Twin Paradox (thorough explanation needed)

[Gulli]

Hi, I'm an undergraduate physics student trying to comprehend why the twin paradox is not a paradox.

The standard reply usually amounts to this: the dude in the spaceship has to turn around at some point to come back to Earth. So he accelerated during his journey (by changing his direction midway) chopping his journey into two parts, each with its own frame of reference (apparently the acceleration required to get to relativistic speed and slow down again when back at Earth doesn't matter, (maybe because it's along the direction of movement) or maybe it does actually matter and it's exactly one of those subtleties I'm looking for here).

Anyway, the above "explanation" seems a bit like a cop out to me: while it's technically true the spaceman has to change to a new frame of reference when he turns around, this alone is a poor proof that the laws of nature will ensure the frame of reference of Earth is precisely right. The same way that saying the derivative of sin(x) is not 39 does not in itself proof the derivative of sin(x) is actually cos(x). I suspect (maybe wrongly) that there is a more thorough explanation.

To help myself gain more understanding I've prepared the following thought experiment:

In 600.000 AD a spaceman sets out in a spaceship from Earth to a star 20 lightyears away, he will travel at 0.5 c so the journey will take 40 years (to an oberver on Earth, 35 years to the spaceman) give or take (you tell me if the "give or take" part because of acceleration to 0.5c, near Earth, and deecceleration to a complete stop near the star, matters or not). Now, and I stress this, the spaceman DOES NOT return to Earth, he will stay near the star.

Now the spaceman tries to find out what year it is on Earth, he does this in 3 different ways:

1) He tunes into his radio dish, does he hear an Earth news broadcast from 600.020 AD, 600.015 AD or 600.025 AD?

2) Before he left Earth he asked someone to send a probe after him with a clock aboard. The probe travels at 30 km/s (c/10.000) which is safely non-relativistic. He waits until the probe arrives at his star. Does he have to wait 400.000 years, 399.995 years or 400.005 years?

3) A group of humans left Earth in the year 100.000 AD, they traveled to the star at 30km/s and established a colony on a planet orbiting the star. The spaceman decides to visit them and asks them what year it is according to their calendar (which is the same as Earth's). Will they answer 600.040 AD, 600.035 AD or 600.045 AD?

I designed this experiment to cut the U-turn from the original twin paradox. In theory (according to the standard explanation) this means the frames of reference of the spaceman and Earth (or the colony, which is pretty much stationary to Earth) are equally valid. So I hope the outcome of this experiment will help me understand the whole thing better.

I understand the math of SR, why Einstein's postulates lead to time dilation and Lorentz contraction and I even understand the solution to the barn and ladder paradox, but I'm having difficulty comprehending the twin paradox.

 

[Janus]

Hi, I'm an undergraduate physics student trying to comprehend why the twin paradox is not a paradox.

The standard reply usually amounts to this: the dude in the spaceship has to turn around at some point to come back to Earth. So he accelerated during his journey (by changing his direction midway) chopping his journey into two parts, each with its own frame of reference (apparently the acceleration required to get to relativistic speed and slow down again when back at Earth doesn't matter, (maybe because it's along the direction of movement) or maybe it does actually matter and it's exactly one of those subtleties I'm looking for here).

What actually matters here is the separation between the Earth and spaceship when the acceleration is made. (Its a relativity of simultaneity thing.)

Anyway, the above "explanation" seems a bit like a cop out to me: while it's technically true the spaceman has to change to a new frame of reference when he turns around, this alone is a poor proof that the laws of nature will ensure the frame of reference of Earth is precisely right. The same way that saying the derivative of sin(x) is not 39 does not in itself proof the derivative of sin(x) is actually cos(x). I suspect (maybe wrongly) that there is a more thorough explanation.

To help myself gain more understanding I've prepared the following thought experiment:

In 600.000 AD a spaceman sets out in a spaceship from Earth to a star 20 lightyears away, he will travel at 0.5 c so the journey will take 40 years (to an oberver on Earth, 35 years to the spaceman) give or take (you tell me if the "give or take" part because of acceleration to 0.5c, near Earth, and deecceleration to a complete stop near the star, matters or not). Now, and I stress this, the spaceman DOES NOT return to Earth, he will stay near the star.

Now the spaceman tries to find out what year it is on Earth, he does this in 3 different ways:

1) He tunes into his radio dish, does he hear an Earth news broadcast from 600.020 AD, 600.015 AD or 600.025 AD?

600,200 AD. You even imagine that he was listening to that broadcast on the whole way out. As he does he will here the broadcast Doppler shifted by the ratio of
$$\sqrt {\frac{1- \beta }{1+ \beta }}$$

where $\beta$ equals v/c

For 0.5c, this comes out to ~0.577. So in the 35 years that he is listening on the way out, he will hear 20 years of "Earth time" broadcasts.

2) Before he left Earth he asked someone to send a probe after him with a clock aboard. The probe travels at 30 km/s (c/10.000) which is safely non-relativistic. He waits until the probe arrives at his star. Does he have to wait 400.000 years, 399.995 years or 400.005 years?

I assume you mean the total time from the moment that he left Earth, to the time the probe catches up to him at the planet. Then the answer is 399,995 years ( the 35 years he spent traveling by his clock and the 399,960 years he waits while at the planet)

3) A group of humans left Earth in the year 100.000 AD, they traveled to the star at 30km/s and established a colony on a planet orbiting the star. The spaceman decides to visit them and asks them what year it is according to their calendar (which is the same as Earth's). Will they answer 600.040 AD, 600.035 AD or 600.045 AD?

600,040 AD. You can reverse the first Doppler shift application for this one too. When he leaves Earth, he is receiving a radio broadcast from the planet that is dated 599,980 AD ( the signal left 20 years before).
During the trip he hears the broadcast Doppler shifted by

$$\sqrt {\frac{1+\beta }{1-\beta }}$$

or ~1.795

meaning that he will hear 1.795 x 35 = ~60 years worth of "planet time" broadcasts on his trip, so that when he arrives he will be receiving signals with no lag time dated 600,040 AD

I designed this experiment to cut the U-turn from the original twin paradox. In theory (according to the standard explanation) this means the frames of reference of the spaceman and Earth (or the colony, which is pretty much stationary to Earth) are equally valid. So I hope the outcome of this experiment will help me understand the whole thing better.

You can use Doppler shift for the round trip also. On the way out, he listens to 20 yrs of Earth broadcasts and on the way back he listens to 60 yrs worth. So in his round trip of 70 years, he gets 80 years worth of broadcasts from the Earth.

If the Earth is listening to him the Earth listens to 60 x .557 = 35 yrs of Red shifted broadcasts ( since the spaceship is 20 ly distant when it turns around, it takes another twenty years for the Earth to get the last signal sent before the ship turns around.)
And since, from the Earth's view, the spaceship is chasing after it own broadcasts on the return trip, the Earth only hears the blue shifted broadcasts for 20 years, so the Earth hears 20* 1.795 = 35 years worth of ship broadcasts for the return leg. So in 80 years, the Earth receives 70 years of broadcasts from the ship

The difference for the Ship and Earth is the fact that the ship, being the one that changes velocity, see the effect of the change on velocity of the Doppler shift immediately, while the Earth has to wait 20 yrs to see the change in Doppler shift.

I understand the math of SR, why Einstein's postulates lead to time dilation and Lorentz contraction and I even understand the solution to the barn and ladder paradox, but I'm having difficulty comprehending the twin paradox.

The Relativity of simultaneity plays a huge role here.

Remember, if it is the same time on Earth and Planet according to either, it will not be so according to the ship traveling between.

 

[Gulli]

=You can use Doppler shift for the round trip also. On the way out, he listens to 20 yrs of Earth broadcasts and on the way back he listens to 60 yrs worth. So in his round trip of 70 years, he gets 80 years worth of broadcasts from the Earth.=

I think this is a major aha moment for me: I thought that in the traditional twin paradox story the trip to the star and back were considered equivalent because gamma has a v^2 in it and (-v)^2 = v^2. But the Doppler shift (that I hadn't considered here) has a v^1 in it so the Doppler shift (redshift) is different for v and -v (which is of course how astronomers know if a galaxy moves away from us or towards us). So when you're moving towards something you get time shrinkage instead of time dilation? If true I wonder why this isn't the standard reply (it seems more like an actual explanation).

So am I correct in saying that:

During the away trip (from Earth to the star system with the colony):

- the spaceman sees time on Earth go slower

- someone on Earth sees time aboard the spaceship go slower (the spaceman can't actually meet the oberver on Earth while separated by distance and simultaneity is relative anyway so this doesn't bother me)

- someone on the colony sees time on the spaceship go faster (this still bothers me somewhat).

During the return trip:

- the spaceman sees time on Earth go faster

- someone on Earth sees time aboard the spaceship go faster

- someone on the colony sees time aboard the spaceship go slower.

The net result is that things add up when the spaceman gets back to Earth, well, sort of: there is a 10 year time difference but at least the spaceman doesn't think the Earthman aged slower than he did. That solves the traditional twin paradox but leaves me to wonder how the colony and spaceman reconcile their time measurements at the end of the away trip.

 

[Mentz114]

Hi, I'm an undergraduate physics student trying to comprehend why the twin paradox is not a paradox.

The standard reply usually amounts to this: the dude in the spaceship has to turn around at some point to come back to Earth. So he accelerated during his journey (by changing his direction midway) chopping his journey into two parts, each with its own frame of reference (apparently the acceleration required to get to relativistic speed and slow down again when back at Earth doesn't matter, (maybe because it's along the direction of movement) or maybe it does actually matter and it's exactly one of those subtleties I'm looking for here).

I don't know where you got this 'standard' answer but it misses the correct explanation. It is a postulate of special relativity that the time on a clock between two points along a worldline is the proper length of the worldline between those points. This makes it possible to calculate the elapsed time on any clock if we know the worldline and the events.

The observation of moving clocks has little to do with it - changing frames, accelerating etc are just aspects of the worldline.

If you are a student of relativity, learn about proper time and all the traveller 'paradoxes' are explained. The Wiki article is quite good

http://en.wikipedia.org/wiki/Proper_time

 

[Gulli]

I don't know where you got this 'standard' answer but it misses the correct explanation. It is a postulate of special relativity that the time on a clock between two points along a worldline is the proper length of the worldline between those points. This makes it possible to calculate the elapsed time on any clock if we know the worldline and the events.

The observation of moving clocks has little to do with it - changing frames, accelerating etc are just aspects of the worldline.

If you are a student of relativity, learn about proper time and all the traveller 'paradoxes' are explained. The Wiki article is quite good

http://en.wikipedia.org/wiki/Proper_time

It's from "Introduction to Electrodynamics" by David J. Griffiths, and I'll look into that wikipedia article.

 

[Mentz114]

It's from "Introduction to Electrodynamics" by David J. Griffiths, and I'll look into that wikipedia article.

That book has a good reputation so I'm surprised proper time is not mentioned. Maybe it's introduced later. The Minkowski spacetime is defined by the infinitesimal relation

ds² = c²dt²-dx²-dy²-dz²

ds is the infinitesimal proper length, and it is invariant under Lorentz transformation ( i.e. change of frame ). The integral of this along a worldline is also invariant, so it is the one quantity that all inertial observers will agree on.

 

[bobc2]

I like Mentz114 comments on this. Follow his lead (follow the world lines) and you will get there. Here are a couple of sketches that may or may not make sense to you. A stay-at-home twin is at rest in the black frame of reference, and his world line (red vertical line) is straight along the X4 axis. The traveling twin moves along the right-slanted world line (blue), then returns along the left slanted world line (blue). The world line distances along the traveler's world line (blue) are calibrated using the hyperbolas (in accordance with Mentz114's distance expression).

The blue dots (proper distance increments) correspond to proper distance marks along the red vertical world line. Note that the traveler blue proper distance traveled is less than the stay-at-home proper distance, even though the line lengths on the computer screen are the opposite. And of course an increment of proper distance along a world line is ds = c(dT) where T is proper time.

Summary: The traveling twin takes a short cut through spacetime (takes a shorter world line path), therefore he has aged less when reuniting with his twin.

 

[yogi]

I don't know where you got this 'standard' answer but it misses the correct explanation. It is a postulate of special relativity that the time on a clock between two points along a worldline is the proper length of the worldline between those points. This makes it possible to calculate the elapsed time on any clock if we know the worldline and the events.

The observation of moving clocks has little to do with it - changing frames, accelerating etc are just aspects of the worldline.

http://en.wikipedia.org/wiki/Proper_time

The standard answer is given over and over again on these forums and in texts and articles on the internet - while it is true actual experiments almost always involve acceleration at some point in the journey - it is not the cause of the age difference but simply a circumstance of the twin returning to his origin - The actual cause of the age difference is due to the fact that the twins have taken different paths through spacetime - once the selection is made as to which fame is to be taken "at rest," then the result determined ...the moving object will necessarily have to travel a distance equal to the quadrature composite of some space interval and some time interval - whereas the stay put twin only travels a temporal interval - so since both intervals are the same - the stay put twin's time interval must be greater than the traveling twins time interval.

 

[JesseM]

while it is true actual experiments almost always involve acceleration at some point in the journey - it is not the cause of the age difference but simply a circumstance of the twin returning to his origin

"Cause" is a vague word, but in SR the fact that one twin accelerates between meetings while the other does not is a necessary and sufficient condition for the twin that accelerated to be younger when they reunite.

once the selection is made as to which fame is to be taken "at rest," then the result determined

It's irrelevant which frame you take to be "at rest", all frames calculate the same answer for the ages of the twins when they reunite.

 

[Gulli]

That book has a good reputation so I'm surprised proper time is not mentioned. Maybe it's introduced later. The Minkowski spacetime is defined by the infinitesimal relation

ds² = c²dt²-dx²-dy²-dz²

ds is the infinitesimal proper length, and it is invariant under Lorentz transformation ( i.e. change of frame ). The integral of this along a worldline is also invariant, so it is the one quantity that all inertial observers will agree on.

Proper time is explained in the book. It just doesn't say it is the answer to the twin paradox. Perhaps I have to explain myself better: I'm not looking for a way to calculate the age difference (I can do that), I'm looking for a way to comprehend the result. Janus' comments on Doppler shift have been really helpful but I'm not quite there yet since I'm still unsure about what happens midway.

 

[Gulli]

I like Mentz114 comments on this. Follow his lead (follow the world lines) and you will get there. Here are a couple of sketches that may or may not make sense to you. A stay-at-home twin is at rest in the black frame of reference, and his world line (red vertical line) is straight along the X4 axis. The traveling twin moves along the right-slanted world line (blue), then returns along the left slanted world line (blue). The world line distances along the traveler's world line (blue) are calibrated using the hyperbolas (in accordance with Mentz114's distance expression).

The blue dots (proper distance increments) correspond to proper distance marks along the red vertical world line. Note that the traveler blue proper distance traveled is less than the stay-at-home proper distance, even though the line lengths on the computer screen are the opposite. And of course an increment of proper distance along a world line is ds = c(dT) where T is proper time.

Summary: The traveling twin takes a short cut through spacetime (takes a shorter world line path), therefore he has aged less when reuniting with his twin.

That's nice but I'm trying to understand why I can't just invert your picture? Why can't I have the spaceman have his world line along the X4-axis and reverse your whole story? Especially for the away trip from Earth towards the midpoint with the star and the colony I should be able to reverse the roles of the observer on the colony and the spaceman as I please.

 

[Gulli]

error

 

[robphy]

That's nice but I'm trying to understand why I can't just invert your picture? Why can't I have the spaceman have his world line along the X4-axis and reverse your whole story? Especially for the away trip from Earth towards the midpoint with the star and the colony I should be able to reverse the roles of the observer on the colony and the spaceman as I please.

No Lorentz transformation will ever straighten out the kink in the spaceman's worldline.

Alternatively,
Suppose the Earth observer constructs a spacetime diagram by stacking his lines [hyperplanes] of simultaneity. If the spaceman tries the same procedure, something funny happens at the turn... his completed "diagram" will have two-sets of coordinates for some events and some other events in the Earth diagram won't appear in the spaceman's diagram. That makes the spaceman distinguishable from the inertial observers (like the Earth observer).

 

[Gulli]

No Lorentz transformation will ever straighten out the kink in the spaceman's worldline.

Alternatively,
Suppose the Earth observer constructs a spacetime diagram by stacking his lines [hyperplanes] of simultaneity. If the spaceman tries the same procedure, something funny happens at the turn... his completed "diagram" will have two-sets of coordinates for some events and some other events in the Earth diagram won't appear in the spaceman's diagram. That makes the spaceman distinguishable from the inertial observers (like the Earth observer).

And what if there's no return? Then there is no kink, though there is still the fact that the spaceman thinks the distance is 17 ly while the people on the colony think the distance is 20 ly (which probably fixes things, exactly how, I am trying to figure out).

 

[I_am_learning]

The easy solution is:
Since the spacecraft isn't in inertia frame of reference (because it has to accelerated/decelerate), you have to take the Earth's frame of reference as the only available inertial frame of reference. If you make calculations on that frame, then the space twins comes younger.
Of course, you can't do the calculation from space twins frame and derive Earth twin younger because space twin isn't in inertial frame.

(there is however, methods to do calculation from space twins frame, by spliting his frame to two inertial frames, 1 going away, and 2 coming back. If you use this method, the space twin again comes younger by equal amount)

So acceleration/deceleration is the key, that creates asymmetry and hence resolves the paradox.

I got this feeling after a long struggle.

 

[Dale]

And what if there's no return?

Then it is not the twin's "paradox".

 

[Mike_Fontenot]

It the traveler changes his velocity (relative to the home twin) during his trip, so that he becomes stationary wrt her, but is still distantly separated from her, then they will be in complete agreement about their respective ages, for as long as their relative velocity remains zero.

The home-twin (who is perpetually inertial) will of course easily determine that the traveler is younger than she is (assuming that they were the same age when the traveler originally left her). And the traveler will completely agree with that (as long as their relative velocity remains zero).

But at any instant during the life of the traveler, when his velocity relative to the home-twin is NOT zero (according to the home-twin), they will disagree about their respective ages (whenever they are not co-located). So, for example, immediately BEFORE the traveler decelerates when he gets to the distant planet, he will say that he is older than the home-twin.

Mike Fontenot

 

[Gulli]

I think it all comes down to the following.

With the distance between Earth and the colony being 20 lightyears according to Earth and the colony, the speed of the spaceship being 0.5c (according to everyone), making the distance 17.32 lightyears according to the spaceman, and the departure date of the spaceman being 600,000 AD (according to Earth).

Take the halfway problem (the spaceman doesn't return to Earth just yet).

The colony looks at it this way. The colony starts receiving the signal of the spaceman's departure from Earth in 600,020 AD (colony years), and then has to wait 20 years before the spaceman is at the same location in space as the colony, during which spaceman signals are "compressed" (Doppler shift), so they receive 1.73x20=35 years of signals, expecting the spaceman to have experienced 35 years. In other words: they think time went slower for the spaceman by a factor 1/gamma. This is the correct result.

The spaceman looks at it this way. He starts receiving signals from the colony's "departure" of 600,000 AD (colony years), in his year 600,017.32 AD (spaceman years). He then waits for another 17.32 years (spaceman years). He receives 1.73x17.32=30 years of signals, so when the colony is at the same location in space as the spaceman, he expects the colony to have experienced 30 years (colony years), or in other words, he thinks time went slower on the colony by a factor 1/gamma, this is wrong, I don't yet understand why.

The twin paradox is basically just the above, and then back again.

If someone can tell me what went wrong in my calculation of the number of years the spaceman thinks went by on the colony (should be 40 instead of 30) I think that would crack the problem for me personally.

 

[Gulli]

But at any instant during the life of the traveler, when his velocity relative to the home-twin is NOT zero (according to the home-twin), they will disagree about their respective ages (whenever they are not co-located). So, for example, immediately BEFORE the traveler decelerates when he gets to the distant planet, he will say that he is older than the home-twin.

Mike Fontenot

But that breaks down when the traveler (spaceman) reaches the colony: he has to agree with the colonists because they never left Earth's frame of reference so the traveler knows for sure colony time is the same as Earth time. That's why I split up the problem in my post directly above.

So are you saying the traveler witnesses a massive warping of time on the colony during his decceleration? A warping that makes him go from being older than a colonist born in the same (Earth/colony year) as him to being younger than that colonist? Because that strikes me as really odd.

 

[Mike_Fontenot]

But that breaks down when the traveler (spaceman) reaches the colony: he has to agree with the colonists because they never left Earth's frame of reference so the traveler knows for sure colony time is the same as Earth time. That's why I split up the problem in my post directly above.

Once the traveler has zero velocity relative to the Earth (and relative to the colony, since the colony is always stationary wrt the earth), he is in complete agreement with BOTH the Earth and the colony, about the correspondence between his age and his home-twin's age (and about the correspondence between his age and any colonist's age).

So are you saying the traveler witnesses a massive warping of time on the colony during his deceleration? A warping that makes him go from being older than a colonist born in the same (Earth/colony year) as him to being younger than that colonist? Because that strikes me as really odd.

It depends on exactly what you mean by "witnesses", or "sees". This is a perpetual source of misunderstandings in special relativity. What the traveler directly sees (for example, through a telescope, or on a TV image) of his home-twin, or of a colonist, does NOT directly tell him how old they currently are. He must make a correction for the finite speed of those signals. Or, he can get the same answer by just using the Lorentz equations.

The traveler will CONCLUDE that his home-twin ages suddenly during his deceleration, but the AMOUNT of that age-swing depends on his distance (according to her) from his home twin during his deceleration. If his deceleration starts when he is fairly close to the colony, he won't conclude that the people in the colony age much during his deceleration. In the simple, idealized case where his velocity changes are instantaneous, he will conclude that the colonists don't age at all during his deceleration, because his separation from them is zero then.

Here is a posting that specifies how separation enters into the disagreements about how ages correspond:

https://www.physicsforums.com/showpost.php?p=2934906&postcount=7 .

(That posting refers to the standard twin "paradox" scenario, but you can probably see how to apply it to the case where the traveler remains at the turnaround location).

Mike Fontenot

 

[bobc2]

That's nice but I'm trying to understand why I can't just invert your picture? Why can't I have the spaceman have his world line along the X4-axis and reverse your whole story? Especially for the away trip from Earth towards the midpoint with the star and the colony I should be able to reverse the roles of the observer on the colony and the spaceman as I please.

I can do that. Bur first, here is the situation where we use a rest frame in which each of the twins starts out moving in opposite directions at relativistic speeds with respect to the black coordinates. The numbers along the world lines for red and for blue represent calibrated proper distances (or proper times). The proper distances on the trip outbound are the same in this reference frame for both red and blue. But, after blue's turnaround, he is taking a shorter path to reunite with red.

I haven't really changed anything, just chose a different reference frame (I should have had the Earth flying off to the left instead of putting the stay-at-home twin in a red rocket--but you get the point). The blue twin is still taking an overall shorter path through spacetime (his short return path) and therefore ages less.

 

[OnlyMe]

But at any instant during the life of the traveler, when his velocity relative to the home-twin is NOT zero (according to the home-twin), they will disagree about their respective ages (whenever they are not co-located). So, for example, immediately BEFORE the traveler decelerates when he gets to the distant planet, he will say that he is older than the home-twin.

But that breaks down when the traveler (spaceman) reaches the colony: he has to agree with the colonists because they never left Earth's frame of reference so the traveler knows for sure colony time is the same as Earth time. That's why I split up the problem in my post directly above.

So are you saying the traveler witnesses a massive warping of time on the colony during his decceleration? A warping that makes him go from being older than a colonist born in the same (Earth/colony year) as him to being younger than that colonist? Because that strikes me as really odd.

The original twin paradox included only two frames of reference. One assumed to be at rest, "the Earth frame" and there other, in motion, "the traveling twin's frame". In this hypothetical, the traveling twin has to have returned home, only so the clocks on Earth and the ship can be once again compared (assuming they were synchronized before the ship departed the earth).

There are two parts to the paradox, assuming only these two frames of reference. The first part is from the perspective of "relative velocity", meaning each twin will see the other as moving. They will both see the other's clock as running slow, or fast depending on the direction of the relative motion. The second part has to do with how the velocity of an object affects length contraction and time dilation. This second part can only be separated from the first by "knowing" which observer is in motion and which is at rest.

There are two ways to determine which is in motion. The first requires that the traveling twin return to the Earth at some point so that the two clocks can be compared. The second is to determine which twin experienced acceleration. The twin on Earth has a hypothetical velocity = 0, while the traveling twin's velocity > 0.

The paradox is that both twins observe the other as moving, length contracted and time dilated and their self as unaffected. Only by introducing some method of knowing which is in motion can it be determined which is length contracted and time dilated.

Adding a third frame of reference, the colony, complicates this thought experiment. While you can assume that their is no relative difference in velocity between the Earth and the colony and that they can have their clocks synchronized, each represents a different perspective of the ship. While the ship is in motion the observers on the Earth and on the colony will not agree, as to their observation of the ship. One will see the ship as time running slow, while the other sees time running fast. The observer on the ship will see the same thing as each observer when viewing that observer.

When the ship stops at the colony, the clock on the ship can be compared to the colony clock. At that point the ship's clock will be found to have run slow. If you could communicate those results back to the earth, the Earth would also see the ships clock as having run slow compared to the colony clock and it's own, when taking the light time difference between the Earth and the colony into consideration.

In both of these thought experiments, acceleration and deceleration only help to determine which observer was in real motion within the hypothetical. Comparing clocks and knowing which is accelerates are just two ways to determine which observe is or was in motion in the hypothetical.

 

[Janus]

But that breaks down when the traveler (spaceman) reaches the colony: he has to agree with the colonists because they never left Earth's frame of reference so the traveler knows for sure colony time is the same as Earth time. That's why I split up the problem in my post directly above.

Not if he doesn't stop. It's the Relativity of Simultaneity again. As long as he is moving at 0.5c relative to the Earth and Planet and along the trajectory he had when traveling from Earth to planet, he will detemine that it is ~10 years later at the planet than it is at the Earth. Thus when he leaves Earth (after accelerating) it will, according to him, be 600,000 AD on Earth and 600,010 AD on the planet. As he travels to the planet, both will advance in age by 30 yrs during his 35 yr trip, so that when he arrives at the planet it will be 600,040 AD on the planet and 600,030 AD on Earth.

So are you saying the traveler witnesses a massive warping of time on the colony during his decceleration? A warping that makes him go from being older than a colonist born in the same (Earth/colony year) as him to being younger than that colonist? Because that strikes me as really odd.

The time the traveler witnesses on the planet does not change from before deceleratio to after deceleration (assuming his movement with respect to the planet is negliable during this time). Thus it will be 600,040 AD on the planet both before and after. During deceleration, the Earth will go from 600,030 AD to 600,040 AD, as the ship transitions from an inertial frame movign at 0.5c relative to the Earth and Planet to one at rest with respect to them.

 

[Gulli]

I can see there are many ways to resolve the paradox, I'm almost there myself, I just need the answer to my question in post #18 (my attempt to make both frames of references as equal as possible).

The time the traveler witnesses on the planet does not change from before deceleratio to after deceleration (assuming his movement with respect to the planet is negliable during this time). Thus it will be 600,040 AD on the planet both before and after. During deceleration, the Earth will go from 600,030 AD to 600,040 AD, as the ship transitions from an inertial frame movign at 0.5c relative to the Earth and Planet to one at rest with respect to them.

You mean, his idea of time on Earth will have to go from 600,020 AD to 600,040 AD during the decceleration? So if decceleration takes a second he'll see Earth grow 20 years older in a second (must be quite a sight)? I guess I can make sense of that by imagining a ray of light (carrying the image of an older Earth) catches up with him when he deccelerates. Could you help me out by answering my question in post #18?

 

[Janus]

I can see there are many ways to resolve the paradox, I'm almost there myself, I just need the answer to my question in post #18 (my attempt to make both frames of references as equal as possible).



You mean, his idea of time on Earth will have to go from 600,020 AD to 600,040 AD during the decceleration? So if decceleration takes a second he'll see Earth grow 20 years older in a second (must be quite a sight)? I guess I can make sense of that by imagining a ray of light (carrying the image of an older Earth) catches up with him when he deccelerates. Could you help me out by answering my question in post #18?

He won't actually "see" Earth aging rapidly, What he'll see is the Doppler shift going away as his speed begins to match the Earth's. He'll "see" the Earth go from aging slowly to aging at normal speed. It is his determination as to what time it is on Earth at any given instant that will "jump forward".

 

[Gulli]

He won't actually "see" Earth aging rapidly, What he'll see is the Doppler shift going away as his speed begins to match the Earth's. He'll "see" the Earth go from aging slowly to aging at normal speed. It is his determination as to what time it is on Earth at any given instant that will "jump forward".

Right, he won't actually see rapid aging but because he now sees the distance to Earth as 20 lightyears (instead of the 17.32 lightyears he saw it as while at 0.5c), his calculations of what year it should be on Earth do go forward (not just age less slowly)?

Can you tell me what went wrong with my calculation in post #18?

 

[OnlyMe]

With the distance between Earth and the colony being 20 lightyears according to Earth and the colony, the speed of the spaceship being 0.5c (according to everyone), making the distance 17.32 lightyears according to the spaceman, and the departure date of the spaceman being 600,000 AD (according to Earth).

Part of the confussion appears to be the difference between distance and length. Distance does not change depending on how fast you are going. The Lorentz transformations apply to the length contraction of a moving object and time dilation resulting from motion.

Below is a link to a calculator for length contraction, time dilation and relativistic mass. Here is a quote from the explanation of the first calculator (for length contraction of a moving object),

The length of any object in a moving frame will appear foreshortened in the direction of motion, or contracted. The amount of contraction can be calculated from the Lorentz transformation. The length is maximum in the frame in which the object is at rest.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

Measurements of distance is between two points and is not time dependent. Any individual measurement of distance is between two points. If there is relative motion between two observers two consecutive measurements of the distance, separated by a given time, can provide the relative velocity.

Time will be experienced by all observers as constant and uniform. Only by comparing once synchronized clocks will a difference be apparent. Two observers will "see" each other equally time dilated and length contracted.

 

[JesseM]

Part of the confussion appears to be the difference between distance and length. Distance does not change depending on how fast you are going. The Lorentz transformations apply to the length contraction of a moving object and time dilation resulting from motion.

On the contrary, Lorentz contraction applies to distance as well. If you have two objects at rest relative to each other and a distance D apart in their mutual rest frame, then to an observer who is moving at speed v relative to those objects (in a direction parallel to the axis between them), in that observer's own frame the distance between them is reduced to $$D * \sqrt{1 - v^2/c^2}$$

 

[Gulli]

Part of the confussion appears to be the difference between distance and length. Distance does not change depending on how fast you are going. The Lorentz transformations apply to the length contraction of a moving object and time dilation resulting from motion.

Below is a link to a calculator for length contraction, time dilation and relativistic mass. Here is a quote from the explanation of the first calculator (for length contraction of a moving object),
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

Measurements of distance is between two points and is not time dependent. Any individual measurement of distance is between two points. If there is relative motion between two observers two consecutive measurements of the distance, separated by a given time, can provide the relative velocity.

Time will be experienced by all observers as constant and uniform. Only by comparing once synchronized clocks will a difference be apparent. Two observers will "see" each other equally time dilated and length contracted.

This is where I get really confused. I thought the traveler had to see the distance contracted (of course he knows the distance is longer in the frame of reference of the planet he took off from), he had to see the distance contracted so he wouldn't get the idea he was going faster than light when flying through space at a relativistic speed.

For example:

Imagine the speed of the ship is 0.9 c and the distance as seen from Earth is 10 lightyears, then someone on Earth (or at the end point) would expect the ship to make the journey in 11.111 years. For v=0.9 gamma is 2.294, so to someone aboard the ship the journey would take only 4.84 years. This means that he would be going faster than light, unless he sees the distance contracted to 4.36 lightyears, than he would conclude he travels at 0.9, like he should.

Distance contraction going hand in hand with time dilation actually makes sense to me, even at an intuitive level (as far as that's possible with relativity), because it ensures the speed of light will be the same to everyone (which is one of Einstein's postulates). So I'm pretty sure that part of my post #18 is correct, but there has to be a flaw somewhere else.

 

[Eli Botkin]

Gulli, first you need to give up thinking of the twin-paradox age difference as resulting from a turn-around acceleration. It’s possible to reform the problem, leaving out the “acceleration” phase, and still get an age difference when the twins meet after initially being the same age. You need to accept that this is how our universe’s space and time are structured.

You likely accept without question the fact that R2 = X2 + Y2 leaves R unchanged under a coordinate-frame rotation of (X, Y). [The 2's are superscripts.] You need also to accept that S2 = (cT)2 – X2 leaves S unchanged under a boost (i.e., a coordinate-frame velocity change in the X direction).

This universal behavior of time (you may be aware from your study of SR) is a result of there being a universal upper limit, c, to relative velocity. That’s really it in a nutshell. The rest is just dressing to help you accept it.

My own favorite description of why time behaves so differently from our expectation is shown in each twin’s plot of the other twin’s clock during the entire trip, from departure to return. Each twin is receiving a televised image of the other twin’s clock. The image-clock time is plotted (on the y-axis) against the receiver’s-clock time (on the x-axis). Do this for both twins (on the same plot-paper) and you will see why the clock times at return CANNOT be the same.

 

[JesseM]

I think it all comes down to the following.

With the distance between Earth and the colony being 20 lightyears according to Earth and the colony, the speed of the spaceship being 0.5c (according to everyone), making the distance 17.32 lightyears according to the spaceman, and the departure date of the spaceman being 600,000 AD (according to Earth).

Take the halfway problem (the spaceman doesn't return to Earth just yet).

The colony looks at it this way. The colony starts receiving the signal of the spaceman's departure from Earth in 600,020 AD (colony years), and then has to wait 20 years before the spaceman is at the same location in space as the colony

An additional 20 years after they start receiving signals from the departure, yes...so they'll see the spaceman arrive in 600,040 AD.

during which spaceman signals are "compressed" (Doppler shift), so they receive 1.73x20=35 years of signals, expecting the spaceman to have experienced 35 years.

Yes, according to the relativistic Doppler shift formula, if the spaceman is sending signals once per year of his own time, they will receive 1.732 signals per year, for a total of 34.64 signals.

The spaceman looks at it this way. He starts receiving signals from the colony's "departure" of 600,000 AD (colony years), in his year 600,017.32 AD (spaceman years).

Yes, assuming his clock reads 600,000 AD at the moment the Earth "departs", and we use the label "E" to refer to the event on the colony's world-line that is simultaneous in his frame with the event of the Earth departing from him (when both his clock and the Earth clock read 600,000 AD), then he won't see the light from event E until 600,017.32 AD. But note that because of the relativity of simultaneity, the event E is not simultaneous with his departure from Earth in the colony's own frame. Instead, the event E on the colony occurs when the colony's own clock reads 600,010 AD.

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.

He then waits for another 17.32 years (spaceman years). He receives 1.73x17.32=30 years of signals, so when the colony is at the same location in space as the spaceman, he expects the colony to have experienced 30 years (colony years), or in other words, he thinks time went slower on the colony by a factor 1/gamma, this is wrong, I don't yet understand why.

No, it's not wrong, it's correct. He sees the event E where the colony clock reads 600,010 AD, then he sees 30 years worth of signals before reaching the colony, at which point he can see that the colony clock does indeed read 600,010 + 30 = 600,040 AD.

 

[JesseM]

Gulli, first you need to give up thinking of the twin-paradox age difference as resulting from a turn-around acceleration. It’s possible to reform the problem, leaving out the “acceleration” phase, and still get an age difference when the twins meet after initially being the same age. You need to accept that this is how our universe’s space and time are structured.

If you have no acceleration phase, then the two twins can't depart from a common location and later reunite at a common location. You could just have two observer approaching inertially without ever having met in the past, but then the relativity of simultaneity would mean that the two twins wouldn't agree on their relative ages prior to the moment they meet, so they wouldn't agree on which one had aged more slowly.

 

[Eli Botkin]

JesseM:

Start with the twins separated, with a closing velocity, and with their clocks synchronized by twin-A to the same reading. When they meet twin-B’s clock will read less than twin-A’s.

If twin-B does the synchronization, then twin-A’s clock will read less than twin-B’s.

 

[Gulli]

@JesseM

So there is an additional boost of 10 years that fixes everything. Am I correct in saying that the invariance (the fact that a similar boost doesn't manifest when we consider the colony stationary, and well, the whole reason the colony and the spaceman experience different time intervals) stems from the fact that we are working with a starting point and ending point which share the same frame of reference, namely that of one of the observers (the one on the colony)?

The 10 year boost then is something the spaceman notices because of the relativity of simultaneity: he's closer to the starting point, Earth, than he is to the colony when he sets out, so he sees a time difference between Earth and the colony, even though Earth and the colony share the same frame of reference and someone located halfway between them would say Earth and the colony have the same time, to which the people of Earth and the colony would agree. Am I right?

 

[JesseM]

JesseM:

Start with the twins separated, with a closing velocity, and with their clocks synchronized by twin-A to the same reading. When they meet twin-B’s clock will read less than twin-A’s.

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.