View Full Version : Twin Paradox
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. Lets 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 lets 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 isnt 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
Apr3-05, 01:07 PM
"Twin 2 travels away from earth at the speed of light. Lets 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 isnt 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).
(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 lets 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.
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.
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?
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.
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?
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.
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 travelled 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.
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 travelled 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
Apr3-05, 06:58 PM
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.
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 whats the actual math/theory behind the issue this thread brought up? Specifically I mean.
hypermorphism
Apr3-05, 07:19 PM
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 Einstein's "Relativity" (http://www.amazon.com/exec/obidos/tg/detail/-/0517884410/qid=1112570235/sr=8-4/ref=sr_8_xs_ap_i4_xgl14/102-4141933-3859329?v=glance&s=books&n=507846), 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 Taylor/Wheeler's Spacetime Physics (http://www.amazon.com/exec/obidos/tg/detail/-/0716723271/qid=1112570268/sr=1-1/ref=sr_1_1/102-4141933-3859329?v=glance&s=books), 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.
Prague,
Suppose the travelling 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?
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 (http://www.physicsforums.com/showthread.php?t=69214).
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: http://www.physicsforums.com/showpost.php?p=515872&postcount=18
Prague,
Suppose the travelling 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 thats correct.
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
Apr4-05, 05:18 PM
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.
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.
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.
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.
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.
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 travelling twin's clock can be running faster after he accelerates and starts getting closer to earth again.
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 travelling 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.
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 travelling twin would be noticeably younger when they met.
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 travelling 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. :uhh:
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. :uhh: 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 travelling 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 travelling 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 travelling 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 travelling 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.
- 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.
- 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
Apr5-05, 05:55 PM
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
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.
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?
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.
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.
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 (http://www.fourmilab.ch/etexts/einstein/specrel/www/), 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?
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.
The hypothenus problem is reversed - in SR the temporal leg and the spatial leg are differenced - so the analogy is confusing
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.
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
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.
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 -
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?
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.
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
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".
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
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.
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.
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).
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.
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? NO. Q IS NOT IN AN INERTIAL REFERENCE FRAME, RULES OF SR LIKE TIME DILATION ONLY WORK IN INERTIAL REFERENCE FRAMES. and will not J measure Q's clock to be slower? Yup, and J is in an inertial reference frame. but in actuality it is Q that is ageing less (if you don't believe this you need to speak with some GPS engineers). There is no objective sense in which Q is aging less at all points on his orbit, although he is aging less over the course of one complete orbit. Suppose that in J's frame, Q is moving to the right at 20,000 miles per hour when he passes the North Pole, and to the left at 20,000 miles per hour when he passes the South Pole. Now consider another observer K who is moving inertially to the right at 5,000 miles per hour in J's frame, so that in K's frame J is moving to the left at 5,000 miles per hour. This means that in K's frame, the satellite Q is moving to the right at 15,000 miles per hour when it crosses the North pole, and it's moving to the left at 25,000 miles per hour when it crosses the South Pole. So you can see that in K's frame, when Q crosses the North Pole its velocity is actually less than J's velocity at that moment, so Q's clock will be ticking faster at that moment. But at other points in the orbit K will see Q's clock ticking slower, and if you do a path integral of the rate that J and Q's clock tick over the course of one complete orbit from the perspective of K's frame, you will see that overall Q does get further and further behind J each time they reunite.
The point is, again, every inertial frame is equally valid in special relativity, so you can look at this problem from either K's or J's perspective here.
Jesse - There is definitely an objective sense to the fact that Q is ageing less at every point in his orbit - we monitor satellite clocks continuously from different ground stations and if the Height is factored out, an uncorrected GPS satellite clock will continually fall behind the ground observer clock - Are you really saying that, relative to the North Pole clock J, Q will be seen it leading and then lagging and that an opposite result would occur at the South Pole... I shutter to think of the state of GPS technology if you had been project engineer.
AS to the fact that the earth is not an inertial system - it is true - but the experiment can always be adjusted to eliminate the slight curvature when making measurments - For example, in the traveling clock example, B can actually straighten out his curve at any number of destinations (if the curvature bothers you so much) where he wants to take note of the clocks in A frame - so long as his velocity remains at v, the observations of B will indicate that clocks at every point in the A frame are running show - they will not be in sync, but that is not the issue - each individual clock will be seen to be running slow.
And yes - one inertial frame is presumed to be as good as another - but that is where you are missing my point - the individual components of the interval are not the same - and that is why, when relativly moving observers only have one piece of data, namely relative velocity, they will both conclude the other clock is running slow (but this is impossible - its an observational fallacy). But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest. But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.
And so has Yogi
- they will not be in sync, but that is not the issue -
that is exactly the issue. A contiuous shift of sync is nothing less than an observed accelerating/decelerating of A´s clock, as seen by B.
- the individual components of the interval are not the same -
SR is all about the individual components of the interval being different from frame to frame.
But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest. But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.
The only permanent thing about those age differences is the way you set up your Gedankenexperimente. You always choose A´s frame as the one where you measure time differences. And you always have in mind that you would let B 'stop' at any point of his journey, i.e. you would match velocities with A and then there would be a residual time difference. That´s true, at least if you would B slowly travel back to A to compare clocks.
But A´s frame is NOT the one and only frame where physically meaningful things happe. If you set up your Gedankenexperiment differently (eg let A join B´s frame), the outcome woul obviously be different, too.
Jesse - There is definitely an objective sense to the fact that Q is ageing less at every point in his orbit - we monitor satellite clocks continuously from different ground stations and if the Height is factored out, an uncorrected GPS satellite clock will continually fall behind the ground observer clock - Are you really saying that, relative to the North Pole clock J, Q will be seen it leading and then lagging and that an opposite result would occur at the South Pole... No, I'm not saying that. Relative to the north pole clock J, Q is always moving at exactly 20,000 miles per hour, so it is always slowed down by the same amount. But relative to the clock K, which is moving inertially at 15,000 mph relative to J (I messed up the numbers in my example last time, if K only saw J moving at 5,000 mph like I said earlier, then K would always see Q moving faster than J), Q's speed varies from between 5,000 mph at the North Pole and 35,000 mph at the South Pole. So at the North Pole K sees Q moving at 5,000 mph to the right and J moving at 15,000 mph to the left, therefore in K's frame at this moment, J's clock is ticking slower than Q's. Do you disagree with this? Do you disagree that K's inertial frame is just as valid for working out the problem as J's inertial frame? AS to the fact that the earth is not an inertial system - it is true I never said that the earth was not an inertial system, over short time intervals it's acceptable to say that the center of the earth (and the poles) are moving inertially. I just said that Q is not an inertial system. but the experiment can always be adjusted to eliminate the slight curvature when making measurments - For example, in the traveling clock example, B can actually straighten out his curve at any number of destinations (if the curvature bothers you so much) where he wants to take note of the clocks in A frame That won't help your argument at all, because the point is that in looking at the entire trip from start to finish, you still can't treat B as if he's in an inertial frame and thus sees A's clock ticking slower throughout the trip. If you analyze the entire problem from within a single inertial frame, at best B can only be at rest in this frame during one of the line segments that make up his trip. so long as his velocity remains at v, the observations of B will indicate that clocks at every point in the A frame are running show Only if you switch inertial frames in the middle of the problem, but that will involve complexities like B's surface of simultaneity suddenly swinging around, and at those moments A's clock may not be running slow, but instead leaping forward very quickly into the future. You can't just assume that the normal formulas of relativity like time dilation apply to the perspective of an observer who does not move inertially throughout his trip, like B.
So again, it's better to analyze the entire problem from start to finish from within a single inertial frame. Then you really can say that at any moment, the rate a clock is ticking is determined solely by its velocity in that frame. And once again, there are perfectly valid inertial frames where B's velocity is slower than A's at certain points in the circle it makes, or where Q's velocity is slower than J's at certain points in its orbit. And yes - one inertial frame is presumed to be as good as another - but that is where you are missing my point - the individual components of the interval are not the same I don't know what you mean by "the individual components of the interval". and that is why, when relativly moving observers only have one piece of data, namely relative velocity, they will both conclude the other clock is running slow (but this is impossible - its an observational fallacy). Do you also think it's "impossible" that two observers in Newtonian mechanics could each see the other observer's velocity as greater than their own, from the perspective of their respective rest frames? If not, I don't see why two observers who each see the other's clock running slower is any more problematic. But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced As long as the clocks start out at the same location, they are synced in all frames. and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest. You are making nonsense of the term "proper distance". Einstein would never have said a path can have a proper length, and neither would any other relativist. Go on, ask some others on this board if you don't believe me. But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.
And so has Yogi Complete nonsense. Einstein would have said it would be just as valid (though less simple mathematically) to analyze the problem from the perspective of a different inertial frame where A is in motion at constant velocity and B's velocity varies as he moves away from A, and if you analyze the problem in this frame you will get exactly the same answer for how far B will have fallen behind A when they reunite, (1/2)t(v/c)^2. That is why the age difference is real, because you will get exactly the same answer for the age difference when they reunite regardless of what frame you use.
Do you agree that we would get the same answer for the age difference when A and B reunite if we used a different frame, and that Einstein and all relativists would say that it would be just as valid to analyze the problem from some other frame besides A's rest frame? Please answer this question yes or no.
yes or no. You are hung up on the idea that the two clocks must return to the same point for there to be real time dilation - and the fact that the moving clock B cannot be changing course in the space defined by the rest frame of A.
As always, our discussions usually turn out to be about as gratifying as shoveling smoke - and I am sure you are equally frustrated.
yes or no. Cute, yogi. Now could you answer the question for real? You are hung up on the idea that the two clocks must return to the same point for there to be real time dilation I'm not "hung up" on this, I'm just telling you something that all relativists, including Einstein, would agree on. If you have an alternate theory, then feel free to present it, but you should stop making the obviously false claim that Einstein's paper supports your position. and the fact that the moving clock B cannot be changing course in the space defined by the rest frame of A. You mean that we can't apply the standard formulas of relativity to B's perspective if B is moving non-inertially? Once again, this is something that all relativists would agree on. If you doubt me on this, please ask some other people on this board, or email some physics professors, you'll see that I'm not just giving you my own interpretation here. As always, our discussions usually turn out to be about as gratifying as shoveling smoke - and I am sure you are equally frustrated. If you would actually address my specific questions, and use the QUOTE function so that I know what comments you're responding to, these discussions would be a lot less frustrating for me. A good place to start would be the yes-or-no question about whether different inertial frames are equally valid, and whether they'll give the same answer for the time delay when the two clocks reunite.
Huckleberry
Apr9-05, 04:23 PM
Maybe this will help you understand what would happen if you brought the twins back together after moving them around at near light speed. If I'm wrong I'm sure someone will correct me.
Twin 1 blasts off to Alpha Centauri 4 light years distant leaving twin 2 standing at the launch pad. Twin 1 would perceive almost no passage of time. He has been travelling along with the same light that was reflected by the Earth. So when he arrives at his destination he pulls out his ultra powerful telescope and looks back at twin 2 standing on the launchpad while the smoke clears. They appear to be the same age.
What twin 1 observes is not reality. Well, it's real enough, but it is light that has been travelling for 4 years from the time it was reflected off twin 2. In reality, while twin 1 is looking at twin 2 on the launch pad, four years has actually passed for twin 2. Twin 1's consciousness would experience no time and as he looked back at Earth his senses would tell him likewise.
Twin 1 turns around and goes home at the same speed he left. He would be running into all the light that has been reflecting from the Earth at an accelerated pace. In 4 light years distance he would experience 8 years worth of reflected light, the 4 from the time he left and 4 more until he returns. Would his consciousness match his senses in this scenario? If it does then both of the twins would be 8 years older than they were when they first seperated.
Twin 2 waits 4 years and looks for twin 1 at Alpha Centauri, but doesn't find him there. He sees him about halfway there. After almost 4 more years he looks again and finds twin 1 has finally arrived and is looking at some point in space where the launch pad was eight years ago. Twin 1 looks like he hasn't aged at all in eight years. A few moments later twin 1 arrives at the launch pad unexpectedly and now both twins are the same age again.
This assumes that just because a person is moving does not guarantee that they have a positive velocity. I imagine their actual velocity would be a vector based on the center of gravity for the universe and taking into account regional irregularities. Any motion requires a 4th dimension.
Ok, you can crucify me now but please be kind enough to educate me instead of just insulting me.
What was the question?
Huck
All right Jesse - I will try to say it in a way that conveys what Einstien said as I see it unambiguously - lets keep it real simple. A and B at rest in the same frame separated by a distance d. Clocks brought to sync. B starts moving toward A at relative velocity v (a short acceleration to get up to speed - then cruises at constant relative v). Einstein says when B arrives the clocks will no longer read the same - B's clock lags (has logged less time - is younger - whatever you want to call it).
I am saying that this is a real difference - it is not an observational apparency - it is a real objective difference in the two clocks.
I think your question is - what if you consider things from B's perspective - once B is in motion, according to the transforms and the fact that B is now in an inertial frame - why can't he say that A's clock is running slow. He will measure it to be running slower if he considers himself at rest and A moving toward him in his own frame.
But if B draws this conclusion, it cannot be a reality. There is only apparent symmetry. When the two clocks arrive, we cannot have a result where B's clock reads less than A's while at the same time A's clock reads less than B's. So if you try to do the problem in the frame of B you would get a different answer if you fail to include the initial conditions But if you take into account the inital conditions (namely - how did the relative motion come about) you will get the correct answer in any frame in which the motions are transformed. So my answer to your question is both yes and no
Huck - didn't mean to ignor your post - Jesse and I have this love affair going (sort of like Roy and the tiger) and so its hard to switch to a new problem when we can't get by first base on this one.
Huckleberry
Apr10-05, 03:08 AM
It's quite alright yogi. I was kind of hoping my post would be ignored anyway. What I wrote feels right, but I have the impression that it is dreadfully wrong and I've made a fool of myself by writing it. I was expecting fire and brimstone and haven't seen it yet. I'm thankful for that. I'm fairly new to this site and don't know what to expect yet.
Huck
What I wrote feels right, but I have the impression that it is dreadfully wrong and I've made a fool of myself by writing it.
Actually, it's mostly more or less correct as a qualitative description, assuming the traveling twin's speed is high enough. The only place you trip up is at the very end of the journey:
A few moments later twin 1 arrives at the launch pad unexpectedly and now both twins are the same age again.
Twin 1 does return home close on the heels of the light by which twin 2 sees him at Alpha Centauri; but they are not the same age when they meet again. Furthermore, if they watch each other continuously through telescopes during the trip, they will see each other aging in such a way that just before twin 2 arrives home, both of them will expect twin 2 to be younger.
For a detailed numeric example using a somewhat lower speed than you're imagining, see posting number 3 in this thread:
http://www.physicsforums.com/showthread.php?t=69214
You might try re-calculating the numbers using, say, v = 0.99c and see what you get.
All right Jesse - I will try to say it in a way that conveys what Einstien said as I see it unambiguously - lets keep it real simple. A and B at rest in the same frame separated by a distance d. Clocks brought to sync. B starts moving toward A at relative velocity v (a short acceleration to get up to speed - then cruises at constant relative v). Einstein says when B arrives the clocks will no longer read the same - B's clock lags (has logged less time - is younger - whatever you want to call it).
I am saying that this is a real difference - it is not an observational apparency - it is a real objective difference in the two clocks.
I think your question is - what if you consider things from B's perspective Not exactly, I mean what if you consider things from the perspective of an inertial frame where B is at rest after B accelerates. In this frame, A and B will initially both be travelling at velocity v, then B will accelerate and come to rest, while I'm not sure if you would say this would be true "from B's perspective" (perhaps you would say that both A and B were initially at rest from B's perspective)--as always, it's tricky to talk about the "perspective" of an observer who accelerates at any point during the time period you're considering, it's better to analyze the whole problem from the point of view of a single inertial frame.
Alternately, if you don't want to talk about "frames", you could talk about the perspective of a third observer C who has been travelling at velocity v relative to A since the beginning of the time period we're looking at, who at first sees B travelling at v and then sees B accelerate and come to rest. once B is in motion, according to the transforms and the fact that B is now in an inertial frame - why can't he say that A's clock is running slow. He will measure it to be running slower if he considers himself at rest and A moving toward him in his own frame.
But if B draws this conclusion, it cannot be a reality. There is only apparent symmetry. When the two clocks arrive, we cannot have a result where B's clock reads less than A's while at the same time A's clock reads less than B's. So if you try to do the problem in the frame of B you would get a different answer if you fail to include the initial conditions I think the problem is that when you say "B's frame" you are considering a perspective where A is initially at rest and synchronized with B, but then after B accelerates A is moving with velocity v and running slow. But again, this isn't a valid frame, since acceleration was involved. If you consider things from the perspective of that third observer C who was always travelling at v relative to A, this observer C will initially see A and B travelling together at velocity v, and will see A's clock ahead of B's by vd/c^2. Then B will accelerate and come to rest, while A continues to move at velocity v in the direction of B. A's clock will be ticking slower than B's, but because A's was already ahead, A's clock will still be ahead when A and B meet. C's prediction about how much A's clock will be ahead of B's when they meet will exactly match the prediction made in A's frame, even though in this frame the explanation for why A's clock was ahead will be different.
So, do you agree that the description from C's perspective is just as valid as the description from A's perspective? If not, do you at least agree that all mainstream physicists, including Einstein, would say that the perspective of one inertial frame is just as good as any other? But if you take into account the inital conditions (namely - how did the relative motion come about) you will get the correct answer in any frame in which the motions are transformed. What if we only start looking at the problem at some time t after B has accelerated? Aren't the initial conditions at t enough to make complete predictions about what will happen in the future, without knowing what happened before t (ie without knowing if it was A or B who accelerated?)
Well - here is where I am probably going to differ from what you refer to as mainstream physics - but I think it is not only consistent with SR, it is also in conformity with all experiments that I am aware of . So, from the standpont of a third frame C, if we consider the velocities of A or B or both after they are up to speed relative to C, we would lose what I consider important, namely how did these velocities come about. Now on the other hand, if A, B and C were all at rest at some instant in the past, and all clocks calibrated and brought into sync, and then A or B or both where later accelerated - then the inital conditions are not lost - namely - which clocks were put into motion relative to C. The transforms would work to yield the actual age differences. For example if A had accelerated toward B we could calculate a spacetime interval for A in the C frame and likewise if B had been accelerated (perhaps a different amount of acceleration than A), then B's interval would have both a temporal component and a spatial component as did A's as observed from the C frame. And a straightforward application of Einstein's formula would give the actual age difference between B and A when the meet.
What i think Einstein did in his 1905 part IV which is significantly entitled "Physical Meaning of the Equations...." was to take the transforms to a different level - from their derivation based upon observations between relatively moving frames where each observer infers the other guys clock is running slow - to a full blown Minkowski space time where it becomes crucial to know which clock moved. Admittedly, once the B clock is in motion, everything looks perfectly symmetrical - but as you have pointed out, if the frames are symmetrical, why would there be a real age difference?
This was the point of my post on the cosmological twin paradox introduced by Garth - I don't see how it is possible to sync clocks on the fly - that is in relative motion - you can read a clock of course - as it passes by, but you have no way of knowing whether it is actually running faster or slower - so on any round trip, whether it be a circumnavigation of the universe or once around the earth, you don't have a way of determining whether the other clock is actually running at an identical speed -
As a matter of academic interest - I think it would be a very interesting to set up a PF pole and ask members if they consider the situation of the two clocks initally in sync and then have B move to A, constitutes a real objective age difference or not.
Are you game?
ps - I know Minkowsk came up with the idea after 1905 - but as I read Einstiens's paper, I see it as implicit in his description.
Well - here is where I am probably going to differ from what you refer to as mainstream physics - but I think it is not only consistent with SR, it is also in conformity with all experiments that I am aware of . What is consistent with SR? The idea that B was objectively aging slower after it accelerated towards A, even though in C's frame it would be A that was aging slower? Or do you disagree that in C's rest frame, A would be aging slower? So, from the standpont of a third frame C, if we consider the velocities of A or B or both after they are up to speed relative to C, What does "up to speed relative to C" mean? C initially sees A and B moving at velocity v, then B accelerates and comes to rest in C's frame. we would lose what I consider important, namely how did these velocities come about. Velocities relative to what? Are you suggesting some concept of absolute velocity here? Why can't C and A have been moving at a constant velocity v relative to one another since the beginning of time? If C and A are just taken as the origins of inertial coordinate systems rather than real physical objects (after all, in relativity you are free to use 'reference frames' which no physical object is permanently at rest in), then the very definition of "inertial reference frames" demands that these two origins always move at constant velocity relative to one another. Now on the other hand, if A, B and C were all at rest at some instant in the past Nope, that's not the scenario I'm asking about. Again, the whole concept of an "inertial reference frame" presupposes that different frames are moving at constant velocity relative to one another for all time--if you want to consider what things look like from the perspective of a physical object which accelerates, you are not talking about inertial reference frames. Please don't duck the question--I'm asking whether the perspectives of different inertial reference frames are equally valid, not the perspectives of different physical observers who don't always move inertially.
By the way, are you saying that if our initial conditions at time t show A and B moving towards each other at constant velocity v, then in order to decide who is aging less we must know who accelerated last before time t? This is in clear contradiction of what you said on this thread (http://www.physicsforums.com/showthread.php?t=51197&page=3), when I asked you: Let me put it this way--do you agree that if we know the complete set of initial conditions at some time t in a particular frame (the position and velocity of both objects at t, the time on their own clocks, etc.) and we want to make some predictions about what will happen later, then since the laws of relativity are completely deterministic, these initial conditions at t are sufficient to make a unique prediction about the future? Do you agree that what happened before t is irrelevant, including the question of which of two clocks accelerated before t? You replied "Yes - i will agree to that." So unless you misunderstood, or unless you are going back and changing your answer, that means if at some time t we see A and C moving towards each other at constant velocity, then the question of whether A or C accelerated at some time before t is not relevant to deciding any physical question, including the question of who is aging slower. and all clocks calibrated and brought into sync, and then A or B or both where later accelerated - then the inital conditions are not lost - namely - which clocks were put into motion relative to C. A was not put into motion relative to C during the period we are considering. A and C have been moving at constant velocity relative to each other since the beginning of the time period we are considering. If you like, you can consider them to have been moving at constant velocity relative to one another since the beginning of time, although as I said above this shouldn't be necessary if you agree that we don't need to know anything about times before our initial conditions to answer the question of who ages slower. What i think Einstein did in his 1905 part IV which is significantly entitled "Physical Meaning of the Equations...." was to take the transforms to a different level - from their derivation based upon observations between relatively moving frames where each observer infers the other guys clock is running slow - to a full blown Minkowski space time where it becomes crucial to know which clock moved. Well, that's just ignorant. Anyone who understands relativity knows that different reference frames disagree about which of two clocks is ticking slower, and that all frames are equally valid, the question of which clock accelerated in the past does not force you to choose one frame over another. This symmetry of different reference frames is a very basic part of "Minksowski space time", if you think that the question of which object "moved" (accelerated) is relevant to deciding which inertial reference frame to use, you're badly confused about how Minkowski space time works. The point of the twin paradox is not that one inertial frame is favored over another depending on who accelerated, it's just that you can't put the perspective of the non-inertial twin on the same footing as an inertial reference frame. Admittedly, once the B clock is in motion, everything looks perfectly symmetrical - but as you have pointed out, if the frames are symmetrical, why would there be a real age difference? How many times do I have to repeat this? The reason there is a real age difference is because different frames define simultaneity differently. Do you not believe me that C's frame will make exactly the same prediction as A's frame about what A and B will read when they meet, in spite of the fact that C's frame says A is ticking slower as they approach each other and A's frame says B is ticking slower as they approach each other? I don't see how it is possible to sync clocks on the fly - that is in relative motion - you can read a clock of course - as it passes by, but you have no way of knowing whether it is actually running faster or slower What do you mean by "sync"? Do you mean to make sure they're reading the same time? That's impossible if they're in relative motion, since they will be ticking at different rates. Or if by "sync" you just mean determining what one clock reads "at the same moment" that another clock reads a given time, like "when clock A read 2:00, clock B read 4:00" (ie the issue of simultaneity), then each frame can do this either using light-signals (taking into account the time the light took to reach you), or by using local readings on a network of clocks which are synchronized in that frame (the clocks are defined to be synchronized in this frame if light emitted at the midpoint of two clocks hits both at the same time according to their readings). This is the procedure that Einstein outlined in his paper, so even if you disagree with it, if you disagree that this is how Einstein defined simultaneity then you're just being ignorant. As a matter of academic interest - I think it would be a very interesting to set up a PF pole and ask members if they consider the situation of the two clocks initally in sync and then have B move to A, constitutes a real objective age difference or not.
Are you game? Only if both the scenario and the question are phrased with enough detail so people don't misunderstand the question. For example, I might phrase it like this: Suppose you have two clocks A and B which are initially at rest relative to one another and synchronized in their own frame, then B accelerates towards A and moves toward it at velocity v until they meet. Do you think it is objectively true that B was aging more slowly as it approached A, even though it is possible to view this situation from a frame where A and B are both initially moving at velocity v, then B comes to rest and A continues to move towards it at velocity v until they meet, so that in this frame A would be ticking more slowly after B comes to rest? Of how about this: Suppose two clocks start out at rest relative to one another and synchronized in their own frame, then one accelerates briefly and after that they move towards one another at constant velocity. If different inertial reference frames disagree about which clock is ticking slower as they move together, can we determine which frame is "objectively" correct by checking which clock was the one that accelerated?
Briefly - rather than go into all your misinterpretations as to what I have said - if you know the complete history of the two clocks - then yes, the predictions as to which is ticking faster, if either, is determined. By complete history you would have to know more than just the fact that the two frames have relative motion.
You can find recogonized authors that interpret Einstein's statement as real one way time dilation (age difference). John D Barrow for example explains the reality based upon the difference in the energy that can be attributed to the change in velocity - anyway - the primary experiments that are available to show time dilation are those dealing with high speed particles, H and K round the world airplane journeys, GPS, all of which are easily explained by taking Einstein's explanation at face value.
So let us see if we can agree upon a correct way to state the poll. I object to the words "objectively correct" I propose the following:
Einstein makes the following statement in part 4 of his 1905 paper under the Heading Physical meaning of the Equations in respect to moving Rigid bodies and moving clocks:
"If at points A and B there are stationary clocks, which viewed in the stationary system are synchronous, and if the clock A is moved with velocity v along the line AB to B, then on its arrival at B, the two clocks no longer synchronize, but the clock moved from A to B, lags behind the other, which has remained at B..."
Question for members - is this a real age difference or would the two clocks actually read the same when compared?
We´ve been through this - The time difference is real, it is the difference in proper time since the clocks of A and B have been synchronized in frame A.
The synchronization is of course frame dependent, and so is the time difference if you chose tho synchronize in a different frame.
You can find recogonized authors that interpret Einstein's statement as real one way time dilation (age difference). John D Barrow for example explains the reality based upon the difference in the energy that can be attributed to the change in velocity - anyway - the primary experiments that are available to show time dilation are those dealing with high speed particles, H and K round the world airplane journeys, GPS, all of which are easily explained by taking Einstein's explanation at face value. No, you are simply misinterpreting them, as usual. Einstein makes the following statement in part 4 of his 1905 paper under the Heading Physical meaning of the Equations in respect to moving Rigid bodies and moving clocks:
"If at points A and B there are stationary clocks, which viewed in the stationary system are synchronous, and if the clock A is moved with velocity v along the line AB to B, then on its arrival at B, the two clocks no longer synchronize, but the clock moved from A to B, lags behind the other, which has remained at B..."
Question for members - is this a real age difference or would the two clocks actually read the same when compared? yogi, this is a totally dishonest strawman on your part--you know perfectly well that I agree the clock A would be behind B when they meet, I just disagree that this means A was "aging slower" during its journey.
Let's say A and B start out 1 light year apart in their rest frame, and synchronized in this frame, then at the moment that both clocks read 2005, A accelerates to 0.5c and moves at constant velocity towards B. In this frame, A will reach B in (1 light year)/(0.5c) = 2 years, in 2007. But A was ticking at (1 - 0.5^2) = 0.866 of B's rate during this trip, so when it meets B it will only read 2005 + 2*0.866 = 2006.73.
So, in B's rest frame:
amount of time that B ages after A accelerates: 2 years
amount of time that A ages after A accelerates: 1.73 years
But in the frame of an inertial observer C who sees A and B initially moving at 0.5c, then sees A decelerate until it's at rest, things look a bit different. In this frame, A and B were initially not synchronized--instead, A and B were initially out-of-sync, with B ahead of A by 0.5 years. So if A read 2005 at the moment it decelerated, B already read 2005.5 at that moment. After that, B was moving towards A at 0.5c, and in this frame the distance between them was only 0.866 light years, so it takes (0.866 light years)/(0.5c) = 1.73 years for B to reach A. Since A was at rest during this time, its clock will elapse 1.73 years during this time, so it will read 2005 + 1.73 = 2006.73 when B arrives. But B is only ticking at 0.866 the normal rate during this time, so itwill tick 1.73*0.866 = 1.5 years in this time, and since it read 2005.5 at the moment A accelerated, it will read 2005.5 + 1.5 = 2007 at the moment it reaches A.
So, in C's rest frame:
amount of time that B ages after A decelerates: 1.5 years
amount of time that A ages after A decelerates: 1.73 years
So, both frames predict that B reads 2007 and A reads 2006.73 at the moment they meet, but in B's rest frame it was A who was aging slower after A accelerated, and in C's frame it was B who was aging slower after A decelerated. The question is, are both these descriptions equally valid? Am I correct in understanding that you would say both descriptions are not equally valid, yogi, but that the description in B's frame is somehow more correct than the description in C's frame? I promise that any relativist, including John D Barrow, would say both descriptions are equally valid.
Jesse - you are weasel wording your way around again - you have previously taken the position that there can be no objectively real age difference in the situation I quoted in post 64.
Jesse - Here is one of my previous statements:
Originally Posted by yogi
Clocks in relative motion do record different absolute times whenever they have been synced at rest and one clock is put into motion - all evidenced by H and F airplane experiments, GPS, and the extended lifetime of high speed muons and pions that are created in the earth reference frame and subsequently move relative thereto.
And your response:
None of these experiments show evidence for an "absolute" truth about which clock is running slow, all of them are perfectly consistent with the idea that each inertial frame sees clocks in other frames running slow, and that there is no way to settle which frame's point of view is the correct one. If you think Einstein would have disagreed, then you are completely misunderstanding his paper.
Did I use the word "correct" or preferred or universal -no as to each
I consistently said that when B moves to A and the two clocks are compared in the frame in which they were synced - A will have logged more time than B - there is a one way age difference, real and objective.
Jesse - you are weasel wording your way around again - you have previously taken the position that there can be no objectively real age difference in the situation I quoted in post 64. No, I explicitly said in many posts (I can quote some if you like) that in the scenario where the two clocks are initially at rest relative to each other and synchronized in their rest frame, and then one accelerates towards the other, then the one that accelerated will show a lesser time reading. My position has always been that although this clock shows a lesster time reading when they meet, that does not mean that the clock that accelerated has "aged less" between the time it accelerated and the time they meet. This is clear in the numerical example I give in my last post, where A accelerates towards B, and all frames agree that when they meet, A will read 2006.73 while B will read 2007. But you see that in B's frame, A ages 1.73 years and B ages 2 years between the time A accelerates and the time they meet; but in C's frame, A ages 1.73 years but B only ages 1.5 years between the time A accelerates and the time they meet. Are you saying that the description of the situation from C's perspective is any less valid than B's, or aren't you? Please give me a simple straight answer to this question. I consistently said that when B moves to A and the two clocks are compared in the frame in which they were synced - A will have logged more time than B - there is a one way age difference, real and objective. "Logged more time" since when? In C's frame, B only logs 1.5 years after A accelerates while A logs 1.73 years, but B still comes out ahead because A read 2005 at the moment it accelerated while B already read 2005.5 at the same moment.
To pin down what you mean here, let's imagine that A and B were out-of-sync in their own rest frame--at the same moment B read 2005, A already read 3164. Then A accelerates, and when they come together, B reads 2007 and A reads 3165.73. Would you say that A has "logged more time" simply because it reads a greater time when they meet, ignoring the question of whether they were in sync to begin with? Or would you say that the question of whether they were initially in sync is relevant to the question of which clock "logged more time"?
Don't add other factors - I am saying that, in the example I proposed as a PF poll question in post 64, between a first spacetime event defined as A and B at rest in the same frame and both set to read zero in the frame in which they are at rest, and a second spacetime event (B's subsequent arrival at A consequent to a short acceleration period followed by constant relative velocity v thereafter), then, if A and B both stop their respective clocks and the hands viewed, the clocks will read different (For example A could read 50 minutes and B could read 40 minutes).
there are 3 spacetime events involved.
Don't add other factors Those "other factors" are what our debate is about, since we both agree that in the scenario you proposed as a poll, A will read a lesser time than B when they meet. So now please address the two questions in my last post, which I think will clarify what it is we actually disagree about: This is clear in the numerical example I give in my last post, where A accelerates towards B, and all frames agree that when they meet, A will read 2006.73 while B will read 2007. But you see that in B's frame, A ages 1.73 years and B ages 2 years between the time A accelerates and the time they meet; but in C's frame, A ages 1.73 years but B only ages 1.5 years between the time A accelerates and the time they meet. Are you saying that the description of the situation from C's perspective is any less valid than B's, or aren't you? and To pin down what you mean here, let's imagine that A and B were out-of-sync in their own rest frame--at the same moment B read 2005, A already read 3164. Then A accelerates, and when they come together, B reads 2007 and A reads 3165.73. Would you say that A has "logged more time" simply because it reads a greater time when they meet, ignoring the question of whether they were in sync to begin with? Or would you say that the question of whether they were initially in sync is relevant to the question of which clock "logged more time"? I am saying that, in the example I proposed as a PF poll question in post 64, between a first spacetime event defined as A and B at rest in the same frame and both set to read zero in the frame in which they are at rest As Ich says, this is not a "spacetime event". A spacetime event is something that happens at a single point in space and a single moment in time--any description of two clocks separated in space cannot qualify as a single event. But this is just an issue of your using incorrect terminology, I understand the scenario you are describing and as I've said practically every time we discussed this scenario, I agree that the clock that accelerates will read a lesser time when they meet. If you ever thought I disagreed about this, you haven't been reading my posts very carefully.
No Ich says there are three events - which is correct - however I am only concerned with the reading of clocks in the rest frame starting at the time B begins to move and the time both clocks are stopped. Tell me Ich - would you agree that it makes no difference whether A and B are initally separated and then B moves to A - or If A and B were initally together and B moves the same distance as he did in the first case - and when he stops at the same point and simultaneously stops his stopwatch (clock) - if he sends a signal back to A - would you that in this case the transmiited signal would indicated the same time - e.g. B would send "my clock reads 40 minutes at the time I reached the destination. And A, Knowing the distance, could then calculate the time that passed on his clock from the time B left up until B signalled - and would he conclude that his clock read 50 minutes as before.
Jesse - the way you frame the problem always leads to diversionary tangents - I know how most SR types like to transform from one frame to another - blindly using the formulas - but not keeping in mind the limitations that - measurements of space and time in another relatively reference frame are illusory - and this leads to the abstruse analysis that colminates in shifting hyper planes and earth clocks jumping ahead by thousand of years because the traveler turns around and so on. When you are all done, that type of analyis leaves one not knowing reality from squat. No wonder Einstein said "Now that the mathematicians have gotten involved in relativity, I am not sure I still understand it myself "
Jesse - the way you frame the problem always leads to diversionary tangents So do you refuse to answer my questions? but not keeping in mind the limitations that - measurements of space and time in another relatively reference frame are illusory Illusory? What the hell does that mean? If A and B are out-of-sync in C's frame, is this somehow more "illusory" than the fact that A and B are in sync in B's frame. If so, then it seems you do indeed think the description in B's frame is more valid than the description in C's frame, even if you are not willing to be straight with me and come out and say it in response to my question about this. and this leads to the abstruse analysis that colminates in shifting hyper planes and earth clocks jumping ahead by thousand of years because the traveler turns around and so on. Not if you are consistent about which inertial frame you use throughout the problem. The shifting hyperplanes only happens if you insist on trying to understand what things look like from the perspective of a non-inertial observer, meaning you have to switch from one frame to another in the middle of the problem. But really, the theory of SR was never meant to tell you anything about the "reference frame" of a non-inertial observer, it is just supposed to tell you how a given physical situation can be understood if you analyze it from beginning to end in one inertial frame or another. In the case of the A-B scenario, if you analyze the whole thing from the perspective of B's frame, or if you analyze the whole thing from the perspective of C's frame, there is no shifting of hyperplanes or sudden jumps in clock readings whatsoever. No wonder Einstein said "Now that the mathematicians have gotten involved in relativity, I am not sure I still understand it myself " Yogi, you sound like a creationist taking quotes by scientists out-of-context (http://www.talkorigins.org/faqs/quotes/mine/project.html) to make it sound like they're supporting positions they obviously don't. If you notice, the only mathematical ideas I'm using in discussing how things would look in B's frame vs. C's frame are the Lorentz transformation equations, which Einstein himself derived in section 3 of his 1905 paper! And the whole concept of the Lorentz transformation is not that they represent some arbitrary mathematical transformation, they are the formulas for relating physical measurements made on systems of measuring-rods and synchronized clocks of different observers. If observer B has a system of measuring-rods spread throughout space which are at rest relative to him, with clocks mounted along them that are synchronized using the assumption that light travels at a constant velocity relative to B, and C also has a system of measuring rods at rest relative to him, with clocks mounted on them that are synchronized using the assumption that light travels at a constant velocity relative to him, then what the position and time coordinates they get by reading off their respective measuring-rod/clock systems will be exactly the ones I gave in my numerical example earlier. For example, if C sees that the clock in his system that was next to A when it accelerated read 2005 at that moment (at which moment A's own clock also read 2005), then if he looks at the moment that B was next to a different clock in his system that also read 2005, he will see that B's own clock read 2005.5 at that moment. This shows the physical meaning of the statement that in C's frame, the event of A accelerating, and of A's clock reading 2005, happened at the same moment that B's clock read 2005.5. So don't give me that nonsense about this being just a matter of abstract mathematics, Einstein explains in detail the physical meaning of the coordinates of different inertial observers in section 1 of his 1905 paper (http://www.fourmilab.ch/etexts/einstein/specrel/www/).
Its not abstract math - it is measurement math - the measurements made in a relatively moving frame are only apparent - they are true in the sense that the measurements are true - but to introduce additional clocks and frames is not necessary for the situation involved in the simple example Einstein gave - at no time in his explanation of the physical meaning of the equations does Einstein make any claim about what the moving (B clock) measures in the A frame. (Its not important) The reciprocity of the frames is not discussed by Einstein in part 4 when he describes the physical meaning. So why consider anything other than the simple fact that the clocks dont agree when the one way journey is completed. As you say "SR is meant to explain what happens from beginning to end in one inertial system" dido to that.
Me a Creationist...In the beginning there was nothing, and God said let there be light. And still there was nothing, but you could see it.
Its not abstract math - it is measurement math - the measurements made in a relatively moving frame are only apparent All frames are "relatively moving" relative to something--for example, the frame where B is at rest is moving relative to C. Again, are you saying there's some reason to believe the measurements in B's frame are any less "illusory" or "apparent" than the ones made in C's frame? but to introduce additional clocks and frames is not necessary for the situation involved in the simple example Einstein gave - at no time in his explanation of the physical meaning of the equations does Einstein make any claim about what the moving (B clock) measures in the A frame. (Its not important) Just a quibble--in Einstein's example it's the B clock that is the inertial one, the A clock is the one that accelerates. But even if you switch the letters, I'm not sure I understand your point--he certainly talks about what the clock that was moved measures, he says that it lags behind the clock that didn't move by (1/2)t(v/c)^2. Maybe that's not what you meant by "what the moving clock measures in the clocks' frame", but if so, what did you mean? Did you mean that Einstein doesn't talk about how things look in the frame of the clock that was moved? Well, I don't either, since the clock that was moved (A) moved non-inertially and thus doesn't have a "frame" in SR. I did talk about how the whole situation would look from the frame of a third observer C who is moving inertially at velocity v relative to the inertial observer B. So is your point just that Einstein didn't talk about how the whole situation would look in a different frame? No, he didn't, but anyone who comprehended the previous sections of the paper would understand that Einstein believed it's an arbitrary choice which frame you use to describe a given physical situation, no inertial frame is more valid than any other. Would you disagree that this is clearly Einstein's view? The reciprocity of the frames is not discussed by Einstein in part 4 when he describes the physical meaning. Yes he does. He first talks about how objects look in a reference frame K (which he has labeled 'the stationary system') when they are moving relative to K, then he says "It is clear that the same results hold good of bodies at rest in the 'stationary' system, viewed from a system in uniform motion". So why consider anything other than the simple fact that the clocks dont agree when the one way journey is completed. Because that's not all that you consider. You also say that the clock which moved "logged less time" than the other clock, or that it "aged less". And you also seem to think it is somehow more valid to say that the two clocks were initially in sync than it is to say they were initially out-of-sync. If you want to renounce all such statements and just say that in Einstein's scenario, the clock that accelerated will be behind the clock that didn't when they meet, then I certainly agree with that much, but if you continue to make statements which go beyond that and seem to implicitly privilege B's frame over other frames, then I still say you're wrong. As you say "SR is meant to explain what happens from beginning to end in one inertial system" dido to that. Yes, but SR also says that the only truly physical facts are the ones that would be true in every inertial frame. The fact that A is behind B when they meet will be predicted in every frame, but the statement that A and B were initially synchronized is not a frame-independent truth, nor is the statement that A's clock was ticking more slowly as A and B approached each other. Me a Creationist...In the beginning there was nothing, and God said let there be light. And still there was nothing, but you could see it. I wasn't accusing you of being a creationist, just of taking quotes out of context like a creationist.
What I have attempted to say - many times - is that the A frame is the frame where the clocks are brought in sync - we never do any syncing in the B frame (I am calling B the traveling clock) - everything is determined in A frame - once B is up to speed it is also a valid inertial frame - but it is different than the A frame
To restate: B is initially at rest in the A frame and in sync with A clock - B leaves the A frame (via acceleration that produces a relative velocity) and returns to a different spatial point in the A frame (or B could return to the same location as A). I am saying that this return does not have to be a physical pulling up to a stop in the A frame adjacent to A - it can be passing any known spatial point that was measured off (a proper distance in the A frame) and B's clock will be read by A either because B physically stops at one of these points or passes by and transmits a radio signal to A.
In contrast to venturing to a distant star and returning to effect real age difference (classical twin paradox) - real age difference is brought about whenever one clock leaves the frame in which it has been calibrated and later returns to any point in the frame in which it was calibrated. That is the essence of Einstien's explanation in part 4
If you do not agree - that is ok - but then show me a single experiment that disproves my interpretation (don't give me the mainstream authority argument - give me an experiment)
yogi and JesseM,
Two clocks, A1 and A2, at rest wrt eachother, separated by a distance L, synchronized as defined by SR.
A third clock, B, moving at v wrt A1 and A2, passes A1 (on his way to A2) and, at the instant he passes A1 he synchs with A1.
Do you guys disagree on what B and A2 read when B passes A2?
What I have attempted to say - many times - is that the A frame is the frame where the clocks are brought in sync - we never do any syncing in the B frame (I am calling B the traveling clock) OK, I have been calling A the travelling clock since that's how Einstein wrote it in his example. But I'll switch from now on so we're using the same convention. - everything is determined in A frame - once B is up to speed it is also a valid inertial frame - but it is different than the A frame
To restate: B is initially at rest in the A frame and in sync with A clock - B leaves the A frame (via acceleration that produces a relative velocity) and returns to a different spatial point in the A frame (or B could return to the same location as A). I am saying that this return does not have to be a physical pulling up to a stop in the A frame adjacent to A - it can be passing any known spatial point that was measured off (a proper distance in the A frame) and B's clock will be read by A either because B physically stops at one of these points or passes by and transmits a radio signal to A. Sure. Or this other point could have a clock that was synchronized with A's clock in A's rest frame, and then you could compare what B's clock reads at the moment it passes this clock, and you'd see B was behind this other clock. This would be the idea described in section 1 of Einstein's paper, where the coordinates of an event (in this case, the event of B's clock reading a certain time) are determined by local readings on a network of measuring-rods and clocks spread throughout space which are all at rest relative to the observer and with the clocks synchronized according to Einstein's procedure. In contrast to venturing to a distant star and returning to effect real age difference (classical twin paradox) - real age difference is brought about whenever one clock leaves the frame in which it has been calibrated and later returns to any point in the frame in which it was calibrated. That is the essence of Einstien's explanation in part 4 What I am saying is that this is not a "real age difference" unless they meet, because nothing is physically "real" in relativity unless it is agreed upon in all frames. It's true that if B reads 2010 at the same moment he's passing a clock in A's network which reads 2015, then at that moment B is five years younger than A, in A's frame. But in C's network of clocks, at the moment B reads 2015 he may be passing a clock in C's network that reads 2020, and at the moment that A passes a clock in C's network that reads 2020, A may read 2014. This would mean that in C's frame, A is one year younger than B. Since the question of whether B is older or younger than A at the moment that B reads 2015 cannot be decided in a way that doesn't depend upon an arbitrary choice of reference frame, there is no physically real age difference between them until they meet at a single point in space. If you do not agree - that is ok - but then show me a single experiment that disproves my interpretation (don't give me the mainstream authority argument - give me an experiment) I am not sure what your interpretation actually is. Why is it, exactly, that your interpretation tells you that the age difference in A's frame is more "real" than the age difference in some other frame? Is it simply because Einstein chose to describe this example from the point of view of this frame? Or is it because they were initially brought into sync in this frame? If they had been initially brought into sync in some frame other than their rest frame, would you say that the "real age difference" would be the one in this other frame, or would you say that their real age difference should still be viewed in terms of their initial rest frame, but that it would no longer match the difference in their clock-readings since they were out-of-sync in their initial rest frame? If you want me to try to find an experiment that shows why your interpretation doesn't make sense, then please answer these questions about how your interpretation works.
yogi and JesseM,
Two clocks, A1 and A2, at rest wrt eachother, separated by a distance L, synchronized as defined by SR. Synchronized in their rest frame, presumably? A third clock, B, moving at v wrt A1 and A2, passes A1 (on his way to A2) and, at the instant he passes A1 he synchs with A1.
Do you guys disagree on what B and A2 read when B passes A2? No, I'm pretty sure we both agree that B would be behind A2. But I think yogi would say this means that B really aged less than A2 did between the time it passed A1 and the time it passed A2, while I would point out that in B's frame, A2 aged less than B between the times of these two events, and I'd say that no inertial frame should be preferred over any other.
So,
Two clocks, A1 and A2, both at rest in S, separated by a distance L, synchronized as defined by SR.
A third clock, B, at rest in S' which is moving at v wrt S, passes A1 (on his way to A2) and, at the instant he passes A1 he synchs with A1.
Suppose you define 3 events, E1, E2, and E3 and use the notation:
event: En[x, t, x', t']
gamma = Y
Then the three events and their S and S' coordinates are:
E1[0, 0, 0, 0] "B passes A1"
E2[0, L, YL, -YLV/C^2] "A2 is set to 0"
E3[L, L/v, 0, (L/v)/Y] "B passes A2"
JesseM, when you say "....in B's frame, A2 aged less than B between the times of these two events" which two events do you mean?
So,
Two clocks, A1 and A2, both at rest in S, separated by a distance L, synchronized as defined by SR.
A third clock, B, at rest in S' which is moving at v wrt S, passes A1 (on his way to A2) and, at the instant he passes A1 he synchs with A1.
Suppose you define 3 events, E1, E2, and E3 and use the notation:
event: En[x, t, x', t']
gamma = Y
Then the three events and their S and S' coordinates are:
E1[0, 0, 0, 0] "B passes A1"
E2[0, L, YL, -YLV/C^2] "A2 is set to 0"
E3[L, L/v, 0, (L/v)/Y] "B passes A2"
JesseM, when you say "....in B's frame, A2 aged less than B between the times of these two events" which two events do you mean? In B's frame, the event of E1 does not happen simultaneously with E2--instead, at the moment B passes A1 and both read 0, A2 reads vL/c^2. So, define the event E2b: "A2 reads vL/c^2". If you want to compare how much A2 ages vs. how much B ages in B's own rest frame, then you must compare how much time elapses on B between events E1 and E3 with how much time elapses on A2 between events E2b and E3. Saying that event E2 is somehow more important than E2b is equivalent to saying that one inertial frame's definition of simultaneity (the A1/A2 rest frame) should be preferred over another inertial frame's definition of simultaneity (B's rest frame).
Ok, how's this?
Two clocks, A1 and A2, both at rest in S, separated by a distance L, synchronized as defined by SR.
A third clock, B, at rest in S' which is moving at v wrt S, passes A1 (on his way to A2) and, at the instant he passes A1 he synchs with A1.
Suppose you define 4 events, E1, E2a, E2b and E3 and use the notation:
event: En[x, t, x', t']
gamma = Y
Then the four events and their S and S' coordinates are:
E1[0, 0, 0, 0] "B passes A1"
E2a[L, 0, YL, -YLv/C^2] "A2 is set to 0"
E2b[L, vL/c^2, L/Y, 0] "A2 reads vL/c^2"
E3[L, L/v, 0, (L/v)/Y] "B passes A2"
djavel - I do not know what is meant by syncing a clock that flies by - we can read its instantaneous value, but we do not have a way of determining that the flyby clock is actually running at the same speed as the two clocks that were in the same rest frame. For example a GPS clock that flies over the North Pole will not be in sync with a clock at the North Pole unless it had been preset to run at the same rate as the clock in the non rotating earth centered reference frame located at the same height as the satellite
Jesse - I do not understand why the B clock has to be adjacent (physically return to the same point occupied by the A clock) for a real time difference to have elapsed.
Jesse - I do not understand why the B clock has to be adjacent (physically return to the same point occupied by the A clock) for a real time difference to have elapsed. As I said, because I'd say the only "real" physical quantities are those that all frames agree on. Do you agree that as long as the two clocks are separated in space, different frames will disagree on the time difference between them? Are you saying the time difference in A's frame is more "real" than the time difference in C's frame?
Then the four events and their S and S' coordinates are:
E1[0, 0, 0, 0] "B passes A1"
E2a[L, 0, YL, -YLv/C^2] "A2 is set to 0"
E2b[L, vL/c^2, L/Y, 0] "A2 reads vL/c^2"
E3[L, L/v, 0, (L/v)/Y] "B passes A2" Yes, this looks fine to me.
Dragongod
Apr14-05, 02:42 PM
the problem is that Einstein never actually proved his special relativity theory. Its based on his assumption that in space-time, which according to him is some kind of forth dimension, time isn't universal. He just merely stated that. He never proved it with logical reasoning. So according to his assumptions, if you have twin at point A and twin at point B if one moves faster than the other over, automatically because time isn't universal, one will have to be older. It doesn't actually make sense. Whether you travel at the speed of light or not, and i can say at the speed of light because it hasn't been proven that the speed of light is actually some kind of cosmic speed limit, the distance you travel over in no way should effect time. Time within a system (the Universe) should be constant. According to Einstein its not but that is irrevelent because who is to say he is right - it is not proven that time isn't constant. And logically, it should be. All that changes, within a system where time is constant, is the distance traveled by the body in that period of time. It doens't destort time itself. Special Relativity Theory is based on Einstein's assumptions about Spacetime.
Dragongod
Apr14-05, 02:47 PM
According to Einstein Time is a subjective thing. Time according to me is not. Time exist whether we do or not and always moves forward at an instanteous rate that we cannot measure. For example, the second you try to measure an instant, you are already too late. Time is always moving forward and cannot be distorted, especially by different frames of reference. Whether i am at point A or you at point B or C or D or F it has no baring on the time that is constantly moving forward regardless of anything.
DaveC426913
Apr14-05, 02:58 PM
Dragongod, are you aware of the countless experiments that have demonstrated beyond a shadow of a doubt that objects travelling at relativistic speeds experience time at a different rate?
We don't need spaceships and distant planets and thought experiments. This is real laboratory experimentation. Particles fired at relativistic speeds in accelarators die at a different rate than if they were not moving. Those rates are predicted with exquisite accuracy, time and time and time again, by Einstien's theory.
Relativity is one of the most tested theories in the history of science, and it has passed spectacularly every time.
Einstein's relativity is not merely the best theory going, it is the only theory that even makes the trip downtown, let alone into the ballpark.
Dragongod
Apr14-05, 03:35 PM
That doesn't prove that the particles are somehow not governed by the same dimension of time. The fact that that happens (particles moving at a faster and rate and living longer than those that are not moving as fast) could be due to the fact that as particles move over a DISTANCE not TIME, they gain more mass and therefore gain more energy. That experiment doesn't necessaritly prove what Einstein was proving - that it is due to different TIMES.
Dragongod
Apr14-05, 04:11 PM
The experiment doesn't necessarily prove that time is relative. It just proves that F=ma and that an increase in force would mean an increase in mass and acceleration. you can't increase acceleration without increasing mass. As something moves over a particular distance it gains more energy. Thats all it shows. If mass and energy could somehow stay constant regardless of acceleration, i don't see how it could, but if it did, neither twin would be older than the other. According to Einstein, one of the twins would still age more. No proof.
My theory (post #88) would still hold up just as Einstein's has. It would be consistant with the experiments thus far because of the reasons I gave in posts #89 and #90. Time exist as dimension that does not exist in the physical realm. The only things we can affect are those that deal with (matter, mass, energy). Time has none of those qualities and therefore cannot be affected. It is not a tangle entity. We can't touch it. It does not change with space, or mass. Time has no mass. It is just a constant that has to exist because we can conceptionally understand it.
Einstein thought Time and Space existed as single entity together. And that is the backbone of all his reasoning for Special Relativity. He arrived at this answer only because he foolishly concluded that space (emptiness according to him) can somehow bend to bodies that have mass or rather large mass. He gave the example of a ball falling on a matress to show this. True the matress bends to the ball but thats only because the matress has mass. Something has no mass cannot bend or curve. Thats like saying that width and depth can exist without taking up space, basically saying without mass. It can't it doesn't make sense. Einstein concluded wrongly.
That doesn't prove that the particles are somehow not governed by the same dimension of time. The fact that that happens (particles moving at a faster and rate and living longer than those that are not moving as fast) could be due to the fact that as particles move over a DISTANCE not TIME, they gain more mass and therefore gain more energy. That experiment doesn't necessaritly prove what Einstein was proving - that it is due to different TIMES.
I would strongly suggest that you actually study Relativity before discussing it or trying to rebut it.
DaveC426913
Apr14-05, 06:32 PM
as particles move over a DISTANCE not TIME, they gain more mass and therefore gain more energy. That experiment doesn't necessaritly prove what Einstein was proving - that it is due to different TIMES.What does gaining more energy have to do them living longer?
The lifespan of particles before decay are known to many decimal places. If, under whatever circumstances you want, they decay in a longer time, they are, in any way you care to define it, experiencing time dilation.
It's not like particles have a rich life happening in there that we can see how fast their tea goes cold. They are created, they last a time, they die. If we record that they take 25% longer time to die, it's because the time they experienced was dilated.
There *isn't* a better definition of time than that. In fact, particle decay is how we measure time. There simply is no more fundamental measurement than subatomic particles by which we could rate this event.
yogi
"djavel - I do not know what is meant by syncing a clock that flies by - we can read its instantaneous value, but we do not have a way of determining that the flyby clock is actually running at the same speed as the two clocks that were in the same rest frame."
Well, I'm beginning to see why you get so confused about relativity!
"Synching" clock B and A1 as B passes A1 simply means that t' = t for the event "B passes A1". Of course at any later (or earlier) time they won't be equal. Just at that one instant. You can do the same thing with the spatial coordinate, setting x = x' for exactly one event. In fact the Lorentz transforms in their familiar, simple form require that t' = t = 0 and x' = x = 0 for the same single event. In other words, the clocks are synched to zero when the spatial origins coincide.
Does that clear up some of your confusion? :wink:
Dragongod
Apr14-05, 09:23 PM
Janus if i am wrong about anything plz correct, i came to this site to learn. Also DaveC426913 that means that u are using a different definition than time than i am. U justify time being relative by saying "They are created, they last a time, they die. If we record that they take 25% longer time to die, it's because the time they experienced was dilated." "There *isn't* a better definition of time than that. In fact, particle decay is how we measure time."
---- this shows that, BY DEFINITION, if I say the electrons LIVE LONGER BECAUSE they gain more mass and energy, you are argreeing with me because you also say that particles can live longer due to particular factors. We have the same definition for two different labels. I define something living longer due to particular factors as having a LONGER LIFESPAN while you define something that lives longer due to particular factors as having MORE TIME. The way I define time has nothing to do with the concept you are speaking concerning time. We are talking about two different concepts using the same label (time). Hopefully i am getting my point accross, i know it sounds a little "iffy." The problem here is semantics as it is in most cases.
djavel - that is exactly what I said - if that is what you are defining as synchronize - ok - you can do that - a single reading - but synchronized can also mean running together - like synchronous motors, See a dictionary: 1) To render synchronous in operation 2) Having the same period 3) Having the same period and phase
DragonGod - it is true Einstein did not prove SR - he did suggest some types of experiments that he felt would reveal the truth of time differences between moving clocks ...and later as experimental technology developed his predictions were found to be in agreement therewith. But Einstein did not proffer a physical reason as to why age differences occur - your questions in this regard are not outrageous - it is natural to want to relate time dilation to some physical source - your approach is very much like that of Curt Renshaw http://renshaw.teleinc.com/index.asp - who has developed a number of equations that are somewhat consistent with the notion that time dilation in SR is energy related. There are also some energy theories that correlate gravitational time dilation with SR. Tom Martin http://www.gravityresearch.org/ and others have developed this notion in connection with inflow theory
Jesse - your post 85 - in theory all observers agree upon the proper length and proper time - distance and time measured in the frame where the clocks are at rest and the rulers are at rest. And yes - if the two clocks are separted in space (e.g. the A frame) - an observer in another frame B moving wrt to A will not see the two clocks as running in sync, nor will this B frame observer be able to correctly measure the distance between the two clocks.
Example: I can agree with the fact that the muon has a 2usec lifetime (a proper time of 2 usec) in every frame, but when it is moving at 0.99c wrt to the earth, I measure it to be greater.
Jesse - your post 85 - in theory all observers agree upon the proper length and proper time - distance and time measured in the frame where the clocks are at rest and the rulers are at rest. And yes - if the two clocks are separted in space (e.g. the A frame) - an observer in another frame B moving wrt to A will not see the two clocks as running in sync, nor will this B frame observer be able to correctly measure the distance between the two clocks.
Example: I can agree with the fact that the muon has a 2usec lifetime (a proper time of 2 usec) in every frame, but when it is moving at 0.99c wrt to the earth, I measure it to be greater.
So you think the Lorentz equations are correct, even with this -vx/c² term?
Jesse - your post 85 - in theory all observers agree upon the proper length and proper time - distance and time measured in the frame where the clocks are at rest and the rulers are at rest. And yes - if the two clocks are separted in space (e.g. the A frame) - an observer in another frame B moving wrt to A will not see the two clocks as running in sync, nor will this B frame observer be able to correctly measure the distance between the two clocks. So are you saying that the "real" answer to the time difference between two clocks separated in space can be found by using length and time measurements in their own rest frame? Does that mean that if two clocks are in motion relative to each other, you think there's no real answer to the time difference between them at any given moment, that only when they come to rest relative to each other can a real time difference be found?
Ich: The LT are correct if the postulates are correct since they follow straightaway - but other postulates like those posed by Selleri recogonize two way isotropy (round trip velocity of c constant) but they do not share Einstein's postulate of one way isotrophy - consequently Selleri transforms predict the same time dilation in terms of the correctness of the interval, but they do not include the vx/c^2 factor. Ironically, because of the methodology used in the physical experiments that seek to measure time dilation, this term doesn't get measured (no clock displacement in the relatively moving frame). So, my above statement is based upon the correctness of the LT, which is why I prefaced it with "In Theory"
Ich: The LT are correct if the postulates are correct since they follow straightaway - but other postulates like those posed by Selleri recogonize two way isotropy (round trip velocity of c constant) but they do not share Einstein's postulate of one way isotrophy - consequently Selleri transforms predict the same time dilation in terms of the correctness of the interval, but they do not include the vx/c^2 factor. Ironically, because of the methodology used in the physical experiments that seek to measure time dilation, this term doesn't get measured (no clock displacement in the relatively moving frame). So, my above statement is based upon the correctness of the LT, which is why I prefaced it with "In Theory"
So those transforms that you prefer are just the LT without vx/c^2?
yogi,
"....consequently Selleri transforms predict the same time dilation in terms of the correctness of the interval, but they do not include the vx/c^2 factor...."
This is great!
I Googled 'Selleri transforms' and the first hit I tried sent me to one of your posts on physicsforums!!!
So I tried the next one. That took me to another one of your posts!!!
At that point I was laughing too hard to try any of the others. :biggrin:
Well - what can I say. I assure you selleri transforms exist independent of yogi and google. Below I will give you a link
http://redshift.vif.com/JournalFiles/V11NO1PDF/V11N1SEL.pdf.
JesseM and yogi,
Sorry about not finishing this; I got side tracked. Let's try it once more.
This time, to avoid confusion about English terminology, let's START by just labeling events (E1, E2....) and giving their coordinates. Then make sure that all events that either of you think are relevant have been included Then make sure we agree that (according to the Lorentz transforms) the coordinates are correct. Only then do we discuss in English what the events correspond to and what their coordinates are telling us.
Terminology:
Two frames S and S' with S' moving at v relative to S
Y = gamma
c2 = c*c
The coordinates of an event En are xn, tn, xn', tn'
So (no earth, no stars, no clocks, no english at all; just some events and their coordinates!)
En..........xn........tn.........xn'........tn'
E1..........0..........0..........0..........0
E2..........L.......Lv/c2.......L/Y........0
E3..........L........L/v..........0.......(L/v)/Y
Are these all the events we need?
Are the coordinates all correct?
JesseM and yogi,
Sorry about not finishing this; I got side tracked. Let's try it once more.
This time, to avoid confusion about English terminology, let's START by just labeling events (E1, E2....) and giving their coordinates. Then make sure that all events that either of you think are relevant have been included Then make sure we agree that (according to the Lorentz transforms) the coordinates are correct. Only then do we discuss in English what the events correspond to and what their coordinates are telling us.
Terminology:
Two frames S and S' with S' moving at v relative to S
Y = gamma
c2 = c*c
The coordinates of an event En are xn, tn, xn', tn'
So (no earth, no stars, no clocks, no english at all; just some events and their coordinates!)
En..........xn........tn.........xn'........tn'
E1..........0..........0..........0..........0
E2..........L.......Lv/c2.......L/Y........0
E3..........L........L/v..........0.......(L/v)/Y
Are these all the events we need? If you don't discuss what these events mean in english, and what you want to find in english, then we can't really say whether these are all the events you "need." (what do you need them for?) But if we say as before that there are two clocks A1 and A2 at rest in S and synchronized in S, and another clock B moving at v relative to them (at rest in S'), and B first passes A1 (with both reading 0 at that moment) and then passes A2, then these three events will tell you how much time elapses on both B and A2 between the moment when B passes A1 and the moment when B passes A2, in the S' frame. If you want to know how much time elapses on both clocks between the times of these events in the S frame, then instead of E2 you need this event:
E2a..........L.......0.......YL........-YLv/C^2
Since yogi now says he doesn't even agree with the Lorentz transform, I don't see him agreeing with the events you chose.
....Since yogi now says he doesn't even agree with the Lorentz transform, I don't see him agreeing with the events you chose.
JesseM,
Well, he's free to pick any additional events he wants. But more to the point, I don't think yogi is interested in discussing relativity with event coordinates, intervals, etc. He wants to discuss things like "time slippage", "length contraction", "time dilation" and "objective differences". They give him the wiggle room he needs to escape whenever you try to pin him down on one of his errors.
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