Twin Paradox (thorough explanation needed)

[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.

 

[JesseM]

@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 spaceman stationary, and well, the whole reason the colony and the spaceman experience different time intervals)

What do you mean "doesn't manifest when we consider the spaceman stationary"? When I said that the station clock would be ahead of the Earth clock by 10 years, I was specifically considering how simultaneity works in the frame where the spaceman is stationary (i.e. in the spaceman's rest frame, the event of the Earth clock reading 600,000 AD is simultaneous with the event of the station clock reading 600,010 AD). In the rest frame of the Earth and station, their clocks are synchronized, by assumption--in their frame, the station clock reads 600,000 AD simultaneously with the Earth clock reading 600,000 AD.

Anyway, if you want a totally symmetrical situation, imagine that behind the spaceman is a second spaceman #2 traveling at the same velocity relative to the Earth/station, and whose distance from spaceman #1 is 20 light-years in the spacemens' rest frame and 17.32 light-years in the Earth/station frame. In that case, assuming the clocks of the spaceman are synchronized in their own rest frame, in the Earth/station frame the clock of spaceman #2 will read 600,010 AD at the same moment the clock of spaceman #1 reads 600,000 AD (also the moment he passes Earth). And to complete the symmetry, when spaceman #2 passes the Earth, spaceman #2's clock reads 600,040 AD while Earth's clock reads 600,034.64 AD, just like how when spaceman #1 passes the station, the station's clock reads 600,040 AD while spaceman #1's clock reads 600,034.64 AD.

 

[Gulli]

What do you mean "doesn't manifest when we consider the spaceman stationary"? When I said that the station clock would be ahead of the Earth clock by 10 years, I was specifically considering how simultaneity works in the frame where the spaceman is stationary (i.e. in the spaceman's rest frame, the event of the Earth clock reading 600,000 AD is simultaneous with the event of the station clock reading 600,010 AD). In the rest frame of the Earth and station, their clocks are synchronized, by assumption--in their frame, the station clock reads 600,000 AD simultaneously with the Earth clock reading 600,000 AD.

Anyway, if you want a totally symmetrical situation, imagine that behind the spaceman is a second spaceman #2, whose distance from spaceman #1 is 20 light-years in the spacemens' rest frame and 17.32 light-years in the Earth/station frame. In that case, assuming the clocks of the spaceman are synchronized in their own rest frame, in the Earth/station frame the clock of spaceman #2 will read 600,010 AD at the same moment the clock of spaceman #1 reads 600,000 AD (also the moment he passes Earth). And to complete the symmetry, when spaceman #2 passes the Earth, spaceman #2's clock reads 600,040 AD while Earth's clock reads 600,034.64 AD, just like how when spaceman #1 passes the station, the station's clock reads 600,040 AD while spaceman #1's clock reads 600,034.64 AD.

I'm sorry, I meant "when we consider the colony stationary". And yes, now that I think of it the distance is also invariant between the two perspectives: the colony thinks it's 20 lightyears, the spaceman thinks it's 17.32 lightyears.

So we have two invariances:

1) 20 lightyears is explicitly defined as the distance in the frame of reference of the colony and Earth, the spaceman measures 17.32 lightyears: it doesn't matter if we consider the spaceman stationary or the planets, the spaceman will always think the distance is 17.32 lightyears, the people on the planets will always think it's 20 lightyears.

2) The starting and ending points are in the frame of reference of the colony and Earth: it doesn't matter if we consider the spaceman stationary or the planets, everyone will agree these two points are in the frame of reference of the planets.

One or both of these lead to the situation not being entirely invertable, preventing a paradox.

 

[JesseM]

I'm sorry, I meant "when we consider the colony stationary".

Well, I'm not sure what you mean when you say "the invariance ... 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)?" What do you mean by "the invariance"? Invariance of what? Does my note about the symmetry of the situation if we add a second spaceman #2 behind spaceman #1 (without changing anything else about the scenario) help answer your question?

edit: I see you added a bit to your post:

So we have two invariances:

1) 20 lightyears is explicitly defined as the distance in the frame of reference of the colony and Earth, the spaceman measures 17.32 lightyears: it doesn't matter if we consider the spaceman stationary or the planets, the spaceman will always think the distance is 17.32 lightyears, the people on the planets will always think it's 20 lightyears.

Yes, since the Earth and station are at rest wrt each other, their distance will be constant in every inertial frame.

2) The starting and ending points are in the frame of reference of the colony and Earth: it doesn't matter if we consider the spaceman stationary or the planets, everyone will agree these two points are in the frame of reference of the planets.

The starting and ending points are events (the event of the spaceman and Earth being at the same position, and the event of the spaceman and station being at the same position), which don't "belong to" any particular frame. The spaceman would say that the starting and ending points are both at the same position but at different times.

One or both of these lead to the situation not being entirely invertable, preventing a paradox.

I don't know what you mean by "invertible", as I said you can just add a second spaceman and then everything is symmetrical under the exchange spaceman #1<-->Earth and spaceman #2<-->station (i.e. if you write an account of what happens, then you do these name-substitutions while leaving every other part of the account unchanged, the account is still accurate)

 

[Gulli]

I don't know what you mean by "invertible", as I said you can just add a second spaceman and then everything is symmetrical under the exchange spaceman #1<-->Earth and spaceman #2<-->station (i.e. if you write an account of what happens, then you do these name-substitutions while leaving every other part of the account unchanged, the account is still accurate)

By being "invertible" I mean the names can be swapped and nothing would change. Obviously this is not the case with the spaceman traveling from Earth to the colony: the spaceman thinks time went faster on the colony, the colony thinks time went faster on the spaceship, but only one of them can be right in the end. So there has to be something in the problem that determines who's right, something that tips the balance.

 

[JesseM]

By being "invertible" I mean the names can be swapped and nothing would change.

Well, obviously the mere fact that the Earth and station are at rest relative to each other while the spaceman is moving relative to both means if you only have those three, you can't swap around the names (this has nothing to do with relativity, it would be true in the Newtonian version of the scenario too). Do you agree that if we add a second spaceman #2 traveling behind spaceman #1, then the scenario does become invertible?

Obviously this is not the case with the spaceman traveling from Earth to the colony: the spaceman thinks time went faster on the colony, the colony thinks time went faster on the spaceship,

You mean "thinks time went slower" in both cases, surely?

but only one of them can be right in the end.

But they are both "right" in the sense that they both make correct predictions, as long as you take into account the relativity of simultaneity. In the colony's frame it took 40 years between the event of the spaceman leaving Earth and the event of the spaceman reaching the colony, but the spaceman only experienced 34.64 years; in the spaceman's frame it took 34.64 years between these events, but the colony only experienced 30 years, since in the spaceman's frame the colony's clock started reading 600,010 AD at the moment the spaceman left Earth and its clock read 600,040 AD when the spaceman reached the colony.

 

[Gulli]

Well, obviously the mere fact that the Earth and station are at rest relative to each other while the spaceman is moving relative to both means if you only have those three, you can't swap around the names (this has nothing to do with relativity, it would be true in the Newtonian version of the scenario too). Do you agree that if we add a second spaceman #2 traveling behind spaceman #1, then the scenario does become invertible?

You mean "thinks time went slower" in both cases, surely?

No, because of the Doppler effect signals will be "compressed" (as Janus explained on page 1 of this thread) so they see each other's clocks run faster.

But they are both "right" in the sense that they both make correct predictions, as long as you take into account the relativity of simultaneity. In the colony's frame it took 40 years between the event of the spaceman leaving Earth and the event of the spaceman reaching the colony, but the spaceman only experienced 34.64 years; in the spaceman's frame it took 34.64 years between these events, but the colony only experienced 30 years, since in the spaceman's frame the colony's clock started reading 600,010 AD at the moment the spaceman left Earth and its clock read 600,040 AD when the spaceman reached the colony.

Of course their math has to work out in the end if they take relativity into effect, otherwise the universe would explode, however the fact remains one of them sees the belief they held "mid-flight" vindicated, the other one won't.

 

[OnlyMe]

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}

Parallel to takes the moving observer out of the coordinate system of the two objects. The observer then can "see" the two objects and the distance between them and the distance would then appear length contracted. That is not the same as being length contracted.

 

[Janus]

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?

Here's a couple of S-T diagrams that might help.

The first shows things according to the spaceship while traveling from Earth to planet, the green line is the Earth worldline, the blue line the ship's and the red line the planet's. The yellow lines represent light signals.

The years start at zero.

The second one shows things from the Earth rest frame (and consequently the frame the ship ends up in after decelerated at the planet. Just imagine that the blue line merges with the red line after they meet)

Notice how the signals from Earth are closed more closely together after deceleration than before and how the light signal sent from Earth at the 20 yr mark not only travels further but had to have left earlier according to the ship after deceleration than it does before acceleration.

 

[JesseM]

No, because of the Doppler effect signals will be "compressed" (as Janus explained on page 1 of this thread) so they see each other's clocks run faster.

OK, but when people talk about time going slower they are normally talking about what's true in a given frame, not what you see visually. After all, even in classical Newtonian mechanics the signals from an oncoming vehicle are compressed in this way, but people wouldn't normally say in a classical context that the vehicle's clock is running faster, it just looks (or sounds, if you're listening to beeps from the clock) like it's running faster. Talk about clocks running slower or faster pretty much always refers to time dilation effects where the clock elapses a different time between two points on its worldline than the coordinate time between those points.

Of course their math has to work out in the end if they take relativity into effect, otherwise the universe would explode, however the fact remains one of them sees the belief they held "mid-flight" vindicated, the other one won't.

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.

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.

 

[OnlyMe]

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.

The traveler will always experience time as being normal. It is not until arriving at one of the destinations that he/she will become aware that time was running slow on his/her clock. The only time the ship is in a position to measure the distance between the Earth and the colony is when it is on the Earth or the colony.

In order to measure the distance between the Earth and the colony, while moving, the ship would have to be in a frame of reference from which the earth, the colony and the distance between them can be treated as a single unit.

None of the observers will know which clocks are running slow until the ship lands and their clocks can be compared.

The ship on Earth measures the distance to the colony at 10 LY. The ship takes off and half way to the colony measures the remaining distance, it turns out to be 5 LY. From within the moving frame of reference you cannot observe the time dilation, length contraction or relativistic mass of the object in motion. Time is experienced as proceeding at a constant and normal rate.., until you land on the colony or back on Earth and find your clocks don't match.

 

[Dale]

the fact remains one of them sees the belief they held "mid-flight" vindicated, the other one won't.

Only if they mistakenly believed something false at the mid point.

 

[JesseM]

Parallel to takes the moving observer out of the coordinate system of the two objects.

What do you mean? Say the two objects use a coordinate system where the Earth is at rest at x=0 and the station is at rest at x=20. Then if the ship is moving parallel to the axis between them, that means it's moving along the x-axis of their coordinate system, for example at t=0 the ship is at x=0, then at t=10 the ship is at x=5, at t=20 the ship is at x=10, etc.

The observer then can "see" the two objects and the distance between them and the distance would then appear length contracted. That is not the same as being length contracted.

No, it's not just about visual appearances, the distance between them really is shorter in the spaceship's own rest frame. Do you know how to use the Lorentz transformation to compare the coordinates of events in different reference frames?

 

[Cleonis]

So there has to be something in the problem that determines who's right, something that tips the balance.

As others have mentioned: what we know is that for the one who travels a longer spatial distance a smaller amount of proper time will elapse.

The two travelers can follow convoluted paths. You can compute the spatial length of each path by integrating along the worldline. When the travelers rejoin you can compare the spatial lengths:Of course, the principle of relativity of inertial motion is prime. I mention the principle of relativity of inertial motion because it may appear as if that principle prevents you from assessing which traveller traveled the longest spatial distance. But that is not the case.

A thought experiment:
Two spaceships, equipped with very accurate clocks and very accurate accelerometers, and the crews of the ships are well aware of the principles of special relativity.
They part from each other, and they agree where and when they will rejoin. Since both crews are good at special relativity calculations they can at all times figure out how much proper time has elapsed for themselves compared to time on a ship that just stays put. So they can agree where and when they will rejoin; it's just that they need to figure in special relativity effects.

The most basic form of moving is to not accelerate at all. Just coasting along. Then your motion is just moving in time. A simple case is that the two spaceship crews agree beforehand that they will rejoin at the point in space and time where a ship would be after one week of proper time of that ship, if it would not accelerate at all.

The onboard accelerometers allow something analogous to http://en.wikipedia.org/wiki/Dead_reckoning" . At all times the accelerometer readings allow you to reconstruct your velocity relative to the point of departure. Your trip will have stages with different acceleration, in different directions.

Two ships can plot different courses, and return to the common point of departure. If the navigators do their math well then two spaceships are able to rejoin, using only the accelerometer reading based "dead reckoning".

When the two spaceships have rejoined you can compare the course plots, and see which one has traveled a longer spatial distance. Note especially that what the comparison yields is a difference. You can figure out who has traveled a longer distance, and how much longer. You make no statements about how much distance has been traveled by each respective ship. By the principle of relativity of inertial motion you cannot make any such statement.
But you are not prevented from saying something about how the spatial distances traveled compare.Special relativity states that in such a scenario the reckoned difference in spatial distance traveled and difference in amount of elapsed proper time follows a law.
Special relativity does not explain why things happen that way. All that special relativity does is describe that that is what will happen.

 

[OnlyMe]

No, it's not just about visual appearances, the distance between them really is shorter in the spaceship's own rest frame. Do you know how to use the Lorentz transformation to compare the coordinates of events in different reference frames?

Yes, in this context the Lorentz transformations can be used to reconcile the perceived contraction of a length or distance, with the same distance as measure from a frame of reference at rest relative to the length or distance.

 

[JesseM]

Yes, in this context the Lorentz transformations can be used to reconcile the perceived contraction of a length or distance, with the same distance as measure from a frame of reference at rest relative to the length or distance.

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.

 

[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.