Can Relativity Explain the Confusion of the Twins Paradox?

[leright]

I am having trouble understanding how the explanation of the paradox solves the problem. What if people in two different frames were moving wrt one another with constant velocity. Say for instance, these people never turn around to meet up once again and are constantly moving at a constant velocity wrt one another.

Clearly, each will percieve the other's frame to be moving more slowly than their own. There is never a change in reference frame so they disagree on who is older. Who is actually older? Are they both younger and both older simultaneously? That seems to be a weird consequence of relativity. What am I missing here?

 

[leright]

Also, I know the speed of light is constant in all inertial frames. However, it does not necessarily have to be constant in accelerated frames, correct? I believe this because the postulate of relativity states that "all laws of physics are the same in all inertial reference frames", so some laws of physics are not the same in accelerated reference frames, and therefore the speed of light is not necessarily constant in all reference frames, but it is only constant in inertial frames?

 

[HallsofIvy]

I am having trouble understanding how the explanation of the paradox solves the problem. What if people in two different frames were moving wrt one another with constant velocity. Say for instance, these people never turn around to meet up once again and are constantly moving at a constant velocity wrt one another.

Clearly, each will percieve the other's frame to be moving more slowly than their own. There is never a change in reference frame so they disagree on who is older. Who is actually older? Are they both younger and both older simultaneously? That seems to be a weird consequence of relativity. What am I missing here?

Each observes the other as being older. Since they are never in the same frame of reference, there is no paradox there. There is no such thing as "actually older" unless they can be compared in the same frame of reference.

Remember that two events being "simultaneous" only applies for two immediately adjacent events in the same frame of reference. You cannot compare the "actual" ages because you cannot compare when each observes the other.

Also, I know the speed of light is constant in all inertial frames. However, it does not necessarily have to be constant in accelerated frames, correct? I believe this because the postulate of relativity states that "all laws of physics are the same in all inertial reference frames", so some laws of physics are not the same in accelerated reference frames, and therefore the speed of light is not necessarily constant in all reference frames, but it is only constant in inertial frames?

Yes, that is true. The fact that the speed of light is the same constant in all inertial frames of reference applies only to inertial frames. In fact, there is no good way to measure the speed of light in a non-inertial frame.

 

[jtbell]

Remember that two events being "simultaneous" only applies for two immediately adjacent events in the same frame of reference.

To clarify: replace "simultaneous" with "absolutely simultaneous." That is, if two events are at the same location, then all observers (no matter how they are moving) will agree about whether those two events are simultaneous or not.

Two events that are not at the same location can be simultaneous in at most one inertial reference frame. In all other inertial reference frames, the events are not simultaneous.

Most of the apparent paradoxes involving length contraction and time dilation can be resolved by taking this relativity of simultaneity into account.

 

[leright]

Each observes the other as being older. Since they are never in the same frame of reference, there is no paradox there. There is no such thing as "actually older" unless they can be compared in the same frame of reference.

Remember that two events being "simultaneous" only applies for two immediately adjacent events in the same frame of reference. You cannot compare the "actual" ages because you cannot compare when each observes the other. Yes, that is true. The fact that the speed of light is the same constant in all inertial frames of reference applies only to inertial frames. In fact, there is no good way to measure the speed of light in a non-inertial frame.

Nobody has proposed a means of calculating what the speed of light would be in a certain non-inertial frame? Yes, there is no way to test these calculations so doing such mat would be meaniningless, but I have seen people do physics that cannot be proven (string theory?) Why is the speed of light any different?

 

[leright]

Each observes the other as being older. Since they are never in the same frame of reference, there is no paradox there. There is no such thing as "actually older" unless they can be compared in the same frame of reference.

Remember that two events being "simultaneous" only applies for two immediately adjacent events in the same frame of reference. You cannot compare the "actual" ages because you cannot compare when each observes the other.


Yes, that is true. The fact that the speed of light is the same constant in all inertial frames of reference applies only to inertial frames. In fact, there is no good way to measure the speed of light in a non-inertial frame.

So, basically, they can think they are both the older twin when they are apart, but when they come together one will in fact be older and the other will be younger.

 

[leright]

btw, I tried to determine the time dilation effect for accelerated frames (assuming light speed is the same for them) by taking the time dilation formula and replacing the deltas with differentials and then I replaced the v in the formula with at for constant acceleration, and simply integrated from 0 to t0. I got delta(t) = (c/a)(inversesin(a*(deltatnaught)/c)).

Clearly, this only works when the argument of the sine is between -1 and +1, so a cannot generally be larger than c, unless the acceleration is for a very small amt of time (one second), and if c=a then the acceleration can only occur for a time arbitrarily smaller than one second.

 

[JesseM]

btw, I tried to determine the time dilation effect for accelerated frames (assuming light speed is the same for them) by taking the time dilation formula and replacing the deltas with differentials and then I replaced the v in the formula with at for constant acceleration, and simply integrated from 0 to t0. I got delta(t) = (c/a)(inversesin(a*(deltatnaught)/c)).

Integrating $$\sqrt{1 - v(t)^2/c^2} \, dt$$ will give you the amount of time elapsed on an accelerating clock with a variable velocity v(t), but as seen in a single inertial frame where the clock has this velocity function v(t), not as seen in an accelerated frame--I'm not sure if that's what you meant by "the time dilation effect for accelerated frames" or not. However "constant acceleration" in relativity usually means the G-force experienced by observers on the ship is constant, which is equivalent to saying that the acceleration measured in the inertial frame where the ship is instantaneously at rest at a given moment will be the same from one moment to another; if this is the case, then the acceleration as seen in a single fixed inertial frame will not be a*t for constant acceleration, see the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page for the correct formulas in this case.

 

[yogi]

So, basically, they can think they are both the older twin when they are apart, but when they come together one will in fact be older and the other will be younger.

Not necessarily - you have to define the experiment so there is an assymetry - if two separated stationary clocks are synchronized and each quickly accelerated to the same velocity so they travel at a contant speed until they meet each other at some point, each will conclude that the other guys clock is running slow during the flight - but when they meet and stop - both clocks will read the same - now compare this with the case where only one clock moves - when it reaches the other clock, the one that has been put in motion will have accumulated less time - what is the difference - the clock put in motion will have undergone both a temporal and spatial increment in his world line - the clock that has not moved will only have experienced a temporal change during the experiment (a straight world line) - in both cases the spacetime interval is the same - but the moving clock has both a temporal and space interval that must be taken into account - so the temporal increment must of necessity be less for the clock that is put in motion

 

[HallsofIvy]

So, basically, they can think they are both the older twin when they are apart, but when they come together one will in fact be older and the other will be younger.

And the point is that, in order for the twins, who have been moving apart, to come back together, one or both must accelerate- so the laws of special relativity no longer apply.

 

[M1keh]

Okay. So Bill stays on the Earth and Bob flies away at 0.6c (180,000kms)for 1 Earth hour. Bob's watch shows a time 12 minutes slower than Bill's after this hour ? Time dilation factor of 0.8 ?

If Bob then turns back and travels at 0.6c for another Earth hour, his watch would be 24 mins behind Bill's when he returns. Is this correct ?

Thanks.

 

[HallsofIvy]

Not necessarily. What happens to his clock during acceleration and decelleration? You would have to specify exactly how the acceleration-decelleration work.

 

[yogi]

Okay. So Bill stays on the Earth and Bob flies away at 0.6c (180,000kms)for 1 Earth hour. Bob's watch shows a time 12 minutes slower than Bill's after this hour ? Time dilation factor of 0.8 ?

If Bob then turns back and travels at 0.6c for another Earth hour, his watch would be 24 mins behind Bill's when he returns. Is this correct ?

Thanks.

You do not have to turn around and return to the starting point to experience time dilation - in fact most of the experiments are one way - e.g. pion created in the lab accelerated to a high velocity will exhibit time dilation as it travels a straight path - if the twin stops at the end of his outbound journey and compares his watch to one on earth, it will be found to have logged less time - the traveling clock can be stopped at any point and compared to a local clock at that point and it will be found to be out of sync therewith - every clock at rest wrt the Earth will read the same as an Earth clock - it is in the Earth rest frame - the twin exercise is best treated as two one way experiments - half the time is lost on the outward journey - the other half is lost on the inward journey - and while you can get the correct result by considering the way each twin observes things during the three phases of the trip, it is easier to comprehend the trip by breaking it into separate events

In Einsteins original publication, he first considers one way motion - then a round trip ... later in 1918 he did a 180 and decided that the twin paradox could best be explained in terms of a pseudo G field that one or the other twin experiences during turn around - this has led several prominent authors to claim general relativity is necessary to explain the twins difference in ages - its of little wonder there are so many different views on the subject

 

[M1keh]

... - the traveling clock can be stopped at any point and compared to a local clock at that point and it will be found to be out of sync therewith -

Thanks. The travel & come-back thing is just so that the two can compare watches when they meet again. They could presumably travel in a circle at 0.6c in opposite directions, never changing their own speed, only the relative speeds ? Presumably, the speed would have to be something other than 0.6c to give the same results, but the exact figures aren't really that important.

Surely describing one of the clocks as 'travelling' is not accurate within the framework of relativity ? Surely both clocks are 'travelling' in relation to each other ? Otherwise one clock's f.o.r. is more important than the other's ?

If Bill stays on Earth, he's still 'travelling' at the same speed as Bob ? His clock should be 24 minutes earlier than Bob's ?

So which is it & why ?

 

[jtbell]

Surely describing one of the clocks as 'travelling' is not accurate within the framework of relativity ? Surely both clocks are 'travelling' in relation to each other ? Otherwise one clock's f.o.r. is more important than the other's ?

For the sake of discussion, let Bob do a simple straight-line out-and-back trip, with very short periods of acceleration at the beginning, turnaround and end, and constant-velocity cruising periods in between

From the relativistic point of view, the essential difference between Bill and Bob is that Bill remains at rest in a single inertial reference frame throughout, whereas Bob is at rest in two different inertial reference frames during the trip: one inertial reference frame for the outbound trip and another one for the inbound trip. To put it another way, if you view Bob's rocket ship as a single reference frame for the entire trip, it cannot be an inertial reference frame.

The difference between inertial and non-inertial reference frames is crucial in relativity. From a relativistic point of view, inertial reference frames are indeed "favored".

The difference between inertial and non-inertial reference frames is not merely mathematical or semantic. There are concrete physical differences which can be perceived by Bill and Bob. If they close their eyes, or equivalently are locked up in a windowless room and spaceship respectively, each one can easily tell whether he remains in an inertial reference frame or not. Bill, in his stationary room, does not perceive anything special. Bob, on the other hand, knows that he is accelerating during takeoff, turnaround, and landing, because he can feel those events taking place, the same way that you can feel it when your car accelerates and your seat pushes against you.

 

[gijeqkeij]

I am having trouble understanding how the explanation of the paradox solves the problem. What if people in two different frames were moving wrt one another with constant velocity. Say for instance, these people never turn around to meet up once again and are constantly moving at a constant velocity wrt one another.

Clearly, each will percieve the other's frame to be moving more slowly than their own. There is never a change in reference frame so they disagree on who is older. Who is actually older? Are they both younger and both older simultaneously? That seems to be a weird consequence of relativity. What am I missing here?

Space and time are just 1 thing in SR and GR. If both can say "you are more distant from me" and being both right why both can't say "you are older than me" and being both right?

gijeqkeij

Universe it's so simple that it's almost impossible for us to understand it

 

[JesseM]

Space and time are just 1 thing in SR and GR. If both can say "you are more distant from me" and being both right why both can't say "you are older than me" and being both right?

That's a little like saying that since all three spatial axes are "1 thing", then being 5 meters away from me on the y-axis is equivalent to being 5 meters away from me on the z-axis, which is not correct. Space and time are both part of a single entity, "spacetime", but they are different directions in spacetime, so an event happening at a later time is not the same as it happening further in some spatial direction. And the time dimension is somewhat different than the spatial dimensions for other reasons--for example, you can't use the regular pythagorean theorem to calculate "distance" in spacetime, instead you must add the squares of the spatial distances and subtract the square of the temporal distance (multiplied by c^2) between two events to get their spacetime separation.

 

[M1keh]

For the sake of discussion, let Bob do a simple straight-line out-and-back trip, with very short periods of acceleration at the beginning, turnaround and end, and constant-velocity cruising periods in between

From the relativistic point of view, the essential difference between Bill and Bob is that Bill remains at rest in a single inertial reference frame throughout, whereas Bob is at rest in two different inertial reference frames during the trip: one inertial reference frame for the outbound trip and another one for the inbound trip. To put it another way, if you view Bob's rocket ship as a single reference frame for the entire trip, it cannot be an inertial reference frame.

The difference between inertial and non-inertial reference frames is crucial in relativity. From a relativistic point of view, inertial reference frames are indeed "favored".

The difference between inertial and non-inertial reference frames is not merely mathematical or semantic. There are concrete physical differences which can be perceived by Bill and Bob. If they close their eyes, or equivalently are locked up in a windowless room and spaceship respectively, each one can easily tell whether he remains in an inertial reference frame or not. Bill, in his stationary room, does not perceive anything special. Bob, on the other hand, knows that he is accelerating during takeoff, turnaround, and landing, because he can feel those events taking place, the same way that you can feel it when your car accelerates and your seat pushes against you.

Good answer ...

But if they compare watches, Bob's watch should now be 24 mintues (approx) behind Bill's ?

If as Bob approaches, Bill accelerates to 0.6c (instantaneously-ish), matching Bob's velocity, his watch won't then be 24 minutes behind Bob's even though he's now in Bob's f.o.r and that's what Bob would expect ?

So why is Bill's f.o.r so 'special' ? From Bob's perspective, Bill raced away at 0.6c, stopped, turned back and returned at 0.6c and when he arrived he stopped. Bill's watch must now be 24 minutes behind Bob's ?

If not, which element is wrong ? At the final second of the journey the times shown on the watches can't surely shift over a range of 48 minutes from -24 to +24 depending on which f.o.r you jump to ? There's no (or close to no) elapsed time in either f.o.r. for this to happen ?

 

[Janus]

Good answer ...

But if they compare watches, Bob's watch should now be 24 mintues (approx) behind Bill's ?

If as Bob approaches, Bill accelerates to 0.6c (instantaneously-ish), matching Bob's velocity, his watch won't then be 24 minutes behind Bob's even though he's now in Bob's f.o.r and that's what Bob would expect ?

No, Bob will expect his clock to be 24 min behind, more as to why latter.

So why is Bill's f.o.r so 'special' ? From Bob's perspective, Bill raced away at 0.6c, stopped, turned back and returned at 0.6c and when he arrived he stopped. Bill's watch must now be 24 minutes behind Bob's ?

No, Bob can not say that. As he watches Bill, he can see that Bill never experiences any acceleration, while during the time that Bill apparently stops and turns around, Bob does experience the effects of acceleration; evidence that is is he who is doing the turning around.

If not, which element is wrong ? At the final second of the journey the times shown on the watches can't surely shift over a range of 48 minutes from -24 to +24 depending on which f.o.r you jump to ? There's no (or close to no) elapsed time in either f.o.r. for this to happen ?

For Bill, Bob's clock always runs slow.
For Bob, Bill's clock runs slow on the outward leg and run slows on the inward leg, but as he (Bob) stops and turns around, form his perspective, Bill's clock will run fast. How fast depends on how hard he acclerates and how far apart Bill and Bob are. It will always work out that the amount of time that Bill's clock gains according to Bob during this period will more than make up for the outgoing and incoming legs and then some. The 'then some', will account for the fact that He will expect his Clock to Be 24 min behind Bill's when they meet up.

The distance factor also explains why that, in your scenerio where Bill accelerates to match Bob's frame of reference when they come back together, Bob's clock is still 24 min slow. Bill doesn't see Bob's clock run fast, because there is no distance( or almost no distance) between them at that time.

 

[M1keh]

No, Bob will expect his clock to be 24 min behind, more as to why latter.
No, Bob can not say that. As he watches Bill, he can see that Bill never experiences any acceleration, while during the time that Bill apparently stops and turns around, Bob does experience the effects of acceleration; evidence that is is he who is doing the turning around.


For Bill, Bob's clock always runs slow.
For Bob, Bill's clock runs slow on the outward leg and run slows on the inward leg, but as he (Bob) stops and turns around, form his perspective, Bill's clock will run fast. How fast depends on how hard he acclerates and how far apart Bill and Bob are. It will always work out that the amount of time that Bill's clock gains according to Bob during this period will more than make up for the outgoing and incoming legs and then some. The 'then some', will account for the fact that He will expect his Clock to Be 24 min behind Bill's when they meet up.

The distance factor also explains why that, in your scenerio where Bill accelerates to match Bob's frame of reference when they come back together, Bob's clock is still 24 min slow. Bill doesn't see Bob's clock run fast, because there is no distance( or almost no distance) between them at that time.

Thanks. You're right of course. The acceleration can be felt by Bob, so his view of things is different. However, nobody seems to be able to explain why this makes a difference. The fact that Bob accelerates can't reverse the time dilation under Relativity as this would require a negative difference in speed ?

In fact, what happens if Bill accelerates to Bob's intial speed in the direction Bob was travelling, stops immediately and then returns at the same speed as Bob, timed so that they both return together ? That eliminates the acceleration completely but the watches still show inconsistent times ?

ie. Bob accelerates in 1 sec to 0.6c and travels for 1hr Earth's time. He stops taking another 1 sec, turns & accelerates to 0.6c, taking another 1sec (lets say turning takes no time at all. could just reverse ). 3 seconds before Bob returns, Bill accelerates towards Bob to 0.6c in 1 sec, stops, taking another 1 sec & accelerates back to Earth, reaching 0.6c in 1 sec. Both experience the same acceleration / deceleration and are in the same place, yet Bill's watch is behind Bob's ?

If instead of Bob being the one to 'travel', both Bill and Bob travel in opposite directions from Earth and then return and compare watches, both have experienced the same acceleration (albeit in opposite directions). I'm assuming there's nothing to suggest that he direction of the acceleration makes a difference ?

So Bob travels one way at 0.6c, Bill travels the other at 0.6c and Suzy stays and watches the clock. When the two adventurers return, they both have to have watches that show 24 mins earlier than Suzy's ? So how are their watches showing the same time ?

 

[Janus]

Thanks. You're right of course. The acceleration can be felt by Bob, so his view of things is different. However, nobody seems to be able to explain why this makes a difference. The fact that Bob accelerates can't reverse the time dilation under Relativity as this would require a negative difference in speed ?

Under Relativity, accelerating towards a clock causes it to run fast from your perspective, This is a separate effect from the dilation caused by relative speed.

In fact, what happens if Bill accelerates to Bob's intial speed in the direction Bob was travelling, stops immediately and then returns at the same speed as Bob, timed so that they both return together ? That eliminates the acceleration completely but the watches still show inconsistent times?

ie. Bob accelerates in 1 sec to 0.6c and travels for 1hr Earth's time. He stops taking another 1 sec, turns & accelerates to 0.6c, taking another 1sec (lets say turning takes no time at all. could just reverse ). 3 seconds before Bob returns, Bill accelerates towards Bob to 0.6c in 1 sec, stops, taking another 1 sec & accelerates back to Earth, reaching 0.6c in 1 sec. Both experience the same acceleration / deceleration and are in the same place, yet Bill's watch is behind Bob's ?

this is little different than the situation you mentioned before. The fact that Bob is three seconds away when Bill accelerates means that the distance between Bob and Bill is much smaller when Bill does his accelerating than when Bob does his. It is not jus the magnitude or dration of the acceleration, it is the distance between them when it occurs.

If instead of Bob being the one to 'travel', both Bill and Bob travel in opposite directions from Earth and then return and compare watches, both have experienced the same acceleration (albeit in opposite directions). I'm assuming there's nothing to suggest that he direction of the acceleration makes a difference ?

So Bob travels one way at 0.6c, Bill travels the other at 0.6c and Suzy stays and watches the clock. When the two adventurers return, they both have to have watches that show 24 mins earlier than Suzy's ? So how are their watches showing the same time ?

If Bob and Bill compared notes at the end of the trip they both agree that each other's clocks accumulated the same amount of time, they would not however agree exactly how each other's clocks reach the final accumulation. For instance, they would not agree as to when each other turned around. (according to Suzy, they turn around at the same time, According to Bob and Bill, one turned around first, and they will disagree as to who did.)

 

[yogi]

There are numerous ways of arriving at the correct amount of age difference - and keeping track of what each twin observes during the 3 phases of the experiment is one of them - but the the underlying question remains as to what is really taking place - in reality, observing a distant clock undergoing turn-around acceleration doesn't have any affect on the accelerating clock - nor does the change in the slope of the planes of simultaneity experienced by the turning aound twin have anything to do with adding a bunch of time to the Earth based twin - nor does acceleration per se cause a clock to speed up or slow down -

So why do these entirely different methods all give the same result - because they each incorporate in some way or another the invariance of the spacetime interval

 

[M1keh]

Under Relativity, accelerating towards a clock causes it to run fast from your perspective, This is a separate effect from the dilation caused by relative speed.this is little different than the situation you mentioned before.

Are you saying it actually runs faster or just appears to ?

The fact that Bob is three seconds away when Bill accelerates means that the distance between Bob and Bill is much smaller when Bill does his accelerating than when Bob does his. It is not jus the magnitude or dration of the acceleration, it is the distance between them when it occurs.

This is news to me. Haven't seen anything stating this before. Are you saying that if two bodies accelerate at the same time, the affects are different depending on the distance between them ?

If Bob and Bill compared notes at the end of the trip they both agree that each other's clocks accumulated the same amount of time, they would not however agree exactly how each other's clocks reach the final accumulation. For instance, they would not agree as to when each other turned around. (according to Suzy, they turn around at the same time, According to Bob and Bill, one turned around first, and they will disagree as to who did.)

This can't be correct ? For an hour Bob & Bill have a difference in speed of over 0.6c. Their watches must differ ? If their watches are the same when they return, there's no time dilation at all ?

 

[Janus]

Are you saying it actually runs faster or just appears to ?

During that period of the trip Bob will determine that Bill's clock really ran faster than his.

This is news to me. Haven't seen anything stating this before. Are you saying that if two bodies accelerate at the same time, the affects are different depending on the distance between them ?

Yes. Put clocks at the nose and tail of an accelerating spaceship and the clock in the nose will run faster than the one in the tail.

This can't be correct ? For an hour Bob & Bill have a difference in speed of over 0.6c. Their watches must differ ? If their watches are the same when they return, there's no time dilation at all ?

Yes. it is.
For each ship, the periods where it determines that the other's clock ran fast (due to its own acceleration) will exactly cancel out the periods where the other clock ran slow. Remember, relativity of simultaneity assures that Bob and Bill will determine that they turned around at different times. If you do the full analysis from Bob's and Bill's perspectives, you will find that relativity predicts they will agree that they accumulated equal times on their clocks during the trip.

 

[M1keh]

During that period of the trip Bob will determine that Bill's clock really ran faster than his.

How does this work. What's it dependent on ? Is it just the acceleration that causes the 'faster' clock or the difference in speed ? Aren't these just the same thing anyway ?

Yes. Put clocks at the nose and tail of an accelerating spaceship and the clock in the nose will run faster than the one in the tail.

Now, originally you said it was just distance, but you've opted for 'front' and 'back' to make it seem more obvious ?

If one is on the left wing and the other on the right wing, is the effect the same ?

Yes. it is.
For each ship, the periods where it determines that the other's clock ran fast (due to its own acceleration) will exactly cancel out the periods where the other clock ran slow. Remember, relativity of simultaneity assures that Bob and Bill will determine that they turned around at different times. If you do the full analysis from Bob's and Bill's perspectives, you will find that relativity predicts they will agree that they accumulated equal times on their clocks during the trip.

I've (previously) mapped out all of the events based on who see's what at what time in their f.o.r. and it all seems to cancel out perfectly ...

but how does that fit in with the 'twins paradox' ? Surely, the whole point is that the times on the watches (ie. their ages) would be significantly different ?

 

[Janus]

How does this work. What's it dependent on ? Is it just the acceleration that causes the 'faster' clock or the difference in speed ? Aren't these just the same thing anyway ?



Now, originally you said it was just distance, but you've opted for 'front' and 'back' to make it seem more obvious ?

If one is on the left wing and the other on the right wing, is the effect the same ?

To put it more clearly, The factors are:
The magnitude of the acceleration
The distance between the clocks as measured on a line parallel to the direction of the acceleration
The direction of the clocks from each other with respect to the acceleration. (if you experience an acceleration towards the other clock, it runs fast, if you experience an acceleration away from the clock it runs slow.

This effect is separate from and in addition to any time dilation you see due to relative velocity alone.
Thus with the clocks in the Nose and Tail, the Tail clock experiences an acceleration towards the nose and determines that the nose clock is running fast. The nose clock experiences an acceleration away from the Tail and thus sees the tail clock as running slow. This is true even though the two clocks have the same speed. (at least as measured by the two clocks).

I've (previously) mapped out all of the events based on who see's what at what time in their f.o.r. and it all seems to cancel out perfectly ...

but how does that fit in with the 'twins paradox' ? Surely, the whole point is that the times on the watches (ie. their ages) would be significantly different ?

In the original Twin paradox only one of the twins ever changes inertial frames while the other does. This means that Bob and Bill take non-symmetrical paths through spacetime, which results in their difference in clock readings at the end.

In the scenerio where they head in opposite directions, they take different, but symmetrical paths through spacetime. The symmetry of their paths results in no time difference at the end.

 

[jtbell]

Ultimately, the cause of this observer-acceleration-related effect on the running-rate of a clock is the Relativity of Simultaneity which is necessary (in addition to length contraction and time dilation) to give a complete account of the effects of the Lorentz transformation between two inertial reference frames.

One way to state an equation for this is as follows. In inertial frame A, let there be two identical clocks at rest, synchronized (in frame A) and separated by a distance L. Let inertial frame B have velocity v with respect to A, along the line joining the two clocks. In frame B, the two clocks run at the same rate (more slowly than in A, of course) but are out of synchronization by an amount

$$\Delta t = \frac{vL}{c^2}$$

The clock that is in front of the other one, with respect to their motion in frame B, reads a time that is behind the other clock by this amount.

From this equation, if the relative velocity of frame B changes (that is, B is now a non-inertial frame), so does the amount of un-synchronization between the two clocks. This shows up as an apparent change in the running-rate of one or both clocks, in addition to the usual time-dilation effect, but only while frame B is changing its relative velocity (i.e. accelerating).

 

[M1keh]

Ultimately, the cause of this observer-acceleration-related effect on the running-rate of a clock is the Relativity of Simultaneity which is necessary (in addition to length contraction and time dilation) to give a complete account of the effects of the Lorentz transformation between two inertial reference frames.

One way to state an equation for this is as follows. In inertial frame A, let there be two identical clocks at rest, synchronized (in frame A) and separated by a distance L. Let inertial frame B have velocity v with respect to A, along the line joining the two clocks. In frame B, the two clocks run at the same rate (more slowly than in A, of course) but are out of synchronization by an amount

$$\Delta t = \frac{vL}{c^2}$$

The clock that is in front of the other one, with respect to their motion in frame B, reads a time that is behind the other clock by this amount.

From this equation, if the relative velocity of frame B changes (that is, B is now a non-inertial frame), so does the amount of un-synchronization between the two clocks. This shows up as an apparent change in the running-rate of one or both clocks, in addition to the usual time-dilation effect, but only while frame B is changing its relative velocity (i.e. accelerating).

Wow. Haven't seen that one before. Obviously need to read more books !

What happens if there are two clocks in B's f.o.r. One in front of and one behind A ? When A get's up to B's speed & f.o.r, he will have to view the two clocks with differing times, whilst B says they're the same ??

ie. B1 & B2 traveling at 0.5c relative to C, but a long distance apart. B1 & B2 are traveling at same speed, relative to A, and agree their clocks are synchronised.

As B1 passes A, A accelerates towards B1, but as B2 hasn't reached A yet, A's accelerating away from B2. B1's clock must therefore travel faster than B2's in A's f.o.r, but remain the same in B1 & B2's.

Now A reaches B1 & B2's speed just as B2 catches up with A. A stops accelerating to match B2's speed. A still sees B1 & B2's clocks as different and B1 & B2 see them as matching.

At no time does A decelerate compared to B1 or B2. At no time does he accelerate towards B2 or away from B1, to reverse the affects. But now the three do not agree on common events in their common f.o.r. ??


What have I missed ?

 

[JesseM]

Wow. Haven't seen that one before. Obviously need to read more books !

What happens if there are two clocks in B's f.o.r. One in front of and one behind A ? When A get's up to B's speed & f.o.r, he will have to view the two clocks with differing times, whilst B says they're the same ??

ie. B1 & B2 traveling at 0.5c relative to C, but a long distance apart. B1 & B2 are traveling at same speed, relative to A, and agree their clocks are synchronised.

You mean B1 and B2 are synchronized in their mutual rest frame, right?

As B1 passes A, A accelerates towards B1, but as B2 hasn't reached A yet, A's accelerating away from B2. B1's clock must therefore travel faster than B2's in A's f.o.r, but remain the same in B1 & B2's.

There's really only a standard way to define "frames of reference" for inertial observers, not accelerating ones like A. What you can do is consider how things look in the inertial reference frame where A is instantaneously at rest at any given moment, although since A is accelerating this will be a different inertial reference frame from moment to moment. But anyway, in A's instantaneous co-moving inertial reference frame at a given moment, both B1 and B2 will have the same velocity in that frame, so their clocks will be ticking at the same rate.

Now A reaches B1 & B2's speed just as B2 catches up with A. A stops accelerating to match B2's speed. A still sees B1 & B2's clocks as different and B1 & B2 see them as matching.

No, they'll all have the same inertial rest frame at this point, and this frame will have a single definition of simultaneity, one which says that B1 and B2 are synchronized. I think you may be confusing yourself by trying to think of the "frame of reference" of a non-inertial observer--although you are free to come up with a non-inertial coordinate system in which such an observer is at rest throughout his journey, you can't assume the laws of physics in this coordinate system will look anything like they do in inertial reference frames (for example, you can't assume that a clock's rate of ticking in a non-inertial coordinate system would depend solely on its coordinate velocity, and you also can't assume that the coordinate velocity of a light beam would always be c).

 

[M1keh]

To put it more clearly, The factors are:
The magnitude of the acceleration
The distance between the clocks as measured on a line parallel to the direction of the acceleration
The direction of the clocks from each other with respect to the acceleration. (if you experience an acceleration towards the other clock, it runs fast, if you experience an acceleration away from the clock it runs slow.

This effect is separate from and in addition to any time dilation you see due to relative velocity alone.
Thus with the clocks in the Nose and Tail, the Tail clock experiences an acceleration towards the nose and determines that the nose clock is running fast. The nose clock experiences an acceleration away from the Tail and thus sees the tail clock as running slow. This is true even though the two clocks have the same speed. (at least as measured by the two clocks).

Ok. So if there are two rockets, R1 & R2, at different distances from Earth. Both rockets agree on the time that a clock on Earth shows as they're both in the same f.o.r and stationary relative to the Earth ?

If they both accelerate away from Earth to 0.5c, starting at the same time, they end up in the same f.o.r., both traveling at 0.5c compared to Earth and stationary compared to each other, but they can't then agree on the time shown by the clock on Earth ?

What have I missed ?

There's a thought. An assumption I'd made without really checking it out. I'm assuming that two observers at opposite sides of the galaxy are in the same frame of reference if they are stationery relative to each other ? The f.o.r. is based on relative speed and independent of distance ? ... must be ??

In the original Twin paradox only one of the twins ever changes inertial frames while the other does. This means that Bob and Bill take non-symmetrical paths through spacetime, which results in their difference in clock readings at the end.

In the scenerio where they head in opposite directions, they take different, but symmetrical paths through spacetime. The symmetry of their paths results in no time difference at the end.

Hmmm. So does that mean that their difference in speed is irrelevant ? The difference in speed has made absolutely no difference to their watches ? Wasn't that the whole point of time dilation ?

Now I'm lost !