

Quote from: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
"The length of any object in a moving frame will appear foreshortened in the direction of motion, or contracted. ..."

Yes, and in the object's own rest frame the object will appear longer, neither frame's perspective is any more "true" than the other's, and length is inherently a frame-dependent matter.

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"

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?

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?

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!

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

Of course it is. Can you give a numerical example? Pick some frame of reference which uses coordinates x,t, and define the velocity of each planet in this frame, as well as their initial positions at t=0. If they are moving "with respect to one another" then it's easy to show that the distance between them changes in this frame, if they both have the same velocity relative to this frame then the distance between them will be constant in this frame, but just using the Lorentz transformation you can show the distance in this frame is shorter than it is in the frame where both planets are at rest. If you don't know enough about the math of SR to give such an example, then you really shouldn't go making such confident statements about a theory whose basic technical details you don't understand!

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.

Sorry but this is nonsense, again if you give a simple numerical example it'll be very easy to show that what you are saying doesn't match with the math of the Lorentz transformation.

 

[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.

 

[Eli Botkin]

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

GrayGhost:
The phrase “twin B can see the twin A clock ticking faster than his own” can be misunderstood. I expect, though, that in this instance your SR expertise is not meaning to interpret that as “seeing through a telescope.” Rather I assume you mean the time that B assigns to events on A’s worldline based on B’s knowledge of the Lorentz transformation equations. If so, then I maintain that an observer will always assign fewer ticks to a moving clock than he experiences on his own clock. This keeps the dilation ratio < 1 even in the instantaneous inertial frames within a period of acceleration. The ratio would rise to 1 as the relative velocity approaches 0, and then fall again.

This assigned time and its resulting dilation rate are not measurable quantities. To see why there is agreement when they are again co-located, I suggest that the more usual meaning of “see” be used. If either observer views the other clock through a telescope, they will first see a slower clock than their own followed by a faster clock (Doppler). [Both of course see the same Doppler shifts.] B (who does the accelerating) will see the two Doppler-shifted periods of A’s clock-images for equal periods on his own (B’s) clock. However, A will see the slower image of B’s clock for a longer time (on A’s clock) than he sees the faster portion. Therefore, at get-together, B’s clock reads less than A’s. And they’ve both been witness to why this has occurred.

Thanks for “listening.”

 

[Mentz114]

If so, then I maintain that an observer will always assign fewer ticks to a moving clock than he experiences on his own clock.

All observers agree on the number of ticks. A tick is an event and events cannot be 'transformed away'. The interval between the ticks is observer dependent. I don't think it affects your argument if you amend the phrase in question.

 

[GrayGhost]

GrayGhost: The phrase “twin B can see the twin A clock ticking faster than his own” can be misunderstood. I expect, though, that in this instance your SR expertise is not meaning to interpret that as “seeing through a telescope.”

Yes, you're right. This is the 2nd time I've stated this poorly in this forum, as it's a bad habit when posting too quickly. Thanx. The faster ticking A-clock (per B) is ascertained from looking at the date/time, track, and navigation data after the fact. The faster ticking clock is "determined" (not seen) by analysis of the data while ignoring the relativistic effects during analysis ... which seems to be the way most folks look it unfortunately (I believe that an incorrect approach).

You're right, the light signals received from twin A will be doppler shifted, and so "the images" of the moving clock can show a clock ticking faster or slower than yor own "due to doppler effects". To know the real relative rate of 2 clocks, the spacetime dilation needs to be determined (and accounted for) from its corresponding relativistic doppler effects for twin B to know the correct A-velocity as it applies to relativity. If NOT DONE, then twin B will determine that twin A moved superluminally and the A-clock ticked (at times) faster than his own per himself. However IF DONE, then the A-velocity will always be luminal (or less) as it should be, and the A-clock will never tick faster than Bs per B.

At any rate, thanks for the correction there Eli.

GrayGhost

 

[Gulli]

@JesseM

I think I understand now.

The Earth and the colony share the same frame of reference and so agree with each other on when an event took place. The colony does not have to agree on simultaneity with the spaceman but they have confirmation from Earth (because Earth and the spaceman were at the same position during the event of the spaceship speeding by) that the spaceship passed them in 600.000. So when the signal from the spaceship speeding past Earth reaches the colony in 600.020 the signal will read "it is now 600.000 and a spaceship just passed us", so no head start. The spaceman does not have to agree on simultaneity with the Earth: the spaceman and Earth agree that the spaceship passed Earth in what they both consider to be 600.000, but the spaceman does not have to agree this event was simultaneous to the year 600.000 on the colony and can therefore infer a head start. The value of the head start depends on the relative velocity between the Earth/colony frame and that of the spaceman, and on the distance between Earth and the colony.

Thanks for this piece of enlightenment!

 

[JesseM]

@JesseM

I think I understand now.

The Earth and the colony share the same frame of reference and so agree with each other on when an event took place. The colony does not have to agree on simultaneity with the spaceman but they have confirmation from Earth (because Earth and the spaceman were at the same position during the event of the spaceship speeding by) that the spaceship passed them in 600.000. So when the signal from the spaceship speeding past Earth reaches the colony in 600.020 the signal will read "it is now 600.000 and a spaceship just passed us", so no head start. The spaceman does not have to agree on simultaneity with the Earth: the spaceman and Earth agree that the spaceship passed Earth in what they both consider to be 600.000, but the spaceman does not have to agree this event was simultaneous to the year 600.000 on the colony and can therefore infer a head start. The value of the head start depends on the relative velocity between the Earth/colony frame and that of the spaceman, and on the distance between Earth and the colony.

Thanks for this piece of enlightenment!

Yup, sounds like you've got it! Glad I could help.

 

[Eli Botkin]

All observers agree on the number of ticks. A tick is an event and events cannot be 'transformed away'. The interval between the ticks is observer dependent. I don't think it affects your argument if you amend the phrase in question.

Mentz114:
You are, of course, correct to say ” All observers agree on the number of ticks” on a worldline.

I should not carelessly assume and then omit clarifying phrases.
I should have written:
If so, then I maintain that an observer will always assign fewer ticks to a moving clock during any sequence of ticks on his own clock. “Fewer ticks” is what is implied by the term “time dilation.

Said another way: Time on a moving clock is “slower “ when compared to the “non-moving” clock. And again: dt/dT < 1 if t is time measured on the moving clock and T is time measured on the “non-moving” clock. Of course, the two clocks must be equivalent in every way (except for where they’re placed).

 

[Gulli]

@JesseM

So then am I correct in assuming the twin paradox has the exact same solution as my spaceman/colony problem because in the twin paradox the traveler returns and so has to turn around somewhere and thus passes a point where his velocity (well, at least the radial component) in relation to Earth is zero and that this point is analogous to the colony?

 

[DTThom]

Everyone is quick to teach and slow to learn. The tens of thousands of documents purporting to explain the Twins Paradox illustrates this as well as anything.

In the process of being so quick to teach, the purveyors of these documents have overlooked the simple and obvious truth about clock rates.

Why do people make things so difficult for themselves? Did it never occur to them that when two reunited clocks (meaning they are now once again at the same place-moment) show an ACTUAL disparity in their recorded time, that there must necessarily have been an ACTUAL difference in clock rates involved while they were in relative motion with each other?

One should never suggest (as they so often do) that there was some sort of "jump in time" involved with the change of inertial frame (meaning at the turn-around point). The simple act of starting a clock as an inbound astronaut passes an outbound astronaut cannot possibly create a "jump in time". (Remember, the outbound astronaut hands off his clock reading to the inbound astronaut.)

The time contraction formula [t' = t * sqr rt of (1 - v^2)] is not linear. That is why the party who changes frames to bring the two parties back together will register the least amount of time on his clock with the symmetry of the situation preserved.

The actual distances and speeds relative to the universe will vary depending on which party changes frames, but the parties involved cannot possibly detect that. That is in keeping with the postulates and deductions of special relativity.

Time-keeping, distance and speed are interminably bound in one equation. Therefore, actual differences in clock rates implies actual length contraction dependent on actual speed relative to the universe. Actual length contraction works in combination with actual time-keeping contraction to preserve the symmetry of measures across inertial frames.

There is clock functioning at every level, dependent on actual light speed, at even the atomic level. Our observations and measuring paradigms of every nature are constrained by the speed of light, as is our "synchronizing" of clocks.

Special relativity can be charted out in actual terms (absolute terms), where light speed is constant in an actual sense. All the results of special relativity, including the consistent measured speed of light, fall naturally into place when charting these actualities against the (experimentally undetectable) rest state of the universe.

Actual time-keeping and length contraction arise naturally from the fact that all phenomena are dependent on the speed of light, which is itself invariant in actuality, being massless.

Consider that A.P. French writes on page 150 of Special Relativity: "Note, though, that we are appealing to the reality of A's acceleration, and to the observability of the inertial forces associated with it. Would such effects as the twin paradox exist if the framework of fixed stars and distant galaxies were not there? Most physicists would say no. Our ultimate definition of an inertial frame may indeed be that it is a frame having zero acceleration with respect to the matter of the universe at large."

And I feel very sorry for any physicist who doesn't understand that.

Michio Kaku states on page 80 of Einstein's Cosmos that bringing the twins together "determines which twin was "really" moving."

Martin Gardner writes on page 114 of Relativity Simply Explained: "There is one all important difference between the relative motion of the astronaut and the relative motion of the stay-at-home. The stay-at-home does not move relative to the universe."

Both Kaku and Gardner were using the simplest of twins paradox scenarios, in which one party is assumed to be at rest with the cosmos. But that need not be the case. There can be any number of "in between" situations, leading to a lesser time differential. It is also not necessary for the twins to reunite to determine which one was "really moving". The noted asymmetry (noted by both parties) in the time-keeping difference builds incrementally, beginning at the moment of inertial change for one party, when radio or light signals are regularly sent forth and back to check on current clock status.

One should do a search on Einstein's clock synchronization, and its bearing on spacetime diagrams. He or she will find that the notorious "jump in time" is built into that clock synchronization, because it is a one way synchronization, which gets instantly replaced with a different synchronization when a new inertial frame is adopted.

There is all the difference of night and day between predicting and explaining. We can use Einstein's clock synchronization and spacetime to predict a time differential, but we must look at relativity in the universal frame of reference to explain not only that time differential, but also all the mutually symmetrical measures made across inertial frames.

The preceding remarks were copied off my copyrighted web document.

By the way, "observations" and "measures" are strictly synonomous, and constrained by light speed. You might think that a "visual observation" is something different than measuring, but the combination of eye to brain processing is precisely a form of measuring. Identically, biological aging is precisely synonomous with clock functioning of all types, right down to the atomic level and things such as the Doppler effect.

Also, the following phrase is a meaningless muddling of terminology: ".. the distance between them really is shorter in the spaceship's own rest frame.

"Really" means "really", a reality independent of anyone's inertial frame.

You may call it the "God's eye view" or the "view from a higher dimension". Such view is the instantaneous view, free from the constraint of light speed, which is finite.


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"First they tell you you're wrong and they can prove it;
then they tell you you're right but it isn't important;
then they tell you it's important but they knew it all along."

Charles Kettering, former head of General Motors
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[DTThom]

I need to clarify one paragraph in the mid portion of my previous post. Here is the first half of that paragraph, better stated:

Both Kaku and Gardner were using the simplest of twins paradox scenarios, in which one party is assumed to be at rest with the cosmos, and the other party both has motion relative to the universe and changes frames. But that need not be the case. There can be any number of "in between" situations -- such as both parties having motion relative to the universe, with still only one party changing frames (same time differential); or both parties changing frames, leading to a lesser time differential.

 

[Dale]

Hi DTThom, welcome to PF!

Why do people make things so difficult for themselves? Did it never occur to them that when two reunited clocks (meaning they are now once again at the same place-moment) show an ACTUAL disparity in their recorded time, that there must necessarily have been an ACTUAL difference in clock rates involved while they were in relative motion with each other?

Consider two cars that travel from Miami to New York, one travels via Washington DC and the other travels via Denver. When the two cars are reunited in New York they show an ACTUAL disparity in their recorded mileage. Does that necessarily imply that there must necessarily have been an ACTUAL difference in their odometer rates? No, both odometers could still have accurately measured one mile per mile and yet obtained different odometer readings.