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## Twin Paradox (thorough explanation needed)

 Quote by 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.
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.

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 Quote by 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 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.

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.

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 Quote by Gulli 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?

 Quote by Gulli 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.
 Quote by Gulli 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.
 Quote by Gulli 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)

 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.

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 Quote by Gulli 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?
 Quote by Gulli 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?
 Quote by Gulli 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.

 Quote by JesseM 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.

 Quote by JesseM 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.

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 Quote by Gulli 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 space ship 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.

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 Quote by Gulli 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.
 Quote by Gulli 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.

 Quote by Gulli 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.

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 Quote by Gulli 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.

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 Quote by OnlyMe 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.
 Quote by OnlyMe 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?

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 Quote by Gulli 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 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.

 Quote by JesseM 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.

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