How Time Passes Differently for Twins Traveling at Light Speed

[Prague]

I am trying to understand the twin paradox, so you have twin 1 and twin 2, both on planet earth. The twins are 23 years old and twin 2 leaves on a ship traveling close to the speed of light and then turns around (with or without a instantaneious turn around time?). On his return home twin 2 finds that twin 1 has aged far more than he has.

Now, why is this? Twin 2 travels away from Earth at the speed of light. Let's say 10 minutes (in a universal time) passes. Even though twin 2 is traveling at the speed of light, isn't he still traveling 10 mintues? And twin 1 would still be waiting for 10 mintues. Now let's say twin 2 turns around, and travels back to earth, this entire trip (from turnaround to landing) takes another 12 minutes. It still is 12 minutes for either twin 1 and 2 isn't it?

Just because he is traveling a distance why should he be younger? Is this just our notion of time (i understand if the times wasn't a 'universal time' it would make them much different in age) but isn't the notion of time false anyways? Our bodies don't slow for time, they always are dieing at an interval. So biologically wouldn't twin 1 and 2 be the same age, but theoretically (if we consider time as we concieve it, a real factor in our aging) there 'age' would be different.

 

[HallsofIvy]

"Twin 2 travels away from Earth at the speed of light. Let's say 10 minutes (in a universal time) passes. "

Right there you have two problems: you can't move at the speed of light (you are welcome to use "99% the speed of light). More importantly there is no such thing as "a universal time" so I don't know what you mean by this.

"It still is 12 minutes for either twin 1 and 2 isn't it?"

No, it isn't! That's the whole point of relativity. Again, there is no "universal time". There is no such thing as "time" except as measured in a particular frame of reference. There is no reason to think that the same amount of time will have passed for both.

And, no, the bodies of two people moving at very different speeds will age at different rates. While that hasn't be done with actual people (the difference in speeds would have to be much greater than anything we can achieve for it to show) it has been shown that elementary particles will have different "life spans" depending upon their speeds (measured relative to the laboratory, of course).

 

[Prague]

(sorry that was a typo, i meant close to the speed)

anyways, what is it that makes our bodies age less? How does speed affect this?

Perhaps my whole view on the times relative to each persons position/speed is wrong. The person moving at the speed of light is aging himself only let's say 10 minutes, but the person on Earth is aging more? I don't understand why this effects age, it's just distance.

Also, is the reason we can't have a universal time because time is distorted my certain objects, like the Earth for example? Time around the Earth is much different then the time around an object much larger than earth.

 

[JesseM]

Also, is the reason we can't have a universal time because time is distorted my certain objects, like the Earth for example?

Even in special relativity, where you ignore gravity, universal time doesn't make sense. For example, as long as two observers are moving at constant velocity (meaning unchanging speed and unchanging direction) relative to each other, there is no absolute truth about who is aging slower--in my reference frame you may be aging at half the speed as I am, but in your reference frame it is me who is aging at half the speed you are. Also, different reference frames disagree about "simultaneity", the question of whether two events at different locations happened "at the same time" or not--if I assign two events the same time-coordinate in my reference frame, then in your reference frame you will assign them two different time-coordinates, saying that one event happened after the other one.

 

[Prague]

Even in special relativity, where you ignore gravity, universal time doesn't make sense. For example, as long as two observers are moving at constant velocity (meaning unchanging speed and unchanging direction) relative to each other, there is no absolute truth about who is aging slower--in my reference frame you may be aging at half the speed as I am, but in your reference frame it is me who is aging at half the speed you are. Also, different reference frames disagree about "simultaneity", the question of whether two events at different locations happened "at the same time" or not--if I assign two events the same time-coordinate in my reference frame, then in your reference frame you will assign them two different time-coordinates, saying that one event happened after the other one.

Yes, but why is it that the observer see's the other aging slower. I am finding it hard to accept that speed and distance result in our aging. If you were to bring the two observers together again, who would be older. One observer saw the person aging slower than him, the other observer saw the same but in reverse. If you bring them together, who was correct?

 

[pervect]

Either one could be "correct". The observer that accelerates will be the younger observer when the re-unite. In order for them to re-unite, one observer has to accelerate.

 

[Prague]

Either one could be "correct". The observer that accelerates will be the younger observer when the re-unite. In order for them to re-unite, one observer has to accelerate.

ok, so why is it that acceleration decreases our age?

 

[robphy]

ok, so why is it that acceleration decreases our age?

It isn't the acceleration that decreases our age. Rather, it is a feature that distinguishes the motions of the two twins.
At the root of the matter, elapsed time is proportional to the arc-length in spacetime.

 

[JesseM]

ok, so why is it that acceleration decreases our age?

It's not that acceleration decreases your age, I'd say it's that the laws of physics have to work equally well in any non-accelerating reference frame, and in each frame a clock moving at velocity v must be ticking at \sqrt{1 - v^2/c^2} times the rate of a clock at rest in that frame. So if you want to know how much time elapses on the clock of an observer who is changing velocities according to some function v(t), between times t_0 and t_1 (with velocity and time defined in terms of that frame) you'd evaluate the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. The value always turns out to be less than the time elapsed on a clock that traveled inertially between the same two points in space in the same time interval. And since the laws of physics should work equally well in any inertial frame, you will get the same answer regardless of which frame you are using to define the time interval and the function v(t).

I don't think there's really an answer of what "causes" clocks to slow down, it's just sort of the nature of spacetime in relativity. But the way spacetime works in relativity does follow uniquely from two postulates, the first being that the laws of physics should work the same in every inertial frame, and the second being that the speed of light should be the same in every inertial frame.

 

[Prague]

It's not that acceleration decreases your age, I'd say it's that the laws of physics have to work equally well in any non-accelerating reference frame, and in each frame a clock moving at velocity v must be ticking at \sqrt{1 - v^2/c^2} times the rate of a clock at rest in that frame. So if you want to know how much time elapses on the clock of an observer who is changing velocities according to some function v(t), between times t_0 and t_1 (with velocity and time defined in terms of that frame) you'd evaluate the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. The value always turns out to be less than the time elapsed on a clock that traveled inertially between the same two points in space in the same time interval. And since the laws of physics should work equally well in any inertial frame, you will get the same answer regardless of which frame you are using to define the time interval and the function v(t).

I don't think there's really an answer of what "causes" clocks to slow down, it's just sort of the nature of spacetime in relativity. But the way spacetime works in relativity does follow uniquely from two postulates, the first being that the laws of physics should work the same in every inertial frame, and the second being that the speed of light should be the same in every inertial frame.

Ok, but we aren't mechanical clocks, we are biological clocks. Does biology agree with this? Sure if twin A was holding a clock and twin B was also holding a clock and they brought them together after B left, then sure I can kind of see now why the time would be different. But what about the twins, would twin B still be a little boy and twin A an old man.

 

[hypermorphism]

Ok, but we aren't mechanical clocks, we are biological clocks. Does biology agree with this?

The results of the postulates of special relativity have nothing to do with the specific construction of time-keeping devices. Simply something that measures position on a time axis from some initial time, or in more everyday language, change.

 

[Prague]

The results of the postulates of special relativity have nothing to do with the specific construction of time-keeping devices. Simply something that measures position on a time axis from some initial time, or in more everyday language, change.

Ok, I cleared up all my doubts, so what's the actual math/theory behind the issue this thread brought up? Specifically I mean.

 

[hypermorphism]

Hmm, I don't think it lends itself to a quick internet exhibition. The resulting equations come from considering two frames in relative motion of uniform velocity with respect to each other, and considering a light signal emitted in one of the frames. A layman's derivation is given in , where one can nowhere see any reference to physical clocks, just relative time and space variables obeying the postulates. There is a light-clock derivation in , but this refers to a specific construction of a clock only as an appeal to the student's geometric intuition. Their book is, however, one of the best introductions to the subject, as it is full of meaningful exercises (some are simplifications of results written in research papers) and intuitive illustrations.

 

[jdavel]

Prague,

Suppose the traveling twin goes to a planet that's 10 light years from Earth and travels at a speed that gets him there in 20 years (hardly close to light speed, but it will do). If each twin watches the other through a telescope throughout the twenty year trip, what will they each see happening? What will they each be seeing at the moment the traveller arrives at the distant planet?

 

[jtbell]

If each twin watches the other through a telescope throughout the twenty year trip, what will they each see happening?

I posted a detailed description of a similar scenario in this forum just a few days ago. See post #3 in this thread.

 

[Prague]

For a much more precise answer, look at this post by JesseM, which helped me understand the twin paradox "when I use it as a reference, I just learned it today afterall. Below is my attempt to explain without the math and without using JesseM's post as a reference.

Link: https://www.physicsforums.com/showpost.php?p=515872&postcount=18

Prague,

Suppose the traveling twin goes to a planet that's 10 light years from Earth and travels at a speed that gets him there in 20 years (hardly close to light speed, but it will do). If each twin watches the other through a telescope throughout the twenty year trip, what will they each see happening? What will they each be seeing at the moment the traveller arrives at the distant planet?

So you have twin A and twin B and Planet X. B departs for X which is 10 light years away. He travels at a speed somewhat near c, and it takes B 20 years to get to X. If B looks back at A during travel A will look as if he is aging slowly. If A looks at B the same affect happens. Now when B looks at a from X which I suppose is traveling relative to A's speed now making B traveling the same speed they would start to age the same speed again.

Now, when B lands on X he will be younger than A because his clock was traveling slower than A. (or perhaps I am wrong on this, I am not sure, becuase occording to the 'usual' twin paradox, B would have to travel back to A to be younger.)

I think that's correct.

 

[Ich]

Yep, B is younger in this frame. But there would still be people who think A is younger, namely those moving near c wrt this rest frame. That´s why they tell B to come back to have an unambigous solution.

 

[TheUnknown]

Yes, but why is it that the observer see's the other aging slower. I am finding it hard to accept that speed and distance result in our aging. If you were to bring the two observers together again, who would be older. One observer saw the person aging slower than him, the other observer saw the same but in reverse. If you bring them together, who was correct?

would they not be the same age when reunited? if observer A flies at close to the speed of light to point C (wich will be farther than point D because of his speed compared to observer B) and observer B flies to point D at half the speed of light... and they both start at F and reunite at F at the same time, wouldn't their ages be the same? ah... i have edited this.. i have caught myself making a huge mistake... because the fact that observer B's travel distance is shorter than A's... he will also have to travel slower back to point F again... BUT... if you add a point G opposite of point F corresponding to points C and D... and they meet there, then they are the same age! because observer B now has to travel at A's initial speed... and A at b's initial speed! ah... time as we know it... it's so misleading. physics are great. :) i love this site, and all you guys are great for doing what you do.

 

[yogi]

Completed round trip journeys always result in the clock which made the trip having logged less time. But there are one way examples of clocks in motion also logging different times relative to the frame in which they were brought to rest - Einstein gives a clear statement of this reality in his 1905 paper - two clocks separated by a distance, both at rest in the same frame, and synchronized so that they run at the same rate. One clock is moved toward the other... when they meet they are found to be out of sync. Jesse doesn't like this experiment - he insists the two clocks can't really read differently - but as between Jesse and Einstein, I am leaning toward the latter.

 

[Ich]

Completed round trip journeys always result in the clock which made the trip having logged less time. But there are one way examples of clocks in motion also logging different times relative to the frame in which they were brought to rest - Einstein gives a clear statement of this reality in his 1905 paper - two clocks separated by a distance, both at rest in the same frame, and synchronized so that they run at the same rate. One clock is moved toward the other... when they meet they are found to be out of sync. Jesse doesn't like this experiment - he insists the two clocks can't really read differently - but as between Jesse and Einstein, I am leaning toward the latter.

Never heard about it - unless you mean "moved" as with a significant fraction of c.

 

[JesseM]

Completed round trip journeys always result in the clock which made the trip having logged less time. But there are one way examples of clocks in motion also logging different times relative to the frame in which they were brought to rest - Einstein gives a clear statement of this reality in his 1905 paper - two clocks separated by a distance, both at rest in the same frame, and synchronized so that they run at the same rate. One clock is moved toward the other... when they meet they are found to be out of sync. Jesse doesn't like this experiment - he insists the two clocks can't really read differently - but as between Jesse and Einstein, I am leaning toward the latter.

Einstein does not say that one has aged less in absolute terms, he just says that one is behind the other when they meet, which of course I agree with. Einstein also says that every reference frame is equally valid, and there is certainly a frame where the clock that accelerates is ticking faster then the one that doesn't as they approach each other (in this frame, they started out out-of-sync, so it will still be true that the clock that accelerated will be behind the other when they meet), so it's you who disagrees with Einstein if you're saying one clock ages less in an absolute sense, not me.

 

[Ich]

Einstein does not say that one has aged less in absolute terms, he just says that one is behind the other when they meet, which of course I agree with. Einstein also says that every reference frame is equally valid, and there is certainly a frame where the clock that accelerates is ticking faster then the one that doesn't as they approach each other (in this frame, they started out out-of-sync, so it will still be true that the clock that accelerated will be behind the other when they meet), so it's you who disagrees with Einstein if you're saying one clock ages less in an absolute sense, not me.

If you make a twin paradox out of this, you will end up with one twin being younger:
A anb B both start moving away from each other, very slowly. If they stop moving relative to another and compare clocks, they will find to be still in sync. Thats yogi´s starting point. When then B accelerates towards A and reaches him, he will have aged less.

 

[JesseM]

If you make a twin paradox out of this, you will end up with one twin being younger:
A anb B both start moving away from each other, very slowly. If they stop moving relative to another and compare clocks, they will find to be still in sync. Thats yogi´s starting point.

But in a frame where the Earth is moving at 0.99c or something, then if the second twin moves away from the Earth "very slowly" (ie if you take the limit as his velocity in the Earth's frame approaches zero), you still find that in this frame his clock gets significantly out-of-sync with the earth-twin's frame as they move a significant distance apart, so in this frame your "starting point" looks quite different, and is still compatible with the idea that the traveling twin's clock can be running faster after he accelerates and starts getting closer to Earth again.

 

[Ich]

But in a frame where the Earth is moving at 0.99c or something, then if the second twin moves away from the Earth "very slowly" (ie if you take the limit as his velocity in the Earth's frame approaches zero), you still find that in this frame his clock gets significantly out-of-sync with the earth-twin's frame as they move a significant distance apart, so in this frame your "starting point" looks quite different, and is still compatible with the idea that the traveling twin's clock can be running faster after he accelerates and starts getting closer to Earth again.

They will be out of sync as long as they are apart. When they rejoin, they will be in sync again except for this little difference one would calculate in the rest frame. Proper time is unambiguous.

 

[JesseM]

They will be out of sync as long as they are apart. When they rejoin, they will be in sync again except for this little difference one would calculate in the rest frame. Proper time is unambiguous.

Yes, if they moved slowly in both directions. But since we were talking about the twin paradox and age difference, I thought you meant that they would move apart very slowly, leading to the initial separation with their clocks still very close to synchronized in the Earth's frame, then reunite at some significant speed so the traveling twin would be noticeably younger when they met.

 

[Ich]

Yes, if they moved slowly in both directions. But since we were talking about the twin paradox and age difference, I thought you meant that they would move apart very slowly, leading to the initial separation with their clocks still very close to synchronized in the Earth's frame, then reunite at some significant speed so the traveling twin would be noticeably younger when they met.

That´s exactly what I meant, provided that they move VERY slowly wrt Earth in the beginning.
I´m afraid I´m starting missing something.

 

[JesseM]

That´s exactly what I meant, provided that they move VERY slowly wrt Earth in the beginning.
I´m afraid I´m starting missing something.

But then why did you say "When they rejoin, they will be in sync again except for this little difference one would calculate in the rest frame"? If they move together quickly, they will not be in sync again (or infinitesimally close to in sync, as they would if their relative motion was very slow in both directions), the clock of the twin who turned around will be significantly behind the clock of the twin who moved inertially.

Anyway, my point is that any inertial reference frame is equally valid, which means you cannot say that the traveling twin objectively "aged less" on just the inbound leg of the trip as yogi would, although you can say he aged less from the moment he departed the earth-twin to the moment they reunited. In a frame where the traveling twin was at rest during the inbound leg and the Earth was moving at high velocity towards him, it is the Earth that ages less during this leg of the trip. But in this frame, the traveling twin aged significantly less during the outbound leg, even though in the Earth's frame his velocity in the outbound leg was close to zero so his age stayed about the same as the earth-twin's. The result is that in this second frame, at the moment the traveling twin accelerates in the direction of the earth, he is already significantly younger than the earth-twin, so even though he ages faster than the earth-twin during the inbound leg in this frame, he will still be younger than the earth-twin when they reunite. But you cannot say that he "aged less" than the earth-twin during the inbound leg, in this frame he aged more during this period because the earth-twin's clock was ticking slower.

 

[Ich]

- that´s what i meant with "little" difference - I did not imagine B to accelerate to .99 c. Anyway - we agree here.
- I think I got your point. Still, the setup of yogi´s gedankenexperiment is such that you have three events:
1) A looks at his clock at time 0, position 0
2) B starts approaching A at time 0, position x
3) B passes A at time t, position 0
All times and position as measured in A´s frame.
It does not matter if B really accelerates or merely flies by - the interval between 2) and 3) is less than the interval between 1) and 3). So B aged less.
However, the situation is in no way symmetric, as yogi may conclude (would he?). If you change to a system which is in motion relative to A (eg B´s system), Events 1) and 2) would no longer be simultaneous. I think that´s what you want to say.

 

[JesseM]

- that´s what i meant with "little" difference - I did not imagine B to accelerate to .99 c. Anyway - we agree here.
- I think I got your point. Still, the setup of yogi´s gedankenexperiment is such that you have three events:
1) A looks at his clock at time 0, position 0
2) B starts approaching A at time 0, position x
3) B passes A at time t, position 0
All times and position as measured in A´s frame.
It does not matter if B really accelerates or merely flies by - the interval between 2) and 3) is less than the interval between 1) and 3). So B aged less.

But B only aged less in A's frame. If you look at things in the frame where A was at rest and B was moving towards him, in this frame A aged less between the time B accelerated and the time A and B passed each other. Yogi doesn't accept that either perspective is equally valid, and says there's some objective, frame-independent sense in which B really aged less between the time he accelerated and the time A and B passed, because B was the one who accelerated.

However, the situation is in no way symmetric, as yogi may conclude (would he?). If you change to a system which is in motion relative to A (eg B´s system), Events 1) and 2) would no longer be simultaneous. I think that´s what you want to say.

Yes, exactly.

 

[Rev Prez]

Just because he is traveling a distance why should he be younger?

Because, if we accept that the "stationary" observer is just that, the only one changing reference frames is the twin we expect would be younger. If you wouldn't describe two objects moving on two "different" paths relative to a stationary as being in the same reference frame, why would you try to do for a single object that moves along two "different" paths?

Rev Prez

 

[Ich]

But B only aged less in A's frame. If you look at things in the frame where A was at rest and B was moving towards him, in this frame A aged less between the time B accelerated and the time A and B passed each other. Yogi doesn't accept that either perspective is equally valid, and says there's some objective, frame-independent sense in which B really aged less between the time he accelerated and the time A and B passed, because B was the one who accelerated.

I think yogi is partially right here. When you set up his experiment exactly as he sais, you obtain those 3 events I described. And the distance between 2) and 3) is in an objective, frame-independent sense less than the distance between 1) and 3).
And so yogi is right in claiming that in his experiment the clocks will be out of sync. But this has nothing to do with acceleration - the frame dependent thing is that he said that the clocks have been synchronized in A´s frame.

 

[JesseM]

I think yogi is partially right here. When you set up his experiment exactly as he sais, you obtain those 3 events I described. And the distance between 2) and 3) is in an objective, frame-independent sense less than the distance between 1) and 3).

But there's no frame-independent sense that 2) and 1) happened at the same moment. We could also imagine event 1b): A looks at a special clock designed to show the correct time in B's frame, and sees that it reads time 0. In this case, it is also a frame-independent truth that the time between 1b) and 3) is less than the time between 2) and 3).

And so yogi is right in claiming that in his experiment the clocks will be out of sync. But this has nothing to do with acceleration - the frame dependent thing is that he said that the clocks have been synchronized in A´s frame.

I agree the clocks will be out-of-sync, of course. But the only question of contention here is, is there any frame-independent sense in which B "aged less" than A from the moment he accelerated towards A to the moment he arrived at the same location as A? Is there any sense in which it is somehow less valid to consider this experiment from the point of view of an inertial frame where, after A accelerates, A is at rest and B is moving towards him?

 

[Ich]

Agreed to all:
yogis experiment has a frame-independent outcome because he gave an exact prescription for the definition of event 1) ("simultaneous to event 2) in A´s frame").
And it makes no sense of talking about who aged less if you leave event 1) somewhat undefined. It all reduces (as always) to the definition of simultaneity.

 

[yogi]

Ok - some interesting comments - let me go back Jesse to the thought that I proposed a few weeks ago - and here is where I stand. We start with A and B at rest in an inertial frame - They can be adjacent - it doesn' matter - they are identical clocks. B Starts out North and then swings East, then South in a large half circle passing Altair and then continues South then bends West and returns to A completing a giant circle. As stated, this is a version of the twin paradox - but there is not abrupt turn around - whatever is going on with B's clock is a continuous happening - the circular path taken by B is continuous and B always travels at a velocity v along the tangent to the circle. Now I think everyone will agree that when B returns to A, B's clock reads less (there is a real objective difference between the time logged on B's clock and the longer time accumulated on A's clock) - The question is when and where does the difference occur - The circular path eliminates those explanations that are based upon shifting slopes of world lines, the lost time at turn around, abruptly shifting hyperplanes and the turn around acceleration as being a factor.

So what do we have - Einstein tells us the clocks will no longer be in sync and that the one which took the trip will lag the other by (1/2)t(v/c)^2 (where t is the total time of the trip) Would you argue that B actually has to reach A to experience a real time difference - what if B pulls up in San Fancisco instead of Los Angles where A is waiting - there is still the same time difference essentially - What if B stop on Mars as he is looping back to Earth and then checks his clock and sends a radio signal to A - well, they won't quite agree - but there will be almost as much age difference between A's clock in Los Angeles and B's clock on Mars as if B had continued on to Los Angeles - and so on - if B stops at Alpha Centuri, A will still be much older that B at the time they communicate by radio, but not as much older as if B had completed the trip because the t factor that is appropriate for stopping on alpha centuri is less that the t factor that would correspond to completing the voyage - what I am saying and what I have always said, is that the situation is proportional - B's clock at every point in the journey is falling proportionately behind A's clock - Einstein tells us this in no uncertain terms - there is no ambiguity in what Einstein says in Part 4 of his 1905 paper -

Where the confusion arises is because of the tendency to hark back to the fact that two clocks in motion always see the other guys clock running slow - that is true - but of course we know that both cannot be running slow relative to the other - the difference in the physical aging experiments is that A's frame and B's frame are not the same - B is moving along a path in A's frame that comprises a real length (yes Jesse I said real - same term Einstein used - we now call it a proper length).

So yes - I do assert that once two clocks are brought to sync in one inertial frame, the clock which is then moved at velocity v relative thereto will always log less time. Of course, to move, B must change its velocity from zero to v - that necessarily involves an acceleration - but it is only incidental to the real understanding of why there is an objectively real age difference that can, and is measured, in every high speed particle experiment.

And I also claim, based upon the above, that, both A's clock and B's clock will read half as much when B passes Altair as they both do When B completes the trip and lands in Los Angeses - and since we know the real distance to altare as measured in A's frame we can verify this with a radio transmission sent from B as he passes Altare.

 

[JesseM]

Ok - some interesting comments - let me go back Jesse to the thought that I proposed a few weeks ago - and here is where I stand. We start with A and B at rest in an inertial frame - They can be adjacent - it doesn' matter - they are identical clocks. B Starts out North and then swings East, then South in a large half circle passing Altair and then continues South then bends West and returns to A completing a giant circle. As stated, this is a version of the twin paradox - but there is not abrupt turn around - whatever is going on with B's clock is a continuous happening - the circular path taken by B is continuous and B always travels at a velocity v along the tangent to the circle.

Well, in a frame where A is not at rest, B's speed will be constantly changing as he moves around the circle, and so the speed his clock runs in this frame will be constantly changing as well, there will be times when his clock is actually running faster than A's. Of course all frames will get the same answer as to what A and B's clocks will read at the moment they reunite.

Now I think everyone will agree that when B returns to A, B's clock reads less (there is a real objective difference between the time logged on B's clock and the longer time accumulated on A's clock) - The question is when and where does the difference occur - The circular path eliminates those explanations that are based upon shifting slopes of world lines, the lost time at turn around, abruptly shifting hyperplanes and the turn around acceleration as being a factor.

The path integral approach works fine as an explanation here. And like I said in another post, all other explanations are really derivative of the path integral approach. For example, as I said in another post, when people use the explanation that the twin who accelerates is the one who'll be younger, this can be thought of as a consequence of the fact that any non-straight path between two points in spacetime will always have a shorter path integral than a straight (non-accelerating) path between them. Someone also mentioned the "triangle paradox", which is the astonishing fact is that a person walking along the hypotenuse of a right triangle will always walk a shorter distance than the person who walks along the other two edges, despite the fact that they start at the same point and end at the same point. You could explain this by integrating the length of each person's path or just by saying that a straight path is always the shortest one, therefore the person who turned around will always walk a greater length. Both answers are equally correct.

So what do we have - Einstein tells us the clocks will no longer be in sync and that the one which took the trip will lag the other by (1/2)t(v/c)^2 (where t is the total time of the trip) Would you argue that B actually has to reach A to experience a real time difference - what if B pulls up in San Fancisco instead of Los Angles where A is waiting - there is still the same time difference essentially

If there are two clocks in San Francisco and Los Angeles which are in sync in their mutual rest frame, then the most that any other frame can see them out-of-sync by is x/c, where x is the distance between San Fran. and LA in their own rest frame. So, I think this means the maximum disagreement you could have about the amount that A and B are out-of-sync would be 2x/c. But yes, I would say this means there is not a single objective time difference between A and B unless they actually meet at the same location, if they don't then I'd say you can only talk about a time difference of t ± x/c, where t is the time difference in the San Francisco/Los Angeles rest frame. But the actual amount that you add or subtract from t will depend on which inertial frame you're using, and Einstein would say that any inertial frame is equally valid.

what I am saying and what I have always said, is that the situation is proportional - B's clock at every point in the journey is falling proportionately behind A's clock

No, that's not true. In a frame where A is in motion, there will be moments along B's journey where B's velocity in this frame is less than A's, agreed? If so, then the formulas of relativity clearly show that in this frame, it would be A's clock that is ticking slower at that moment.

Einstein tells us this in no uncertain terms - there is no ambiguity in what Einstein says in Part 4 of his 1905 paper -

I just looked over section 4 of the paper, I saw him correctly state that the clock that went in the circle would be behind by (1/2)tv^2/c^2 when they reunited, but nowhere did I see anything about B's clock falling behind A's clock at the same rate throughout the journey. If you think he did, please post the quote you're thinking of.

Where the confusion arises is because of the tendency to hark back to the fact that two clocks in motion always see the other guys clock running slow - that is true - but of course we know that both cannot be running slow relative to the other - the difference in the physical aging experiments is that A's frame and B's frame are not the same - B is moving along a path in A's frame that comprises a real length (yes Jesse I said real - same term Einstein used - we now call it a proper length).

But a path isn't an object, it can't have a proper length since it doesn't have a rest frame.

So yes - I do assert that once two clocks are brought to sync in one inertial frame, the clock which is then moved at velocity v relative thereto will always log less time.

But what do you mean by "log less time"? I would certainly agree that when the clock that is accelerated meets up with the other one, it will be behind it. But do you agree that in an inertial frame where both clocks start out in motion, and the clock that accelerates actually decreases in velocity, then before it accelerates it will start out significantly behind the other clock, and then it will tick faster after it acelerates, just not by enough to catch up with the other clock when they meet?

And I also claim, based upon the above, that, both A's clock and B's clock will read half as much when B passes Altair

But to ask what A's clock reads "when B passes Altair" depends on your definition of simultaneity. Do you agree that according to Einstein, no inertial frame's definition of simultaneity is preferred over any other's? Do you agree that in a frame where B is at rest while A and Altair are in motion, it would be A's clock that's behind B's clock at the moment B passes Altair?

 

[yogi]

Jesse - I will take your statements one at a time - no - B's velocity does not change - your forgetting what Einstein said - any pologonal path will do - I chose a circle - and Einstein gave the formula (read it again (1/2)t(v/c)^2 - your again trying to confuse the simple unambiguous explanation given by Einstein. As long as his velocity does not change we can use exactly what Einstein prescribed. Velocity does not change for an object that is undergoing centripetal acceleration - there is no change in the energy of the B spaceship - we consider it teathered by a long string with a central anchor point midway between Earth and Altair. I will take up each oint separately.

 

[yogi]

The hypothenus problem is reversed - in SR the temporal leg and the spatial leg are differenced - so the analogy is confusing

 

[JesseM]

Jesse - I will take your statements one at a time - no - B's velocity does not change

Think about this more carefully, yogi. Relativity is not even important here--even in Newtonian physics, if you travel in a circle at a constant speed relative to the center of the circle, then in a frame where the center of the circle is in motion, your speed as you move around the circle will constantly be changing. For example, if you're moving clockwise, so that at the bottom of the circle your velocity is totally to the left and at the top of the circle your velocity is totally to the right, then in a frame where the center of the circle is moving to the right, your speed will be higher at the top of the circle than at the bottom of the circle.

and Einstein gave the formula (read it again (1/2)t(v/c)^2 - your again trying to confuse the simple unambiguous explanation given by Einstein.

Yes, I have already said I agree that when the two clocks reunite at time t, the one that went around the circle will have gotten behind by (1/2)t(v/c)^2. But Einstein does not say that at some earlier time t', like when the clock has only gone around half the circle, the clock will be behind by (1/2)t'(v/c)^2--the amount that it's behind at any given moment before they reunite depends on your choice of reference frames.

 

[yogi]

3rd point - There is no common time between the two frames - but we can still know shat Bis clock reads at any point in his journey - because his spatial path is totally known in advance - it is a measured circle in the A frame which includes a flyby of Altair, Alpha Centuri, Mars, whatever. At any of thiese known objects B can hold up a sign indicating his clock's reading and the transmitter on these remote points will send a signal to A - so even theough there is no common time - A can still know figure out how far behind B's clock is lagging - you can't really believe that they have to get back together to determine that there is a real physical difference at every point in the trip - What if B returns but he is an inch away from A - could they compare --come on.
udes a

 

[JesseM]

The hypothenus problem is reversed - in SR the temporal leg and the spatial leg are differenced - so the analogy is confusing

Just exchange "path of shortest length" with "worldlines of greatest proper time" and the analogy works fine. Remember that in general relativity, objects follow geodesics, which are the worldlines that maximize the proper time, just as a geodesic along a curved spatial surface is the path of minimal spatial distance.

 

[yogi]

Your 4th point is not even possible given the initial conditions - there is only one velocity involved - and this is where we differ - you want to say that because part of the journey (where B passes Altair) A will believe the velocity is different - or when B is heading home - its a negative velocity - no - it makes no difference - these are observational aspects that would lead to apparent differences where each clock observes the other to be running slow - but what we are talking about is object age difference - any pologonal path (Again I quote Einstein) and that means the velocity is constant during the entire trip but the trip can wander all over the place -

 

[JesseM]

3rd point - There is no common time between the two frames - but we can still know shat Bis clock reads at any point in his journey

Sure we can. But if we know B's clock reads time t when it's at a particular point on the circle, the only way to say what A's clock reads "at the same time" that B is at that position is to pick a reference frame. Are you denying that different frames would disagree about this according to the rules of relativity? If not, are you denying that Einstein would say that no frame's definition of simultaneity can be preferred over any other?

because his spatial path is totally known in advance - it is a measured circle in the A frame which includes a flyby of Altair, Alpha Centuri, Mars, whatever. At any of thiese known objects B can hold up a sign indicating his clock's reading and the transmitter on these remote points will send a signal to A - so even theough there is no common time - A can still know figure out how far behind B's clock is lagging

How far it is lagging IN WHOSE FRAME?

- you can't really believe that they have to get back together to determine that there is a real physical difference at every point in the trip - What if B returns but he is an inch away from A - could they compare --come on.
udes a

If B was an inch away from x, then the quantity 2x/c, which is the upper bound on the amount that different frames can disagree about how far the clocks are out-of-sync, would be a very tiny amount--about 0.00000000017 seconds. Still, as long as they are not at exactly the same position, there will be at least some disagreement between different frames about how far they are out-of-sync, and the greater the distance, the greater the potential disagreement. Again, do you deny that different inertial reference frames in relativity will disagree about the amount that two separated clocks are out-of-sync? Do you deny that relativity (and Einstein) say that no inertial reference frame is preferred over any other?

 

[JesseM]

Your 4th point is not even possible given the initial conditions - there is only one velocity involved

If you really believe that an object moving at constant speed in a circle in one frame would be moving at a constant speed in other frames, then you don't even understand Newtonian mechanics.

- and this is where we differ - you want to say that because part of the journey (where B passes Altair) A will believe the velocity is different

I never said anything of the sort, obviously the speed is constant in A's frame. But there's no reason we have to analyze the problem in A's frame, is there? You'd get the same answer for how far B is behind A at the moment they reuinite no matter which frame you used.

 

[yogi]

5th point - A path certainly can have a proper length because it is defined in the A frame - I have described a path that B will follow - it is an itenery that A and B work out while they are both at rest in A frame - and all points are measured in the A frame - just because it is a circle doesn't mean it doesn't have a length

 

[JesseM]

5th point - A path certainly can have a proper length because it is defined in the A frame

Nonsense, you're misusing the term "proper length". Proper length is the length of an object in its own rest frame, not just whatever frame we choose to define its length in for the purpose of writing down a problem. I could say that in A's frame there's a rod moving at 0.866c that's 1 meter long, but that's not its proper length, its proper length is 2 meters because that's the length in its own rest frame. A path has no rest frame, and again, the frame we arbitrarily choose to use when writing down a particular problem has jack squat to do with "proper length".

 

[yogi]

I don't understand your 6th point - but to the extent I do - The situation is only defined per Einstein when the two clocks are moving together or stationary and brought into sync - whether one slows or not is of no moment - there is just an acceleration in a different direction - we only refere to the situation when they were last in sync and take all readings from there

 

[JesseM]

I don't understand your 6th point

What sentences in my post are you calling my "6th point"? I've said this before, but I would prefer that you would use the "quote" function instead of just responding to what you think my points were.

- but to the extent I do - The situation is only defined per Einstein when the two clocks are moving together or stationary and brought into sync - whether one slows or not is of no moment - there is just an acceleration in a different direction - we only refere to the situation when they were last in sync and take all readings from there

You'll have to be more specific, I don't understand what a single one of these phrases ('only defined per Einstein', 'whether one slows or not is of no moment', 'an acceleration in a different direction', 'take all readings from there') refers to or means.

 

[yogi]

Your 7th point hits the nail on the head as to where we disagree - Yes as to there being no preferred frame - and Yes as to the question as to What B would measure as to A and Altair - B would at say that both A's clock and the clock on Altair are both running slower than his - but this is an apparent effect - in fact by the same line of reasoning B would get all the way back home and if he made a number of observations on the trip considering his own frame as stationary - every one of those observations made by B would yield the same result - of course he would be very certain that his brother A was aging less - but here is the difference Einstein claims when he explains the physical consequences of the transforms - what he says in Part 4 - there is a real age difference that accumulates in accordance with how far you have traveled in the frame where the two clocks were initailly at rest and in sync - when you add in this factor - the intervals are still equal, but the components that make up the interval for the traveler (B) are different that the single component (time) that comprises the totality of the interval for the A clock. Age differences are real and you don't have to turn around to measure them - The interval for B has a spatial and temporal component, the interval for A has only a temporal component.

 

[yogi]

Here is another way to sharpen the issues - Einstein tells us that a clock at the equator will run slower than a clock at the North Pole (actually there are oblateness issues that cancel - but we will ignor that) - now suppose we have a high tower on the North Pole and we put a clock J on top - we launch another clock Q in polar orbit at the same height as the top of the tower. Since Q and J are at the same height we do not need to consider GR effects - so according to Einstein the orbiting clock Q will run slower - but if Q considers himself stationary for a short time each time he passes J, will he not measure J's clock to be slower? - and will not J measure Q's clock to be slower? - but in actuality it is Q that is ageing less (if you don't believe this you need to speak with some GPS engineers).

 

[JesseM]

Your 7th point hits the nail on the head as to where we disagree

Arrgghh! Please quote my post, I don't know what "point" you're replying to!

Yes as to there being no preferred frame - and Yes as to the question as to What B would measure as to A and Altair - B would at say that both A's clock and the clock on Altair are both running slower than his - but this is an apparent effect - in fact by the same line of reasoning B would get all the way back home and if he made a number of observations on the trip considering his own frame as stationary - every one of those observations made by B would yield the same result

I don't remember saying anything about what things would look like from B's perspective, since B doesn't even have a valid inertial frame. Maybe I did say something like that, but I don't feel like going back and rereading all my posts--again, please quote when responding so I can have some idea of the context.

And no, as a matter of fact I wouldn't say that B would see A's cock running slower throughout the journey. If you're talking about what B would actually "see" using light-signals, at times he would see A's clock running slower and at other times he'd see it running faster. If you're talking about how fast A's clock would be running at any given time in B's coordinate system, this depends on how you define B's coordinate system, since B is not moving inertially I don't think there's any "standard" way of doing it.

When I talked about looking at the same problems in different frames, I meant looking at things from the frame of a separate inertial observer (let's call him 'C') moving at constant velocity relative to A (and remember, 'constant velocity' means both constant speed and constant direction, so B doesn't count even though his speed relative to A is constant).

of course he would be very certain that his brother A was aging less

No he wouldn't. Do you understand that the rules of special relativity do not apply to non-inertial coordinate systems?

but here is the difference Einstein claims when he explains the physical consequences of the transforms - what he says in Part 4 - there is a real age difference that accumulates in accordance with how far you have traveled in the frame where the two clocks were initailly at rest and in sync

If you're saying that Einstein ever said that one inertial reference frame's view of a problem is more physically real than any other's, then you are totally and utterly confused about the most basic ideas of relativity.