Understanding Time Dilation: How Fast Do You Have to Go?

[Sherlock]

I would like to point out to Sherlock that it's possible to conduct the experiment without anyone undergoing any acceleration.

Bob is on Earth and Alice flies past at a constant velocity. As Alice passes Bob she synchronizes her clock with Bob's.

After a while Alice meets Carol traveling at a constant velocity towards Earth. As Carol passes Alice she synchronizes her clock with Alice's.

When Carol passes Bob on Earth (without decelerating) she compares her clock against Bob's and finds hers is ahead of his.

Alice, Bob and Carol are all moving at constant velocities relative to each other throughout the whole experiment. So whose motion is the most "natural"? Of course they're all equally natural.

In order to do your experiment we have to accelerate Alice and Carol initially away from Bob and each other, and then into states of motion where they are heading toward each other at a uniform rate, and each is heading toward Bob at a different uniform rate.

So, it's not possible to do the experiment without anyone undergoing any acceleration.

Both Alice's and Carol's clocks are running slower than Bob's (ie., slower than their clocks would run on Earth next to Bob's -- we're assuming that the clocks are identical and run at the same rate while next to each other in the same state of motion), so even though Alice set's her time to Bob's as she passes Bob, and then Carol sets her time to Alice's (as Carol is heading toward, and Alice away from, Bob) as she passes Alice, then as Carol passes Bob her accumulated time should be less than Bob's.

In this scenario, Alice and Carol would be the anomalous movers.

Of course there could be something to this that I just don't understand yet. So, I'll be interested to read why you have Carol's clock accumulating more time than Bob's.

 

[JesseM]

Or the traveller could keep heading away from Earth or whatever. But the question I was responding to involved the round trip scenario.

If they pass each other at a single point in space, that is a round trip scenario, because there can only be a single objective answer to what their clocks read at the moment they pass, it can't be different for different reference frames.

Ok, but wrt my reply to the original questioner, we were talking about the scenario where clock A is eventually returned to it's starting position next to clock B.

What you're not understanding is that "passing at a single point in space" is the same as "returning to its original starting location" (at least, it would be the 'original starting location' in the Earth's frame). Whether the traveler changes velocity and comes to rest relative to the earthbound observer at that location, or simply passes by the earthbound observer instantaneously, makes no difference to the problem, because we're only interested in what their clocks read at the moment they first reunite.

The questioner was apparently of the opinion that the landing approach would cancel the effects of the takeoff, not realizing that even on slowing down wrt the earth, though the timekeeping rates of the traveller's clocks (biological or artificial) are becoming ever closer to what they were while on the earth, the periods of the oscillators are still larger than what they were on earth.

Or the period of the oscillators are smaller than what they were on earth, it depends what frame you use.

If the period was smaller than on earth, then clock A would record *more* time during it's trip than clock B.

All you can say is that the average period of clock A's ticks throughout the trip is larger than period of clock B's ticks. But since clock A changes velocities, the period of its ticks will not be constant in most frames. You referred specifically to what would happen during the return portion of the trip as clock A changed velocity to come to rest relative to clock B; there are certainly frames where clock A's ticks during the inbound leg of the trip are shorter than those of clock B, but this is balanced by the fact that in these frames, clock A's ticks during the outbound leg are longer than those of clock B. So at the end of the inbound leg when clock A changes velocity to come to rest relative to clock B (assuming this actually happens, as I said earlier you could also just have them pass each other instantaneously and compare clocks then), in these frames clock A's ticks are shorter than those of clock B but are getting longer as it gets closer to B's velocity.

Since accelerating the Earth isn't too plausible, let's change the earth-twin to a twin on a space station, but otherwise the situation is the same--the ship and the station are initially at rest relative to each other at a single location, then they begin to move apart, then the ship fires its rockets and turns around, and finally they reunite and compare clocks. Now, note that when I described this scenario, I just said they "begin to move apart", I didn't say whether this was because the ship fired its rockets to move away from the space station or because the space station fired its rockets to move away from the ship. Do you agree that I don't need that piece of information (assuming the initial acceleration was near-instantaneous), that either way the ship's clock will be the one that's behind when they reunite? If so, how can the answer to the question possibly have any relevance as a "physical explanation" to why the ship's clock is behind when they reunite?

The way you describe this we don't have the information about whether the ship or the station, or both, accelerated to produce the initial separation. So, if we want to make some inferences about the affects of acceleration, then we would have to describe the reference state as being the state of the system at the instant the ship fired its rockets to return to the station.

I still don't know what you mean by "reference state", this is a term that you have made up without defining, and which makes no sense to me. Why would we have to describe the reference state as the one where ship fired its rockets to return to the station? Do you agree or disagree that if we know their initial relative velocity as they begin to move apart on the outbound leg, and we know when the ship fired its rockets to turn around, and we know the relative velocity of the ship and the station on the inbound leg after the rocket has been fired (assume the acceleration due to the rocket firing was arbitrarily brief), then this is enough information to tell us which clock will be behind when they reunite, and by how much? Do you agree, in other words, that the answer to the question of whether it was the ship or the station that initially accelerated to get them moving apart has no effect whatsoever on the final time difference when they reunite? Please answer this question yes or no, because it's not clear to me whether you understand this.

But I don't know how that would work, because we want to first compare the clocks while they are in the same state of motion

Why is it important compare them when they are in the same state of motion? All that's important is that you compare them when they are at the same spatial location, that way there will be no disagreement between different reference frames as to the time on each clock at that moment (in contrast to when the clocks are at different locations, and the question of what the two clocks read 'at a single moment' depends on your definition of simultaneity).

and the easiest way to do this is to have them sitting next to each other. Then we accelerate *one* clock and note any differences in timekeeping between the two.

I agree that if you start them out sitting next to each other and then accelerate one, then you can have an objective answer to which one has aged less when they reunite. But again, which one you accelerate initially will have no effect on the final outcome.

As for physical explanations, a pretty simple one is that the laws of physics are Lorentz-symmetric. Lorentz-symmetry is just a mathematical property of a given equation--it can be defined with no reference to physics whatsoever--and it so happens that the equations describing the laws of electromagnetism have this property, which guarantees that any clocks based on electromagnetic phenomena must appear to slow down when they are moving in your frame, regardless of which frame you're in.

Yes, but as is demonstrated by the phenomenon of differential aging or timekeeping, the affects of acceleration aren't just matters of appearance or perspective due to a finite c.

I didn't say anything about it being a matter of appearance or perspective. Again, if you figure out what the correct equations of physics are in a single frame, and then you check the equation you found and see that it has the mathematical property of Lorentz-symmetry, this means you know that observers in different coordinate systems related to yours by the Lorentz transform would see the laws of physics obeying the exact same equations, and this guarantees that clocks based on these laws must be measured by each coordinate system to slow down as their velocity in that frame increases. It is not logically possible to have a universe where the equations representing the laws of physics as seen by an inertial observer have the mathematical property of Lorentz-symmetry but where the physical phenomenon of differential aging seen in the twin paradox would not be observed. Do you agree with this? If you like I can give a little more explanation about what I mean when I talk about "Lorentz-symmetry" in a purely mathematical sense, as a property of certain equations, as opposed to a more physical sense.

I'm not sure what you mean by "valid".

I mean saying that one frame's perspective on which clock slows down should be preferred in some way over any other's, as you seem to be arguing with your "anomalous motion" vs. "natural motion" distinction.

We're asking whether acceleration to a different state of motion (eg., a different velocity wrt some reference state) produces physical changes in the accelerated body.

Maybe that's what you're asking, but I still am not clear on what you mean by "reference state", you have never really defined this term. All I'm saying is that if two clocks start at the same location so that there is a single objective answer to what each one reads "at the same moment", then move apart, then reunite so their readings are compared again and we can see which clock elapsed more time, it will always be whichever clock had to accelerate to turn around that elapsed less time.

Apparently it does. But in order to see this we have to know which body is accelerating wrt some reference state, don't we? Otherwise, if we just have two clocks with different readings and no other information, then either, or both, could have been accelerated wrt the reference state.

As long as we know the accelerations and velocities throughout the period when they are apart, that's all you need to figure out what the readings will be when they reunite. Knowing which accelerated initially to begin to move them apart in the first place is irrelevant.

So, we make some assumptions about the experimental situation. In the case of clock A moving around the solar system while clock B remains on earth, we're assuming that clock A's motion is the anomalous motion wrt the regular motion of the solar system and not the earth's.

"Anomalous motion" is another term you made up and didn't define. And again, in the ship/space station example, if the question of whose motion is "anomalous" depends on whether the ship or the station accelerated initially, then obviously this question is irrelevant to the problem, since either way you'll get the same answer to which clock is behind when they reunite, and by how much (it'll be the ship's clock that's behind, because the ship was the one that accelerated to turn around at the midpoint of the journey).

And we *can* say with a pretty high degree of certainty that it's clock A that has accelerated and not clock B.
But it does make a difference. If it was clock B rather than clock A that traveled around the solar system, then clock B would show less time for the trip interval than clock A. Which clock we choose to accelerate is an arbitrary decision of course, since they're identical clocks, keeping time at exactly the same rate while next to each other on earth. But once we've made the choice, and accelerated one or the other, then it makes a difference, wrt our experimental hypothesis, which one was accelerated and which one wasn't.

It makes a difference which clock accelerated to turn around at the midpoint of the trip, it doesn't make a difference which clock accelerated initially to get the two clocks moving apart after they had been at rest relative to each other in a single location. Agreed?

The point of a 'reference state' is simply to be able to make some objective statements about differential timekeeping.

And any such "objective statement" would totally violate the spirit of relativity.

I don't think so. Especially since we're using Relativity to predict the results. And the results do seem to agree very well with the predictions of Relativity.

Yes, and the predictions of relativity about objective physical questions like what two clocks read at the moment they reunite will be the same regardless of what reference frame you use to analyze the problem. But these frames will disagree on other less "objective" issues, like which clock was ticking slower during the inbound leg of the trip. From what I understand, you're using the concept of a "reference state" to pick a particular reference frame's answers to these sorts of questions (the Earth's frame, say) and label them more "objectively true" then the answers in other frames--that's what I meant when I said it would totally violate the spirit of relativity. Am I misunderstanding what you are saying about reference states here?

So, I'm just exploring what sort of inferences about the physical causes of differential timekeeping can be made. The conjecture is that acceleration produces physical changes in accelerated bodies -- eg., the periods of oscillators are altered, due to physical changes in the oscillators, as they move from one state of motion to another. If you don't think that the results of relativistic experiments lead to this idea, then you still haven't explained satisfactorily why? Aren't length contraction and mass increase as prescribed by Relativity real physical changes?

Something is only a "real physical change" if it is the same in every frame. Length contraction and time dilation depend on the frame you choose--they're more analogous to the slope of a curve drawn on a piece of paper at a particular point, which depends on what angle you place your x and y axes. Nevertheless, you can calculate the length of the curve in different coordinate systems by integrating a function of the slope in that coordinate system, and you'll get the same answer regardless of how you orient your axes. Similarly, you can integrate a function of the time dilation in different frames in relativity to get the total time elapsed on a clock which follows a given path between two points in spacetime, and you'll get the same answer in every frame.

We're using a *reference state* (earth-clock A-clock B) which we disassemble and then reassemble by accelerating *one* component (clock A) of that state.

Who's "we"? Physicists don't use "reference states" when analyzing these problems. Perhaps we could look at a numerical example of a twin-paradox-like situation, and I could show you how physicists would analyze it, then you could try to explain how it could be analyzed in terms of your own concepts?

 

[DrGreg]

In order to do your experiment we have to accelerate Alice and Carol initially away from Bob and each other, and then into states of motion where they are heading toward each other at a uniform rate, and each is heading toward Bob at a different uniform rate.

But that acceleration takes place before the experiment begins and you seem to be assuming that Bob's presence on Earth makes him different than Alice and Carol.

Consider a variation of the experiment where Alice is on Earth, Bob flies past the Earth at a constant velocity, and later Carol flies past the Earth chasing after Bob at an even higher constant velocity. Carol's speed is such that her speed relative to Bob (towards) is the same as Alice's speed relative to Bob (away). The result of this experiment would be identical to the experiment I previously described, because the relative motions of Alice, Bob and Carol during each experiment are identical.

The presence of the Earth is a red herring - it doesn't matter that Alice, Bob and Carol, as human beings, must once have been born on Earth. In my previous post, Alice and Carol could be aliens from other planets who have spent their entire lives traveling at constant speed, the argument would still hold.

So, I'll be interested to read why you have Carol's clock accumulating more time than Bob's.

Whoops, that was a bad choice of words I used which I have now corrected. You are right that Carol's clock shows a time earlier than Bob's clock.

 

[Janus]

Sherlock,
I would like you to consider another scenerio:
Assume we have two spaceships headed towards Earth at the same constant velocity but separated from each other by a lightyear along the line of travel. At some point, the first ship "A" passes a marker buoy and radios to the second ship as it does so that it is setting its clock to zero.
Some time latter, the second ship "B" passes the same marker. Knowing both the their speed and the distance to ship to ship A, ship B sets its clock such that it reads the same as Ships A's. Thus the two clocks read exactly the same and tick at the same rate at this point since they are motionless with respect to each other.
When ship A reaches Earth it decelerates and stops. Ship B continues until it reaches the Earth and then stops.
By your understanding which clock, A or B, will read less once they both reach Earth and why?
Please do not respond that this is not the scenerio we are discussing, as any explanation/understanding of Relativity must be consistant with all given scenerios.
Also, it does not matter that Ships A and B might have had to accelerate at some time in the past in order to attain the velocity they have, as we are only concerned with the period of time after A and B synchonize their clocks.

 

[Sherlock]

Whether the traveler changes velocity and comes to rest relative to the earthbound observer at that location, or simply passes by the earthbound observer instantaneously, makes no difference to the problem, because we're only interested in what their clocks read at the moment they first reunite.

I'm interested in returning clock A to its original state of motion, next to clock B on earth.

All you can say is that the average period of clock A's ticks throughout the trip is larger than period of clock B's ticks.

That's right, and in the experiment I'm talking about that's all I'm interested in.

I still don't know what you mean by "reference state", this is a term that you have made up without defining, and which makes no sense to me.

I'm using reference state to mean some state of motion -- defined by some group of objects' positions and velocities wrt each other. The earth-clock A-clock B reference state is defined as the state where these objects are contiguous and not moving wrt each other.
Clock A and clock B sitting next to each other and stationary wrt the Earth is the easiest to set up and the easiest to reproduce.

Why is it important compare them when they are in the same state of motion?

Because the hypothesis is that changing an oscillator's state of motion changes its period. By keeping clock B (which is identical to and running the same as clock A) in the original (reference) state we're effectively comparing clock A to itself (ie., to what it would have read had we not accelerated it).

I agree that if you start them out sitting next to each other and then accelerate one, then you can have an objective answer to which one has aged less when they reunite. But again, which one you accelerate initially will have no effect on the final outcome.

If we accelerate both clock A and clock B then I think that would unnecessarily complicate the experiment.

It is not logically possible to have a universe where the equations representing the laws of physics as seen by an inertial observer have the mathematical property of Lorentz-symmetry but where the physical phenomenon of differential aging seen in the twin paradox would not be observed. Do you agree with this?

Yes. At least wrt a universe that has the electrodynamic properties that our universe apparently has.

All I'm saying is that if two clocks start at the same location so that there is a single objective answer to what each one reads "at the same moment", then move apart, then reunite so their readings are compared again and we can see which clock elapsed more time, it will always be whichever clock had to accelerate to turn around that elapsed less time.

Ok.

I don't understand why you're bringing this up ... so much. :-) Why would you want to accelerate both clocks if the question you're asking has to do with comparing an accelerated clock to an unaccelerated clock? The way you're talking about doing it is difficult to set up in the first place and even more difficult to reproduce.

"Anomalous motion" is another term you made up and didn't define.

An anomaly is a deviation from the normal, or a certain, order. I forget why I used this term in the first place. I wouldn't worry about it.

From what I understand, you're using the concept of a "reference state" to pick a particular reference frame's answers to these sorts of questions (the Earth's frame, say) and label them more "objectively true" then the answers in other frames--that's what I meant when I said it would totally violate the spirit of relativity. Am I misunderstanding what you are saying about reference states here?

As you've noted, the experiment might be done in a number of different ways, but it's more convenient to use an earth-based reference state. We're talking about real experiments, right? Iirc, the specific experiment that I'm talking about has been done a few times already.

Physicists don't use "reference states" when analyzing these problems.

They might not talk about it in exactly those terms. But in experiments where two clocks are compared while next to each other, then one is accelerated away from and back to the other clock (which hasn't been accelerated) and the two clocks are compared again -- they're using a 'reference state'.

 

[Sherlock]

Sherlock,
I would like you to consider another scenerio:
Assume we have two spaceships headed towards Earth at the same constant velocity but separated from each other by a lightyear along the line of travel. At some point, the first ship "A" passes a marker buoy and radios to the second ship as it does so that it is setting its clock to zero.
Some time latter, the second ship "B" passes the same marker. Knowing both the their speed and the distance to ship to ship A, ship B sets its clock such that it reads the same as Ships A's. Thus the two clocks read exactly the same and tick at the same rate at this point since they are motionless with respect to each other.
When ship A reaches Earth it decelerates and stops. Ship B continues until it reaches the Earth and then stops.
By your understanding which clock, A or B, will read less once they both reach Earth and why?
Please do not respond that this is not the scenerio we are discussing, as any explanation/understanding of Relativity must be consistant with all given scenerios.
Also, it does not matter that Ships A and B might have had to accelerate at some time in the past in order to attain the velocity they have, as we are only concerned with the period of time after A and B synchonize their clocks.

A would read less. A was set to 0 at the buoy. B was set to 0 + 1 light year's worth of ticks of it's clock.

 

[JesseM]

I'm interested in returning clock A to its original state of motion, next to clock B on earth.

But why? Do you agree that, regardless of whether A comes to rest again next to B or just passes next to B instantaneously without slowing down, we'll get the same answer to the question of what each clock reads at the moment they meet?

All you can say is that the average period of clock A's ticks throughout the trip is larger than period of clock B's ticks.

That's right, and in the experiment I'm talking about that's all I'm interested in.

So do you agree that although we can say the average period of A's ticks is slower over the course of the whole trip, there is no "objective" truth about whether A's ticks are slower or faster than B's during a particular portion of the trip, like the inbound leg?

I'm using reference state to mean some state of motion -- defined by some group of objects' positions and velocities wrt each other. The earth-clock A-clock B reference state is defined as the state where these objects are contiguous and not moving wrt each other.

Yes, but what are you doing with the reference state? What part does it play in your analysis of which clock is "objectively" running slower or which frame's perspective is to be preferred?

Clock A and clock B sitting next to each other and stationary wrt the Earth is the easiest to set up and the easiest to reproduce.

That's why I asked you to consider clock A being on a space station rather than the earth, so it would be easier to imagine either the station or the rocket accelerating initially.

Because the hypothesis is that changing an oscillator's state of motion changes its period.

But is your hypothesis that changing the state of motion always slows it down, or do you accept that, depending on your reference frame, changing velocity can either slow an oscillator down or speed it up, and that there is no reason to prefer one frame over another, thus there is no "objective truth" about whether any given oscillator is slowed down or sped up during a given acceleration?

If we accelerate both clock A and clock B then I think that would unnecessarily complicate the experiment.

Even if you think it would complicate it on a conceptual level, you didn't answer my question about whether the outcome of the experiment would be changed. Please answer this question:

Do you agree or disagree that if we know their initial relative velocity as they begin to move apart on the outbound leg, and we know when the ship fired its rockets to turn around, and we know the relative velocity of the ship and the station on the inbound leg after the rocket has been fired (assume the acceleration due to the rocket firing was arbitrarily brief), then this is enough information to tell us which clock will be behind when they reunite, and by how much? Do you agree, in other words, that the answer to the question of whether it was the ship or the station that initially accelerated to get them moving apart has no effect whatsoever on the final time difference when they reunite? Please answer this question yes or no, because it's not clear to me whether you understand this.

It is not logically possible to have a universe where the equations representing the laws of physics as seen by an inertial observer have the mathematical property of Lorentz-symmetry but where the physical phenomenon of differential aging seen in the twin paradox would not be observed. Do you agree with this?

Yes. At least wrt a universe that has the electrodynamic properties that our universe apparently has.

You're hedging. When I said "not logically possible", I meant there is no way to avoid the conclusion of differential aging in any possible universe with Lorentz-symmetric laws, regardless of whether the universe "has the electrodynamic properties that our universe apparently has". In other words, if you were creating a computer simulation of a universe, and you first established some coordinate system for identifying locations in space and moments of time within the simulation, then you wrote down some equations defining the universe's "laws of physics" in terms of this coordinate system, then regardless of what arbitrary equations you came up with, regardless of whether they had any relation to the equations of physics in the real world, if all your equations had the mathematical property of Lorentz-invariance then it would be absolutely guaranteed you'd see differential aging in any sort of regular oscillators that could exist in the simulated universe. Do you agree with this?

All I'm saying is that if two clocks start at the same location so that there is a single objective answer to what each one reads "at the same moment", then move apart, then reunite so their readings are compared again and we can see which clock elapsed more time, it will always be whichever clock had to accelerate to turn around that elapsed less time.

Ok.
I don't understand why you're bringing this up ... so much. :-) Why would you want to accelerate both clocks if the question you're asking has to do with comparing an accelerated clock to an unaccelerated clock?

Because I'm only interested in accelerations that occurred during the two clock's journeys between the point they departed each other and the point they reunited, accelerations that occurred before or at these two endpoints are irrelevant to your conclusion. If you had two clocks sitting next to each other and at rest relative to each other, would you think it was important to know that one clock was driven to this location 3 days ago and that the trip involved lots of acceleration? If not, then what I'm saying is that the question of which clock accelerates initially to get them moving apart at constant velocity is equally irrelevant, assuming we can treat the acceleration as instantaneously brief so that as soon as they move apart from their common spatial location, they are both moving at constant velocity.

As you've noted, the experiment might be done in a number of different ways, but it's more convenient to use an earth-based reference state. We're talking about real experiments, right?

Not particularly, I was just talking about analyzing thought-experiments like the twin paradox--obviously this experiment of sending one twin on a journey and having him come back noticeably younger than his stay-at-home twin has never been done.

They might not talk about it in exactly those terms. But in experiments where two clocks are compared while next to each other, then one is accelerated away from and back to the other clock (which hasn't been accelerated) and the two clocks are compared again -- they're using a 'reference state'.

No they aren't, because again, it's irrelevant to the problem whether the two clocks were at rest with respect to each other initially or if they just passed each other by at a single location moving at constant velocity, and if they were at rest with respect to each other, it's irrelevant whether the first clock accelerated instantaneously or the second one did to get them moving apart at constant velocity. These issues simply make no difference to the analysis of the problem, any more than knowing the color of the clocks. Again, if you want I can show you a simple numerical example so you can see why these issues are irrelevant.

 

[Janus]

A would read less. A was set to 0 at the buoy. B was set to 0 + 1 light year's worth of ticks of it's clock.

And?
When B is set to "0 + 1 light year's worth of ticks of it's clock" as it passes the bouy, A had continued to run until it read "0 + 1 light year's worth of ticks of it's clock" IOW at this instant, according to A & B, Both A & B read the same time and are ticking at the same rate. So your answer doe not address the question.

The question was what comparative time shows on A and B after A stops at Earth and B catches up to it.

 

[Sam Woole]

DrGreg, thanks for your effort to help me. I have spent many hours on your demonstration and I should say it made very good sense to me, except one more difficulty.
This is what Alice sees:

Alice’s   Alice’s
own       view of
clock     Bob’s clock
12:00     12:00
12:15     12:10
12:30     12:20
12:45     12:30
13:00     12:40

The above shows, after each 15 minutes of travel according to Alice's clock, a 5 minute delay is created. But now, let us compare to the following.

Ted’s   Ted’s     Ted’s
own     view of   view of
clock   Alice’s   Bob’s
clock     clock
17:20    12:00    12:00
17:30    12:15    12:10
17:40    12:30    12:20
17:50    12:45    12:30
18:00    13:00    12:40

This one shows, after each 10 minutes of travel according to Ted's clock, the 5 minute delay remains there. It is also there according to Alice's clock showing 10 minutes of travel, in this:

Alice’s   Alice’s
own       view of
clock     Bob’s clock
13:00     12:40
13:10     12:55
13:20     13:10
13:30     13:25
13:40     13:40
13:50     13:55
14:00     14:10

So when Alice and Bob reunite, Alice’s clock reads 14:00, yet Bob’s clock reads 14:10. This is the twins “paradox”.

My difficulty can be better seen by the following figure:
Bob, A-----B-----C-----D-----E, Ted, while Alice travels from point to point and return.

Each segment above represents 15 minutes of travel.  When Alice returns, she can only cut the 5-minute delay by finishing each segment, which must take 15 minutes.
But your demonstration gave only 10 minutes in the return trip.

I gues, maybe it has something to do with the principle of relativity that allowed you switching between 3/2 and 2/3. But I could not figure it how.

 

[Sam Woole]

That guy is a typical anti-relativity crackpot. You need to be very careful about learning relativity from Web sites, because there are a lot of crackpots out there. I wouldn't waste any time on sites like that until after you already understand relativity fairly well, in which case it can be a useful exercise to spot the flaws in their arguments.

Thanks. I will be careful.

 

[Sherlock]

And?
When B is set to "0 + 1 light year's worth of ticks of it's clock" as it passes the bouy, A had continued to run until it read "0 + 1 light year's worth of ticks of it's clock" IOW at this instant, according to A & B, Both A & B read the same time and are ticking at the same rate. So your answer doe not address the question.
The question was what comparative time shows on A and B after A stops at Earth and B catches up to it.

Yes, I see. I was assuming that A's clock was stopped on landing.

 

[Sherlock]

Do you agree that, regardless of whether A comes to rest again next to B or just passes next to B instantaneously without slowing down, we'll get the same answer to the question of what each clock reads at the moment they meet?

The different scenarios would produce different average relative velocities for the trip interval, wouldn't they? Not that that really matters. I agree that a comparison can be done whether you put them back next to each other or not.

So do you agree that although we can say the average period of A's ticks is slower over the course of the whole trip, there is no "objective" truth about whether A's ticks are slower or faster than B's during a particular portion of the trip, like the inbound leg?

Yes, I agree that anywhere during the trip, there is some frame that would see A's clock ticking faster than B's. There's apparently no way to distinguish one frame from another, and all frames see things differently, so no reason to prefer one rather than another. So, I guess the idea of different states of motion is out.

I've been reading that differential timekeeping (the twin clock effect) has nothing to do with acceleration, and this seems to be supported by what I read about the Moessbauer and muon experiments. So I guess the idea that differential timekeeping is due to some physical change due to acceleration is out. On the other hand, there are the effects of gravitational fields on clocks, and gravitation is acceleration. And, then there is Keating saying that differential timekeeping is *only* due to acceleration. So, I don't know how to think about this.

Thanks for your (and others) help. I have lots of reading to do now, and will probably post some questions.

 

[Janus]

Yes, I see. I was assuming that A's clock was stopped on landing.

Okay, Now that you realize that clocks A & B continue to run while clock B catches up to A; By your understanding, how do the clocks compare after B has caught up to A?

 

[Sam Woole]

"Difference" means what you get when you subtract one time from another. This is not constant, as you can see:
4 - 2.4 = 1.6
5 - 3 = 2
6 - 3.6 = 2.4
When we say a clock is "behind" another by a constant amount, we mean that the difference between the times is constant. For example, if one clock was always 0.6 seconds behind another, that would mean that (time on first clock) - (time on second clock) was always 0.6 seconds--when the first clock read 5 seconds the second would read 4.4, when the first clock read 6 seconds the second would read 5.4, and so on. Clearly that is not the case here, one clock is ticking slower than the other clock, which is different from being behind by a constant amount. And clearly they are not "ticking at the same rate" if the gap in times between the two clocks is continually increasing! Each is ticking at a constant rate, but one clock's rate is faster than the other clock's rate in any given frame. Well, even if you didn't understand every aspect of what I was saying in the text, did you understand that the pictures just show the readings on the clocks set along the two rulers at different times in each ruler's reference frame? Could you follow the progress of a particular clock--say, the one with the red hand--from one time to another? OK, I'm glad you said that and are not just assuming this page shows a contradiction because it fits your personal views. Unfortunately the images on that page don't load, so I can't tell what equations he's citing. But look, if this were some other debate about math that had nothing to do with relativity, who would you be inclined to trust more--the entire community of highly-trained practicing mathematicians, or some guy with a webpage saying everyone else was wrong? Please consider my comment from the other post:

JesseM, I am sorry I did make a mistake. But do you think there still is a constant difference if I change my language. That is, when one clock accumulates 1 second, the other clock constantly accumulates 0.6 seconds. The difference between the two accumulations is constantly 40%.

According to you, this difference is symmetrical. Namely from the other observer's view, my clock was accumulating 0.6 seconds while his was doing 1 second. Since this difference is only an illusion created by light, both would have accumulated the same thing when they come together.

I do know that you were trying to show otherwise, not accumulating the same thing. I am trying to learn this.

As to the web page I introduced to you, it allows people to communicate back. I have just done so. We shall see what will happen next. Maybe he has removed his math challenge after he realized his own fault.

As to "who would you be inclined to trust more--the entire community of highly-trained practicing mathematicians, or some guy with a webpage saying everyone else was wrong?", of course I am more inclined to trust the highly-trained practicing mathematicians than others. My story is this. I became a dissident against Relativity because I read too much relativity, about 5 books. One Australian guy became a dissident after he has tried 19. Gradually I discovered that relativity dissidents were many, all over the world, thanks to the internet. As a result I think there might be somthing wrong in relativity. As far as I am concerned, I believe time dilation is the one.

 

[Sherlock]

Okay, Now that you realize that clocks A & B continue to run while clock B catches up to A; By your understanding, how do the clocks compare after B has caught up to A?

After B passes the buoy, A and B read the same and are running at the same rate. A and B are in the same inertial (nonaccelerating) frame of reference and are synchronized. Then A accelerates from this frame to another inertial frame (earth) moving toward B. During A's acceleration to the Earth frame, the clocks can't be compared. [1]

After landing, A has, in effect, turned around and is now heading back toward B. After A lands, A and B are no longer synchronized and their tick rate is different (A wrt B, and B wrt A) as they move toward each other. From A's frame, B's tick rate is dilated. From B's frame A's tick rate is dilated. The magnitude of the dilation is proportional to their relative velocity.

The differential timekeeping starts when A lands on earth. It doesn't matter whether B eventually accelerates to land on Earth or not since no comparison can be made during the acceleration interval. But since B does land on earth, then what we're interested in is the interval between the instant that A lands on Earth and the instant that B lands on earth, and, in the absence of any other information, A and B will experience an average symmetric time dilation proportional to their average relative velocity during this interval.

During this interval both A and B have undergone the same (I'm assuming) acceleration. So, I don't know how to analyse this. It seems sort of like the twin paradox, which in its usual form is easy because only one twin accelerates during the interval considered. In the current problem, I'm not really sure what interval to consider.


[1] This isn't the same thing as saying that acceleration has no effect on, or doesn't produce, differential timekeeping -- for if neither A nor B accelerates, then they remain synchronized and ticking at the same rate.

 

[DrGreg]

My difficulty can be better seen by the following figure:
Bob, A-----B-----C-----D-----E, Ted, while Alice travels from point to point and return.

Rather than use your letters ABCDE I’d prefer to use abcdefghijklm. Each of my letters represents 5 minutes of Alice’s travel time.

B           T
o           e
b           d
|           |
abcdefghijklm
A--B--C--D--E

Ted’s     Ted’s     Ted’s
own       view of   view of
clock     Alice’s   Bob’s
          clock     clock
17:20(m)  12:00(a)  12:00(a)
17:30(m)  12:15(d)  12:10(a)
17:40(m)  12:30(g)  12:20(a)
17:50(m)  12:45(j)  12:30(a)
18:00(m)  13:00(m)  12:40(a)

Each row of this table represents a ray of light traveling from Bob to Ted. For example in the second row, a ray of light leaves Bob, at (a), at 12:10 Bob-time, passes Alice, at (d), at 12:15 Alice-time, and arrives at Ted, at (m), at 17:30 Ted-time. If you ignore Alice’s column and just look at the rays leaving Bob and arriving at Ted, each ray arrives at Ted(m) at a Ted-time that is 5h20m later than the Bob-time that it left Bob(a).

The first diagram I have attached to this post may make things clearer. Each spot represents a 5-minute interval. The yellow lines are light rays.

The conclusion to draw from this table is that if you travel away from somebody and observe an apparent 2/3 “red-shift” change of clock rates (Alice observing Bob), then it must follow that if you travel towards somebody at the same speed, you must observe an apparent 3/2 “blue-shift” change of clock rates (Ted observing Alice).

For the return journey my table shows, not EDCBA (mjgda) but mkigeca. That was my choice.

Alice’s   Alice’s
own       view of
clock     Bob’s clock
13:00(m)  12:40(a)
13:10(k)  12:55(a)
13:20(i)  13:10(a)
13:30(g)  13:25(a)
13:40(e)  13:40(a)
13:50(c)  13:55(a)
14:00(a)  14:10(a)

The reason for the 3/2 rate is because the speed of Bob towards Alice, during the return, is exactly the same as was the speed of Alice towards Ted on the outward journey. Einstein’s postulates imply that any effects you can measure should depend only on the relative velocity and on nothing else. So the same rate of 3/2 applies in both cases.

The second diagram I have attached to this post shows Alice’s whole journey.

 

[JesseM]

JesseM, I am sorry I did make a mistake. But do you think there still is a constant difference if I change my language. That is, when one clock accumulates 1 second, the other clock constantly accumulates 0.6 seconds. The difference between the two accumulations is constantly 40%.

Yes, in a given frame the ratio between the length of each clock's ticks will be constant as long as their velocities are constant. However, the ratio will be different in different frames--in another frame the ratio might be 60%, or 200%.

According to you, this difference is symmetrical. Namely from the other observer's view, my clock was accumulating 0.6 seconds while his was doing 1 second. Since this difference is only an illusion created by light, both would have accumulated the same thing when they come together.

Where did you get the idea that it's an "illusion created by light"? In fact, the numbers I gave were supposed to be the times you calculate after correct for the delays in the signal due to the finite speed of light. For example, suppose at a time of 7.2 seconds you look through your telescope and see an image of my clock reading 2.4 seconds, and at that moment I am 3.2 light-seconds away. Then you can calculate that this image must have taken 3.2 seconds to reach you, so the actual event took place 7.2 - 3.2 = 4 seconds in your frame, and since you know that both our clocks read 0 seconds at the moment we departed each other, you conclude that after 4 ticks of your clock my clock had only ticked 2.4 seconds. Likewise, when I look through my own telescope at a time of 7.2 seconds according to my clock, I will see an image of your clock reading 2.4 seconds at a distance of 3.2 light-seconds away from me, so I will do the same sort of calculation to conclude that when my clock read 4 seconds, yours only read 2.4 seconds.

As to "who would you be inclined to trust more--the entire community of highly-trained practicing mathematicians, or some guy with a webpage saying everyone else was wrong?", of course I am more inclined to trust the highly-trained practicing mathematicians than others. My story is this. I became a dissident against Relativity because I read too much relativity, about 5 books. One Australian guy became a dissident after he has tried 19. Gradually I discovered that relativity dissidents were many, all over the world, thanks to the internet. As a result I think there might be somthing wrong in relativity. As far as I am concerned, I believe time dilation is the one.

You may have read a number of books, but you have said yourself that you don't understand basic concepts like that of different "reference frames" (a concept which is important in Newtonian mechanics, not just relativity) so isn't your position just based on the "argument from incredulity", much like the creationist argument against evolution? Also, have you read about the different experiments demonstrating time dilation? For example, were you aware that there have been experiments where extremely precise atomic clocks were placed aboard space shuttles, and when the shuttle returned the clocks were slightly behind clocks on earth, by the amount relativity predicts?

 

[Janus]

They would read the same.

And your reasoning for this answer?

 

[Sherlock]

They would read the same.

And your reasoning for this answer?

I don't know if that's correct. I wrote it last night when I was depressed over the idea that my hypothesis about acceleration and differential timekeeping didn't seem tenable.

I've since edited it. So, you can check my reasoning. Thanks.

 

[Janus]

I don't know if that's correct. I wrote it last night when I was depressed over the idea that my hypothesis about acceleration and differential timekeeping didn't seem tenable.
I've since edited it. So, you can check my reasoning. Thanks.

First off, your assertion that the clocks cannot be compared during the acceleration phases is not correct. You can deal with acceleration in SR if you are careful. One thing that must be kept in mind is whether you comparing from or to an accelerating frame. If you are comparing clocks from an inertial frame to an accelerating frame all you need to concern yourself with is the time dilation due to the Relative velocity at any given instant. If you are comparing the same clocks from the accelerating frame you have to take an additional effect into account. In this case the time rate on the other clock(the non accelerating one) will depend on the magnitude of your acceleration, the distance between you and the other clock (as measured on a line parallel to the line of acceleration), and the direction it lies from you compared to the direction of acceleration.

Example, if you are accelerating away from the other clock it will run slow compared to your clock (In addition to the regular time dilation), the greater the distance, the slower it will run. If you are accelerating towards the other clock, it will run fast compared to your clock, the further the clock is away, the faster it runs.

Now let's see how this works in our example.

First from the perspective of B.

A & B start off sychronized after B passes the Buoy. As A slows to a stop at Earth, it slowly increases its relative speed to B and undergoes progressive Time dilation. After A come to a complete stop it undergoes a constant time dilation and run slow while the distance between A and B closes. When B reaches Earth it begins to undergo its own acceleration which is away from A, The Relative velocity between A and B will decrease and the time dilation factor will decrease until the two clocks once again run at the same rate. There will be an additional slowing of A due to B's own acceleration, but since we will assume a fairly short acceleration period, this happens when the distance between A and B is small it won't make much difference. Since A ran slow for the entire period that B caught up to A, from B's perspective, A will show less time.

From the perspective of A.

A & B start off sychronized after B passes the Buoy. A begins its acceleration to stop at Earth. The realtive velocity between A and B begins to increase as doe the standard time dilation Factor. The acceleration is however towards B and B starts at 1 ly from B, therefore this will cause the effect of B clock running very fast compared to A. So fast, that it overshadows the time dilation. The acceleration period stops, and A & B maintain a constant velocity and clock B runs slow while the distance closes(But by this time clock B has gained a lot of time during the acceleration period). B begins its acceleration and the relative velocity between the two decreases, as doe the time dilation factor. Taking into account the time gained by B during A's acceleration, and the time dilation together, A determines that B's clock has accumulated more time then A's clock

From the Earth:

A passes the buoy and set it clock to zero. It then continues on towards the Earth at a constant velocity and its clock undergoes time dilation and runs slow. B then passes the buoy ands sets itself to the time that it determines A is reading at that instant (for instance if the velocity of A & B is .866c relative to the Earth B will set it clock to 1.15 years, as this is the amount of time that it takes between A and B passing the Bouy according to A & B).
But, This is not the same amount of time that A accumlates between its passing of the Buoy and B passing the Buoy according to the Earth. For one, due to length contraction, the distance between A and B is only .5 ly in our example and two, Clock A is running at half speed. Thus, according to the Earth, A will read 1.15/4 =.2875 years. This means that according to the Earth, B will lead A by 0.8625 yrs. A & B continue towards the Earth, each running slow by the same amount and maintaining the same diference in time.
A begins its acceleration period and matches speed with the Earth, after which it Runs at the same time rate as the Earth. B continues on, its clock running at half speed. B reaches the Earth and accelerates, matching speed with the Earth, after which its clock run at the same speed as the Earth's and B's. Now, even though it ran slow between the time that A stopped and it reached the Earth, it isn't enough to overcome the time it lead Clock A and it B will read more time than A.

Thus, from all three perspectives involved, A ends up reading less time than B, but each perspective comes to this conclusion for different reasons.

 

[Sam Woole]

Rather than use your letters ABCDE I’d prefer to use abcdefghijklm. Each of my letters represents 5 minutes of Alice’s travel time.

B           T
o           e
b           d
|           |
abcdefghijklm
A--B--C--D--E

Ted’s     Ted’s     Ted’s
own       view of   view of
clock     Alice’s   Bob’s
clock     clock
17:20(m)  12:00(a)  12:00(a)
17:30(m)  12:15(d)  12:10(a)
17:40(m)  12:30(g)  12:20(a)
17:50(m)  12:45(j)  12:30(a)
18:00(m)  13:00(m)  12:40(a)

Each row of this table represents a ray of light traveling from Bob to Ted. For example in the second row, a ray of light leaves Bob, at (a), at 12:10 Bob-time, passes Alice, at (d), at 12:15 Alice-time, and arrives at Ted, at (m), at 17:30 Ted-time. If you ignore Alice’s column and just look at the rays leaving Bob and arriving at Ted, each ray arrives at Ted(m) at a Ted-time that is 5h20m later than the Bob-time that it left Bob(a).
The first diagram I have attached to this post may make things clearer. Each spot represents a 5-minute interval. The yellow lines are light rays.
The conclusion to draw from this table is that if you travel away from somebody and observe an apparent 2/3 “red-shift” change of clock rates (Alice observing Bob), then it must follow that if you travel towards somebody at the same speed, you must observe an apparent 3/2 “blue-shift” change of clock rates (Ted observing Alice).
For the return journey my table shows, not EDCBA (mjgda) but mkigeca. That was my choice.

Alice’s   Alice’s
own       view of
clock     Bob’s clock
13:00(m)  12:40(a)
13:10(k)  12:55(a)
13:20(i)  13:10(a)
13:30(g)  13:25(a)
13:40(e)  13:40(a)
13:50(c)  13:55(a)
14:00(a)  14:10(a)

The reason for the 3/2 rate is because the speed of Bob towards Alice, during the return, is exactly the same as was the speed of Alice towards Ted on the outward journey. Einstein’s postulates imply that any effects you can measure should depend only on the relative velocity and on nothing else. So the same rate of 3/2 applies in both cases.
The second diagram I have attached to this post shows Alice’s whole journey.

I would say I have learned why and how the Doppler effect was applied. But I have also found more difficulties.

The first was, when Alice moved toward Ted, the 3/2 rate was applied to the moving clock of Alice, not to the stationary clock of Ted; while Alice moved toward Bob, the 3/2 rate was applied to the stationary clock of Bob. I could not understand this unbalanced treatment.

The second was, when Alice was moving away from Ted, the stationary observer Ted did the observation, applying the 2/3 rate to Alice's clock. This was the reverse of Alice moving away from Bob, the stationary observer. So what will happen if we let Bob observe Alice's journey both ways and apply the 2/3 and 3/2 to Alice's clock?

When I tried, I got the following table, supposing the starting time be 00 minutes, and eliminating the 3rd observer.

At start both clocks show 00 minutes:
00.......00, at start. After every 15 minutes on Bob's clock, he applies 2/3 to Alice's clock,

Bob's view of Alice clock...Bob's view of his own.
10........15,
20........30,
30........45,
40........60, In Bob's view, Alice has traveled 60 minutes, and she returns at this moment. Now Bob will observer every 10 minutes, and apply the 3/2rate.
55........70,
70........80,
85........90,
100.......100,
115.......110,
130.......120.

My table showed, after 2 hours of travel by Alice from Bob's view, Alice's clock should have accumulated 10 minutes more than his. If my table were sound, we could deduce that the earthbound twin got younger than his traveled brother.

Let me repeat my difficulties. When Alice was moving toward Ted, why the 3/2 rate was not applied to Ted's clock? Second, can Bob do the observation?

 

[Sherlock]

First off, your assertion that the clocks cannot be compared during the acceleration phases is not correct.

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)

This is much more complicated than I thought it was going to be. :-)

Regarding the assertion that clocks can't be compared during acceleration phases -- I forget where I got that. I was Googling a lot last night and read something like that somewhere. It wasn't from a quack site (at least I don't think it was). I also remember reading somewhere that a clock's tick rate is independent of any acceleration that the clock might experience.

 

[Janus]

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)
This is much more complicated than I thought it was going to be. :-)
Regarding the assertion that clocks can't be compared during acceleration phases -- I forget where I got that. I was Googling a lot last night and read something like that somewhere. It wasn't from a quack site (at least I don't think it was).

What might have been meant is that the clocks cannot be compared in an absolute way. IOW, If you compare clocks from the perspective of each clock they might both determine that is the other clock that is running slow and there is no way to say that one of them is absolutely right and the other wrong

I also remember reading somewhere that a clock's tick rate is independent of any acceleration that the clock might experience.

That is correct, a clocks rate is independent of any acceleration it experiences. What is not indedependent of that acceleration is how that clock measures the time rate of other clocks separated from it by a distance measured along the line of acceleration.

 

[DrGreg]

I would say I have learned why and how the Doppler effect was applied. But I have also found more difficulties.
The first was, when Alice moved toward Ted, the 3/2 rate was applied to the moving clock of Alice, not to the stationary clock of Ted; while Alice moved toward Bob, the 3/2 rate was applied to the stationary clock of Bob. I could not understand this unbalanced treatment.

The basic principle of Relativity is that the concepts of "stationary" and "moving" are relative terms. Something is stationary relative to something else, or something is moving relative to something else.

When Alice moved towards Ted, Alice was moving relative to Ted with a Doppler shift, relative to Ted, of 3/2.

When Alice moves towards Bob, it is also true to say that Bob is moving relative to Alice. From Alice's point of view, Bob is moving towards her. And the speed at which he is moving is exactly the same speed that, before, Alice was moving towards Ted. That is why, in both cases, the 3/2 Doppler shift applies.

This is the crux of what Relativity is all about. If you are sat in a moving train, you can think of yourself as being stationary and the rail tracks as moving relative to you. As long as the train moves at a constant speed in a straight line, any experiments you carry out inside the train carriage will give exactly the same results as if the train were stopped.

If A moves at speed v relative to B, then B moves at speed v relative to A. And if A observes B with a Doppler shift of 3/2, then B observes A with a Doppler shift of 3/2.

The second was, when Alice was moving away from Ted, the stationary observer Ted did the observation, applying the 2/3 rate to Alice's clock. This was the reverse of Alice moving away from Bob, the stationary observer. So what will happen if we let Bob observe Alice's journey both ways and apply the 2/3 and 3/2 to Alice's clock?

In my original post https://www.physicsforums.com/showpost.php?p=783491&postcount=68", the last table I gave was for Bob's view of Alice. (I never gave a table for Ted's view of Alice on her return journey.)

When I tried, I got the following table, supposing the starting time be 00 minutes, and eliminating the 3rd observer.
At start both clocks show 00 minutes:
00.......00,
at start. After every 15 minutes on Bob's clock, he applies 2/3 to Alice's clock,
Bob's view of Alice clock...Bob's view of his own.
10........15,
20........30,
30........45,
40........60,

Yes, that is correct.

In Bob's view, Alice has traveled 60 minutes, and she returns at this moment. Now Bob will observer every 10 minutes, and apply the 3/2 rate.
55........70,
70........80,
85........90,
100.......100,
115.......110,
130.......120.

No. Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:

50.......75
60.......90

Now, Bob sees Alice return and the factor 3/2 applies:

60.......90
75.......100
90.......110
105........120
120........130

Another diagram attached...

 

[jtbell]

[at the 60-minute mark, according to Bob's clock] Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:

To make this more explicit, in Bob's reference frame, when Alice turns around she is 25 light-minutes away. It takes 25 minutes for light from this event to travel back to Bob, and that's when he actually sees her turn around. In the meantime Bob continues to watch Alice's clock running at the 2/3 rate.

 

[Sam Woole]

No. Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:
50.......75
60.......90
Now, Bob sees Alice return and the factor 3/2 applies:
60.......90
75.......100
90.......110
105........120
120........130
Another diagram attached...

Here I might have to disagree with you. Alice turned around when she saw 13:00 on her clock. If Bob observed Alice's clock from (a), he would see 12:40. Therefore, I believe Alice's return trip should begin when Alice saw 60 on her clock, namely when Bob saw 40 on Alice's clock.

If you insist, you have to explain 2 things. First, where was the 20 minutes delay? When Bob saw 40 on Alice's clock, Alice has traveled 60 minutes already due to the 20 minutes delay.

Secondly, why the equal row disappeared. Your table on post #86 had an equal row at (e), 13:40 = 13:40, and mine had an equal row 100=100.
This new table above of yours had no equal row.

Your table on post #86:
Alice’s... Alice’s
own... view of
clock..... Bob’s clock
13:00(m)... 12:40(a)
13:10(k)... 12:55(a)
13:20(i)... 13:10(a)
13:30(g)... 13:25(a)
13:40(e)... 13:40(a)
13:50(c)... 13:55(a)
14:00(a)... 14:10(a)

 

[jtbell]

I believe Alice's return trip should begin when Alice saw 60 on her clock, namely when Bob saw 40 on Alice's clock.

Imagine that in order to initiate the turnaround, Alice has to push a button on her ship's control panel, and the button is right next to her clock. From her point of view, she pushes the button when her clock reads 60 minutes after departure. Do you really believe that Bob is going to see Alice pushing the button when her clock (which is right next to the button!) reads 40?

 

[Sam Woole]

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)

Since you have the book by Halliday, Resnick and Walker, would you please pay attention to the language used in the chapter on Relativity. By this I meant, try to find out whether there is a lie in the said chapter.

 

[Sherlock]

Since you have the book by Halliday, Resnick and Walker, would you please pay attention to the language used in the chapter on Relativity. By this I meant, try to find out whether there is a lie in the said chapter.

What did you read that causes you to say this? I'll look for it.

 

[Sherlock]

... a clock's rate is independent of any acceleration it experiences.

I don't think I understand what this means.

A and B are on Earth with identical, synchronized clocks. A accelerates away and eventually assumes a uniform speed wrt B and earth.

B sees A's clock as ticking at a slower rate than his. How is this independent of the acceleration that brought A to the relative speed that is causing the time dilation?

 

[Janus]

I don't think I understand what this means.
A and B are on Earth with identical, synchronized clocks. A accelerates away and eventually assumes a uniform speed wrt B and earth.
B sees A's clock as ticking at a slower rate than his. How is this independent of the acceleration that brought A to the relative speed that is causing the time dilation?

In that acceleration itself does not cause any time dilation effect on a clock outside of that caused by a change in relative speed. Remember, you can have acceleration without any change of relative speed, as acceleration is a change of velocity and velocity consists of both speed and direction.

For example, you can put a clock on a centrifuge and spin it at a constant RPM, and will be both constantly accelerating and maintaining a constant speed. In this case, an observer watching the centrifuge from a frame at rest with respect to the centrifuge axis will only measure a time dilation for the clock that is due to its constant relative speed.

 

[jtbell]

Another way to put it: time dilation does not depend directly on acceleration. It depends only indirectly on acceleration insofar as acceleration produces a change in an object's speed. It is not necessary to know an object's "acceleration history" in order to calculate the amount of time dilation, only the object's current speed.

 

[DrGreg]

Sam,

You must remember that each row in my tables does not show two things happening at the same time. Each row represents a ray of light traveling from Bob to Alice in one table, or from Alice to Bob in the other table. There is always a delay between light being sent and being received.

First, where was the 20 minutes delay? When Bob saw 40 on Alice's clock, Alice has traveled 60 minutes already due to the 20 minutes delay.

The “20 minute delay” is for light sent from Bob at (a) to Alice at (m). There is also a “20 minute delay” for light sent from Alice at (i) (not (m)) to Bob at (a). When Bob sees 40 on Alice’s clock, Bob is seeing Alice when she was at (i). As Alice traveled from (i) to (m) she was still traveling away from Bob, so when Bob sees Alice traveling from (i) to (m) the delay is still increasing and the Doppler rate of 2/3 still applies. The fact that Bob’s own clock has gone past 60 is irrelevant – he is still seeing light which left Alice before her clock reached 60 so the delay from Alice to Bob is still increasing, not decreasing.

Secondly, why the equal row disappeared. Your table on post #86 had an equal row at (e), 13:40 = 13:40, and mine had an equal row 100=100. This new table above of yours had no equal row.

This is because these are different tables. The table with 13:40(e)…13:40(a) that you quoted was Alice’s view of Bob. The table in my last message was Bob’s view of Alice.

It is wrong to say that the event of Alice’s clock showing 13:40 occurs “at the same time” as the event of Bob’s clock showing 13:40. The table with 13:40(e)…13:40(a) shows that light leaving (a) at 13:40 on Bob’s clock arrives at (e) at 13:40 on Alice’s clock. If these two events occurred at the same time, light would have traveled instantly from (a) to (e) at infinite speed. This is impossible.

In Relativity, the concept of “at the same time” is a relative concept, which is determined by a convention. Using that convention, two different observers can disagree whether two separated events occur “at the same time”.

However, in my argument, I make no assumptions about simultaneity. Clocks are compared only by sending light from one clock to another, which takes time.

It might avoid confusion if I redesign the experiment so that Alice and Bob do not synchronize their clocks at the start. Let’s say that we start with Bob’s clock showing 12:00 but Alice’s clock showing 15:00.

The two tables now look like this.


Alice’s view of Bob

Bob...abcdefghijklm.Alice
12:00 aa ...15:00
12:10 a>>>d ...15:15
12:20 a>>>>>>g ...15:30
12:30 a>>>>>>>>>j ...15:45
12:40 a>>>>>>>>>>>>m 16:00
12:40 a>>>>>>>>>>>>m 16:00
12:55 a>>>>>>>>>>k ..16:10
13:10 a>>>>>>>>i ...16:20
13:25 a>>>>>>g ...16:30
13:40 a>>>>h ...16:40
13:55 a>>c ...16:50
14:10 aa ...17:00
Bob...abcdefghijklm.Alice


Bob’s view of Alice

Bob...abcdefghijklm.Alice
12:00 aa ...15:00
12:15 a<<c ...15:10
12:30 a<<<<e ...15:20
12:45 a<<<<<<g ...15:30
13:00 a<<<<<<<<i ...15:40
13:15 a<<<<<<<<<<k ..15:50
13:30 a<<<<<<<<<<<<m 16:00
13:30 a<<<<<<<<<<<<m 16:00
13:40 a<<<<<<<<<j ...16:15
13:50 a<<<<<<g ...16:30
14:00 a<<<d ...16:45
14:10 aa ...17:00
Bob...abcdefghijklm.Alice


The only times that we can sensibly compare Bob’s clock directly with Alice’s clock is when Bob and Alice are at the same place (a). At the beginning there is a difference of 3 hours but at the end the difference is 3 hours 10 minutes. We cannot say what the difference is in between.

 

[Sam Woole]

Sam,
You must remember that each row in my tables does not show two things happening at the same time. Each row represents a ray of light traveling from Bob to Alice in one table, or from Alice to Bob in the other table. There is always a delay between light being sent and being received.
The “20 minute delay” is for light sent from Bob at (a) to Alice at (m). There is also a “20 minute delay” for light sent from Alice at (i) (not (m)) to Bob at (a). When Bob sees 40 on Alice’s clock, Bob is seeing Alice when she was at (i). As Alice traveled from (i) to (m) she was still traveling away from Bob, so when Bob sees Alice traveling from (i) to (m) the delay is still increasing and the Doppler rate of 2/3 still applies. The fact that Bob’s own clock has gone past 60 is irrelevant – he is still seeing light which left Alice before her clock reached 60 so the delay from Alice to Bob is still increasing, not decreasing.
This is because these are different tables. The table with 13:40(e)…13:40(a) that you quoted was Alice’s view of Bob. The table in my last message was Bob’s view of Alice.
It is wrong to say that the event of Alice’s clock showing 13:40 occurs “at the same time” as the event of Bob’s clock showing 13:40. The table with 13:40(e)…13:40(a) shows that light leaving (a) at 13:40 on Bob’s clock arrives at (e) at 13:40 on Alice’s clock. If these two events occurred at the same time, light would have traveled instantly from (a) to (e) at infinite speed. This is impossible.
In Relativity, the concept of “at the same time” is a relative concept, which is determined by a convention. Using that convention, two different observers can disagree whether two separated events occur “at the same time”.
However, in my argument, I make no assumptions about simultaneity. Clocks are compared only by sending light from one clock to another, which takes time.
It might avoid confusion if I redesign the experiment so that Alice and Bob do not synchronize their clocks at the start. Let’s say that we start with Bob’s clock showing 12:00 but Alice’s clock showing 15:00.
The two tables now look like this.
Alice’s view of Bob
Bob...abcdefghijklm.Alice
12:00 aa ...15:00
12:10 a>>>d ...15:15
12:20 a>>>>>>g ...15:30
12:30 a>>>>>>>>>j ...15:45
12:40 a>>>>>>>>>>>>m 16:00
12:40 a>>>>>>>>>>>>m 16:00
12:55 a>>>>>>>>>>k ..16:10
13:10 a>>>>>>>>i ...16:20
13:25 a>>>>>>g ...16:30
13:40 a>>>>h ...16:40
13:55 a>>c ...16:50
14:10 aa ...17:00
Bob...abcdefghijklm.Alice
Bob’s view of Alice
Bob...abcdefghijklm.Alice
12:00 aa ...15:00
12:15 a<<c ...15:10
12:30 a<<<<e ...15:20
12:45 a<<<<<<g ...15:30
13:00 a<<<<<<<<i ...15:40
13:15 a<<<<<<<<<<k ..15:50
13:30 a<<<<<<<<<<<<m 16:00
13:30 a<<<<<<<<<<<<m 16:00
13:40 a<<<<<<<<<j ...16:15
13:50 a<<<<<<g ...16:30
14:00 a<<<d ...16:45
14:10 aa ...17:00
Bob...abcdefghijklm.Alice
The only times that we can sensibly compare Bob’s clock directly with Alice’s clock is when Bob and Alice are at the same place (a). At the beginning there is a difference of 3 hours but at the end the difference is 3 hours 10 minutes. We cannot say what the difference is in between.

I do not think your tables are honest. Your first table showed the turn around took place when the moving clock has accumulated 60 minutes, 60 vs 40 (40 was the illusion in the telescope. Actual number was 60 due to 20 minutes delay). Your second table showed the turn around took place when the moving clock has accumulated 90 minutes, 90 vs 60 (60 is the illusion in the telescope. Actual number is 80 due to 20 minutes delay.). This was the point I could not understand. I did not find any solution to it in your post.

My understanding of your original tables was, it was a two hour return journey, one hour each way. I do not understand how could the turn around take place when one clock has accumulated 90 minutes.

I did not challenge the truthfulness of the equal row, nor the definition of simultaneity. What I did challenge was your consistency. You changed rules of the game by prolonging the one way journey to 90 minutes which made the equal row disappear, and as a result the turn around took place 30 minutes later. To be consistent, we should have kept that equal row, we should have the turn around to take place when the moving clock has accumulated 60 minutes, meaning the journey's one end.

Although you changed the time on Alice clock to be 15:00 hours at start, but the equal row was still there in the first table above, 13:40 and 16:40, accumulation of 100 minutes on each clock, or 1:40 hours, meaning light is instantaneous. How could it be so?

I had an inkling that people cannot make the time dilation idea stand unless they contradict themselves, such as the equal row. I believe it was all an illusion. You said so in the beginning; you knew it was.

 

[JesseM]

IYour second table showed the turn around took place when the moving clock has accumulated 90 minutes, 90 vs 60 (60 is the illusion in the telescope.

You're misunderstanding the second table. The second table shows that Alice's clock read 16:00 at the time of the turnaround, but the light from that event didn't reach Bob until his clock read 13:30. The first table, on the other hand, says that Alice's clock read 16:00 at the time of the turnaround, and that at that moment she was seeing light from Bob's clock which read 12:40. So in both tables it's true that the turnaround happened at 16:00 according to Alice's clock, so both tables say the moving clock accumulated 60 minutes.