Can Relativity Explain the Confusion of the Twins Paradox?

[selfAdjoint]

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

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

 

[M1keh]

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

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

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

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

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

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

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

Are there any others ?

 

[JesseM]

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

 

[M1keh]

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

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

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

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

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


Sorry. This is where it all started.

 

[MeJennifer]

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

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

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

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

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

 

[JesseM]

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

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

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

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

 

[JesseM]

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

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

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

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

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

 

[MeJennifer]

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

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

 

[JesseM]

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

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

 

[Emanresu]

Very clever people are rare.

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

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

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

E.

 

[MeJennifer]

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

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

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

 

[JesseM]

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

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

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

 

[MeJennifer]

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

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

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

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

 

[M1keh]

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

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

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

( Long weekend. A bit grumpy this morning. )

 

[JesseM]

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

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

 

[M1keh]

... 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]

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 traveling 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 traveling 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 traveling 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 traveling 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]

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 traveling 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 traveling 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 traveling 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 traveling 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]

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 traveling 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 traveling twin's outbound rest frame, the time on Earth might be 2010; but at the instant of turnaround in the traveling 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 traveling 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 traveling 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 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 traveling 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 traveling triplets move at 0.6c, and turn around after 10 years. In the Earth's frame, each traveling 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]

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 traveling clock and the Earth clock during each unit. No, because again, different inertial frames define simultaneity differently. At the instant of turnaround in the traveling twin's outbound rest frame, the time on Earth might be 2010; but at the instant of turnaround in the traveling 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 traveling 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 traveling 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 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 traveling 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 traveling triplets move at 0.6c, and turn around after 10 years. In the Earth's frame, each traveling 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]

...

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 traveling twins (we won't 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]

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 traveling twins (we won't 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]

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 won't he view E's as having elapsed 8 years ? He would have traveled 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 traveled 10 local years representing 8 of E's years ? He won't travel 8 years and look at E as having traveled 6.4 years ?

Did we introduce one two many ajustments somewhere earlier ?

 

[JesseM]

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 traveling triplets, while 20 years have passed on the clock of the earthbound triplet.

If the two traveling twins (we won't 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]

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]

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]

...

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 traveled 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 traveled 0.88235*8 = 7.06 ly.


Am I messing this up somewhere ?

 

[JesseM]

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 traveled 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]

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

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]

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 traveled 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 traveled 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 Earth's--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 traveling 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 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 Earth's 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 Earth's 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 Earth's' rest frame.

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

 

[pess5]

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]

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]

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 truly 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]

This scenario is called "the handoff technique". Minkowski diagrams are truly 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]

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.