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, becuase 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, becuase 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 !

 

[M1keh]

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

Yes.

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. 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).

I sort of get this point. It was my understanding earlier today.

My query was in response to an earlier post suggesting ( if I understood it correctly ? ) that the acceleration of an observer towards a clock would result in the clock speeding up and that the acceleration of an observer away from a clock would result in the clock slowing down (not just appearing to speed up / slow down) in addition to any time dilation affects resulting from the difference in speeds ?

Does accelerating towards / away from a clock affect the time dilation (apart from the fact that it changes the difference in speed) or is it purely the difference in speed that affects time dilation ? ie. can we effectively ignore acceleration / deceleration unless we're trying to come up with EXACT timings / figures ?

 

[JesseM]

My query was in response to an earlier post suggesting ( if I understood it correctly ? ) that the acceleration of an observer towards a clock would result in the clock speeding up and that the acceleration of an observer away from a clock would result in the clock slowing down (not just appearing to speed up / slow down) in addition to any time dilation affects resulting from the difference in speeds ?

Ah, I see. Well, probably the most common way to define a coordinate system for a non-inertial observer is just to define it in such a way that the observer's definition of simultaneity and distance at any given moment will match that of her instantaneous inertial reference frame at that moment (see the diagram here, from this section of the Twin Paradox FAQ, to see how the lines of simultaneity for an accelerating observer would look at different points on their path); if you do things this way, I think it'd be true that clocks would speed up when you accelerate towards them, and slow down when you accelerate away from them (in some cases they could actually run backwards when you accelerate away from them, which is why this method of defining a coordinate system for a non-inertial observer can create problems). But it's best not to think of these as being "in addition to" time dilation effects, since time dilation effects are a property of inertial reference frames...it's really just a matter of the laws of physics looking completely different in non-inertial coordinate systems, including the fact that time dilation does not work the same way.

Anyway, if we think of how things will look in this sort of non-inertial coordinate system for the accelerating observer A in your example, what we conclude is that before A accelerates he'll observe clock B2 in the back being significantly ahead of clock B1 in the front (remember that B1 and B2 are synchronized in their own rest frame but will appear out-of-sync in other frames), then as A accelerates towards B1 and away from B2 he'll observe B1 speed up and B2 slow down, narrowing the gap between them, until when he finally comes to rest with respect to B1 and B2, the gap will have dwindled to zero and they'll now be in sync.

Does accelerating towards / away from a clock affect the time dilation (apart from the fact that it changes the difference in speed) or is it purely the difference in speed that affects time dilation ? ie. can we effectively ignore acceleration / deceleration unless we're trying to come up with EXACT timings / figures ?

Like I said, the laws of physics look fundamentally different in an accelerating coordinate system. Usually when dealing with acceleration, the normal practice is to consider the accelerating object from the point of view of an inertial coordinate system, so you can still use the usual rules of SR in analyzing how its clock slows down; since at any given instant a clock moving at velocity v in an inertial coordinate system will only be ticking at \sqrt{1 - v^2/c^2} the normal rate in that coordinate system, if you want to find the time elapsed on a clock whose velocity as a function of time is v(t) in your inertial coordinate system, between two coordinate times t_0 and t_1, you just do the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt and that should give you the right answer without ever having to use any non-inertial coordinate systems.

 

[M1keh]

Ah, I see. Well, probably the most common way to define a coordinate system for a non-inertial observer is just to define it in such a way that the observer's definition of simultaneity and distance at any given moment will match that of her instantaneous inertial reference frame at that moment (see the diagram here, from this section of the Twin Paradox FAQ, to see how the lines of simultaneity for an accelerating observer would look at different points on their path); if you do things this way, I think it'd be true that clocks would speed up when you accelerate towards them, and slow down when you accelerate away from them (in some cases they could actually run backwards when you accelerate away from them, which is why this method of defining a coordinate system for a non-inertial observer can create problems). But it's best not to think of these as being "in addition to" time dilation effects, since time dilation effects are a property of inertial reference frames...it's really just a matter of the laws of physics looking completely different in non-inertial coordinate systems, including the fact that time dilation does not work the same way.

Anyway, if we think of how things will look in this sort of non-inertial coordinate system for the accelerating observer A in your example, what we conclude is that before A accelerates he'll observe clock B2 in the back being significantly ahead of clock B1 in the front (remember that B1 and B2 are synchronized in their own rest frame but will appear out-of-sync in other frames), then as A accelerates towards B1 and away from B2 he'll observe B1 speed up and B2 slow down, narrowing the gap between them, until when he finally comes to rest with respect to B1 and B2, the gap will have dwindled to zero and they'll now be in sync. Like I said, the laws of physics look fundamentally different in an accelerating coordinate system. Usually when dealing with acceleration, the normal practice is to consider the accelerating object from the point of view of an inertial coordinate system, so you can still use the usual rules of SR in analyzing how its clock slows down; since at any given instant a clock moving at velocity v in an inertial coordinate system will only be ticking at \sqrt{1 - v^2/c^2} the normal rate in that coordinate system, if you want to find the time elapsed on a clock whose velocity as a function of time is v(t) in your inertial coordinate system, between two coordinate times t_0 and t_1, you just do the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt and that should give you the right answer without ever having to use any non-inertial coordinate systems.

Thanks for the reply. I think you're confirming, as I suspected, that there's no suggestion that acceleration affects time dilation, just the measured time dilation ? The clocks only appear to speed up / slow down with acceleration / deceleration ?


Thanks.

 

[JesseM]

Thanks for the reply. I think you're confirming, as I suspected, that there's no suggestion that acceleration affects time dilation, just the measured time dilation ? The clocks only appear to speed up / slow down with acceleration / deceleration ?


Thanks.

I don't understand the distinction you're making between acceleration affecting "time dilation" vs. "measured time dilation". If you travel away from a twin on earth, and return to find that you have aged less than your twin and everyone else on earth, would you say that less time has elapsed for you, or that you only "appeared" to age less? Would you say this is actual time dilation or only "measured" time dilation? Physics doesn't deal with philosophical statements about what is "really" going on, only with the results of empirical tests and measurements.

By the way, time dilation in a particular inertial frame is technically a function of velocity, not acceleration, although of course a clock's velocity must be changing if it is accelerating, so it will be experiencing time dilation as it accelerates. Note that in the integral I gave above for calculating the total time elapsed on an accelerating clock, you're integrating the clock's velocity as a function of time v(t), not its acceleration a(t).

 

[M1keh]

I don't understand the distinction you're making between acceleration affecting "time dilation" vs. "measured time dilation". If you travel away from a twin on earth, and return to find that you have aged less than your twin and everyone else on earth, would you say that less time has elapsed for you, or that you only "appeared" to age less? Would you say this is actual time dilation or only "measured" time dilation? Physics doesn't deal with philosophical statements about what is "really" going on, only with the results of empirical tests and measurements.

The distinction I was trying to make was between the difference in the time that distant events happen and the time that you view them. It's never very clear from examples what people are referring to.

When two events happen at the same point in space, there's no issue, but there's a complication when there's a large distance between the two becuase of the time taken for the light from the event to reach the observer. This creates a difference between the 'actual' time of the event and the 'observed' time of the event.

I'd say that both twins would be the same age and also 'appear' to be the same age, but that's just me.

In the context of the discussions, I'd say the traveling twin both was and appeared to be younger. I wasn't disagreeing with the context of the original example ... honestly.

By the way, time dilation in a particular inertial frame is technically a function of velocity, not acceleration, although of course a clock's velocity must be changing if it is accelerating, so it will be experiencing time dilation as it accelerates. Note that in the integral I gave above for calculating the total time elapsed on an accelerating clock, you're integrating the clock's velocity as a function of time v(t), not its acceleration a(t).

Hate to be a bore, but it's speed not velocity ? An important distinction ?

I was trying to agree with you above. Time dilation is a function of speed not acceleration. It's just that somebody else had suggested that accelerating / decelerating changed all of the 'rules'.

I'd provided an example and asked how it worked and the response was that the acceleration / deceleration at the end of the journey 'affected' the results.

So basically, when discussing examples of time dilation, we can ignore the 'observed' time of events and concentrate on the 'actual' time of events and we can ignore acceleration / deceleration and assume 'instant' changes in velocity - to make things easier ?


eg. If the twin travels away from Earth at 0.6c for one Earth hour, he'll reach his destination in 60 Earth mins, 48 local mins ( 0.8 time dilation factor ? ), but the twin on Earth won't see him arrive for another 60 minutes, when the light at the time & place of his arrival gets back to Earth. The 'actual' time dilation is -12 mins, but the 'observed' time dilation is -72mins, the -12mins shown on the traveling twins watch, taking 60 mins to reach the twin on Earth ?

 

[selfAdjoint]

When two events happen at the same point in space, there's no issue, but there's a complication when there's a large distance between the two becuase of the time taken for the light from the event to reach the observer. This creates a difference between the 'actual' time of the event and the 'observed' time of the event.

The mere distance a light beam has to travel doesn't affect clocks or simultaneity. It is relative speed between observers that does that, because contra to intuition they both experience the SAME speed of light though they each see the other as moving relative to themselves.

 

[M1keh]

The mere distance a light beam has to travel doesn't affect clocks or simultaneity. It is relative speed between observers that does that, because contra to intuition they both experience the SAME speed of light though they each see the other as moving relative to themselves.

Thanks for confirming. That was my understanding of the theory before this all started.

I can't say it works for me yet, but it was my understanding of the theory.

Just out of interest. Is there an experiment somewhere that shows that they do experience the same speed of light ? Something less complicated than the Maxwell experiments ?

The results of the Maxwell experiments are written in Greek or Klingon and I'm reluctant to learn either.

The Michelson-Morley experiments seem to prove the opposite ... to the untrained eye.

The 'All frequencies of light from a distant object turning up at the same time / speed' examples don't appear to prove much at all ?

Are there any others ?

 

[JesseM]

The distinction I was trying to make was between the difference in the time that distant events happen and the time that you view them. It's never very clear from examples what people are referring to.

When two events happen at the same point in space, there's no issue, but there's a complication when there's a large distance between the two becuase of the time taken for the light from the event to reach the observer. This creates a difference between the 'actual' time of the event and the 'observed' time of the event.

It's conventional in SR to distinguish between when an observer "sees" an event and when they "observe" it. Seeing is when the light signal from the event actually reaches you, so it is affected by the light-signal delays you're talking about, but observing is based on retroactively assigning a time-coordinate to the event, taking into account the light signal delay based on knowledge of the distance. For example, suppose in 2006 I look through my telescope and notice an explosion happening 5 light years away according to my measurements. Then I would say I see the explosion in 2006, but I would say that in my coordinate system I "observed" the explosion's time coordinate to be 2006-5 = 2001.

When physicists say that a clock moving at speed v will be slowed down by a factor of \sqrt{1 - v^2/c^2}, they are talking about what is "observed" after you factor out the different distances that light beams from successive ticks had to travel to reach you (this different distance of successive ticks is the explanation for Doppler shift), not what you actually see with your eyes. In fact, because of the Doppler shift, if the clock is moving away from you at speed v you will see it ticking even slower than that, and if it is moving towards you at speed v you will see it ticking faster, faster than your own clocks in fact.

I'd say that both twins would be the same age and also 'appear' to be the same age, but that's just me.

When would you say that? Would you still say it even if one twin returned to Earth and stood next to the other, and he was visibly younger-looking? Or would you just say it while they were moving away from each other at constant velocity or something?

In the context of the discussions, I'd say the traveling twin both was and appeared to be younger. I wasn't disagreeing with the context of the original example ... honestly.

Well, before either of them accelerates, it would depend what frame you were using, there would be frames where it was the earth-twin that aged less during the constant-velocity phase before the traveling twin turned around.

Hate to be a bore, but it's speed not velocity ? An important distinction ?

Yeah, you're right of course. The direction of the velocity vector doesn't affect the time dilation, only its magnitude (the speed) is important.

I was trying to agree with you above. Time dilation is a function of speed not acceleration. It's just that somebody else had suggested that accelerating / decelerating changed all of the 'rules'.

Both are true, though. Accelerating does change all the rules, because you can only apply the rules of relativity from within an inertial reference frame, so you can't apply these rules in the non-inertial "frame" of the traveling twin. However, from within any given inertial frame, the amount of time dilation experienced by a moving clock is just a function of the clock's speed in that frame. It works out so that even though different frames disagree on the relative rate of the twins' clocks during the different phases of the journey--for example, there would be inertial frames where the traveling twin's clock was ticking faster than the Earth twin's clock before he turned around, but was then ticking even slower than the Earth twin's clock after the turnaround--they will all agree on the value of the total time elapsed on each clock between the times the two twins depart and reunite, with the total time calculated in each frame using the integral \int \sqrt{1 - v(t)^2/c^2} \, dt of each twin's speed as a function of time v(t) in that frame.

So basically, when discussing examples of time dilation, we can ignore the 'observed' time of events and concentrate on the 'actual' time of events and we can ignore acceleration / deceleration and assume 'instant' changes in velocity - to make things easier ?


eg. If the twin travels away from Earth at 0.6c for one Earth hour, he'll reach his destination in 60 Earth mins, 48 local mins ( 0.8 time dilation factor ? ), but the twin on Earth won't see him arrive for another 60 minutes, when the light at the time & place of his arrival gets back to Earth. The 'actual' time dilation is -12 mins, but the 'observed' time dilation is -72mins, the -12mins shown on the traveling twins watch, taking 60 mins to reach the twin on Earth ?

Like I said, usually in relativity when you talk about what is "observed" you have already factored out the light-signal delays. But in your example, since the traveling twin doesn't turn around and reunite with the earth-twin at a single point in space and time, you also have simultaneity issues to worry about--the two twins disagree about what tick of the earth-twin's clock happened at the "same time" that the traveling twin was reaching his destination, so that the traveling twin would say the earth-twin's clock had only elapsed 38.4 minutes (0.8 * 48) at the moment that he reached his destination and his own clock read 48 minutes, while the earth-twin would say his clock had elapsed 60 minutes at the moment the traveling twin reached his destination and his clock read 48 minutes. So with no acceleration involved, the situation is symmetrical, each twin observes the other one's clock to be slowed down by a factor of 0.8.

 

[M1keh]

It's conventional in SR ... traveling twin would say the earth-twin's clock had only elapsed 38.4 minutes (0.8 * 48) at the moment that he reached his destination and his own clock read 48 minutes, while the earth-twin would say his clock had elapsed 60 minutes at the moment the traveling twin reached his destination and his clock read 48 minutes. So with no acceleration involved, the situation is symmetrical, each twin observes the other one's clock to be slowed down by a factor of 0.8.

Apologies for cutting out most of the reply. Your reply was quite concise and confirmed my understanding of how it's supposed to work.

One question. If the 'travelling' twin returns at the same speed and meets his twin. What time will their watches show ? If Twin A expects Twin B's time to be 24 minutes behind (2*12?) and Twin B expects Twin A's to be 19.2 mins behind (2*9.6?), who is right ?

If it depends on which f.o.r you use, then either twin can switch to the other's f.o.r at the last moment. So what would the watches show in both ?

Don't they have to regain or lose time instantly to agree on what the watches show ? If not, how do they get back into sync ?


Sorry. This is where it all started.

 

[MeJennifer]

It's conventional in SR to distinguish between when an observer "sees" an event and when they "observe" it. Seeing is when the light signal from the event actually reaches you, so it is affected by the light-signal delays you're talking about, but observing is based on retroactively assigning a time-coordinate to the event, taking into account the light signal delay based on knowledge of the distance.

Yes this difference between "seeing" and "observing" an event adds a lot of confusion.

While it is true that in flat (and non expanding) space one can calculate the difference by using only on the distance, for curved space or expanding space one cannot.

Frankly I do not see any advantage in considering the "observed" over the "seeing" event. What's the use?

Furthermore, I personally am not happy with "observing" and 'seeing" as terms in scientific context. Rather than using "seeing" we could be more exact and talk about receiving a light signal. And "observing" well again what is the point of such a reconstruction of reality?

 

[JesseM]

One question. If the 'travelling' twin returns at the same speed and meets his twin. What time will their watches show ? If Twin A expects Twin B's time to be 24 minutes behind (2*12?) and Twin B expects Twin A's to be 19.2 mins behind (2*9.6?), who is right ?

But the traveling twin has no right to expect the earth-twin's clock to have elapsed (time elapsed on traveling twin's clock)*(\sqrt{1 - v^2/c^2), because the traveling twin knows he did not remain in a single inertial frame, and the time dilation equation only works in inertial frames. As long as you pick a single inertial frame--it doesn't have to be the earth-twin's rest frame, it could also be the frame where the traveling twin was at rest during the outbound leg but not the inbound leg, or the frame where the traveling twin was at rest during the inbound leg but not the outbound leg--then you will always get the same answer to what the two twins' clocks will read when they reunite.

If it depends on which f.o.r you use

As long as you use an inertial frame, it doesn't. If you try to use a non-inertial frame, you cannot assume the laws of physics (and thus the rules for calculating the elapsed time on a given clock) would look anything like the standard laws of special relativity.

 

[JesseM]

Yes this difference between "seeing" and "observing" an event adds a lot of confusion.

While it is true that in flat (and non expanding) space one can calculate the difference by using only on the distance, for curved space or expanding space one cannot.

Frankly I do not see any advantage in considering the "observed" over the "seeing" event. What's the use?

Furthermore, I personally am not happy with "observing" and 'seeing" as terms in scientific context. Rather than using "seeing" we could be more exact and talk about receiving a light signal. And "observing" well again what is the point of such a reconstruction of reality?

I think the short answer would be that coordinate systems are a basic feature of both special relativity and general relativity, and it's useful to have a shorthand for "the time and position coordinates that were assigned to an event in your chosen coordinate system", and the general agreement is to use the word "observed" for this. The laws of physics are always stated in terms of equations that describe the coordinate paths of objects and the values of different fields as a function of coordinates, this is true in GR as well as SR--to try to rewrite the laws of physics in terms of the times a particular observer actually receives the light from different events, with no reference to a coordinate system at all, would probably be both very complicated mathematically and totally impractical for solving problems, not to mention you'd have to rewrite the laws every time you picked a new observer with a different worldline.

 

[MeJennifer]

I think the short answer would be that coordinate systems are a basic feature of both special relativity and general relativity, and it's useful to have a shorthand for "the time and position coordinates that were assigned to an event in your chosen coordinate system", and the general agreement is to use the word "observed" for this. The laws of physics are always stated in terms of equations that describe the coordinate paths of objects and the values of different fields as a function of coordinates, this is true in GR as well as SR--to try to rewrite the laws of physics in terms of the times a particular observer actually receives the light from different events, with no reference to a coordinate system at all, would probably be both very complicated mathematically and totally impractical for solving problems, not to mention you'd have to rewrite the laws every time you picked a new observer with a different worldline.

There are alternative ways of considering paths or light. Think of twistor theory or even theories like the Feynman-Wheeler theory.

 

[JesseM]

There are alternative ways of considering paths or light. Think of twistor theory or even theories like the Feynman-Wheeler theory.

I would think that calculating anything in these theories would still involve the use of coordinate systems (although I think points in twistor theory represent entire light paths rather than events), I don't think this is equivalent to phrasing the laws of physics solely in terms of the time that light from various events hits your worldline.

 

[Emanresu]

Very clever people are rare.

Very clever people who can explain things so that stupid people like me can understand them are even rarer.

Pervect is one of the very rare ones (if you pester him enough !)

Here is a another one : http://sheol.org/throopw/sr-ticks-n-bricks.html

E.

 

[MeJennifer]

I would think that calculating anything in these theories would still involve the use of coordinate systems (although I think points in twistor theory represent entire light paths rather than events), I don't think this is equivalent to phrasing the laws of physics solely in terms of the time that light from various events hits your worldline.

Just for the good order there is nothing wrong by using coordinate systems, however to insist that we gain much by interpreting relativity by projecting it endlessly onto a 3D hyperplane is something I disagree with. On the contrary it causes more confusion.

In my views in special relativity we have to understand that if two or more objects are in relative motion we cannot have an understanding of the total situation if we insist in creating a coordinate system from one perspective only. For general relativity this complexity "blows up", it is simply next to useless to project what is happening onto a 3D hyperplane if one wants to gain an understanding of what is going on.

 

[JesseM]

Just for the good order there is nothing wrong by using coordinate systems, however to insist that we gain much by interpreting relativity by projecting it endlessly onto a 3D hyperplane is something I disagree with. On the contrary it causes more confusion.

In my views in special relativity we have to understand that if two or more objects are in relative motion we cannot have an understanding of the total situation if we insist in creating a coordinate system from one perspective only. For general relativity this complexity "blows up", it is simply next to useless to project what is happening onto a 3D hyperplane if one wants to gain an understanding of what is going on.

But I'm not talking about gaining a conceptual understanding, I'm talking about how you calculate the answers to specific problems in specific situations. I don't see a way of doing this without using coordinate systems, although of course you can make use of more than one coordinate system in the course of solving a problem (and juggling multiple coordinate systems is probably the best way to gain the sort of conceptual understanding you're talking about).

 

[MeJennifer]

But I'm not talking about gaining a conceptual understanding, I'm talking about how you calculate the answers to specific problems in specific situations. I don't see a way of doing this without using coordinate systems, although of course you can make use of more than one coordinate system in the course of solving a problem (and juggling multiple coordinate systems is probably the best way to gain the sort of conceptual understanding you're talking about).

Agreed.
Perhaps from now on we should explain everything using twistor space, that will clear up everything

Now by the way, do you think we "see" or only "observe" length contraction or neither?
From a twistor-space perspective we see or observe no contraction, since Lorentz transformations are shape preserving in twistor space.

It gets more fantastic if we would include Feynman-Wheeler like theories in SR. There a free photon does not exist, only the closed path between an "emitting" and "absorbing" photon constitutes an event.

 

[M1keh]

But the traveling twin has no right to expect the earth-twin's clock to have elapsed (time elapsed on traveling twin's clock)*(\sqrt{1 - v^2/c^2), because the traveling twin knows he did not remain in a single inertial frame, and the time dilation equation only works in inertial frames. As long as you pick a single inertial frame--it doesn't have to be the earth-twin's rest frame, it could also be the frame where the traveling twin was at rest during the outbound leg but not the inbound leg, or the frame where the traveling twin was at rest during the inbound leg but not the outbound leg--then you will always get the same answer to what the two twins' clocks will read when they reunite. As long as you use an inertial frame, it doesn't. If you try to use a non-inertial frame, you cannot assume the laws of physics (and thus the rules for calculating the elapsed time on a given clock) would look anything like the standard laws of special relativity.

Ok. Great ! So we can only work out what will happen as long as nobody accelerates anywhere ?

Appologies if I misunderstood, but are you saying that the theory only works in a universe where acceleration isn't possible ?

( Long weekend. A bit grumpy this morning. )

 

[JesseM]

Ok. Great ! So we can only work out what will happen as long as nobody accelerates anywhere ?

No, what I'm saying is that if you want to use the laws of SR, you must use a reference frame that doesn't accelerate--but you can certainly analyze the paths of accelerating objects from within this reference frame! For example, in the twin paradox it is often assumed that the acceleration is instantaneous, so that the traveling twin's worldline just consists of two joined straight line segments, an outbound leg and an inbound leg. In this case you can calculate the total time elapsed on the traveling twin's clock by picking a single inertial reference frame, figuring out how long the outbound and inbound leg last as measured by this frame, then figuring out the velocity during the outbound leg and the velocity during the inbound leg in this frame, and from this you can predict the time elapsed on the traveling twin's own clocks by calculating (time of outbound leg in your frame)*(time dilation factor based on velocity during outbound leg in your frame) + (time of inbound leg in your frame)*(time dilation factor based on velocity during inbound leg in your frame). For example, if the outbound leg lasts T_1 and the inbound leg lasts T_2 and the outbound velocity is v_1 while the inbound velocity is v_2 in your chosen inertial frame, then the time elapsed on the traveling twin's clock will be T_1 * \sqrt{1 - v_1^2 /c^2} + T_2 * \sqrt{1 - v_2^2 / c^2}. More generally, if the acceleration is not instantaneous, then if you want to know the time elapsed on an accelerating clock between two events on its worldline with time-coordinates t_1 and t_2 in your inertial frame, and the accelerating clock's changing velocity as a function of time as measured in your frame is given by some function v(t), then the total time elapsed on the clock would be calculated by doing the integral \int_{t_1}^{t_2} \sqrt{1 - v(t)^2 / c^2} \, dt. So in both cases, you have calculated the time elapsed on the accelerating clock using the coordinate system of an inertial frame, not an accelerating frame. And if you do the same calculations from the perspective of a different inertial frame, you will get exactly the same answer for the total time elapsed on the accelerating clock, despite the fact that details like the time and velocity during the outbound leg or the velocity as a function of time v(t) will be different in this frame--the total time elapsed on a moving clock (known as the 'proper time') between two events on its own worldline is a frame-invariant quantity, it won't depend on which frame you use to calculate it. If you'd like to see an example of this, I could come up with a simple one so you could see the math.