Can Relativity Explain the Confusion of the Twins Paradox?

[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.

 

[JesseM]

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]

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.

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 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.

That'll take a while to sink in !

 

[pess5]

JesseM,

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

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

pess

 

[pess5]

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]

JesseM,

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

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

pess

Cool. Thanks. It's take some time to decypher & digest. Gives me something to consider over the weekend.

 

[M1keh]

JesseM,

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

https://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]

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]

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 ...

https://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 possesses 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]

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]

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.

Well, I'd have to say you are correct on this matter. I therefore will post an attachment here for the benefit of all regarding the ...

the Relativity of Simultaniety​

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.

jtBell ... does this suffice?

 

[robphy]

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]

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. The story behind this is that physicists had always imagined that light only traveled 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 Earth's' 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]

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 website 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, thanks 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]

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]

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]

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]

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.

Ok.

And all observers in a common f.o.r, no matter how far apart, agree on the time in their f.o.r ?

ie. If I'm one side of the Universe, stationery compared with another observer on the other side of the Universe, we will agree on the current time ?

 

[jtbell]

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.

 

[M1keh]

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 traveling 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]

Ok. Here goes.

If A1, A2 & A3 are all in the same f.o.r, but 6ly apart, and B1, B2, B3 are all traveling 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]

... 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]

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]

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]

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 traveling 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]

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.

Thanks. Top diagrams. Like the way you've scanned them in and attached them.

One question first. 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 ?

In B's f.o.r the A's are traveling at 0.6c, exactly the same as the A's view of the B's in their f.o.r ?

Why the difference ?

 

[jtbell]

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]

If the A's see the B's as 6ly apart and traveling 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]

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]

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 traveling 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]

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 traveling 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 traveling 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 traveled 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 ?