twins paradox confusion [Archive]

M1keh

Sep13-06, 06:04 AM

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 wont 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

Sep13-06, 09:29 AM

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 wont 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

Sep13-06, 09:58 AM

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

Sep13-06, 11:51 AM

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

Sep13-06, 10:52 PM

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

Sep14-06, 02:39 AM

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

Sep14-06, 06:33 AM

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

Sep14-06, 07:14 AM

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

Sep14-06, 09:14 AM

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

Sep14-06, 10:34 AM

M1keh

Sep15-06, 05:50 AM

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 travelling at 0.5c relative to C, but a long distance apart. B1 & B2 are travelling 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

Sep15-06, 09:07 AM

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 travelling at 0.5c relative to C, but a long distance apart. B1 & B2 are travelling 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

Sep15-06, 09:09 AM

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 travelling 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

Sep15-06, 09:26 AM

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

Sep15-06, 09:46 AM

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 (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/gr.gif), from this section (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html) of the Twin Paradox FAQ (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html), 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

Sep28-06, 06:58 AM

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 (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/gr.gif), from this section (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html) of the Twin Paradox FAQ (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html), 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

Sep28-06, 10:50 AM

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

Sep29-06, 04:08 AM

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 travelling 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 wont 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 travelling twins watch, taking 60 mins to reach the twin on Earth ?

selfAdjoint

Sep29-06, 08:19 AM

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

Sep29-06, 08:59 AM

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

Sep29-06, 09:47 AM

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 travelling 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 travelling 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 travelling 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 travelling 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 wont 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 travelling 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 travelling 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 travelling twin was reaching his destination, so that the travelling 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 travelling 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

Sep29-06, 10:05 AM

It's conventional in SR ... travelling 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 travelling 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

Sep29-06, 12:11 PM

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

Sep29-06, 01:34 PM

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 travelling twin has no right to expect the earth-twin's clock to have elapsed (time elapsed on travelling twin's clock)*(\sqrt{1 - v^2/c^2), because the travelling 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 travelling twin was at rest during the outbound leg but not the inbound leg, or the frame where the travelling 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

Sep29-06, 01:44 PM

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

Sep29-06, 02:03 PM

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

Sep29-06, 03:02 PM

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

Sep29-06, 03:20 PM

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

Sep29-06, 03:30 PM

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. :smile:

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

Sep29-06, 03:41 PM

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. :smile:

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

Sep29-06, 03:47 PM

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 :tongue:

Now by the way, do you think we "see" or only "observe" length contraction or neither? :smile:
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

Oct2-06, 01:48 AM

But the travelling twin has no right to expect the earth-twin's clock to have elapsed (time elapsed on travelling twin's clock)*(\sqrt{1 - v^2/c^2), because the travelling 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 travelling twin was at rest during the outbound leg but not the inbound leg, or the frame where the travelling 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 ? :confused:

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

( Long weekend. A bit grumpy this morning. :cry: )

JesseM

Oct2-06, 11:42 AM

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

M1keh

Oct3-06, 02:29 AM

... If you'd like to see an example of this, I could come up with a simple one so you could see the math.

JesseM. Nice answer. You've obviously done this all before ?

I think I follow the answer and it's pretty much as I believed the case to be. So the acceleration / deceleration, although they change the figures because of the change in difference in speed between the observers, have no other affect on the results ? ie. there's nothing special about an accelerating / decelerating frame of reference ?

An example would be great, but I've seen the twins paradox example.

What I'm looking for is an explanation of what happens if there are triplets and one triplet travels away from Earth at 0.6c in one direction for an Earth hour and another triplet travels in the opposite direction at 0.6c for an Earth hour. Both triplets then returning at 0.6c.

When they return and compare watches with the Earth triplet, what do the watches show.

This is the one I'm really struggling with. My understanding of the rules, briefly is :

* Time dilation is related to speed not velocity.
* All frames of reference are equal.
* There are no 'special' rules for acceleration / deceleration. ie. when changing between f.o.r's.

Are these correct ?

JesseM

Oct3-06, 09:00 AM

I think I follow the answer and it's pretty much as I believed the case to be. So the acceleration / deceleration, although they change the figures because of the change in difference in speed between the observers have no other affect on the results ? I'm not sure I understand your question. I was talking about analyzing the same accelerating path from the point of view of two different inertial frames, each of which would have a different view of the travelling twin's velocity during the inbound and outbound legs of the trip (and also of how long each leg lasted), but both nevertheless calculating the same answer to how much time will have elapsed on the traveling twin's clock. ie. there's nothing special about an accelerating / decelerating frame of reference ? No, as I've said many times, you can't use accelerating reference frame and expect the laws of physics to remotely resemble the laws of SR! Do you understand the difference between 1) trying to use an accelerating reference frame, and 2) analyzing the behavior of an accelerating object using the coordinate system of an inertial reference frame? An example would be great, but I've seen the twins paradox example. But have you seen it analyzed from two different inertial reference frames, showing how they both get the same prediction for the total time elapsed on the travelling twin's clock? That was what I was talking about in my last post. Let me know if you'd like to see this. What I'm looking for is an explanation of what happens if there are triplets and one triplet travels away from Earth at 0.6c in one direction for an Earth hour and another triplet travels in the opposite direction at 0.6c for an Earth hour. Both triplets then returning at 0.6c.

When they return and compare watches with the Earth triplet, what do the watches show.

This is the one I'm really struggling with. My understanding of the rules, briefly is :

* Time dilation is related to speed not velocity.
* All frames of reference are equal. All inertial frames, yes.
* There are no 'special' rules for acceleration / deceleration. ie. when changing between f.o.r's. OK, I think I might see what the problem is. When I talk about using different frames of reference, I am talking about analyzing the problem from beginning to end in one inertial frame, then picking a different inertial frame and once again analyzing the problem from beginning to end in this second frame. I am not talking about trying to take the point of view of the twin who changes speeds, and using one frame during the outbound leg and then switching to a different frame in mid-problem when he begins the inbound leg, somehow patching together these two frames to analyze the whole problem. "Changing reference frames" has nothing to do with when any physical observer changes speeds, it's just a question of us, the omniscient observers of the problem, using different coordinate systems to analyze the problem. So I could certainly analyze that triplets problem from the point of view of different reference frames, but each frame would be used to analyze the whole problem from start to finish, not to try to take the "point of view" of a twin who changes speeds by starting out with his rest frame during the inbound leg and then switching in mid-problem to his rest frame during the outbound leg (this wouldn't make sense because the frames define simultaneity differently--the time on the earth's clock that is simultaneous with the time on the travelling twin's clock at the moment before he changes speeds in his outbound rest frame would be completely different from the time on the earth's clock that is simultaneous with the time on the travelling twin's clock at the moment after he changes speeds in the inbound reference frame). Let me know if you want to see this sort of analysis.

M1keh

Oct3-06, 09:26 AM

I'm not sure I understand your question. I was talking about analyzing the same accelerating path from the point of view of two different inertial frames, each of which would have a different view of the travelling twin's velocity during the inbound and outbound legs of the trip (and also of how long each leg lasted), but both nevertheless calculating the same answer to how much time will have elapsed on the traveling twin's clock.

Unless I misunderstood, we just break the acceleration down into small units of static reference frames and sum them all up ?

No, as I've said many times, you can't use accelerating reference frame and expect the laws of physics to remotely resemble the laws of SR!

Why not ? As you say next ...

Do you understand the difference between 1) trying to use an accelerating reference frame, and 2) analyzing the behavior of an accelerating object using the coordinate system of an inertial reference frame?

Hmmm. I see the difference, but my problem is I'm not sure why it makes a difference. However, if we assume 'instant' acceleration / deceleration we can pretty much ignore these ? Or does this seriously affect any example where someone turns around ?

But have you seen it analyzed from two different inertial reference frames, showing how they both get the same prediction for the total time elapsed on the travelling twin's clock? That was what I was talking about in my last post. Let me know if you'd like to see this.

Yes please. That would help.

All inertial frames, yes.
OK, I think I might see what the problem is. When I talk about using different frames of reference, I am talking about analyzing the problem from beginning to end in one inertial frame, then picking a different inertial frame and once again analyzing the problem from beginning to end in this second frame. I am not talking about trying to take the point of view of the twin who changes speeds, and using one frame during the outbound leg and then switching to a different frame in mid-problem when he begins the inbound leg, somehow patching together these two frames to analyze the whole problem. "Changing reference frames" has nothing to do with when any physical observer changes speeds, it's just a question of us, the omniscient observers of the problem, using different coordinate systems to analyze the problem. So I could certainly analyze that triplets problem from the point of view of different reference frames, but each frame would be used to analyze the whole problem from start to finish, not to try to take the "point of view" of a twin who changes speeds by starting out with his rest frame during the inbound leg and then switching in mid-problem to his rest frame during the outbound leg (this wouldn't make sense because the frames define simultaneity differently--the time on the earth's clock that is simultaneous with the time on the travelling twin's clock at the moment before he changes speeds in his outbound rest frame would be completely different from the time on the earth's clock that is simultaneous with the time on the travelling twin's clock at the moment after he changes speeds in the inbound reference frame). Let me know if you want to see this sort of analysis.

YES PLEASE ! That would be ideal. The only point I would make, probably clumsily, is that the twin who travels out at 0.6c and comes back at 0.6c, doesn't 'change speed' ??? only velocity ???

The difference in speed between the two twins, ignoring the 'instant' change in direction, is a constant 0.6c ?

JesseM

Oct3-06, 11:25 AM

Unless I misunderstood, we just break the acceleration down into small units of static reference frames and sum them all up ? You break down the path into units of constant-velocity motion, but you use a single inertial frame to calculate how much time elapses on both the travelling clock and the earth clock during each unit. Hmmm. I see the difference, but my problem is I'm not sure why it makes a difference. However, if we assume 'instant' acceleration / deceleration we can pretty much ignore these ? No, because again, different inertial frames define simultaneity differently. At the instant of turnaround in the travelling twin's outbound rest frame, the time on earth might be 2010; but at the instant of turnaround in the travelling twin's inbound rest frame, the time on earth might be 2020. So, you'd get the wrong answer if you tried to figure out the total time elapsed on earth between departure and return by saying something like "the ship left earth in 2005, and in the outbound rest frame only 5 years passed on earth between the moment of departure and the moment of turnaround, then in the inbound rest frame only 5 more years passed on earth between the moment of turnaround and the moment of return to earth, therefore earth's clock will read 2015 at the moment of return." This would miss the discontinuous gap between the earth's time at the moment of turnaround in the outbound frame and the earth's time at the moment of turnaround in the inbound frame, due to the two frames defining simultaneity differently--because of this gap, the actual time when the ship returned would be 2020 + 5 = 2025, not 2015 as in the naive calculation. If you stick to a single inertial frame for adding each segment you won't run into this sort of problem. Yes please. That would help. OK, let's say our ship departs earth and travels at 0.6c for 10 years in the earth's frame, then turns instantaneously and travels back at 0.6c for another 10 years in the earth's frame. In the earth's frame, the time dilation factor is \sqrt{1 - 0.6^2} = 0.8, so the travelling twin's clock will only tick (10 years)*(0.8) = 8 years during the outbound leg, and (10 years)*(0.8) = 8 years during the inbound leg, so when they reunite the earth's clock will show 20 years have elapsed while the ship's clock only shows 16. Note that both the 10 years and the 0.8 time dilation factor were calculated solely from the perspective of the earth's inertial frame.

Now let's analyze the problem in a different inertial frame--the frame where the travelling twin is at rest during the outbound leg. In this frame, the earth will be moving away at a constant speed of 0.6c, while the ship will first be at rest for 8 years, during which time the earth has moved away a distance of (8 years)*(0.6c) = 4.8 light years, and its clock has advanced by (8 years)*(0.8) = 6.4 years, while the ship's clock has advanced forward by 8 years since it is at rest. Then after 8 years the ship will accelerate instantaneously, and using the velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) we know its speed in this frame after acceleration will be (0.6c + 0.6c)/(1 + 0.6^2) = 1.2c/1.36 = 0.88235c. Since the earth is 4.8 light years away but only moving at 0.6c, the time for the ship to catch up can be found by solving for t in 4.8 + 0.6t = 0.88235t, which gives t = 17 years in this frame. During this time, the earth's time dilation factor will still be 0.8, while the ship's will be \sqrt{1 - 0.88235^2} = 0.4706, so the earth's clock will have advanced forward by (17 years)*(0.8) = 13.6 years while the ship's clock will have advanced forward by (17 years)*(0.4706) = 8 years. So, by the time the ship catches up to the earth, the earth's clock will have advanced forward by the 6.4 years of the outbound leg plus the 13.6 years of the inbound leg, while the ship's clock will have advanced by the 8 years of the outbound leg plus the 8 years of the inbound leg. So, we find that the earth's clock had advanced 20 years while the ship's clock has advanced only 16, just as we found in the earth's frame.

We could also analyze everything from the point of view of the frame where the ship is at rest during the inbound leg. This will just be the mirror image of how things looked in the outbound rest frame--during the outbound leg, the earth is moving at 0.6c while the ship is moving at 0.88235c in the same direction, which lasts for 17 years, then the ship instantaneously accelerates and comes to rest while the earth continues to approach it at 0.6c from a distance of 4.8 light years, catching up to it after 8 years in this frame. And again, in a mirror image of the analysis in the outbound rest frame, the ship's clock elapses (17)*(0.4706) = 8 years during the outbound leg and (8)*(1) = 8 years in the inbound leg, while the earth's clock elapses (17)*(0.8) = 13.6 years in the outbound leg and (8)*(0.8) = 6.4 years in the inbound leg, so you again predict the ship's clock elapses a total of 16 years while the earth's clock elapses a total of 20 years.

Now look what would happen if you tried to switch frames in mid-problem, without worrying about simultaneity issues. Start out using the outbound rest frame during the outbound leg, and you'll conclude that at the moment before the ship does its instantaneous acceleration, the ship's clock reads 8 years and the earth is 4.8 light years away, its clock reading 6.4 years. Now if you switch to the inbound rest frame, and you don't realize that because of the different definition of simultaneity the earth should "jump" to reading 13.6 years, then you'll mistakenly continue to think that at the beginning of the inbound leg the earth's clock reads 6.4 years, and since the inbound frame says the ship is at rest during the inbound leg while the earth is approaching it at 0.6c from 4.8 light years away, in the inbound frame the ship's clock must advance by 8 years during the inbound leg and the earth's clock must advance by (8)*(0.8) = 6.4 years, so you would conclude the earth's clock only read 6.4 + 6.4 = 12.8 years when the ship returned? But this answer contradicts the answer you get when you stick to a single inertial frame throughout the problem, whether the earth rest frame, the inbound rest frame or the outound rest frame; you get the wrong answer if you try to "patch together" different frames in mid-problem this way, without taking into account discontinuous jumps in clock time due to the frames defining simultaneity differently. YES PLEASE ! That would be ideal. The only point I would make, probably clumsily, is that the twin who travels out at 0.6c and comes back at 0.6c, doesn't 'change speed' ??? only velocity ??? Although the time dilation factor as measured in a given inertial frame is only a function of a clock's speed in that frame, not its velocity, a frame can only qualify as "inertial" in the first place if it is moving at constant velocity, not just constant speed. So even though the instantaneous acceleration means the travelling twin never changed speeds in the earth's inertial frame (although he did change speeds in every other inertial frame, like the outbound rest frame), he did not stay at rest in a single inertial frame throughout the journey.

Analyzing the triplets problem wouldn't differ much from the twins problem. Let's again assume that in the earth's frame, the two travelling triplets move at 0.6c, and turn around after 10 years. In the earth's frame, each travelling triplet ages (10 years)*(0.8) = 8 years during their outbound legs, and another 8 years during their inbound legs, so that they have both aged 16 years when they return while the earth twin has aged 20 years.

Now look at things from the perspective of an inertial frame where one of the triplets is at rest during the oubound leg. In this frame, for the first 8 years this triplet (call him A) will be at rest, while the earth-triplet (call him B) is moving away at 0.6c and the third triplet (call him C) is moving away at (0.6c + 0.6c)/(1 + 0.6^2) = 0.88235c. Since C's clock is ticking slow by a factor of 0.4706 in this frame, and he won't accelerate until 8 years have passed on his own clock, he won't turn around until 8/0.4706 = 17 years have passed in this frame; when he does turn around, he'll be at rest in this frame, and at this moment the distance between him and the earth is (17 years)*(0.88235c - 0.6c) = 4.8 light years. The earth will continue to approach him at 0.6c after this point, reaching him after an additional 4.8/0.6 = 8 years in this frame. And when A turns around after 8 years, he is now moving in the direction of the earth at 0.88235c, with the earth having moved a distance of (8 years)*(0.6c) = 4.8 light years in those 8 years. The time for A to catch up to the earth is given by 4.8 + 0.6t = 0.88235t, or 17 years. So in this frame A starts out at rest, accelerates after 8 years, then takes another 17 years to catch up to earth, while C starts out at high speed, accelerates to rest after 17 years, and then the earth takes an another 8 years to catch up to him, meaning all three twins reunite after 25 years have passed in this frame.

So now let's figure out how much time has passed on each twin's clock after 8 years, after 17 years, and after 25 years in this frame, using this frame's values of the speeds and time dilation factors.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

M1keh

Oct4-06, 01:31 AM

You break down the path into units of constant-velocity motion, but you use a single inertial frame to calculate how much time elapses on both the travelling clock and the earth clock during each unit. No, because again, different inertial frames define simultaneity differently. At the instant of turnaround in the travelling twin's outbound rest frame, the time on earth might be 2010; but at the instant of turnaround in the travelling twin's inbound rest frame, the time on earth might be 2020. So, you'd get the wrong answer if you tried to figure out the total time elapsed on earth between departure and return by saying something like "the ship left earth in 2005, and in the outbound rest frame only 5 years passed on earth between the moment of departure and the moment of turnaround, then in the inbound rest frame only 5 more years passed on earth between the moment of turnaround and the moment of return to earth, therefore earth's clock will read 2015 at the moment of return." This would miss the discontinuous gap between the earth's time at the moment of turnaround in the outbound frame and the earth's time at the moment of turnaround in the inbound frame, due to the two frames defining simultaneity differently--because of this gap, the actual time when the ship returned would be 2020 + 5 = 2025, not 2015 as in the naive calculation. If you stick to a single inertial frame for adding each segment you won't run into this sort of problem. OK, let's say our ship departs earth and travels at 0.6c for 10 years in the earth's frame, then turns instantaneously and travels back at 0.6c for another 10 years in the earth's frame. In the earth's frame, the time dilation factor is \sqrt{1 - 0.6^2} = 0.8, so the travelling twin's clock will only tick (10 years)*(0.8) = 8 years during the outbound leg, and (10 years)*(0.8) = 8 years during the inbound leg, so when they reunite the earth's clock will show 20 years have elapsed while the ship's clock only shows 16. Note that both the 10 years and the 0.8 time dilation factor were calculated solely from the perspective of the earth's inertial frame.

Now let's analyze the problem in a different inertial frame--the frame where the travelling twin is at rest during the outbound leg. In this frame, the earth will be moving away at a constant speed of 0.6c, while the ship will first be at rest for 8 years, during which time the earth has moved away a distance of (8 years)*(0.6c) = 4.8 light years, and its clock has advanced by (8 years)*(0.8) = 6.4 years, while the ship's clock has advanced forward by 8 years since it is at rest. Then after 8 years the ship will accelerate instantaneously, and using the velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) we know its speed in this frame after acceleration will be (0.6c + 0.6c)/(1 + 0.6^2) = 1.2c/1.36 = 0.88235c. Since the earth is 4.8 light years away but only moving at 0.6c, the time for the ship to catch up can be found by solving for t in 4.8 + 0.6t = 0.88235t, which gives t = 17 years in this frame. During this time, the earth's time dilation factor will still be 0.8, while the ship's will be \sqrt{1 - 0.88235^2} = 0.4706, so the earth's clock will have advanced forward by (17 years)*(0.8) = 13.6 years while the ship's clock will have advanced forward by (17 years)*(0.4706) = 8 years. So, by the time the ship catches up to the earth, the earth's clock will have advanced forward by the 6.4 years of the outbound leg plus the 13.6 years of the inbound leg, while the ship's clock will have advanced by the 8 years of the outbound leg plus the 8 years of the inbound leg. So, we find that the earth's clock had advanced 20 years while the ship's clock has advanced only 16, just as we found in the earth's frame.

We could also analyze everything from the point of view of the frame where the ship is at rest during the inbound leg. This will just be the mirror image of how things looked in the outbound rest frame--during the outbound leg, the earth is moving at 0.6c while the ship is moving at 0.88235c in the same direction, which lasts for 17 years, then the ship instantaneously accelerates and comes to rest while the earth continues to approach it at 0.6c from a distance of 4.8 light years, catching up to it after 8 years in this frame. And again, in a mirror image of the analysis in the outbound rest frame, the ship's clock elapses (17)*(0.4706) = 8 years during the outbound leg and (8)*(1) = 8 years in the inbound leg, while the earth's clock elapses (17)*(0.8) = 13.6 years in the outbound leg and (8)*(0.8) = 6.4 years in the inbound leg, so you again predict the ship's clock elapses a total of 16 years while the earth's clock elapses a total of 20 years.

Now look what would happen if you tried to switch frames in mid-problem, without worrying about simultaneity issues. Start out using the outbound rest frame during the outbound leg, and you'll conclude that at the moment before the ship does its instantaneous acceleration, the ship's clock reads 8 years and the earth is 4.8 light years away, its clock reading 6.4 years. Now if you switch to the inbound rest frame, and you don't realize that because of the different definition of simultaneity the earth should "jump" to reading 13.6 years, then you'll mistakenly continue to think that at the beginning of the inbound leg the earth's clock reads 6.4 years, and since the inbound frame says the ship is at rest during the inbound leg while the earth is approaching it at 0.6c from 4.8 light years away, in the inbound frame the ship's clock must advance by 8 years during the inbound leg and the earth's clock must advance by (8)*(0.8) = 6.4 years, so you would conclude the earth's clock only read 6.4 + 6.4 = 12.8 years when the ship returned? But this answer contradicts the answer you get when you stick to a single inertial frame throughout the problem, whether the earth rest frame, the inbound rest frame or the outound rest frame; you get the wrong answer if you try to "patch together" different frames in mid-problem this way, without taking into account discontinuous jumps in clock time due to the frames defining simultaneity differently. Although the time dilation factor as measured in a given inertial frame is only a function of a clock's speed in that frame, not its velocity, a frame can only qualify as "inertial" in the first place if it is moving at constant velocity, not just constant speed. So even though the instantaneous acceleration means the travelling twin never changed speeds in the earth's inertial frame (although he did change speeds in every other inertial frame, like the outbound rest frame), he did not stay at rest in a single inertial frame throughout the journey.

Analyzing the triplets problem wouldn't differ much from the twins problem. Let's again assume that in the earth's frame, the two travelling triplets move at 0.6c, and turn around after 10 years. In the earth's frame, each travelling triplet ages (10 years)*(0.8) = 8 years during their outbound legs, and another 8 years during their inbound legs, so that they have both aged 16 years when they return while the earth twin has aged 20 years.

Now look at things from the perspective of an inertial frame where one of the triplets is at rest during the oubound leg. In this frame, for the first 8 years this triplet (call him A) will be at rest, while the earth-triplet (call him B) is moving away at 0.6c and the third triplet (call him C) is moving away at (0.6c + 0.6c)/(1 + 0.6^2) = 0.88235c. Since C's clock is ticking slow by a factor of 0.4706 in this frame, and he won't accelerate until 8 years have passed on his own clock, he won't turn around until 8/0.4706 = 17 years have passed in this frame; when he does turn around, he'll be at rest in this frame, and at this moment the distance between him and the earth is (17 years)*(0.88235c - 0.6c) = 4.8 light years. The earth will continue to approach him at 0.6c after this point, reaching him after an additional 4.8/0.6 = 8 years in this frame. And when A turns around after 8 years, he is now moving in the direction of the earth at 0.88235c, with the earth having moved a distance of (8 years)*(0.6c) = 4.8 light years in those 8 years. The time for A to catch up to the earth is given by 4.8 + 0.6t = 0.88235t, or 17 years. So in this frame A starts out at rest, accelerates after 8 years, then takes another 17 years to catch up to earth, while C starts out at high speed, accelerates to rest after 17 years, and then the earth takes an another 8 years to catch up to him, meaning all three twins reunite after 25 years have passed in this frame.

So now let's figure out how much time has passed on each twin's clock after 8 years, after 17 years, and after 25 years in this frame, using this frame's values of the speeds and time dilation factors.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

Ouch ! I mean thanks. It'll take a while to go through ... but I will. This is the first time anyone's come back with a real example and some figures. Thankyou.

M1keh

Oct4-06, 03:57 AM

...

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

JesseM. That seems like a very detailed and accurate representation of the theory. The solutions presented for these problems are very elegant. Almost an art form ?

At first I though this solved my problem as the figures do work out exactly right. Albeit difficult to get your head around at first.

But then ...

My problem is what the twins watches would show at each stage of the journey. The twins & their watches jump between f.o.r.'s at regular points so they dont' have the luxury of measuring time from one f.o.r. ?

If the two travelling twins (we wont call them triplets ?) 'stop' relative to the Earth when they reach their turning points, what are the times on the three twins' watches ?

They will have jumped back into the Earth's f.o.r, relative speed now zero ?

Janus

Oct4-06, 06:28 AM

My problem is what the twins watches would show at each stage of the journey. The twins & their watches jump between f.o.r.'s at regular points so they dont' have the luxury of measuring time from one f.o.r. ?

If the two travelling twins (we wont call them triplets ?) 'stop' relative to the Earth when they reach their turning points, what are the times on the three twins' watches ?

They will have jumped back into the Earth's f.o.r, relative speed now zero ?

That depends on what frame of reference you are asking the question from. From the frame of reference in which they stopped, the Traveling twins will show the same time on their respective clocks and this will be less than that shown on the Earth Clock. (Though you have to remember that even though the two travelers stop at the same time according to the Earth clock, they do not according to their own clocks)

From other frames of reference, the three clocks will show different times after they stop. For instance, from a frame that continues to travel away from the Earth at the same speed as one of the Traveling twins, the twin nearest to him will show the greatest time on his clock, the Earth clock will show less, and the other twin even less.

M1keh

Oct4-06, 08:13 AM

That depends on what frame of reference you are asking the question from. From the frame of reference in which they stopped, the Traveling twins will show the same time on their respective clocks and this will be less than that shown on the Earth Clock. (Though you have to remember that even though the two travelers stop at the same time according to the Earth clock, they do not according to their own clocks)

From other frames of reference, the three clocks will show different times after they stop. For instance, from a frame that continues to travel away from the Earth at the same speed as one of the Traveling twins, the twin nearest to him will show the greatest time on his clock, the Earth clock will show less, and the other twin even less.

Hold on ... rewind ... back to Earth twin E & Travelling twin T ...

E's time elapses 10 years with T's appearing as 8 years, from E's f.o.r.

In T's f.o.r. his time elapses 8 years, & E's elapses 6.4 years ...

Is this correct ? Wont T's local time also elapse 10 years and wont he view E's as having elapsed 8 years ? He would have travelled for 8 Earth years but they would have appeared as 8/0.8 = 10 elapsed years to T ??

The questions still stand, but aren't 10, 8 (E) -> 10, 8 (T) the correct values rather than 10, 8 (E) -> 8, 6.4 (T) ?

After all. If E says to T, travel that way at 0.6c for 10 years, won't T then have travelled 10 local years representing 8 of E's years ? He wont travel 8 years and look at E as having travelled 6.4 years ?

Did we introduce one two many ajustments somewhere earlier ?

JesseM

Oct4-06, 10:37 AM

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.JesseM. That seems like a very detailed and accurate representation of the theory. The solutions presented for these problems are very elegant. Almost an art form ?

At first I though this solved my problem as the figures do work out exactly right. Albeit difficult to get your head around at first.

But then ...

My problem is what the twins watches would show at each stage of the journey. The twins & their watches jump between f.o.r.'s at regular points so they dont' have the luxury of measuring time from one f.o.r. ? They don't have the luxury of calculating things as if they have a single rest frame that stays the same throughout the journey, but when I say things above like "x years have passed on his clock" I am talking about the actual time on each triplet's clocks, which is different from the amount of coordinate time that has passed in any particular frame of reference. The three coordinate times were 8 years, 17 years and 25 years; from the above, you can see that at the points on each triplet's worldlines that are assigned a time-coordinate of t=8 years in this frame (which includes the point on A's worldline where he turns around), triplet A's clock reads 8 years, triplet B's clock reads 7.2 years, and triplet C's clock reads 3.765 years; at the points on each triplet's worldlines that are assigned a time-coordinate of t=17 years in this frame (which includes the point on C's worldline where he turns around), triplet A's clock reads 8 + 4.235 = 12.235 years, triplet B's clock reads 7.2 + 6.4 = 13.6 years, and triplet C's clock reads 3.765 + 4.235 = 8 years; and at the points on each triplet's worldlines that are assigned a time-coordinate of t=25 years in this frame (which is the point on each triplet's worldline where they reunite at a single place and time on earth), triplet A's clock reads 12.235 + 3.765 = 16 years, triplet B's clock reads 13.6 + 6.4 = 20 years, and triplet C's clock reads 8 + 8 = 16 years. So, these numbers do verify that each triplet's clock reads 8 years at the point they turn around, and that at the point all three reunite 16 years have passed on the clocks of the two travelling triplets, while 20 years have passed on the clock of the earthbound triplet. If the two travelling twins (we wont call them triplets ?) Yes, my mistake, I should have said "triplet" rather than "twin" tin the section you quoted above. 'stop' relative to the Earth when they reach their turning points, what are the times on the three twins' watches ? At the point that a given triplet turns around, his own clock will read 8 years. But if you ask a question like "what time did triplet B's clock read at the same moment that triplet A turned around", the answer is different in different reference frames, because they define simultaneity differently. They will have jumped back into the Earth's f.o.r, relative speed now zero ? I think you're confusing yourself with this way of speaking--nobody "jumps into" one reference frame or another, a reference frame is just a coordinate system that we, the omniscient observer thinking about the problem as a whole, use to calculate things. A change in velocity may result in a ship having a different rest frame, but that doesn't mean its clocks will somehow reset to match the coordinate time of that frame or anything, and we have no obligation to use the ship's rest frame to calculate things like the amount of time that elapses on its own clock.

Again, the time between two given events on a clock's worldline, like the event of the clock leaving earth and the event of it instantaneously accelerating at some distance from earth, is known as the "proper time", and it's a frame-invariant quantity, meaning the answer you get for the proper time between two events along a particular worldline won't depend in any way on which reference frame you use to calculate it. If the part of the worldline between these two events consists of constant-velocity motion, as in this example, then you can calculate the proper time in whatever reference frame you (the omniscient observer) are using by multiplying (time between the events in terms of the time-coordinates of the f.o.r. you're using)*(time dilation factor based on clock's speed in the f.o.r. you're using). If the clock changed velocities between the two events you want to find the proper time between, then if the accelerations were instantaneous you can just repeat the above procedure for each constant-velocity segment to find the proper time between the beginning and end of each segment, and then add up all these individual proper times to find the total proper time; if the acceleration was non-instantaneous, with velocity as a function of time changing continuously according to some function v(t), then you'd have to do the integral I mentioned earlier, \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. The important thing is that when you pick two specific events on a clock's worldline, then when you calculate the proper time along the worldline between those two events, you will get the same answer regardless of what reference frame you use to calculate it.

JesseM

Oct4-06, 10:56 AM

Hold on ... rewind ... back to Earth twin E & Travelling twin T ...

E's time elapses 10 years with T's appearing as 8 years, from E's f.o.r.

In T's f.o.r. his time elapses 8 years, & E's elapses 6.4 years ...

Is this correct ? No, assuming E moved inertially while T accelerated at some point (even if instantaneously), there will be no inertial reference frame where the second is true.

M1keh

Oct5-06, 05:47 AM

No, assuming E moved inertially while T accelerated at some point (even if instantaneously), there will be no inertial reference frame where the second is true.

JesseM. Apologies. I'm getting confused now.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.

In the example, when we look at A's f.o.r, "A stays stationery for 8 years and 6.4 years have passed on B's watch" ?

Isn't that what we said earlier ?

So if we look at just A & B,

B sees himself stay stationery for 10 years and A's watch move 8 years.
A sees himself stay stationery for 8 years and B's watch move 6.4 years.

Is this right ?

Ah. Ok. I've worked through the figures from each of the three f.o.r's and answered the question for myself. I was just thinking out aloud ?

It all works perfectly well and exactly as you detailed. Thanks.

I need some thinking time to absorb this ... hurts between the ears, a bit.

M1keh

Oct5-06, 10:30 AM

...

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

Ok. Finally. After much head scratching and working this out from start to finish, I come up with exactly the same figures and I think I understand why.

It took a while as I couldn't believe it worked correctly. Something still didn't seem right, although the numbers proved it was.

But I've come up with another example ? This one's a slant on the original that I believe doesn't work ? I'm sure you'll prove me wrong ?

Suppose we have 3 Earths, separated by 6 ly (12 ly total). B is on E2, the Middle-Earth :-), A is on E1 and C is on E3.

Now all observers are synchronised in Earth's f.o.r, the same as the original example.

Both A & C leave at the same time and head towards E2 at 0.6c. When they get there they stop and compare watches.

How does the earlier method of explaining events work for this example ?

If you look at A's view for the 10 years, don't you get :

View From A’s Starting F.O.R.

First and only 8 years.

A remains stationery.

B[123] will travel at 0.6c and their watches will show 0.8*8years = 6.4 years. They will have travelled 0.6*8 = 4.8 ly.

C will travel towards A at (0.6+0.6)/(1+0.6^2) = 0.88235c, and his watch will show 0.4706*8years = 3.765 years. He will have travelled 0.88235*8 = 7.06 ly.

Am I messing this up somewhere ????

JesseM

Oct5-06, 10:30 AM

No, assuming E moved inertially while T accelerated at some point (even if instantaneously), there will be no inertial reference frame where the second is true. JesseM. Apologies. I'm getting confused now. Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock. In the example, when we look at A's f.o.r, "A stays stationery for 8 years and 6.4 years have passed on B's watch" ? Yes, but this was during a single section of the trip where neither A nor B accelerated, so they both had a single inertial rest frame. When you were talking about E and T before, I thought you were talking about a twin-paradox type situation where T travelled away, then turned around and returned to E so they could compare clocks at a single location...if you weren't talking about this scenario, then forget that comment. So if we look at just A & B,

B sees himself stay stationery for 10 years and A's watch move 8 years.
A sees himself stay stationery for 8 years and B's watch move 6.4 years.

Is this right ? Yup, that's what each one would observe in his own inertial rest frame, up until the moment A turned around.

pess5

Oct5-06, 03:20 PM

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

The relativistic contractions are both see'able and observable. Although, it wouldn't be easy by any stretch for tiny bodies. All normally spherical planets would appear ellipsoidal after accelerating to near light speeds.

One's proper length never changes with changes in one's own state of motion, but those of relative motion >0 will see contractions of you. They are real, even though a body never sees its own proper length (or tick of tock) change.

JesseM

Oct5-06, 05:09 PM

Ok. Finally. After much head scratching and working this out from start to finish, I come up with exactly the same figures and I think I understand why.

It took a while as I couldn't believe it worked correctly. Something still didn't seem right, although the numbers proved it was.

But I've come up with another example ? This one's a slant on the original that I believe doesn't work ? I'm sure you'll prove me wrong ?

Suppose we have 3 Earths, separated by 6 ly (12 ly total). B is on E2, the Middle-Earth :-), A is on E1 and C is on E3.

Now all observers are synchronised in Earth's f.o.r, the same as the original example.

Both A & C leave at the same time and head towards E2 at 0.6c. When they get there they stop and compare watches.

How does the earlier method of explaining events work for this example ?

If you look at A's view for the 10 years, don't you get :

View From A’s Starting F.O.R.

First and only 8 years.

A remains stationery.

B[123] will travel at 0.6c and their watches will show 0.8*8years = 6.4 years. They will have travelled 0.6*8 = 4.8 ly.

C will travel towards A at (0.6+0.6)/(1+0.6^2) = 0.88235c, and his watch will show 0.4706*8years = 3.765 years. He will have travelled 0.88235*8 = 7.06 ly.

Am I messing this up somewhere ???? Yes, in this case you need to take into account the fact that the frame where A is at rest as he approaches Earth2 defines simultaneity differently than does the rest frame of the three earths--in A's rest frame, Earth2's clock is ahead of the clock on Earth1 where he departed from, and Earth3's clock is further ahead still, and C departed from Earth3 well before A departed from Earth1. In general, if two clocks are at rest with respect to each other and synchronized in their rest frame, and the distance between them in their rest frame is L, then if you have an observer moving at velocity v with respect to them in the same direction as the line between the two clocks, in this observer's rest frame the back clock's time will be ahead of the front clock's time by vL/c^2. So in your example, A will observe E2's clock as being ahead of E1's by (6 l.y.)*(0.6c) = 3.6 years. So even though he only observe the clock of E2 advance by (8 years)*(0.8) = 6.4 years during the 8 years it took him to move from E1 to E2 according to his own clock, since E2's clock started out at 3.6 years at the beginning of the journey, at the end it would read 3.6 + 6.4 = 10 years.

Meanwhile, E3's clock already read 7.2 years at the moment A left E1, and since it is only ticking at 0.8 the normal rate, it would have read zero 7.2/0.8 = 9 years before A's departure. Since C left E3 when E3's clock read zero, this means that in the rest frame of A, C departed E3 9 years earlier than A left E1, which means C was travelling for 9 + 8 = 17 years before they finally met at E2. To figure out how fast C was moving in A's frame, we can use the velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) which tells us that in this frame C must have been moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, which gives a time dilation factor of 0.47059. So, in the 17 years between the time C departed and the time he met with A, his clock would have advanced forward by (17)*(0.47059) = 8 years.

If you don't want to have to make use of a bunch of specialized formulas (time dilation formula, length contraction formula, formula for how much moving clocks will appear out-of-sync, velocity addition formula), a more general way to approach any relativity problem is to use the Lorentz transformation, which transforms between the coordinates of an event in one inertial frame's coordinate system and the coordinates of the same event in other frame's system. If we have one frame which labels events with spatial coordinates x,y,z and time coordinate t, and another frame which uses spatial coordinates x',y',z' and time coordinate t', and the spatial origin of the second frame is moving along the first frame's x-axis with velocity v, with both origins coinciding at time t=t'=0, then the coordinate transform would be:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

In your problem, we can ignore the y and z directions and just concentrate on one space coordinate. Let's have the x,t coordinate system be the one in which all three earths are at rest, with x=0 being the position of E1 and t=0 being the time when A and C departed E1 and E3 in this frame. Then the x',t' coordinate system can be the one where A is at rest during the journey, and it will also have its origin at the event of A departing E1. So in the coordinate transform, we'd have v=0.6c, and \gamma = 1.25.

Now, it's easy to figure out the coordinates of various important events in the x,t system where the earths are at rest:

A leaves E1: x=0 l.y., t=0 years
C leaves E3: x=12 l.y., t=0 years
E2's clock reads 3.6 years: x=6 l.y., t=3.6 years
E3's clock reads 7.2 years: x=12 l.y., t=7.2 years

Now, use the formulas I gave above to find the corresponding x',t' coordinates of each event:

A leaves E1: x'=0 l.y., t'=0 years
C leaves E3: x'=15 l.y., t'=-9 years
E2's clock reads 3.6 years: x'=4.8 l.y., t'=0 years
E3's clock reads 7.2 years: x'=9.6 l.y., t'=0 years

So, figuring things out explicitly in terms of coordinates gives the same answers that were found earlier: in A's frame, C departed E3 nine years before A departed E1, and at the same moment that A departed E1, E2's clock read 3.6 years and E3's clock read 7.2 years. Note that I confirmed the times on E2 and E3 in a sort of backwards way, by already knowing the times they should read and then confirming that they read these times at t'=0 in A's frame, but a more logical way to do it would be to figure out the positions of E2 and E3 at t'=0 in A's frame (since they were 6 light years and 12 light years away in their own rest frame, Lorentz contraction tells us they'd be 6*0.8 = 4.8 light years and 12*0.8 = 9.6 light years away in A's frame at the moment he left E1), and then plug those coordinates into the reverse version of the coordinate transformation:

x = \gamma (x' + vt')
t = \gamma (t' + vx'/c^2)

That way, plugging in (x'=4.8 l.y., t'=0 years) and (x'=9.6 l.y., t'=0 years) would tell us that these events happened at t=3.6 years and t=7.2 years, respectively, in the earths' rest frame.

This was pretty involved, so as always, let me know if you have questions about any of the steps here.

pess5

Oct5-06, 11:14 PM

JesseM,

Is one able to post PDFs or MS WORD docs in this forum environment offhand? Reason I ask is because a Minkowski spacetime diagram paints a picture of a 1000 verbal posts. Is this possible?

jtbell

Oct5-06, 11:31 PM

Yes, it is possible to attach pictures etc., with restrictions on size. See the "Additional Options" section below the message-composition field when you reply or start a thread.

pess5

Oct5-06, 11:49 PM

For the benefit of all.

Here's a portable document file (PDF) I drafted a number of years back. It's a Minkowski worldline diagram of the twins scenario. A spacetime diagram paints a 1000 words.

This scenario is called "the handoff technique". Minkowski diagrams are truely designed to represent inertial reference frames, however the effects of acceleration may be extrapolated.

This scenario has 2 figures, O stationary and A stationary ...

1. Observer O & A flyby event, where they align their clocks at flyby.
2. Observer A & C flyby, whereby C aligns his clock to A's.
3. Clock comparison at final C & O flyby.

It represents the equivalent of an observer A departing from observer O with clocks aligned, then immediately returning at some point back to O for clock comparison, where accelerations are considered instantaneous.

I also show an observer B who represents a mirrored image of observer A, although this may be ignored far as the twins scenario is concerned. I use v=0.866c outbound & inbound since gamma is conveniently = 2.

JesseM

Oct6-06, 12:05 AM

This scenario is called "the handoff technique". Minkowski diagrams are truely designed to represent inertial only scenarios, however the effects of acceleration may be extrapolated. I'd modify that to say Minkowski diagrams are designed to represent inertial reference frames, but within such a frame you can certainly draw in the non-straight worldline of an accelerating object.

Would it be possible for you to translate the diagrams into jpgs or gifs? I have a mac and I don't know what application to use for the doc you attached. But if it'd be much trouble don't worry about it, presumably most other people don't have this problem.

pess5

Oct6-06, 12:37 AM

I'd modify that to say Minkowski diagrams are designed to represent inertial reference frames, but within such a frame you can certainly draw in the non-straight worldline of an accelerating object.

done.

Would it be possible for you to translate the diagrams into jpgs or gifs? I have a mac and I don't know what application to use for the doc you attached. But if it'd be much trouble don't worry about it, presumably most other people don't have this problem.

done, but I reposted it in PDF. I figure most folks have the Adobe Acrobat Reader, however the MS WORD 97 doc sure looks cleaner. But PDF isn't bad though.

JesseM

Oct6-06, 12:43 AM

done, but I reposted it in PDF. I figure most folks have the Adobe Acrobat Reader, however the MS WORD 97 doc sure looks cleaner. But PDF isn't bad though. Thanks! Very nice (and detailed) diagrams there.

M1keh

Oct6-06, 06:43 AM

Yes, in this case you need to take into account the fact that the frame where A is at rest as he approaches Earth2 defines simultaneity differently than does the rest frame of the three earths--in A's rest frame, Earth2's clock is ahead of the clock on Earth1 where he departed from, and Earth3's clock is further ahead still, and C departed from Earth3 well before A departed from Earth1.

Ouch. Didn't see that one coming. How can that be ?

If A starts on E1 and C starts on E3 and E1-3 are 'stationery', A & C start at exactly the same f.o.r as E1-3 and all of their clocks are sync'd ?

When A accelerates (instantly) to 0.6c. His time will dilate by a factor of 0.8 compared with all 3 ? The distance will also compress by a factor of 1.25 compared with all 3 ?

In general, if two clocks are at rest with respect to each other and synchronized in their rest frame, and the distance between them in their rest frame is L, then if you have an observer moving at velocity v with respect to them in the same direction as the line between the two clocks, in this observer's rest frame the back clock's time will be ahead of the front clock's time by vL/c^2. So in your example, A will observe E2's clock as being ahead of E1's by (6 l.y.)*(0.6c) = 3.6 years. So even though he only observe the clock of E2 advance by (8 years)*(0.8) = 6.4 years during the 8 years it took him to move from E1 to E2 according to his own clock, since E2's clock started out at 3.6 years at the beginning of the journey, at the end it would read 3.6 + 6.4 = 10 years.

Ok. Now I'm struggling to keep up.

Before the journey starts, tA = tE1 = tE2 = tE3 = tC ?

At what point does A see E2's time jump ahead by 3.6 years ? and E3's time jump ahead by 7.2 years ?

The only 'rules' I've seen so far state that time slows down, it doesn't jump ?

Am I missing a 'rule' somewhere ? Or misunderstanding what you're saying ?

Meanwhile, E3's clock already read 7.2 years at the moment A left E1, and since it is only ticking at 0.8 the normal rate, it would have read zero 7.2/0.8 = 9 years before A's departure. Since C left E3 when E3's clock read zero, this means that in the rest frame of A, C departed E3 9 years earlier than A left E1, which means C was travelling for 9 + 8 = 17 years before they finally met at E2. To figure out how fast C was moving in A's frame, we can use the velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) which tells us that in this frame C must have been moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, which gives a time dilation factor of 0.47059. So, in the 17 years between the time C departed and the time he met with A, his clock would have advanced forward by (17)*(0.47059) = 8 years.

If you don't want to have to make use of a bunch of specialized formulas (time dilation formula, length contraction formula, formula for how much moving clocks will appear out-of-sync, velocity addition formula), a more general way to approach any relativity problem is to use the Lorentz transformation, which transforms between the coordinates of an event in one inertial frame's coordinate system and the coordinates of the same event in other frame's system. If we have one frame which labels events with spatial coordinates x,y,z and time coordinate t, and another frame which uses spatial coordinates x',y',z' and time coordinate t', and the spatial origin of the second frame is moving along the first frame's x-axis with velocity v, with both origins coinciding at time t=t'=0, then the coordinate transform would be:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

In your problem, we can ignore the y and z directions and just concentrate on one space coordinate. Let's have the x,t coordinate system be the one in which all three earths are at rest, with x=0 being the position of E1 and t=0 being the time when A and C departed E1 and E3 in this frame. Then the x',t' coordinate system can be the one where A is at rest during the journey, and it will also have its origin at the event of A departing E1. So in the coordinate transform, we'd have v=0.6c, and \gamma = 1.25.

Now, it's easy to figure out the coordinates of various important events in the x,t system where the earths are at rest:

A leaves E1: x=0 l.y., t=0 years
C leaves E3: x=12 l.y., t=0 years
E2's clock reads 3.6 years: x=6 l.y., t=3.6 years
E3's clock reads 7.2 years: x=12 l.y., t=7.2 years

Now, use the formulas I gave above to find the corresponding x',t' coordinates of each event:

A leaves E1: x'=0 l.y., t'=0 years
C leaves E3: x'=15 l.y., t'=-9 years
E2's clock reads 3.6 years: x'=4.8 l.y., t'=0 years
E3's clock reads 7.2 years: x'=9.6 l.y., t'=0 years

So, figuring things out explicitly in terms of coordinates gives the same answers that were found earlier: in A's frame, C departed E3 nine years before A departed E1, and at the same moment that A departed E1, E2's clock read 3.6 years and E3's clock read 7.2 years. Note that I confirmed the times on E2 and E3 in a sort of backwards way, by already knowing the times they should read and then confirming that they read these times at t'=0 in A's frame, but a more logical way to do it would be to figure out the positions of E2 and E3 at t'=0 in A's frame (since they were 6 light years and 12 light years away in their own rest frame, Lorentz contraction tells us they'd be 6*0.8 = 4.8 light years and 12*0.8 = 9.6 light years away in A's frame at the moment he left E1), and then plug those coordinates into the reverse version of the coordinate transformation:

x = \gamma (x' + vt')
t = \gamma (t' + vx'/c^2)

That way, plugging in (x'=4.8 l.y., t'=0 years) and (x'=9.6 l.y., t'=0 years) would tell us that these events happened at t=3.6 years and t=7.2 years, respectively, in the earths' rest frame.

This was pretty involved, so as always, let me know if you have questions about any of the steps here.

That'll take a while to sink in !

pess5

Oct6-06, 10:08 AM

JesseM,

I've reposted the Handoff Scenario rev c (PDF attachment) with a few cleanup tweeks at the original post ...

http://www.physicsforums.com/showpost.php?p=1104227&postcount=68

pess

pess5

Oct6-06, 10:15 AM

Where discussing the twins scenario gets real interesting, is when one addresses the what the non-inertial twin sees of the inertial twin, especially over periods of active acceleration. Also, how the spacetime grid of one twin frame maps into the spacetime grid of the other.

M1keh

Oct6-06, 10:17 AM

M1keh

Oct6-06, 10:35 AM

JesseM,

I've reposted the Handoff Scenario rev b (PDF attachment) with a few cleanup tweeks at the original post ...

http://www.physicsforums.com/showpost.php?p=1104227&postcount=68

pess

pess. What about the scenario I mentioned earlier ?

If the twins start from opposite sides of the Earth, 6ly away (Earth distance) and stationary compared with the Earth. They then accelerate to 0.6c and travel to Earth. What are the time lines ?

This way they have the perfect hand-off as they all meet at Earth at the same time, or not, or they do, or they don't ???

jtbell

Oct6-06, 10:36 AM

Ouch. Didn't see that one coming. How can that be ?

Don't feel too bad about it. You're yet another victim of incomplete presentations of basic relativistic phenomena in popular-level science books, and even in many general-physics textbooks at the college/university level.

They all talk about length contraction and time dilation, but most of them either don't mention relativity of simultaneity at all, or present it in a way that doesn't make clear from the beginning that it is at the same level of importance as length contraction and time dilation.

Start with a line of clocks, all stationary in inertial reference frame S, and all synchronized in S. In reference frame S' which is moving along that line with respect to S, three things about these clocks are different:

1. The clocks are closer together than in S: length contraction.

2. The clocks tick more slowly than an identical clock which is at rest in S': time dilation.

3. The clocks are not synchronized in S': relativity of simultaneity.

In general, you need to take all three of these phenomena into account, not just the first two.

pess5

Oct6-06, 11:02 AM

Mikeh,

I just retweeked it again, so please reprint the updated (newer) rev b version I just posted. You got the 1st rev b. See ...

http://www.physicsforums.com/showpost.php?p=1104227&postcount=68

In this 2nd rev c version, I added a line-of-simultaneity in the bottom figure. This in relation to JesseM's point regarding synchronized clocks appearing to lag each other in clock readouts per a moving observer vantage. You might want to print this post out when looking at the figure printout...

The top figure shows the O stationary perspectve. O's line-of-simultaneity is horizontal, and 12:34am-C, 1:09am-O, 12:34am-B all ly on the horizontal black line. A horizontal line-of-simultaneity is a line representing NOW across space for the stationary observer perspective at some moment. So in this case, O always sees B & C clocks aligned in time readout at any moment, ie they appear synchronized per O.

The bottom figure shows the A stationary perspective. O's line-of-simultaneity is slanted because O is in motion with slanted worldline, and 12:34am-C, 1:09am-O, 12:34am-B all ly on the slanted gray line. This represents O's sense of NOW at 12:09am-O. A slanted line-of-simultaneity is a line representing NOW across space for the moving observer perspective. Note that the observers C,O,B times which intersect this slanted line-of-simultaneity match that of the upper figure!

To determine a moving observer's line-of-simultaneity, draw his x'-axis such that the angle it makes with the 45 deg light beam is identical to the angle his slanted worldline makes with the 45 deg light beam, but on the other side of the light beam. This is to say ...

the faster an observer moves, the more slanted his worldline becomes. Light paths are always 45 deg on Minkowski worldline diagrams, since they travel one unit space per 1 unit time, and units of space & time are set equal (normalized), and so speed c=1. The more a moving observer's worldline tilts away from your own stationary vertical worldline (ie your time axis), the more his x'-axis tilts upward toward the 45 deg ray path from your x-axis. That x'-axis is his line-of-simultaneity, his sense of NOW!

Note that in the top diagram, O stationary, the observer A & B clocks read the same in that O instant of 1:09am-O. In the bottom A stationary figure, observers B & C never possess the same time readouts in any observer A moment, because B & C are in motion. The most aft clock (C's) reads well ahead of the foremost clock (B's). A's line-of-simultaneity is of course horizontal in the lower figure, since that's an A stationary vantage. Lines-of-simultaneity are your spatial axis, so x-axis for one guy, x'-axis for the other guy. Minkowski spacetime diagrams portray 1-space vs time.

pess

pess5

Oct6-06, 12:11 PM

pess. What about the scenario I mentioned earlier ?

If the twins start from opposite sides of the Earth, 6ly away (Earth distance) and stationary compared with the Earth. They then accelerate to 0.6c and travel to Earth. What are the time lines ?

This way they have the perfect hand-off as they all meet at Earth at the same time, or not, or they do, or they don't ???

Mikeh,

Ahhh. Assuming all clocks are in sync from the start, and accelerate/decelerate identically, the twins will arrive on earth with identical clock readings. However their clocks will lag behind the earth clock. If we assume the magical instant accelrations, each twin traverses 0.8(6ly)=4.8ly, since a v=0.6 produces a contraction of 0.8. So 4.8ly/0.6c=8yr, thus the twins age 8 yrs in their trek to earth.

Earthlings see it differently of course. They have to await the twins to transcend 6ly at 0.6c, so 6ly/0.6c=10 yr. So although the twins only age 8 yr over the trek, the earthlings age 10 yr. Personally, I'd rather age 10 yr on earth with my friends than age 8 yr alone in a tin can alone.

I know you been thru this with JesseM already with all the math, but I'm just shooting from the cuff here.

PS ... my twins handoff scenario PDF is now at a rev c. That's the best one you'll want to look at. thanx.

pess

pess5

Oct6-06, 01:17 PM

robphy

Oct6-06, 01:56 PM

Please see the following PDF attachment. Goes back a number of years. It's a scenario depicting vulcano eruptions on Mt Hood & Mt Ranier and investigates how a seismologist & passing spaceship capt witness the events.

Hmmm... sounds like something from the Physics Education Group at U. Washington.

JesseM

Oct6-06, 03:17 PM

Ouch. Didn't see that one coming. How can that be ?

If A starts on E1 and C starts on E3 and E1-3 are 'stationery', A & C start at exactly the same f.o.r as E1-3 and all of their clocks are sync'd ? The difference in synchronizations shouldn't exactly be thought of as a physical effect--if you accelerate and then begin moving inertially again, your rulers and clocks can generally assumed to now measure length and time the same as rulers and clocks that have always been at rest in your current inertial rest frame, but clocks on different parts of your ship will not automatically by synchronized the same way as your current rest frame's definition of what it means for different clocks to be "synchronized".

Ultimately, all the special equations we've used--lorentz contraction equation, time dilation equation, velocity addition equation, and that formula for how clocks will be out-of-sync in frames where they're moving if they're in-sync in their own rest frame--are derived from the general coordinate transformation I gave you, which is known as the "Lorentz transformation". So it might help if I gave you an idea of what assumptions are needed to derive the Lorentz transformation in the first place. Basically, Einstein started out by assuming each inertial observer should define the coordinates of any event in terms of local measurements made on a system of rulers and synchronized clocks which are at rest with respect to that observer. So, for example, if I look through my telescope and see a firecracker going off in the distance, then if I see that it happened right next to the 300-meter mark on a ruler laid out parallel to my x-axis, I'll assign the event an x-coordinate of 300 meters, and if I see that as the explosion was happening the clock at the 300-meter mark was reading 25 seconds, I'll assign the event a t-coordinate of 25 seconds. The fact that I'm relying only on local measurements made in the immediate area of the event means I don't have to worry about the delay between the time an event happens and the time I actually see it, due to the finite speed of light.

But this brings up a new problem--what does it mean for clocks at different positions in my system of rulers and clocks to be "synchronized"? The most obvious method of synchronizing distant clocks is to move them to a common position, synchronize them right next to each other, and then move them apart again. But by the time Einstein was working on SR, it had already been suggested by Lorentz that moving clocks might slow down, in order to account for the results of the Michelson-Morley experiment (http://en.wikipedia.org/wiki/Michelson-Morley_experiment). The story behind this is that physicists had always imagined that light only travelled at exactly c in the rest frame of the "ether", which was supposed to be the medium that light waves were vibrations in just like sound waves are vibrations in the air, and that if you were moving at velocity v relative to the ether, you'd measure light to move at c+v in one direction and c-v in the other, just as would be true if you measured sound waves while in motion relative to the rest frame of the air around you. But the Michelson-Morley experiment had shown that light was measured to move at c in all directions at different points in the earth's orbit, when its velocity relative to any inertial frame like the frame of the ether should have been noticeably different, and Lorentz found that this could be accounted for if you imagine that rulers shrunk when moving relative to the ether, and clocks slowed down.

So back to clock synchronization: if clocks change their rate of ticking depending on how they move, this causes problems for the obvious method of synchronizing clocks at a common location and then moving them apart, because they could get out of sync during the process of moving them. Einstein suggested a different method: what if each observer synchronizes different clocks in their system using light-signals, making the assumption that light travels at the same speed in all directions in their own rest frame? (This assumption wouldn't make any sense if you believed in the ether, but the results of making this assumption led to some interesting results which made Einstein think the ether should be abandoned altogether.) If each inertial observer makes this assumption, they could each synchronize their own clocks by setting off a light flash at the midpoint of two clocks, then setting both clocks to read the same time at the moment the light from the flash reaches them.

But a little thought shows that this method must lead different inertial observer to disagree about simultaneity. Suppose I am in a ship which is moving forward in your rest frame, and I decide to synchronize two clocks at the front and back of my ship using this method of setting off a flash at the center and setting them to read the same time when the light reaches them. In your frame, both clocks were still the same distance from the point where the flash was set off, but the clock at the front of the ship is moving away from that point as the ship moves forward, while the clock at the back is moving towards that point, so naturally if you assume that both light beams move at the same speed relative to yourself, you must conclude that the light beam moving towards the back clock caught up with that clock at an earlier time than the other light beam caught up with the front clock. But if I set both clocks to read the same time when the light reached them, naturally this leads you to conclude that my front clock is running behind my back clock! As long as each observer assumes light moves at the same speed in all directions in their own rest frame, there is no way to avoid this sort of disagreement about simultaneity.

Note that this definition of what it means for two clocks at rest in a given frame to be "synchronized" in that frame is not really forced on you by nature--you'd be free to use a different method of synchronization if you wanted, like having one special observer synchronize his clocks using this light-signal method, but then having every other observer synchronize their clocks so that they'd agree with the special observer's definition of simultaneity. For this reason, the standard synchronization method in special relativity is known as the "Einstein synchronization convention." But it's not arbitrary either, there is in fact a good reason to prefer this method of synchronization to any other. If two inertial observers both define their coordinate systems in terms of measurements made on the sort of ruler-clock system described by Einstein, then if the first observer figures out the correct equations to describe the laws of physics in terms of his coordinates x,y,z,t, and the second observer figures out the correct equations for the laws of physics in terms of his own coordinates x',y',z',t', they will find that they are both using identical equations, save for the replacement of x with x' and y with y' and z with z' and t with t'. This would not be true if they had used a different clock synchronization convention in defining their coordinate systems. It's a particular feature of the laws of physics in our universe that they all have the property of looking the same in the different coordinate systems constructed in this way, known as "Lorentz-invariance" (and you can show that these coordinate systems will be related by the Lorentz transformation which I mentioned before). It's this Lorentz-invariance of all the known fundamental laws which makes this the most "natural" way to define the coordinate systems of inertial observers, including each system's differing definition of simultaneity.

As always, let me know if you have questions about any of these ideas. Conceptually, it may also help you to learn more about "Minkowski diagrams", which show the x and t axes of different reference frames in a single diagram (lines of simultaneity, or constant t-coordinate, for different frames will appear slanted with respect to one another, for example), along with the worldlines of different objects. The ones pess5 provided are good examples, and you might want to find some general tutorial as well. When A accelerates (instantly) to 0.6c. His time will dilate by a factor of 0.8 compared with all 3 ? The distance will also compress by a factor of 1.25 compared with all 3 ? By "all 3" you mean the 3 planets, not including C whose velocity relative to A is different from that of the planets, right? If so, yes, in the rest frame of the planets A's clock will be compressed to 0.8 the length they were previously, and the length of his rulers will be divided by 1.25 (or shrunk by a factor of 0.8, equivalently). Ok. Now I'm struggling to keep up.

Before the journey starts, tA = tE1 = tE2 = tE3 = tC ?

At what point does A see E2's time jump ahead by 3.6 years ? and E3's time jump ahead by 7.2 years ? Again, it's best not to think of the differences in simultaneity as this sort of physical effect. If you imagine two networks of rulers and synchronized clocks moving past each other inertially, one at rest in the earths' frame and one which will be at rest with respect to A after his acceleration, then you can just imagine A switching which network of rulers/clocks he uses to measure distances and times before and after he accelerates, since he always wants to make measurements on a system that is currently at rest with respect to him.

As I've said in earlier posts, I think the best course is to avoid trying to think about the "point of view" of a non-inertial observer altogether, and just analyze the problem from start to finish in a single inertial frame (after which you can of course go back and analyze it from start to finish in a different frame, the point is not to try to switch frames in the middle).

pess5

Oct6-06, 04:07 PM

Hmmm... sounds like something from the Physics Education Group at U. Washington.

Sounds like you didn't like the scenario much? I thought it was pretty decent.

I just looked up the web site for Physics Education Group at U. Washington here, and you were right of course. I debated this scenario online long ago, but couldn't remember what university had it posted. At the time, we debated about 8 or 9 presumed paradoxes of relativity theory. As it turned out, none were paradoxes after all. The error always arises from injecting absolute simultaneity midstream into a relativistic scenario, whereby the debates always began assuming the postulates true. Anyways, thanx much. It's an excellent scenario for grasping relative simultaneity. Here's the link...

http://arxiv.org/ftp/physics/papers/0207/0207081.pdf

See Section VI ... Assessing Student Understanding of Simultaniety

pess

robphy

Oct7-06, 05:06 AM

Sounds like you didn't like the scenario much? I thought it was pretty decent.

I just made the comment to note a similarity.
It wasn't intended to complain about the scenario.
(I haven't had a chance to really look at it in detail...been a little busy.)

M1keh

Oct9-06, 01:28 AM

Don't feel too bad about it. You're yet another victim of incomplete presentations of basic relativistic phenomena in popular-level science books, and even in many general-physics textbooks at the college/university level.

They all talk about length contraction and time dilation, but most of them either don't mention relativity of simultaneity at all, or present it in a way that doesn't make clear from the beginning that it is at the same level of importance as length contraction and time dilation.

Start with a line of clocks, all stationary in inertial reference frame S, and all synchronized in S. In reference frame S' which is moving along that line with respect to S, three things about these clocks are different:

1. The clocks are closer together than in S: length contraction.

2. The clocks tick more slowly than an identical clock which is at rest in S': time dilation.

3. The clocks are not synchronized in S': relativity of simultaneity.

In general, you need to take all three of these phenomena into account, not just the first two.

Thanks for the reply.

The sqrt( 1 - (v**2/c**2) ) accounts for 1 & 2. I think JesseM gave me the formula for 3. I'll check.

Now, I'm probably wrong, but my understanding was that two observers in different f.o.r's can synchronise their watches as they pass each other, using light singals ? So they can confirm that their watches show the same times ?

This presumably also confirms that they ARE at the same place ?

jtbell

Oct9-06, 06:33 AM

Yes, two observers (each carrying a clock with her) that pass each other closely (ideally two pointlike observers passing infinitesimally close to each other) can unambiguously synchronize their clocks at the instant that they pass. Of course, the clocks go out of sync again immediately, because they run at different rates in any inertial reference frame.

Similarly, even if the two observers don't synchronize their clocks as described above, they always agree on the readings of each others' clocks at the instant that they pass.

For that matter, if any two moving clocks pass each other infinitesimally closely, all observers agree on the readings of those two clocks at the instant that they pass.

M1keh

Oct9-06, 06:45 AM

jtbell

Oct9-06, 09:42 AM

M1keh

Oct11-06, 12:30 AM

Yes, all observers who are at rest in the same inertial reference frame, and therefore at rest with respect to each other, can synchronize their clocks so that they all agree on what time it is, in that inertial reference frame.

In any other inertial reference frame, those clocks will not be synchronized.

Ok. Here goes.

If A1, A2 & A3 are all in the same f.o.r, but 6ly apart, and B1, B2, B3 are all travelling at 0.6c relative to the A's but, locally, 4.8ly apart, they could reach B1, B2 & B3 respectively, simultaneously. B1, B2 & B3 could then synchronise their watches with A1, A2 & A3 respectively.

Now A1, A2 & A3 must agree on their local time & B1, B2 & B3 must agree on their local time. If A3 agree's with B3's time (now synchronised) A1 must agree with B3's time ?

What am I missing ?

JesseM

Oct11-06, 12:55 AM

Ok. Here goes.

If A1, A2 & A3 are all in the same f.o.r, but 6ly apart, and B1, B2, B3 are all travelling at 0.6c relative to the A's but, locally, 4.8ly apart, they could reach B1, B2 & B3 respectively, simultaneously. Only in the B frame will they pass the A clocks simultaneously. In the A frame, the B clocks are 4.8*0.8 = 3.84 ly apart while the A clocks are 6 ly apart, so each B clock will pass its respective A clock at a different time. B1, B2 & B3 could then synchronise their watches with A1, A2 & A3 respectively. If they do that, then their watches will no longer be synchronized in their own rest frame. Your definition of simultaneity is not based on what your clocks actually happen to read, since your clocks could very well be out of sync, it's based on what an ideal network of clocks at rest in your frame and synchronized according to Einstein's clock synchronization procedure would define as simultaneous. Now A1, A2 & A3 must agree on their local time & B1, B2 & B3 must agree on their local time. If A3 agree's with B3's time (now synchronised) A1 must agree with B3's time ? In the A frame, the B clocks pass their corresponding A clocks at different times, as I mentioned above, so even if each B clock sets itself to the same time as the A clock at the moment it passes it, since the B clocks are still ticking slow in A's frame, the B clocks won't be synchronized (first B1 passes A1 and sets itself to the same time as all the A clocks, but then 3.6 years later in the A frame, B2 passes A2 and sets itself to the same time as the current reading on all the A clocks, but B1 has only advanced by 3.6*0.8 = 2.88 years in this time so it's 3.6-2.88 = 0.72 years behind B2, and likewise B2 will be 0.72 years behind B3 when it sets itself to the same time as B3). Meanwhile, in the B frame all three B clocks pass their A clocks simultaneously, but the three A clocks are out of sync in this frame--as I mentioned before, if two clocks are x distance apart and synchronized in their own rest frame, then in a frame where they're moving at v along the axis joining them, the back clock's time will be ahead of the front ones by vx/c^2. So since the A clocks are 6 ly apart and moving at 0.6c in the B frame, in this frame each one will be out of sync with the one ahead of it by (0.6 ly/year)*(6 ly)/(1 ly/year)^2 = 3.6 years. So when the B clocks synchronize with the A clocks as they pass, they'll now each be 3.6 years out of sync from their nearest neighbor in their own rest frame.

M1keh

Oct11-06, 01:32 AM

... if two clocks are x distance apart and synchronized in their own rest frame, then in a frame where they're moving at v along the axis joining them, the back clock's time will be ahead of the front ones by vx/c^2. So since the A clocks are 6 ly apart and moving at 0.6c in the B frame, in this frame each one will be out of sync with the one ahead of it by (0.6 ly/year)*(6 ly)/(1 ly/year)^2 = 3.6 years. So when the B clocks synchronize with the A clocks as they pass, they'll now each be 3.6 years out of sync from their nearest neighbor in their own rest frame.

Ok. I'm failing to tie these together somewhere ?

Let's check it step by step ? Maybe my original question is wrong ?

What are the distances viewed by each of the two sets of observers ?

If the A's see each other as 6ly apart, then the B's will see the A's as 4.8 ly apart, their measure of A's distances being contracted by a factor of 1.25 ?

What will the B's measure the distance between the A's as ?

Length contraction is a constant factor ? ie. 1.25 at 0.6c ? It's not affected by distance from the object ?

No. This must be wrong. Where am I making the mistake ?

JesseM

Oct11-06, 02:34 AM

What are the distances viewed by each of the two sets of observers ?

If the A's see each other as 6ly apart, then the B's will see the A's as 4.8 ly apart, their measure of A's distances being contracted by a factor of 1.25 ?

What will the B's measure the distance between the A's as ? As you say, it'll be 6 ly/1.25, or 4.8 ly. That's why, as I said, in the B frame each B clock will pass its corresponding A clock simultaneously--if the Bs do not reset their clocks, each will read the same time at the moment it passes its A clock. But in this frame the A clocks are out of sync, so despite the fact that they each pass their B clock at the same time in this frame, each A clock reads a different time at the moment it passes its B clock (in the A frame, the explanation for this is that the three passing events actually happen at different times, not simultaneously). Length contraction is a constant factor ? ie. 1.25 at 0.6c ? It's not affected by distance from the object ? No, it's not affected by distance. No. This must be wrong. Where am I making the mistake ? Why do you think you are making a mistake? Do you disagree with any part of my description of how the three passing events look in the two different frames?

jtbell

Oct11-06, 03:02 PM

For what it's worth, here are a couple of diagrams of the situation being discussed. The simpler one is frameB.gif, which shows the two sets of clocks passing each other simultaneously in frame B. I've let this be time 0 on the B clocks, and I've arranged things so clocks A1 and B1 both read zero. A2 and A3 have nonzero readings because I've assumed that the A clocks are synchronized in frame A.

In frameA.gif, I show the two sets of clocks at three different times in frame A, namely the points in time at which each of the three pairs of A & B clocks pass each other.

I think this is the same situation that M1keh and JesseM are discussing, except that I have not synchronized B1, B2 and B3 with A1, A2 and A3 when they pass each other in frame B, because that would put B1, B2 and B3 out of sync with each other. The A clocks are synchronized in A, and the B clocks are synchronized in B, as is usual in thought experiments like this.

M1keh

Oct11-06, 11:24 PM

As you say, it'll be 6 ly/1.25, or 4.8 ly. That's why, as I said, in the B frame each B clock will pass its corresponding A clock simultaneously--if the Bs do not reset their clocks, each will read the same time at the moment it passes its A clock. But in this frame the A clocks are out of sync, so despite the fact that they each pass their B clock at the same time in this frame, each A clock reads a different time at the moment it passes its B clock (in the A frame, the explanation for this is that the three passing events actually happen at different times, not simultaneously). No, it's not affected by distance. Why do you think you are making a mistake? Do you disagree with any part of my description of how the three passing events look in the two different frames?

Ok. Penny's dropped. I see how this should work. But ...

If the A's see the B's as 6ly apart and travelling at 0.6c, in A's f.o.r surely the B's must reach the next 6ly stage in 10 years ? Each arriving at the same time ?

And ...

If, as they reach the A's the B's 'stop', or effectively accelerate to match the A's velocities, they'd be at the same place as the A's, but now within the A's f.o.r ? They'd see each other as at their respective A's ?

What would the A's see ?

M1keh

Oct11-06, 11:30 PM

jtbell

Oct12-06, 12:09 AM

Why are the diagrams for A's f.o.r and B's f.o.r so different? What's so 'special' about B's f.o.r ?

You specified that the "A clocks" and the "B clocks" should have equal spacings in frame B (4.8 ly), so that in frame B at one particular moment, which I chose to be time 0 in frame B, the following three events would occur simultaneously:

A1 passes B1;
A2 passes B2; and
A3 passes B3.

Therefore I could draw a single diagram that shows all three events happening at time 0 in frame B.

Because of relativity of simultaneity, these three events do not occur simultaneously in frame A. Therefore I had to make three diagrams for frame A, representing three different times in frame A:

time 0, when A1 passes B1;
time 3.6 years, when A2 passes B2; and
time 7.2 years, when A3 passes B3.

JesseM

Oct12-06, 12:17 AM

If the A's see the B's as 6ly apart and travelling at 0.6c, They don't. You said the B's were 4.8 ly apart in their rest frame, so in the A rest frame they are 4.8/1.25 = 3.84 ly apart. Remember, different observers don't agree on the ratios between their clock ticks/ruler lengths--if you measure my rulers as shrunk relative to yours, that doesn't mean I measure your rulers expanded relative to mine, I measure yours as shorter than my own. The laws of physics have to work the same way in every frame, so each observer must measure rulers shrunk by a factor of \sqrt{1 - v^2/c^2}, where v is the velocity of the ruler in the observer's own rest frame. If, as they reach the A's the B's 'stop', or effectively accelerate to match the A's velocities, they'd be at the same place as the A's, but now within the A's f.o.r ? They'd see each other as at their respective A's ?

What would the A's see ? Again, in the A frame the Bs are only 3.84 ly apart while the As are 6 ly apart, so when B1 arrives at A1, B2 will still be 6 - 3.84 = 2.16 ly away from A2, and at 0.6c it'll take 2.16/0.6 = 3.6 more years to reach A2. Likewise, when B2 catches up with A2, B3 is still 2.16 ly away from B3, and it takes 3.6 more years to catch up with B3. If the Bs stop when they reach their corresponding As, each B will read the same time at the moment it reaches its A, since in the A frame each B's clock is ahead of the next one by 3.6 years (again, using the formula vx/c^2, where x is the distance between the Bs in their own rest frame and v is their velocity in the As' frame). Of course, in the frame where the Bs were at rest before they matched velocities with the As, the reason they all read the same time at the moment they reach their respective A is because each B reaches its A at the same moment, and the B clocks are synchronized in this frame.

It doesn't really make sense to ask what the B's "see" from the beginning to the end of this problem if they stop in the middle (except in a purely local sense of what each B will see on the A right next to it), because they change velocities and thus they don't have a single inertial rest frame throughout the problem. You can talk about what is measured in the frame where they are at rest before stopping in A's frame, and you can talk about what is measured in the frame where they are at rest after stopping in A's frame, but in each case what you're really talking about is what would be measured by a grid of rulers and clocks that were at rest in that frame for all time (and synchronized using the Einstein clock synchronization convention).

M1keh

Oct12-06, 12:58 AM

They don't. You said the B's were 4.8 ly apart in their rest frame, so in the A rest frame they are 4.8/1.25 = 3.84 ly apart. Remember, different observers don't agree on the ratios between their clock ticks/ruler lengths--if you measure my rulers as shrunk relative to yours, that doesn't mean I measure your rulers expanded relative to mine, I measure yours as shorter than my own. The laws of physics have to work the same way in every frame, so each observer must measure rulers shrunk by a factor of \sqrt{1 - v^2/c^2}, where v is the velocity of the ruler in the observer's own rest frame.

Ok. I sort of get that, I think.

However, if the laws of physics have to be the same in both f.o.r's why does A's estimate of B's measurement of distance exactly match B's measurement, ie. both come to 4.8ly, but B's estimate of A's measurement of distance is nowhere near A's measurement, ie. 3.84ly & 6ly ?

Isn't this a difference in the laws of physics between the two f.o.r's ?

Again, in the A frame the Bs are only 3.84 ly apart while the As are 6 ly apart, so when B1 arrives at A1, B2 will still be 6 - 3.84 = 2.16 ly away from A2, and at 0.6c it'll take 2.16/0.6 = 3.6 more years to reach A2. Likewise, when B2 catches up with A2, B3 is still 2.16 ly away from B3, and it takes 3.6 more years to catch up with B3.

Ok. Yes that fits.

If the Bs stop when they reach their corresponding As, each B will read the same time at the moment it reaches its A, since in the A frame each B's clock is ahead of the next one by 3.6 years (again, using the formula vx/c^2, where x is the distance between the Bs in their own rest frame and v is their velocity in the As' frame). Of course, in the frame where the Bs were at rest before they matched velocities with the As, the reason they all read the same time at the moment they reach their respective A is because each B reaches its A at the same moment, and the B clocks are synchronized in this frame.

It doesn't really make sense to ask what the B's "see" from the beginning to the end of this problem if they stop in the middle (except in a purely local sense of what each B will see on the A right next to it), because they change velocities and thus they don't have a single inertial rest frame throughout the problem. You can talk about what is measured in the frame where they are at rest before stopping in A's frame, and you can talk about what is measured in the frame where they are at rest after stopping in A's frame, but in each case what you're really talking about is what would be measured by a grid of rulers and clocks that were at rest in that frame for all time (and synchronized using the Einstein clock synchronization convention).

Ok.

1. what is measured in the frame where they are at rest before stopping in A's frame ?

2. what is measured in the frame where they are at rest after stopping in A's frame ?

3. what do the watches on the wrists of the 6 observers show at both 1 & 2 ?

Presumably, 3 is a reasonable question ? It does mean that an explanation is required of what happens as the B's move from their original f.o.r to A's f.o.r, but we're not saying that's impossible ?

JesseM

Oct12-06, 01:16 AM

Ok. I sort of get that, I think.

However, if the laws of physics have to be the same in both f.o.r's why does A's estimate of B's measurement of distance exactly match B's measurement, ie. both come to 4.8ly, but B's estimate of A's measurement of distance is nowhere near A's measurement, ie. 3.84ly & 6ly ?

Isn't this a difference in the laws of physics between the two f.o.r's ? No, the fact that the distances match in one frame is just a property of the particular physical setup of the objects you're analyzing, the laws of physics are general equations that aren't specific to any particular physical setup, they describe things like how an arbitrary clock will slow down as a function of velocity. To take an analogy from Newtonian physics, if two cars are travelling at 100 mph in opposite directions in my frame, their speeds are the same in my frame, but in the frame of someone moving at 50 mph relative to me, one car is moving at 50 mph and the other at 150 mph; you wouldn't say the fact that the speeds are identical in one frame but different in the other means the laws of physics work differently in the two frames, would you? Ok.

1. what is measured in the frame where they are at rest before stopping in A's frame ? In this frame, all the Bs reach their corresponding As simultaneously (because both As and Bs are spaced 4.8 ly apart in this frame), and then they accelerate to match the velocity of the As. But the As are all out of sync by vx/c^2 = (0.6 ly/year)*(6 ly)/(1 ly/y)^2 = 3.6 years, so at the moment when the Bs line up with the As, A1 reads 0 years, A2 reads 3.6 years and A3 reads 7.2 years. 2. what is measured in the frame where they are at rest after stopping in A's frame ? This is just the A rest frame, I discussed what happens in this frame in the last post. 3. what do the watches on the wrists of the 6 observers show at both 1 & 2 ? All frames must agree on local events, so once any A and B pair are lined up all frames agree on their respective times. When B1 and A1 first line up, B1 reads 0 years and A1 reads 0 years; when B2 and A2 first line up, B2 reads 0 years and A2 reads 3.6 years; and when B3 and A3 first line up, B3 reads 0 years and A3 reads 7.2 years. From then on they tick at the same rate since they are at rest relative to one another, so B1 always reads the same time as A1, B2 is always 3.6 years behind A2, and B3 is always 7.2 years behind A3. Presumably, 3 is a reasonable question ? It does mean that an explanation is required of what happens as the B's move from their original f.o.r to A's f.o.r, but we're not saying that's impossible ? Clocks don't naturally jump times when they change velocities, if the acceleration is instantaneous then the reading on a clock will be the same immediately after acceleration as it was immediately before. Again, just because the B clocks come to rest in the A frame that doesn't mean they are now in sync in the A frame, they'd have to be reset in order to be synchronized in their new rest frame.

M1keh

Oct12-06, 02:17 AM

No, the fact that the distances match in one frame is just a property of the particular physical setup of the objects you're analyzing, the laws of physics are general equations that aren't specific to any particular physical setup, they describe things like how an arbitrary clock will slow down as a function of velocity. To take an analogy from Newtonian physics, if two cars are travelling at 100 mph in opposite directions in my frame, their speeds are the same in my frame, but in the frame of someone moving at 50 mph relative to me, one car is moving at 50 mph and the other at 150 mph; you wouldn't say the fact that the speeds are identical in one frame but different in the other means the laws of physics work differently in the two frames, would you?

No. I agree that you wouldn't all agree on each other's speeds.

However, you would all agree on what each would measure his own speed as ?

For the two travelling at 100 mph. A would estimate B's measure of his own speed to be 100mph and B would estimate A's measure of his own speed to be 100mph. Over an hour, both would estimate the other's measurement of their distance travelled as the same value. The 'laws' of physics would agree.

Unless I'm missing something, the example above isn't "just a property of the particular physical setup of the objects" ? It's impossible to come up with an example where the two measurements are the same as the rules always say B = A * 0.8 and A = B * 0.8. They will NEVER match ?

In this frame, all the Bs reach their corresponding As simultaneously (because both As and Bs are spaced 4.8 ly apart in this frame), and then they accelerate to match the velocity of the As. But the As are all out of sync by vx/c^2 = (0.6 ly/year)*(6 ly)/(1 ly/y)^2 = 3.6 years, so at the moment when the Bs line up with the As, A1 reads 0 years, A2 reads 3.6 years and A3 reads 7.2 years. This is just the A rest frame, I discussed what happens in this frame in the last post. All frames must agree on local events, so once any A and B pair are lined up all frames agree on their respective times. When B1 and A1 first line up, B1 reads 0 years and A1 reads 0 years; when B2 and A2 first line up, B2 reads 0 years and A2 reads 3.6 years; and when B3 and A3 first line up, B3 reads 0 years and A3 reads 7.2 years. From then on they tick at the same rate since they are at rest relative to one another, so B1 always reads the same time as A1, B2 is always 3.6 years behind A2, and B3 is always 7.2 years behind A3. Clocks don't naturally jump times when they change velocities, if the acceleration is instantaneous then the reading on a clock will be the same immediately after acceleration as it was immediately before. Again, just because the B clocks come to rest in the A frame that doesn't mean they are now in sync in the A frame, they'd have to be reset in order to be synchronized in their new rest frame.

Hmmm. Ok. This looks okay for a snapshot of a fixed view from B's f.o.r, ie. when the B's see themselves passing the A's.

Lets step back to a point before they're all sync'd in B's f.o.r. The first event will be B3 passing A1 ?

__________ A1 <- A2 <- A3
B1 -> B2 -> B3___________

B3 & A1 sync watches to A's agreed time. Zero time. The A's already agree on this time ?

B3 passes a signal back to B2 & B1 to sync to the same time, adjusting for the length of time the signal will take to reach them.

Effectively, then all A's & all B's will show the same zero time ? All sync'd to a common event in the two f.o.r's when B3 passes A1.

When B2 passes A1 and B3 passes A2. What will their respective watches show ?

jtbell

Oct12-06, 04:23 AM

Why are the diagrams for A's f.o.r and B's f.o.r so different? What's so 'special' about B's f.o.r ?You specified that the "A clocks" and the "B clocks" should have equal spacings in frame B (4.8 ly),

Another consequence of your initial condition is that the A-clocks have to be further apart in their rest-frame (A) than the B-clocks are in their rest-frame (B), so that length contraction can can make the distances equal in frame B.

If we had started out with both sets of clocks being spaced 6 ly apart in their own rest frames, then the two diagrams would have been much more symmetrical.

M1keh

Oct12-06, 04:55 AM

Another consequence of your initial condition is that the A-clocks have to be further apart in their rest-frame (A) than the B-clocks are in their rest-frame (B), so that length contraction can can make the distances equal in frame B.

If we had started out with both sets of clocks being spaced 6 ly apart in their own rest frames, then the two diagrams would have been much more symmetrical.

Yes. An omission. I didnt' specify that the B-Clocks were 4.8ly apart in their f.o.r, because I believed that they had to be ? There's no option ?

Presumably, if A's are 6ly apart, B's have to be 4.8ly apart ? They can't be 6ly apart in both f.o.r's ? A time dilation factor of 0.8 (0.6c) results in a length contraction factor of 1.25 ??

jtbell

Oct12-06, 05:26 AM

Presumably, if A's are 6ly apart, B's have to be 4.8ly apart?

If you want corresponding A's and B's to pass simultaneously in frame B, then yes, you have to set things up that way. It also prevents any "symmetry" between the way things happen in the two frames, because there's no way that this can be true in both frames.

If you remove the requirement that corresponding A's and B's pass simultaneously in frame B, then you can make things "symmetrical" between the two frames, by making the spacing of the A's and B's equal in their own respective rest frames. In this case, corresponding A's and B's pass simultaneously in neither frame A nor frame B, although I think there is a frame in which this is possible. This would be the frame that is sort of "halfway between" frame A and frame B, in which frame A and frame B move with equal speeds in opposite directions. I'd have to work this out in detail to make sure.

By the way, going completely off-topic, why do you put a space between the last word in a question, and the question mark? (like this ? and not like this?) Sometimes it causes the question mark to end up at the beginning of the next line all by itself, which looks strange.

JesseM

Oct12-06, 04:23 PM

No. I agree that you wouldn't all agree on each other's speeds.

However, you would all agree on what each would measure his own speed as ? In each observer's own rest frame, their speed is 0.
For the two travelling at 100 mph. A would estimate B's measure of his own speed to be 100mph and B would estimate A's measure of his own speed to be 100mph. Over an hour, both would estimate the other's measurement of their distance travelled as the same value. The 'laws' of physics would agree. Both will agree on their distance travelled relative to the road in an hour, in Newtonian physics. But that's just a particular feature of Newtonian physics where all rulers and clocks agree, it's not in general true that if "the laws of physics agree", then observers in different frames must agree on the distance A moves relative to B in a given amount of time. The laws of physics agreeing just means the same equations are used in every frame to calculate things, that's all. Unless I'm missing something, the example above isn't "just a property of the particular physical setup of the objects" ? It's impossible to come up with an example where the two measurements are the same as the rules always say B = A * 0.8 and A = B * 0.8. They will NEVER match ? Yes, as a general rule, the distances can only be equal in one frame. But this is no different from the fact that in Newtonian physics two objects travelling in opposite directions can have equal speeds in only one frame.

Also, note that if the distances are the same in the rest frame of each group of clocks, then they will view each other symmetrically--if they're both 6 ly apart in their own rest frames, then each will observe the other group to be 4.8 ly apart. In this frame, all the Bs reach their corresponding As simultaneously (because both As and Bs are spaced 4.8 ly apart in this frame), and then they accelerate to match the velocity of the As. But the As are all out of sync by vx/c^2 = (0.6 ly/year)*(6 ly)/(1 ly/y)^2 = 3.6 years, so at the moment when the Bs line up with the As, A1 reads 0 years, A2 reads 3.6 years and A3 reads 7.2 years. This is just the A rest frame, I discussed what happens in this frame in the last post. All frames must agree on local events, so once any A and B pair are lined up all frames agree on their respective times. When B1 and A1 first line up, B1 reads 0 years and A1 reads 0 years; when B2 and A2 first line up, B2 reads 0 years and A2 reads 3.6 years; and when B3 and A3 first line up, B3 reads 0 years and A3 reads 7.2 years. From then on they tick at the same rate since they are at rest relative to one another, so B1 always reads the same time as A1, B2 is always 3.6 years behind A2, and B3 is always 7.2 years behind A3. Clocks don't naturally jump times when they change velocities, if the acceleration is instantaneous then the reading on a clock will be the same immediately after acceleration as it was immediately before. Again, just because the B clocks come to rest in the A frame that doesn't mean they are now in sync in the A frame, they'd have to be reset in order to be synchronized in their new rest frame. Hmmm. Ok. This looks okay for a snapshot of a fixed view from B's f.o.r, ie. when the B's see themselves passing the A's.

Lets step back to a point before they're all sync'd in B's f.o.r. The first event will be B3 passing A1 ? What do you mean, "before they're all sync'd in B's f.o.r."? In the frame where the B's are at rest before reaching the As and accelerating, they are all in sync throughout the process. For example, when B1 has time t=-2 years (t=0 years is defined as the time the B's reach the A's), B2 also has time t=-2 years according to this frame's definition of simultaneity, as does B3.

The first event will be B3 passing A1 ?

__________ A1 <- A2 <- A3
B1 -> B2 -> B3___________ Yes, B3 passes A1 before any of the others pass each other. B3 & A1 sync watches to A's agreed time. Zero time. The A's already agree on this time ? Wait, you're changing the scenario now, you never said anything about the B's adjusting the times on their clocks before. All my and jtbell's previous explanations were based on the assumption that everyone is letting their clocks run naturally after the 3 Bs are synchronized in their rest frame and the three As are synchronized in theirs. Also, it was assumed that the Bs read t=0 at the moment each reaches their own corresponding A (ie when B3 reaches A3 and so forth, not when B3 reaches A1), and A1 also reads 0 when B1 reaches it, although if the three As are synchronized in their own rest frame then A2 and A3 won't read 0 when their own corresponding Bs reach them.

Would you like to change the scenario now, or were you not understanding that this was what jtbell and I had been assuming? B3 passes a signal back to B2 & B1 to sync to the same time, adjusting for the length of time the signal will take to reach them. Sync them in whose frame, the rest frame of the three Bs (before they accelerate to stop relative to the As, if you're still assuming that) or the rest frame of the As? Effectively, then all A's & all B's will show the same zero time ? Only if one of the two sets of clocks have been set so that they are out-of-sync in their own rest frame. When B2 passes A1 and B3 passes A2. What will their respective watches show ? Under the original assumption I described above, or making some new assumptions about the B clocks being reset when B3 passes A1? If the latter you need to be clear about what frame you want the Bs to be synchronized in.

M1keh

Oct12-06, 10:54 PM

If you want corresponding A's and B's to pass simultaneously in frame B, then yes, you have to set things up that way. It also prevents any "symmetry" between the way things happen in the two frames, because there's no way that this can be true in both frames.

If you remove the requirement that corresponding A's and B's pass simultaneously in frame B, then you can make things "symmetrical" between the two frames, by making the spacing of the A's and B's equal in their own respective rest frames. In this case, corresponding A's and B's pass simultaneously in neither frame A nor frame B, although I think there is a frame in which this is possible. This would be the frame that is sort of "halfway between" frame A and frame B, in which frame A and frame B move with equal speeds in opposite directions. I'd have to work this out in detail to make sure.

By the way, going completely off-topic, why do you put a space between the last word in a question, and the question mark? (like this ? and not like this?) Sometimes it causes the question mark to end up at the beginning of the next line all by itself, which looks strange.

The space after the ? is just a habit. It looks clearer ... provided it doesn't wrap around to the next line.

Now back to the examples.

Unless I misunderstood, earlier discussions went along the lines of 'If A measures distance as 6ly, B, travelling at 0.6c relative to A, must measure distance as 1 / \sqrt{1 - v^2/c^2} ? ie. 6/1.25 = 4.8 ly ?

Is this true or not ?

M1keh

Oct12-06, 11:54 PM

In each observer's own rest frame, their speed is 0.
Both will agree on their distance travelled relative to the road in an hour, in Newtonian physics. But that's just a particular feature of Newtonian physics where all rulers and clocks agree, it's not in general true that if "the laws of physics agree", then observers in different frames must agree on the distance A moves relative to B in a given amount of time. The laws of physics agreeing just means the same equations are used in every frame to calculate things, that's all. Yes, as a general rule, the distances can only be equal in one frame. But this is no different from the fact that in Newtonian physics two objects travelling in opposite directions can have equal speeds in only one frame.

Whilst that's true, it's now what I was saying ? In Newtonian Physics both objects would measure the other's speed as exactly the same (assuming they measure themselves as stationery). Therefore the rules are the same ?

Also, note that if the distances are the same in the rest frame of each group of clocks, then they will view each other symmetrically--if they're both 6 ly apart in their own rest frames, then each will observe the other group to be 4.8 ly apart. What do you mean, "before they're all sync'd in B's f.o.r."? In the frame where the B's are at rest before reaching the As and accelerating, they are all in sync throughout the process. For example, when B1 has time t=-2 years (t=0 years is defined as the time the B's reach the A's), B2 also has time t=-2 years according to this frame's definition of simultaneity, as does B3.

Ok. Lets try to clear this one up. It may be I misunderstood the earlier discussions.

If A's are 6ly apart, B's are travelling at 0.6c relative to A's, therefore their measure of the distance must be 4.8ly, (using 1 / \sqrt{1 - v^2/c^2} ).

Is this one of the 'rules of physics' that we're using or did I misread it ?

All of the earlier examples were specified on this basis ? I earlier suggested that both should be 6ly and measure the other as 4.8ly but everyone said this was incorrect ?

( Must stop typing so many ???? )

Yes, B3 passes A1 before any of the others pass each other. Wait, you're changing the scenario now, you never said anything about the B's adjusting the times on their clocks before. All my and jtbell's previous explanations were based on the assumption that everyone is letting their clocks run naturally after the 3 Bs are synchronized in their rest frame and the three As are synchronized in theirs. Also, it was assumed that the Bs read t=0 at the moment each reaches their own corresponding A (ie when B3 reaches A3 and so forth, not when B3 reaches A1), and A1 also reads 0 when B1 reaches it, although if the three As are synchronized in their own rest frame then A2 and A3 won't read 0 when their own corresponding Bs reach them.

Apologies. Yes I changed it. The example seemed to have been reduce into a snapshot of time when the B's see themselves as level with the A's.

This restricts the example somewhat.

I didn't think there'd be a problem with syncing watches at the first event and then playing the events forward ?

Would you like to change the scenario now, or were you not understanding that this was what jtbell and I had been assuming? Sync them in whose frame, the rest frame of the three Bs (before they accelerate to stop relative to the As, if you're still assuming that) or the rest frame of the As? Only if one of the two sets of clocks have been set so that they are out-of-sync in their own rest frame. Under the original assumption I described above, or making some new assumptions about the B clocks being reset when B3 passes A1? If the latter you need to be clear about what frame you want the Bs to be synchronized in.

Interesting question.

If A1 & B3 are passing each other. They can sync their times across f.o.r's ? A simultaneous event ?

Presumably there's no question that the A's can then sync their times effectively instantaneously, in their local f.o.r, as can the B's ? I agree that by the time the B's sync their watches locally, they'd be out of sync with the A's but they would have, effectively, been in sync when A1 & B3 passed ?

I may have confused the conversation. Maybe it's worth listing out the rules we're working to ? Which of these are correct and which are not ?

* Time dilation is a factor of relative speed, which at 0.6c is 0.8.

* Length contraction is, effectively, the inverse of time dilation, 1.25.

* At 0.6c, a local observer must expect the remote observer to measure time at 0.8 of his time.

* At 0.6c, a local observer must expect the remote observer to measure distance at 0.8 of his distance.

* Time dilation is a function of relative speed, not relative velocity.

* Two passing observers can agree on the 'simultaneous event' of them passing each other ( & can therefore sync watches at that point ).

Now, I'm not sure how you specify the lenght between observer's function that you listed earlier. How would you put that into words ?

Thanks.

JesseM

Oct13-06, 12:31 AM

Whilst that's true, it's now what I was saying ? In Newtonian Physics both objects would measure the other's speed as exactly the same (assuming they measure themselves as stationery). Therefore the rules are the same ? Yes, their view of each other's speed is symmetrical. Ok. Lets try to clear this one up. It may be I misunderstood the earlier discussions.

If A's are 6ly apart, B's are travelling at 0.6c relative to A's, therefore their measure of the distance must be 4.8ly, (using 1 / \sqrt{1 - v^2/c^2} ).

Is this one of the 'rules of physics' that we're using or did I misread it ? The length contraction factor of 1 / \sqrt{1 - v^2/c^2} is a general equation that holds in all inertial frames. But this equation tells you the amount that the length is contracted from the length in the rest frame--and since the A clocks and B clocks have different separations in their rest frames, then the separation each measures the other ones to have will also be different. The A clocks are 6 ly apart in their own rest frame and therefore 4.8 ly apart in the B frame, while the B clocks are 4.8 ly apart in their own rest frame and therefore 3.84 ly apart in the A frame. So applying the same laws of physics in both frames results in each getting a different result for the other's separation, since their separation in their own rest frames is different. All of the earlier examples were specified on this basis ? I earlier suggested that both should be 6ly and measure the other as 4.8ly but everyone said this was incorrect ? Yes, it's incorrect. In the A frame, the B clocks are 3.84 ly apart. Apologies. Yes I changed it. The example seemed to have been reduce into a snapshot of time when the B's see themselves as level with the A's.

This restricts the example somewhat.

I didn't think there'd be a problem with syncing watches at the first event and then playing the events forward ? I'm fine with changing the scenario, but you still haven't explained what you mean by "syncing watches". In which frame do the three B clocks synchronize their watches? Under the original assumption I described above, or making some new assumptions about the B clocks being reset when B3 passes A1? If the latter you need to be clear about what frame you want the Bs to be synchronized in. Interesting question.

If A1 & B3 are passing each other. They can sync their times across f.o.r's ? A simultaneous event ? At the moment A1 and B3 pass each other, if they each set their clock to read 0, then it will be true in all frames that they both read 0 at the same moment, since this is an event which happens at a single location. But that's not what I was asking. I was asking what the other two B clocks, B1 and B2, are doing--if you want B1 and B2 to also set themselves to 0 "at the same time" that B3 passes A1, you need to specify which frame's definition of simultaneity they should use, since if B1 and B2 set themselves to read 0 simultaneously with B3 passing A1 in one frame, in other frames they will set themselves to read zero earlier than or later than the moment B3 passes A1. Presumably there's no question that the A's can then sync their times effectively instantaneously, in their local f.o.r, as can the B's ? I don't understand your question here--what is a "local f.o.r.", for example? "Local" in relativity usually refers to events that happen at a single place and time like two clocks passing arbitrarily close to each other, while a frame of reference is a system of coordinates covering a large region of space and time, and giving a definition of simultaneity for distant (i.e. not local) events.

And what does it mean for the A's to sync their times effectively instantaneously? Sync their times with what, each other (in which case you need to pick one frame's definition of simultaneity) or with a B that's passing them (in which case, if you say that A1 sets itself to the same time as B3 at the moment they pass, you need to specify which Bs the other two As set themselves to match the time of as they pass) or something else? I agree that by the time the B's sync their watches locally What does sync their watches locally mean? Sync with what? Can you be explicit about what event triggers each of the Bs--B1, B2, and B3--to reset their watch, and if it's a distant event (like B2 setting its watch to 0 at the 'same time' that B3 passes A1) then specify which frame's definition of simultaneity you're using? Without these specifics I can't really follow your way of talking here. I may have confused the conversation. Maybe it's worth listing out the rules we're working to ? Which of these are correct and which are not ?

* Time dilation is a factor of relative speed, which at 0.6c is 0.8.

* Length contraction is, effectively, the inverse of time dilation, 1.25. Yes. * At 0.6c, a local observer must expect the remote observer to measure time at 0.8 of his time. Again, I don't understand how you're using the word "local" here. Do you just mean that if I see a clock moving at 0.6c in my own rest frame, I will measure it to tick at 0.8 the rate of my own clocks? If so, then that's correct. * At 0.6c, a local observer must expect the remote observer to measure distance at 0.8 of his distance. Again, if you just mean that a ruler moving at 0.6c in my frame will be measured to be shrunk to 0.8 the size of my ruler in my frame, then yes. * Time dilation is a function of relative speed, not relative velocity. Yes. * Two passing observers can agree on the 'simultaneous event' of them passing each other ( & can therefore sync watches at that point ). I would use the word "local" for these events, since there is a negligible spatial separation between them. "Simultaneous events" can mean events that happen at a large distance from each other and are simultaneous in one particular reference frame. Of course, local events will be simultaneous in every reference frame, they'll be no disagreement about what each clock read at the moment they passed next to each other. Now, I'm not sure how you specify the lenght between observer's function that you listed earlier. How would you put that into words ? What do you mean by "the length between observer's function"? Can you quote from the post where I gave this function so I'll know what you're referring to?

M1keh

Oct13-06, 05:51 AM

Apolgies. My post wasn't clear and I'm struggling with some of the terminology and where it applies.

Yes, their view of each other's speed is symmetrical. The length contraction factor of 1 / \sqrt{1 - v^2/c^2} is a general equation that holds in all inertial frames. But this equation tells you the amount that the length is contracted from the length in the rest frame--and since the A clocks and B clocks have different separations in their rest frames, then the separation each measures the other ones to have will also be different. The A clocks are 6 ly apart in their own rest frame and therefore 4.8 ly apart in the B frame, while the B clocks are 4.8 ly apart in their own rest frame and therefore 3.84 ly apart in the A frame.

I'm struggling with this. The formula is 'symmetrical' in both f.o.r's, but it's use or the interpretation of the results isn't ?

If you look from A's f.o.r, we start with A measures the distance as 6ly. We apply the formula and say, therefore B must see the distance as 4.8ly. Then we jump to B's f.o.r and say 'B MUST therefore measure the distance as 4.8 ly'.

But if you then apply this from B's f.o.r, we start with B measures the distance as 4.8ly. We apply the formula and say, therefore A must see the distance as 3.84 ly, but when we jump back to A's f.o.r, it isn't.

Now if the rules apply equally well from both f.o.r's, why do we get two different results from applying the rules ? Bearing in mind, we could just as easily have started from B's f.o.r and set the initial rules from there ?

So applying the same laws of physics in both frames results in each getting a different result for the other's separation, since their separation in their own rest frames is different. Yes, it's incorrect. In the A frame, the B clocks are 3.84 ly apart. I'm fine with changing the scenario, but you still haven't explained what you mean by "syncing watches". In which frame do the three B clocks synchronize their watches?

By local, I meant local to their own f.o.r. So the B's can instantaneously synchronise their watches to the event where B3 passes A1, as can the A's ?

At the moment A1 and B3 pass each other, if they each set their clock to read 0, then it will be true in all frames that they both read 0 at the same moment, since this is an event which happens at a single location.

Ok.

But that's not what I was asking. I was asking what the other two B clocks, B1 and B2, are doing--if you want B1 and B2 to also set themselves to 0 "at the same time" that B3 passes A1, you need to specify which frame's definition of simultaneity they should use, since if B1 and B2 set themselves to read 0 simultaneously with B3 passing A1 in one frame, in other frames they will set themselves to read zero earlier than or later than the moment B3 passes A1.

?? If the B's can observe B3 passing A1 and the A's can observe A1 passing B3. If they all synchroise their watches to the time this event happens in their own f.o.r, at the instant that they observed the event, don't they all agree on the exact time their watches show when the event happened ??

If the A1 synchs with B3 and then the A's synch their watches with A1 in their f.o.r and the B's synch their watches with B3 in their f.o.r, aren't all of the watches synch'd to each other in both f.o.r's at that moment, but then the times on A's and B's watches drift apart afterwards ?

I don't understand your question here--what is a "local f.o.r.", for example? "Local" in relativity usually refers to events that happen at a single place and time like two clocks passing arbitrarily close to each other, while a frame of reference is a system of coordinates covering a large region of space and time, and giving a definition of simultaneity for distant (i.e. not local) events.

So 'locally' means at the same time & place and 'simultaneously' means a single event viewed from our super-human viewing point, but potentially viewed as happening at different times from different f.o.r's ?

Is there a term for a single event witnessed within a single f.o.r by multiple observers at different places, ie. All the B's viewing B3 pass by A1 ? An f.o.r-local event ?

And what does it mean for the A's to sync their times effectively instantaneously? Sync their times with what, each other (in which case you need to pick one frame's definition of simultaneity) or with a B that's passing them (in which case, if you say that A1 sets itself to the same time as B3 at the moment they pass, you need to specify which Bs the other two As set themselves to match the time of as they pass) or something else? What does sync their watches locally mean? Sync with what? Can you be explicit about what event triggers each of the Bs--B1, B2, and B3--to reset their watch, and if it's a distant event (like B2 setting its watch to 0 at the 'same time' that B3 passes A1) then specify which frame's definition of simultaneity you're using? Without these specifics I can't really follow your way of talking here.

Again, apologies. All 6 sync their watches to when they observe A1 & B3 pass each other. As the event is local to A1 & B3 it's also a common event for A2, A3, B2 & B3 ? The A's will all agree on an event in their f.o.r and the B's will all agree on an event in their f.o.r ?

Yes. Again, I don't understand how you're using the word "local" here. Do you just mean that if I see a clock moving at 0.6c in my own rest frame, I will measure it to tick at 0.8 the rate of my own clocks? If so, then that's correct. Again, if you just mean that a ruler moving at 0.6c in my frame will be measured to be shrunk to 0.8 the size of my ruler in my frame, then yes. Yes. I would use the word "local" for these events, since there is a negligible spatial separation between them. "Simultaneous events" can mean events that happen at a large distance from each other and are simultaneous in one particular reference frame. Of course, local events will be simultaneous in every reference frame, they'll be no disagreement about what each clock read at the moment they passed next to each other. What do you mean by "the length between observer's function"? Can you quote from the post where I gave this function so I'll know what you're referring to?

Lets ignore that last bit for now. It doesn't help. It'll come back later.

Are the rules :

* Time dilation @ 0.6c = 0.8.
* Length contraction @ 0.6c = 1.25
* 'Local' events are events happening at the same point in space and are viewed as happening simultaneously by anyone at that point in space, in all f.o.r's.
* Simultaneous events are events that happen at the same time but may be viewed as happening at different times from different f.o.r's ?

* All observers within a single f.o.r will observe an event happening at the same time. (f.o.r-local event ?).

* All rules must work the same in any f.o.r.

jtbell

Oct13-06, 06:52 AM

If you look from A's f.o.r, we start with A measures the distance as 6ly.

I assume you mean the distance between, say, clocks A1 and A2 in our example.

We apply the formula and say, therefore B must see the distance as 4.8ly. Then we jump to B's f.o.r and say 'B MUST therefore measure the distance as 4.8 ly'.

Right.

But if you then apply this from B's f.o.r, we start with B measures the distance as 4.8ly. We apply the formula and say, therefore A must see the distance as 3.84 ly, but when we jump back to A's f.o.r, it isn't.

If you're still referring to clocks A1 and A2, we cannot apply the length contraction formula in the same way here, to switch from frame B back to frame A, because A1 and A2 are moving in frame B.

However, we can apply the length contraction formula to the distance between clocks B1 and B2 in our example, to switch from frame B back to frame A, because B1 and B2 are at rest in frame B.

Explanation: in the length contraction formula

L = L_0 \sqrt {1 - v^2 / c^2}

L_0 is always the length of an object, or the distance between two objects at rest with respect to each other, in the reference frame in which the objects are at rest. We often call this the proper length of the object.

In the example we're discussing here, the distance between clocks A1 and A2, as measured in frame B, is not the proper length between A1 and A2, because A1 and A2 are not at rest in frame B. Therefore we cannot use it as L_0 in the length contraction formula, in switching from frame B to frame A.

On the other hand, the distance between B1 and B2, as measured in frame B, is the proper length between B1 and B2, because B1 and B2 are at rest in frame B. Therefore we can use it as L_0 in the length contraction formula, in switching from frame B to frame A.

M1keh

Oct13-06, 07:33 AM

I assume you mean the distance between, say, clocks A1 and A2 in our example.

Right.

If you're still referring to clocks A1 and A2, we cannot apply the length contraction formula in the same way here, to switch from frame B back to frame A, because A1 and A2 are moving in frame B.

However, we can apply the length contraction formula to the distance between clocks B1 and B2 in our example, to switch from frame B back to frame A, because B1 and B2 are at rest in frame B.

Explanation: in the length contraction formula

L = L_0 \sqrt {1 - v^2 / c^2}

L_0 is always the length of an object, or the distance between two objects at rest with respect to each other, in the reference frame in which the objects are at rest. We often call this the proper length of the object.

In the example we're discussing here, the distance between clocks A1 and A2, as measured in frame B, is not the proper length between A1 and A2, because A1 and A2 are not at rest in frame B. Therefore we cannot use it as L_0 in the length contraction formula, in switching from frame B to frame A.

On the other hand, the distance between B1 and B2, as measured in frame B, is the proper length between B1 and B2, because B1 and B2 are at rest in frame B. Therefore we can use it as L_0 in the length contraction formula, in switching from frame B to frame A.

Hold on. That's a new 'rule'. Not seen that one before.

We can't use length contraction for B's measurement of the distance between A1 and A2, because they're moving relative to B ?

If A measures the distance between two objects as 6ly whether or not they're moving in his f.o.r, presumably B, moving at 0.6c relative to A will measure the distance between the objects as 4.8 ly, whether or not they are moving ?

The length contraction is a function of the difference in speed between A & B, not the speed of the objects ?

jtbell

Oct13-06, 08:05 AM

If A measures the distance between two objects as 6ly whether or not they're moving in his f.o.r, presumably B, moving at 0.6c relative to A will measure the distance between the objects as 4.8 ly, whether or not they are moving?

No. B will measure the distance between the objects as 4.8 ly, only if the objects are at rest in A's reference frame. Only then is L_0, the proper distance between the objects, equal to 6 ly (the distance that A measures).

JesseM

Oct13-06, 10:43 AM

I'm struggling with this. The formula is 'symmetrical' in both f.o.r's, but it's use or the interpretation of the results isn't ?

If you look from A's f.o.r, we start with A measures the distance as 6ly. We apply the formula and say, therefore B must see the distance as 4.8ly. Then we jump to B's f.o.r and say 'B MUST therefore measure the distance as 4.8 ly'.

But if you then apply this from B's f.o.r, we start with B measures the distance as 4.8ly. We apply the formula and say, therefore A must see the distance as 3.84 ly, but when we jump back to A's f.o.r, it isn't. As jtbell said, it is understood that the lorentz contraction formula tells you how much a ruler's length will be shrunk from its length in its own rest frame. Now, you could come up with a general formula that tells you "if an object's length in your frame is L and its speed in your frame is v1, what is its length in a frame which is moving at speed v2 in the same/the opposite direction", and this formula would work the same way in every frame, but the lorentz contraction formula is not it. By local, I meant local to their own f.o.r. So the B's can instantaneously synchronise their watches to the event where B3 passes A1, as can the A's ? OK, as I said that isn't the way "local" is used in relativity, but I think I understand what you're saying now. In the B rest frame, B1 and B2 will read 0 at the same time B3 passes A1, and in the A rest frame, A2 and A3 will read 0 at the same time B3 passes A1, is that correct? But that's not what I was asking. I was asking what the other two B clocks, B1 and B2, are doing--if you want B1 and B2 to also set themselves to 0 "at the same time" that B3 passes A1, you need to specify which frame's definition of simultaneity they should use, since if B1 and B2 set themselves to read 0 simultaneously with B3 passing A1 in one frame, in other frames they will set themselves to read zero earlier than or later than the moment B3 passes A1.?? If the B's can observe B3 passing A1 and the A's can observe A1 passing B3. If they all synchroise their watches to the time this event happens in their own f.o.r, at the instant that they observed the event, don't they all agree on the exact time their watches show when the event happened ?? No, because the Bs are using a different frame's definition of simultaneity than the As, are they not? If I'm understanding you, B1 and B2 set their watches to 0 "simultaneously" with B3 passing A1 as defined in the B rest frame, while A2 and A3 set their watches to 0 "simultaneously" with B3 passing A1 as defined in the A rest frame. Tell me if I got that wrong, but if not, you understand that the A rest frame and the B rest frame define simultaneity differently, right? So it should be no surprise that in the B rest frame, A2 and A3 do not set their watches to 0 at the same moment B3 passes A1, and likewise in the A rest frame, B1 and B2 do not set their watches to 0 at the same moment B3 passes A1. If the A1 synchs with B3 and then the A's synch their watches with A1 in their f.o.r and the B's synch their watches with B3 in their f.o.r, aren't all of the watches synch'd to each other in both f.o.r's at that moment, but then the times on A's and B's watches drift apart afterwards ? No, again because what distant clocks read "at that moment" is a question of simultaneity, and the B rest frame defines simultaneity differently from the A rest frame. So 'locally' means at the same time & place yup. and 'simultaneously' means a single event viewed from our super-human viewing point, but potentially viewed as happening at different times from different f.o.r's ? "Simultaneously" refers to the question of whether two different events happening at different points in space happened "at the same time" or not. And if "our super-human viewing point" is taken to mean a view of spacetime as a whole, there is no "objective" answer from this viewing point as to whether two events happened simultaneously, any more than a person viewing a piece of paper with two dots on it would say there's an objective answer to whether both dots share the same y-coordinate, it all depends on where you place your coordinate axes (assume no coordinate axes are drawn on the paper, the observer has to lay coordinate axes on himself, so it's up to him to decide how they're oriented). The super-human view of spacetime sees that there are different ways you could lay space and time axes on spacetime, corresponding to different reference frames--for two events with a spacelike separation, there would be one orientation that would give the events the same t-coordinate (so they happen 'simultaneously'), while other orientations give them different t-coordinates. A lot of this would probably be more intuitive to you if you learned the basics of using minkowski diagrams in SR (this page (http://en.wikibooks.org/wiki/Special_Relativity:_Simultaneity,_time_dilation_an d_length_contraction) seems to have a decent tutorial), since they actually show the orientation of the space and time axes for different frames. Is there a term for a single event witnessed within a single f.o.r by multiple observers at different places, ie. All the B's viewing B3 pass by A1 ? An f.o.r-local event ? What do you mean by "viewing"? We aren't talking about the time they actually see the light from the event of B3 passing A1, are we? I assumed that B1 and B2 would set their clocks to 0 "at the same time" in their rest frame that B3 passed A1, which would be well before the light from this event would actually reach them and they could see it happening. So talking about them "witnessing" the event doesn't really make sense to me, they've just calculated what time on their watches will be simultaneous with that event in their rest frame, and at that moment they reset their watches to 0. Again, apologies. All 6 sync their watches to when they observe A1 & B3 pass each other. Again, by "observe" I'm presuming you don't mean the time they actually see the light from this event, but just that they set their watches to 0 at the time that is simultaneous with this event in their own rest frames (with the As using a different definition of 'simultaneous' than the Bs)? As the event is local to A1 & B3 it's also a common event for A2, A3, B2 & B3 ? What do you mean by "also a common event"? It's true that no matter what frame you use, the event of A1 setting his clock to 0 and the event of B3 setting his clock to 0 are simultaneous, since these two events are local. But the event of any other observer like A2, A3, B2 or B3 setting their own clock to 0 is a separate event happening at a different position in space, so different frames disagree on whether, for example, the event of B2 setting his watch to 0 happens simultaneously with the event of A1 passing B3. The A's will all agree on an event in their f.o.r and the B's will all agree on an event in their f.o.r ? Yes, but three clocks at different locations setting their time to 0 is not "an event", it is three different events. An event is something that happens at a single point in time and space, like a single watch reading a particular time...a single dot in a minkowski diagram. The A's all agree that A1 and B3 set their time to 0 at the moment they passed, since these two events happen locally, but they don't agree that B1 and B2 set their times to 0 at the same moment, since those two events happened at different times in the A rest frame. Are the rules :

* Time dilation @ 0.6c = 0.8.
* Length contraction @ 0.6c = 1.25 Yes, assuming you're talking about dilation from the rate the clock is ticking in its own rest frame, and contraction from the object's length in its own rest frame.
* 'Local' events are events happening at the same point in space and are viewed as happening simultaneously by anyone at that point in space, in all f.o.r's. Basically yes, although you don't have to be "at that point in space" to say the events happened simultaneously--your judgement about simultaneity has nothing to do with where you are located, it only depends on what frame you are using (usually we adopt the convention that 'your' frame means your rest frame, although of course you are actually free to calculate things from the perspective of some other frame). * Simultaneous events are events that happen at the same time but may be viewed as happening at different times from different f.o.r's ? No, there is no objective answer to whether two events "happen at the same time". Simultaneity is a completely frame-dependent concept, all you can say is whether two events happen simultaneously (same time-coordinate) or non-simultaneously (different time coordinates) in a particular f.o.r. * All observers within a single f.o.r will observe an event happening at the same time. (f.o.r-local event ?). Provided by "observe" you just mean the time-coordinate that will be assigned to that event, and not the time the light from the event actually reaches them, then yes, if they're all using the same f.o.r. then they will all assign the same time-coordinate to a given event. But I wouldn't use the term "f.o.r.-local event" for this, I don't think it really needs a term at all, it's just a tautology that if two people use the same coordinate system, they'll assign the same coordinates to every event. * All rules must work the same in any f.o.r. Yes.

M1keh

Oct17-06, 02:44 AM

As jtbell said, it is understood that the lorentz contraction formula tells you how much a ruler's length will be shrunk from its length in its own rest frame. Now, you could come up with a general formula that tells you "if an object's length in your frame is L and its speed in your frame is v1, what is its length in a frame which is moving at speed v2 in the same/the opposite direction", and this formula would work the same way in every frame, but the lorentz contraction formula is not it.

Is that true ? I'm still relatively new to all this, but the Lorentz transformations are used for conversions from one rest frame to another, nothing more ?

The fact that two objects are moving in A's rest frame doesn't change the way we calculate the distance between them in B's rest frame ? The formulae are dependent on the relationship between the two rest frames, nothing more ?

If A measures two 'stationary' objects as 6ly apart and two 'moving' objects as 6ly apart, both must appear as 4.8ly apart to B (moving at 0.6c). Both distances are contracted by a factor of 1.25 ?

Or have I missed something ?

OK, as I said that isn't the way "local" is used in relativity, but I think I understand what you're saying now. In the B rest frame, B1 and B2 will read 0 at the same time B3 passes A1, and in the A rest frame, A2 and A3 will read 0 at the same time B3 passes A1, is that correct? No, because the Bs are using a different frame's definition of simultaneity than the As, are they not? If I'm understanding you, B1 and B2 set their watches to 0 "simultaneously" with B3 passing A1 as defined in the B rest frame, while A2 and A3 set their watches to 0 "simultaneously" with B3 passing A1 as defined in the A rest frame. Tell me if I got that wrong, but if not, you understand that the A rest frame and the B rest frame define simultaneity differently, right? So it should be no surprise that in the B rest frame, A2 and A3 do not set their watches to 0 at the same moment B3 passes A1, and likewise in the A rest frame, B1 and B2 do not set their watches to 0 at the same moment B3 passes A1. No, again because what distant clocks read "at that moment" is a question of simultaneity, and the B rest frame defines simultaneity differently from the A rest frame.

Hmmm. If A1 & B3 can agree on the exact time and B1 & B2 can agree with B3 on the exact time, surely B1 & B2 are agreeing with A1 ? Albeit only for that single moment ?

The same should follow for A2 & A3, meaning that all 6 agree on the exact time of the event where A1 & B3 meet ?

(Yes. let's stick with 'observed' in the GR sense, not allowing for the time light takes to travel in each rest frame).

yup. ... --your judgement about simultaneity has nothing to do with where you are located, it only depends on what frame you are using

So all of the B's would agree that B3 is at A1 and all of the A's would agree that A1 is at B3 ?

(usually we adopt the convention that 'your' frame means your rest frame, although of course you are actually free to calculate things from the perspective of some other frame). No, there is no objective answer to whether two events "happen at the same time". Simultaneity is a completely frame-dependent concept, all you can say is whether two events happen simultaneously (same time-coordinate) or non-simultaneously (different time coordinates) in a particular f.o.r. Provided by "observe" you just mean the time-coordinate that will be assigned to that event, and not the time the light from the event actually reaches them, then yes, if they're all using the same f.o.r. then they will all assign the same time-coordinate to a given event. But I wouldn't use the term "f.o.r.-local event" for this, I don't think it really needs a term at all, it's just a tautology that if two people use the same coordinate system, they'll assign the same coordinates to every event. Yes.

When you move from A's rest frame, measuring the distance between two objects as 6ly, to B's rest frame, travelling at 0.6c relative to A, B MUST measure the distance as 4.8ly ? 6.0ly * 0.8 ??

Is this correct ?

Is this one of the necessary outcomes of the 'laws of physics' as we now know them ?

jtbell

Oct17-06, 09:50 AM

Is that true ? I'm still relatively new to all this, but the Lorentz transformations are used for conversions from one rest frame to another, nothing more ?

The Lorentz transformation equations relate the position and time of an event in one inertial reference frame to the position and time of the same event in another inertial reference frame:

x^{\prime} = \gamma (x - v t)

t^{\prime} = \gamma (t - v x / c^2)

They always apply, between any two inertial reference frames.

The familiar length contraction equation relates the length of an object (or distance between two objects) in its (their) own rest frame to its (their) length in another inertial reference frame. It is more restricted in application than the Lorentz transformation equations, in that one of the frames must be the object's rest frame.

You might be confusing the length contraction and time dilation formulas with the Lorentz transformation equations. They are different, although related in that one can derive the length contraction and time dilation formulas from the Lorentz transformation equations.

If A measures two 'stationary' objects as 6ly apart and two 'moving' objects as 6ly apart, both must appear as 4.8ly apart to B (moving at 0.6c). Both distances are contracted by a factor of 1.25 ?

No, they are not. The objects that are stationary (and 6 ly apart) in frame A are indeed 4.8 ly apart in frame B. The objects that are moving (and 6 ly apart) in frame A are a different distance apart in frame B; that distance depends on how fast they are moving in frame A, as well as on the relative velocity of frames A and B.

The difference in the two situations can be explained using relativity of simultaneity. In order to measure the distance between two moving objects in any frame, you must measure their positions simultaneously, whereas if the objects are stationary, you don't have to measure their positions simultaneously. I have to go out for the rest of the day soon, but tonight or tomorrow I can probably come up with a worked-out example, unless JesseM beats me to it.

JesseM

Oct17-06, 02:09 PM

As jtbell said, it is understood that the lorentz contraction formula tells you how much a ruler's length will be shrunk from its length in its own rest frame. Now, you could come up with a general formula that tells you "if an object's length in your frame is L and its speed in your frame is v1, what is its length in a frame which is moving at speed v2 in the same/the opposite direction", and this formula would work the same way in every frame, but the lorentz contraction formula is not it.Is that true ? I'm still relatively new to all this, but the Lorentz transformations are used for conversions from one rest frame to another, nothing more ? Sorry about the confusion there, Lorentz contraction is just a synonym for length contraction, it refers to the formula L = L_0 \sqrt{1 - v^2/c^2}, while the Lorentz transformation is a general set of equations for transforming the coordinates of one frame to the coordinates of another. In the above quote I was talking about the Lorentz contraction formula, not the Lorentz transformation. The fact that two objects are moving in A's rest frame doesn't change the way we calculate the distance between them in B's rest frame ? The formulae are dependent on the relationship between the two rest frames, nothing more ? No, the Lorentz contraction formula is based on an object's speed in the frame you're using, and on the object's rest length in its own rest frame. As I said above, though, it would be relatively simple to come up with a new formula that tells you "if an object's speed in my frame is v1 and its length in its direction of motion in my frame is L, what is its length in another frame which is moving at speed v2 in the same/opposite direction as the object's direction of motion", but this would be a different formula from the Lorentz contraction formula L = L_0 \sqrt{1 - v^2/c^2}. If A measures two 'stationary' objects as 6ly apart and two 'moving' objects as 6ly apart, both must appear as 4.8ly apart to B (moving at 0.6c). Both distances are contracted by a factor of 1.25 ? No, again, the contraction factor in the Lorentz contraction formula is contraction from the object's rest length. Hmmm. If A1 & B3 can agree on the exact time and B1 & B2 can agree with B3 on the exact time, surely B1 & B2 are agreeing with A1 ? Albeit only for that single moment ? Yes, if A1 and B3 both read 0 at the same moment ('at the same moment' in all frames, since these events are local), then in the B-frame, B1, B2, B3 and A1 all read 0 at the same moment in the B-frame (but aside from B3 and A1, these events do not all happen simultaneously in other frames, since they occur at distinct locations in space). The same should follow for A2 & A3 Yes, if by "the same" you mean that in the A-frame, B3, A1, A2 and A3 all read 0 at the same moment in the A-frame (but again, aside from B3 and A1, these events do not all happen simultaneously in other frames). meaning that all 6 agree on the exact time of the event where A1 & B3 meet ? No--again, you have to keep track of which frame you're talking about. In the B-frame, at the "same moment" that B1, B2, B3 and A1 all read 0, A2 reads 3.6 years and A3 reads 7.2 years. And in the A-frame, at the "same moment" that B3, A1, A2 and A3 all read 0, B2 reads 2.88 years and B1 reads 5.76 years. There is no inertial frame where all 6 read 0 simultaneously.So all of the B's would agree that B3 is at A1 and all of the A's would agree that A1 is at B3 ? In both the B-frame and the A-frame, there would be a single moment when B3 and A1 are passing and their clocks both read 0. But if you're trying to compare the times in the two frames, asking whether the A's see A1 meeting B3 "at the same time" that the B's see them meeting, that question wouldn't really make sense (aside from the question of whether both frames assign the same time-coordinate to these events, but that can be changed just by moving the temporal origin of one frame's coordinate system, which is no more significant than moving the spatial origin of your x, y and z axes). The two frames aren't like two movies playing in parallel where we can ask what's happening on the two screens "at the same time" from our perspective in the theater, the phrase "at the same time" is only used for talking about simultaneity within a particular frame. When you move from A's rest frame, measuring the distance between two objects as 6ly, to B's rest frame, travelling at 0.6c relative to A, B MUST measure the distance as 4.8ly ? 6.0ly * 0.8 ??

Is this correct ? No, see above on how the Lorentz contraction formula works.

Thrice

Oct17-06, 02:50 PM

Jesse are you the same IIDB jesse? If so, will you marry me? :p

JesseM

Oct17-06, 03:13 PM

Jesse are you the same IIDB jesse? If so, will you marry me? :p Yup, one and the same. But marriages based solely on a shared love of physics rarely work out ;) Are you on IIDB yourself?

Thrice

Oct17-06, 03:29 PM

I'm on IIDB too, yeah. Come to think of it, I may have gotten the link to this forum from you many months ago. Physics people are few & far between. I'd spend all my time here if these forums were more active.

M1keh

Oct18-06, 06:42 AM

The Lorentz transformation equations relate the position and time of an event in one inertial reference frame to the position and time of the same event in another inertial reference frame:

x^{\prime} = \gamma (x - v t)

t^{\prime} = \gamma (t - v x / c^2)

Ouch. More maths. :-( ... inevitable I guess.

Ah. Okay. So the \sqrt{1 - v^2/c^2} is just the start of it and a specialisation of the real formula ?

Ok. Apologies for my ignorance, but the notation used isn't familiar.
t^{\prime}
\gamma (t - v x / c^2) ? Is the \gamma just a shorthand for \sqrt{1 - v^2/c^2} ?

They always apply, between any two inertial reference frames.

The familiar length contraction equation relates the length of an object (or distance between two objects) in its (their) own rest frame to its (their) length in another inertial reference frame. It is more restricted in application than the Lorentz transformation equations, in that one of the frames must be the object's rest frame.

But not if the distance between moving objects is measured in their own rest frame ? This only applies to stationery objects ?

You might be confusing the length contraction and time dilation formulas with the Lorentz transformation equations. They are different, although related in that one can derive the length contraction and time dilation formulas from the Lorentz transformation equations.

No. But I had made assumptions about the application of the formulae above. Having not understood them properly.

No, they are not. The objects that are stationary (and 6 ly apart) in frame A are indeed 4.8 ly apart in frame B. The objects that are moving (and 6 ly apart) in frame A are a different distance apart in frame B; that distance depends on how fast they are moving in frame A, as well as on the relative velocity of frames A and B.

The difference in the two situations can be explained using relativity of simultaneity. In order to measure the distance between two moving objects in any frame, you must measure their positions simultaneously, whereas if the objects are stationary, you don't have to measure their positions simultaneously. I have to go out for the rest of the day soon, but tonight or tomorrow I can probably come up with a worked-out example, unless JesseM beats me to it.

Oh, bugger. Missed this completely.

I'll need to understand how these formulae work. Are they complicated or relatively straight forward ?

jtbell

Oct18-06, 09:05 AM

Following up on posting #114... I wrote all of this offline and pasted it into the reply window. I haven't found any problems with the LaTex equations or other typos yet, but don't take that as a guarantee!

Here's a worked-out example that uses the Lorentz transformation equations to calculate the distance between two objects in frame B, (a) when the objects are at rest in frame A, and (b) when the objects are moving in frame A. It shows that the well-known length contraction formula gives the correct result for (a) but not for (b).

The setup: in frame A we have two objects, 1 and 2. Alternatively, they could be the two ends of a single object. Assume that in frame A at time 0 they have coordinates x_1 = 0 and t_1 = 0 at one end and x_2 = 6 and t_2 = 0 at the other end. Imagine little flashbulbs going off on both objects at time 0, if you like.

Frame B is moving to the right at 0.6c with respect to frame A, so \gamma = 1 / \sqrt {1 - v^2 / c^2} = 1.25.

The Lorentz transformation equations are

x^\prime = \gamma (x - v t)

t^\prime = \gamma (t - v x / c^2)

We'll use units of light-years for distance and years for time, so c = 1 and we can omit c from our equations to simplify them.

Case (a): The objects are stationary in frame A

We use the Lorentz transformation to find the coordinates of the two objects in frame B. First let's try

x_1^\prime = \gamma (x_1 - v t_1) = 1.25 (0 - 0.6 \cdot 0) = 0

t_1^\prime = \gamma (t_1 - v x_1) = 1.25 (0 - 0.6 \cdot 0) = 0

x_2^\prime = \gamma (x_2 - v t_2) = 1.25 (6 - 0.6 \cdot 0) = 7.5

t_2^\prime = \gamma (t_2 - v x_2) = 1.25 (0 - 0.6 \cdot 6) = -4.5

We expect the distance between the two objects in frame B to equal x_2^\prime - x_1^\prime = 7.5 - 0 = 7.5. But this can't be right, because it should be contracted to L_0 / \gamma = 6 / 1.25 = 4.8, according to the length contraction equation!

The problem is that t_1^\prime and t_2^\prime are different. In frame B, the two objects are moving, so we have to measure their positions at the same time in order to calculate the distance between them using x_2^\prime - x_1^\prime. Because of the relativity of simultaneity, two events at different locations that are simultaneous in frame A are not simultaneous in frame B, and vice versa.

We need to find x_2^\prime when t_2^\prime = 0, not when t_2^\prime = -4.5. This means we can't use t_2 = 0 in the Lorentz transformation, but must leave it unknown instead. Nevertheless, we can still use x_2 = 6 because object 2 is stationary in frame A, that is, it's always at that position no matter what time it is. (I've put this in boldface because it's a crucial assumption that we're going to return to later.)

So our two Lorentz transformation equations for object 2 now become (substuting t_2^\prime = 0 and leaving t_2 unknown)

x_2^\prime = 1.25 (6 - 0.6 t_2)

0 = 1.25 (t_2 - 0.6 \cdot 6)

Solving these gives us x_2^\prime = 4.8 and t_2 = 3.6. Now we can calculate the distance between the two objects in frame B as x_2^\prime - x_1^\prime = 4.8 - 0 = 4.8, which agrees with the result from the length contraction equation.

Case (b): The objects are moving in frame A

Now, let's suppose the two objects are both moving to the right in frame A, at speed 0.8c. At time 0 they have the same coordinates that we assumed before. The only difference is that now, they're moving.

For object 1, we can use the Lorentz transformation just like before, to get x_1^\prime = 0 and t_1^\prime = 0.

For object 2, we can follow similar reasoning as before, up to the assumption that I put in boldface earlier. We cannot now use x_2 = 6 because the object is moving. Object 2 is at position x_2 = 6 only at time t_2 = 0, in frame A. But now t_2 is unknown, so x_2 is also unknown, and the Lorentz transformation equations for object 2 now look like this:

x_2^\prime = 1.25 (x_2 - 0.6 t_2)

0 = 1.25 (t_2 - 0.6 x_2)

We have two equations for three unknowns, which are impossible to solve, as is. We need another equation! Fortunately, object 2 moves at constant velocity 0.8 ly/yr in frame A, starting from x_2 = 6 at t_2 = 0, so we can easily write down an equation which tells where object 2 is at any time, in frame A (its equation of motion):

x_2 = 6 + 0.8 t_2

We now have three equations in three unknowns. We can easily solve them to get t_2 = 6.923, x_2 = 11.539, and x_2^\prime = 9.231.

Therefore the distance between the two objects (which are moving in frame A) is x_2^\prime - x_1^\prime = 9.231 - 0 = 9.231 ly in frame B, whereas we got 4.8 ly in case (a).

Appendix

There's a way we can verify our answer for case (b) without using the Lorentz transformation equations. I'll describe the steps and show the results here, and leave it as an exercise for the reader to perform the calculations.

Step 1: The distance between the moving objects in frame A is already length-contracted, so use the length-contraction equation "backwards" to find the proper distance between them. That is, find the distance between the objects in the reference frame in which they are at rest. I'll call this frame R. The proper distance turns out to be 10 ly.

Step 2: Use the relativistic "velocity addition" equation to find the velocity of frame B relative to frame R, given that the velocity of frame R relative to frame A is 0.8 ly/yr and the velocity of frame B relative to frame A is 0.6 ly/yr. This velocity turns out to be 0.3846 ly/yr.

Step 3: Use the length contraction equation with the velocity found in step 2, to find the distance between the objects in frame B. This turns out to be 9.231 ly. (Surprise! :smile: )

JesseM

Oct18-06, 11:05 AM

Ouch. More maths. :-( ... inevitable I guess.

Ah. Okay. So the \sqrt{1 - v^2/c^2} is just the start of it and a specialisation of the real formula ?

Ok. Apologies for my ignorance, but the notation used isn't familiar.
t^{\prime}
\gamma (t - v x / c^2) ? Is the \gamma just a shorthand for \sqrt{1 - v^2/c^2} ? \gamma is the inverse of that, \frac{1}{\sqrt{1 - v^2/c^2}}. Note that I discussed the use of the Lorentz transformation formula, and gave an example, back in the second half of my post #65 (http://www.physicsforums.com/showpost.php?p=1103939&postcount=65) earlier on the thread. And as jtbell shows, the length contraction formula can be derived from the Lorentz transformation, as can the other specialized formulas we've discussed like time dilation and the equation that shows how out-of-sync moving clocks will be if they are synchronized in their rest frame.

M1keh

Oct19-06, 02:38 AM

Wow. Thanks.

Obviously, this will take some time to go through. Should keep me quiet for a while. :-)

M1keh

Oct19-06, 02:47 AM

\gamma is the inverse of that, \frac{1}{\sqrt{1 - v^2/c^2}}. Note that I discussed the use of the Lorentz transformation formula, and gave an example, back in the second half of my post #65 (http://www.physicsforums.com/showpost.php?p=1103939&postcount=65) earlier on the thread. And as jtbell shows, the length contraction formula can be derived from the Lorentz transformation, as can the other specialized formulas we've discussed like time dilation and the equation that shows how out-of-sync moving clocks will be if they are synchronized in their rest frame.

Thanks folks.

I thought there must have been some details I was missing.

These will take a bit of time to digest. A bit of light reading for tonight :-)

M1keh

Jan18-07, 11:40 AM

You break down the path into units of constant-velocity motion, but you use a single inertial frame to calculate how much time elapses on both the travelling clock and the earth clock during each unit. No, because again, different inertial frames define simultaneity differently. At the instant of turnaround in the travelling twin's outbound rest frame, the time on earth might be 2010; but at the instant of turnaround in the travelling twin's inbound rest frame, the time on earth might be 2020. So, you'd get the wrong answer if you tried to figure out the total time elapsed on earth between departure and return by saying something like "the ship left earth in 2005, and in the outbound rest frame only 5 years passed on earth between the moment of departure and the moment of turnaround, then in the inbound rest frame only 5 more years passed on earth between the moment of turnaround and the moment of return to earth, therefore earth's clock will read 2015 at the moment of return." This would miss the discontinuous gap between the earth's time at the moment of turnaround in the outbound frame and the earth's time at the moment of turnaround in the inbound frame, due to the two frames defining simultaneity differently--because of this gap, the actual time when the ship returned would be 2020 + 5 = 2025, not 2015 as in the naive calculation. If you stick to a single inertial frame for adding each segment you won't run into this sort of problem. OK, let's say our ship departs earth and travels at 0.6c for 10 years in the earth's frame, then turns instantaneously and travels back at 0.6c for another 10 years in the earth's frame. In the earth's frame, the time dilation factor is \sqrt{1 - 0.6^2} = 0.8, so the travelling twin's clock will only tick (10 years)*(0.8) = 8 years during the outbound leg, and (10 years)*(0.8) = 8 years during the inbound leg, so when they reunite the earth's clock will show 20 years have elapsed while the ship's clock only shows 16. Note that both the 10 years and the 0.8 time dilation factor were calculated solely from the perspective of the earth's inertial frame.

Now let's analyze the problem in a different inertial frame--the frame where the travelling twin is at rest during the outbound leg. In this frame, the earth will be moving away at a constant speed of 0.6c, while the ship will first be at rest for 8 years, during which time the earth has moved away a distance of (8 years)*(0.6c) = 4.8 light years, and its clock has advanced by (8 years)*(0.8) = 6.4 years, while the ship's clock has advanced forward by 8 years since it is at rest. Then after 8 years the ship will accelerate instantaneously, and using the velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) we know its speed in this frame after acceleration will be (0.6c + 0.6c)/(1 + 0.6^2) = 1.2c/1.36 = 0.88235c. Since the earth is 4.8 light years away but only moving at 0.6c, the time for the ship to catch up can be found by solving for t in 4.8 + 0.6t = 0.88235t, which gives t = 17 years in this frame. During this time, the earth's time dilation factor will still be 0.8, while the ship's will be \sqrt{1 - 0.88235^2} = 0.4706, so the earth's clock will have advanced forward by (17 years)*(0.8) = 13.6 years while the ship's clock will have advanced forward by (17 years)*(0.4706) = 8 years. So, by the time the ship catches up to the earth, the earth's clock will have advanced forward by the 6.4 years of the outbound leg plus the 13.6 years of the inbound leg, while the ship's clock will have advanced by the 8 years of the outbound leg plus the 8 years of the inbound leg. So, we find that the earth's clock had advanced 20 years while the ship's clock has advanced only 16, just as we found in the earth's frame.

We could also analyze everything from the point of view of the frame where the ship is at rest during the inbound leg. This will just be the mirror image of how things looked in the outbound rest frame--during the outbound leg, the earth is moving at 0.6c while the ship is moving at 0.88235c in the same direction, which lasts for 17 years, then the ship instantaneously accelerates and comes to rest while the earth continues to approach it at 0.6c from a distance of 4.8 light years, catching up to it after 8 years in this frame. And again, in a mirror image of the analysis in the outbound rest frame, the ship's clock elapses (17)*(0.4706) = 8 years during the outbound leg and (8)*(1) = 8 years in the inbound leg, while the earth's clock elapses (17)*(0.8) = 13.6 years in the outbound leg and (8)*(0.8) = 6.4 years in the inbound leg, so you again predict the ship's clock elapses a total of 16 years while the earth's clock elapses a total of 20 years.

Now look what would happen if you tried to switch frames in mid-problem, without worrying about simultaneity issues. Start out using the outbound rest frame during the outbound leg, and you'll conclude that at the moment before the ship does its instantaneous acceleration, the ship's clock reads 8 years and the earth is 4.8 light years away, its clock reading 6.4 years. Now if you switch to the inbound rest frame, and you don't realize that because of the different definition of simultaneity the earth should "jump" to reading 13.6 years, then you'll mistakenly continue to think that at the beginning of the inbound leg the earth's clock reads 6.4 years, and since the inbound frame says the ship is at rest during the inbound leg while the earth is approaching it at 0.6c from 4.8 light years away, in the inbound frame the ship's clock must advance by 8 years during the inbound leg and the earth's clock must advance by (8)*(0.8) = 6.4 years, so you would conclude the earth's clock only read 6.4 + 6.4 = 12.8 years when the ship returned? But this answer contradicts the answer you get when you stick to a single inertial frame throughout the problem, whether the earth rest frame, the inbound rest frame or the outound rest frame; you get the wrong answer if you try to "patch together" different frames in mid-problem this way, without taking into account discontinuous jumps in clock time due to the frames defining simultaneity differently. Although the time dilation factor as measured in a given inertial frame is only a function of a clock's speed in that frame, not its velocity, a frame can only qualify as "inertial" in the first place if it is moving at constant velocity, not just constant speed. So even though the instantaneous acceleration means the travelling twin never changed speeds in the earth's inertial frame (although he did change speeds in every other inertial frame, like the outbound rest frame), he did not stay at rest in a single inertial frame throughout the journey.

Analyzing the triplets problem wouldn't differ much from the twins problem. Let's again assume that in the earth's frame, the two travelling triplets move at 0.6c, and turn around after 10 years. In the earth's frame, each travelling triplet ages (10 years)*(0.8) = 8 years during their outbound legs, and another 8 years during their inbound legs, so that they have both aged 16 years when they return while the earth twin has aged 20 years.

Now look at things from the perspective of an inertial frame where one of the triplets is at rest during the oubound leg. In this frame, for the first 8 years this triplet (call him A) will be at rest, while the earth-triplet (call him B) is moving away at 0.6c and the third triplet (call him C) is moving away at (0.6c + 0.6c)/(1 + 0.6^2) = 0.88235c. Since C's clock is ticking slow by a factor of 0.4706 in this frame, and he won't accelerate until 8 years have passed on his own clock, he won't turn around until 8/0.4706 = 17 years have passed in this frame; when he does turn around, he'll be at rest in this frame, and at this moment the distance between him and the earth is (17 years)*(0.88235c - 0.6c) = 4.8 light years. The earth will continue to approach him at 0.6c after this point, reaching him after an additional 4.8/0.6 = 8 years in this frame. And when A turns around after 8 years, he is now moving in the direction of the earth at 0.88235c, with the earth having moved a distance of (8 years)*(0.6c) = 4.8 light years in those 8 years. The time for A to catch up to the earth is given by 4.8 + 0.6t = 0.88235t, or 17 years. So in this frame A starts out at rest, accelerates after 8 years, then takes another 17 years to catch up to earth, while C starts out at high speed, accelerates to rest after 17 years, and then the earth takes an another 8 years to catch up to him, meaning all three twins reunite after 25 years have passed in this frame.

So now let's figure out how much time has passed on each twin's clock after 8 years, after 17 years, and after 25 years in this frame, using this frame's values of the speeds and time dilation factors.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

Just noticed something.

If, from A's f.o.r, in the first 8 years A is stationery and B is 0.6c, then the furthest distance between A and B, according to A is 4.8 ly ? According to B this is 8 ly ? So did A reach his destination 8 ly from B or not ?

He would have had to measure his distance from B as 6.4 ly to have reached the destination ?

JesseM

Jan18-07, 11:53 AM

Just noticed something.

If, from A's f.o.r, in the first 8 years A is stationery and B is 0.6c, then the furthest distance between A and B, according to A is 4.8 ly ? According to B this is 8 ly ? So did A reach his destination 8 ly from B or not ?

He would have had to measure his distance from B as 6.4 ly to have reached the destination ? Remember that in the frame of B (the Earth-twin) the travelling twin A was moving away for 10 years instead of 8. So, with their relative speed being 0.6c, the B sees A move 6 light years away before reaching the destination, not 8 light years. In A's frame, this distance is Lorentz-contracted to 6*0.8 = 4.8 light years, so it's consistent.

M1keh

Jan19-07, 03:31 AM

Doh. Fools rush in !! Thanks.

M1keh

Jan19-07, 09:13 AM

Remember that in the frame of B (the Earth-twin) the travelling twin A was moving away for 10 years instead of 8. So, with their relative speed being 0.6c, the B sees A move 6 light years away before reaching the destination, not 8 light years. In A's frame, this distance is Lorentz-contracted to 6*0.8 = 4.8 light years, so it's consistent.

Ok. I'm not giving up ...

Instead of journeys out from Earth, how about the same measurements but two travellers travelling towards Earth at 0.6c from two satellites 6ly away from Earth ?

How do the figures work then ??

Now all three start in Earth time so they can synchronise watches ? The travellers instantly accelerate to 0.6c and measure the distances, in their new f.o.r.'s

A will see B as 4.8 ly away and C as 9.6 ly away ?
C will see B as 4.8 ly away and A as 9.6 ly away ?
B will see A as 6 ly away and C as 6 ly away ?

All three view themselves as remaining stationery in their new f.o.r ?

From A's f.o.r :

First 8 years.
* A will remain stationery for 8 years.
* B will travel at 0.6 and his watch will show 0.8 * 8years = 6.4 years and he will have travelled 0.6*8 = 4.8ly.
* C will travel at (0.6+0.6)/(1+0.6^2) = 0.88235c and will travel 0.88235*8 = 7.06 ly. Watch will have moved 3.765 years.

Now, B's watch must be 1.6 years earlier than A's, C's watch must be 4.235 years before A's and C won't have reached A ?

What's wrong ?

JesseM

Jan20-07, 12:16 AM

Ok. I'm not giving up ...

Instead of journeys out from Earth, how about the same measurements but two travellers travelling towards Earth at 0.6c from two satellites 6ly away from Earth ?

How do the figures work then ??

Now all three start in Earth time so they can synchronise watches ? The travellers instantly accelerate to 0.6c and measure the distances, in their new f.o.r.'s

A will see B as 4.8 ly away and C as 9.6 ly away ?
C will see B as 4.8 ly away and A as 9.6 ly away ?
B will see A as 6 ly away and C as 6 ly away ? That's right.

edit: sorry, that's actually not quite right. In the frame of reference where A is at rest after acceleration, it's true that C was 9.6 light years from A before either C or A accelerated. However, in this frame C accelerated well before A (see below), so at the moment A accelerates C has already been travelling towards B for some time, so the distance is smaller. Something analogous is true in the frame where C is at rest after accelerating. All three view themselves as remaining stationery in their new f.o.r ?

From A's f.o.r :

First 8 years.
* A will remain stationery for 8 years.
* B will travel at 0.6 and his watch will show 0.8 * 8years = 6.4 years and he will have travelled 0.6*8 = 4.8ly. Yes.
* C will travel at (0.6+0.6)/(1+0.6^2) = 0.88235c Right. and will travel 0.88235*8 = 7.06 ly. Watch will have moved 3.765 years. This part is wrong, you're forgetting that in this frame, although C was 4.8 ly from B at the moment before it accelerated, after it accelerates B is still moving away from it at 0.6c, so C is only gaining on B at a speed of 0.88235c - 0.6c = 0.28235c. Also, because of the relativity of simultaneity, C accelerated much earlier than A in this frame.

The main thing you're forgetting is that even though all their clocks were synchronized at the moment of acceleration in the Earth's inertial rest frame, in the inertial rest frame where A is at rest after the acceleration, the three clocks were not synchronized at the moment A accelerated, because of the relativity of simultaneity. If you don't want to use the full Lorentz transform, you can use the formula vx/c^2 to figure out how out-of-sync the clocks are before any of them accelerate, in the frame where A is at rest after acceleration: here, x is the distance between two synchronized clocks in their own rest frame, and v is their velocity in your frame. Since A and B are 6 ly apart in their mutal rest frame before acceleration, in a frame moving at 0.6c relative to them, B will be ahead of A by (0.6c)(6 ly)/c^2 = 3.6 years. Likewise, C will be ahead of B by a constant factor of 3.6 years until it accelerates, in this frame. So since C accelerates when its own clock reads 0, in this frame A's clock read -7.2 years at the moment C accelerated. And since A's clock is ticking at 0.8 the normal rate until it accelerates, it takes 7.2/0.8 = 9 years in this frame before A's clock reads 0, at which point B's clock reads 3.6 years. Since C has been moving at 0.88235 c during these nine years, C's clock has been moving at 0.4706 the normal rate, so it only moved by 9*0.4706 = 4.235 years. It read 0 at the beginning of the 9 years, so it now reads 4.235 years.

So, in this frame, when A accelerates and comes to rest, A reads 0, B reads 3.6 years, and C reads 4.235 years. It takes 8 years for B to reach A in this frame, so during that time A ticks at the normal rate and advances to 0 + 8 = 8 years, B ticks at 0.8 times the normal rate and advances to 3.6 + 0.8*8 = 10 years, and C ticks at 0.4706 times the normal rate and advances to 4.235 + 0.4706*8 = 8 years. This is the point where they all meet, and the predictions about each one's clock-reading when they meet is the same as what you'd predict in the Earth's rest frame.