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

[JesseM]

Here you were repeating the other experts' idea and experiments (by Hafele-Keating). But my difficulty with time dilation was on the symmetrical aspect. As long as it is symmetrical, time dilation cannot be justified because both clocks are producing equal time intervals.

No, it is "symmetrical" because the second clock runs slow in the first clock's frame, while the first clock runs slow in the second clock's reference frame.

Now supporters of time dilation changed the game rules by introducing acceleration. I said this because when the twin paradox was first introduced to give support to the time dilation idea, acceleration was not specified. Acceleration came into the game only after some dissitents attacked the weird idea by means of uniform motion. (I read about it but I could not quote the source.)

You don't know your history. Einstein specified each reference frame must be in a state of uniform linear motion (ie no acceleration) in his original 1905 paper, and the twin paradox was certainly not thought up until after that.

Even if we accept the change of rules, I believe time dilation still cannot be justified, especially when spaceships were meant to return to earth. For the spaceship to return, it must undergo deceleration, such as NASA's spacecraft re-entering the earth. Theoretically the spaceship's acceleration should equal its deceleration. Hence your "oscillator dilated" would be cancelled. Don't you think so? Are the twins of the same age?

There is no distinction between "acceleration" and "deceleration" in physics--whether an accelerating object is increasing in velocity or decreasing in velocity depends on your reference frame, this is true in Newtonian physics as well.

Look, whether or not you believe relativity is physically correct (and all the experimental evidence says it is), if you're trying to argue that it doesn't make mathematical sense you're just being foolish. The Lorentz transformation is a perfectly consistent way to transform between different coordinate systems, and if a clock moving at velocity v is ticking at \sqrt{1 - v^2/c^2} the normal rate in a given frame, then if you know some basic calculus it's clear that given the clock's velocity as a function of time v(t), the total time elapsed between two times t_0 and t_1 must be given by the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can prove mathematically that the answer to the total time elapsed between two points on the clock's worldline (a 'worldline' is a path through spacetime) in one frame will be equal to the answer in another reference frame if you use the Lorentz transform to describe the worldline in the two different frames. And you can also prove that the path between two points with the greatest elapsed time will always be the straight one, so any non-straight worldline between those same points will have elapsed less time, and straight worldline = no acceleration while non-straight worldline = at least some acceleration. This is analogous to the fact that the shortest distance between two points in 2D space is always a straight line, any non-straight path between the two points will have a greater length.

 

[Sherlock]

Sherlock, when you say "reference state", do you mean the same thing as a "reference frame" or something different? If it's something different, can you explain, maybe with an example?

If I'm understanding what reference frame means, then the reference state includes 3 reference frames -- the earth, clock A, and clock B. But, since clock B doesn't move from it's location on earth, then the reference frame that is clock B is essentially the same as the reference frame that is it's location on earth.

According to relativity time dilation is a function of velocity rather than acceleration. But if two clocks are moving apart at constant velocity, then in each clock's own rest frame it will be the other clock that is running slower--the only situation where you get an objective answer to which clock is "really" behind is the one where one clock turns around (accelerates) and moves back towards the first clock, so they can meet at a single location in space and see which clock is behind when they meet.

As long as B remains in the original location on Earth while A is moving wrt the original configuration (the reference state), and as long as we can continue to track A's motion, then I think we get an objective answer to which clock is really behind, and by how much, whether A and B are reunited.

You mention that time dilation is a function of velocity rather than acceleration. Acceleration is velocity per unit of time.

 

[JesseM]

If I'm understanding what reference frame means, then the reference state includes 3 reference frames -- the earth, clock A, and clock B. But, since clock B doesn't move from it's location on earth, then the reference frame that is clock B is essentially the same as the reference frame that is it's location on earth.

As long as B remains in the original location on Earth while A is moving wrt the original configuration (the reference state), and as long as we can continue to track A's motion, then I think we get an objective answer to which clock is really behind, and by how much, whether A and B are reunited.

Just because A and B start out in a single rest frame, why should that mean that this frame's answer to which clock is later ticking faster is more correct than any other? Why does the "original state" matter, especially since the choice of what moment to pick as the beginning is totally arbitrary? If at an even earlier time A and B had been traveling through space at a high velocity relative to earth, then they came to rest on Earth and stayed there for a while, then B took off from the Earth and returned to its original velocity, which frame would you use to decide which clock is really ticking faster?

You mention that time dilation is a function of velocity rather than acceleration. Acceleration is velocity per unit of time.

Acceleration is change in velocity per unit time--if an object isn't changing velocity, its acceleration is zero. Anyway, what's your point?

 

[pervect]

The results of such comparisons are fairly clear. The clock whose state of motion has changed wrt the reference state accumulates time at a different rate. So, the average period of it's oscillator is different during the interval in which its average velocity wrt the reference state is different.

Ok, but we specify a reference state, so naturally we need to reassemble that state in order to make any meaningful comparisons between the clocks.

EDIT: I changed my mind about this. Because we specify a reference state involving a frame of reference in addition to the two clocks, then we don't need to bring the clocks back together in order to make meaningful statements about which clock is moving slower.

We of course do need to keep one clock in the reference state, and keep them both running, and continue tracking them both during the separation interval.

*IF* you make the additional assumption to use a specific reference frame's defintion of simultaneity, your logic is sound. But this additional assumption is required, and leads to a seeming paradox.

The seeming paradox is commonly called the "twin paradox", where A concludes that B's clocks are running slow, and B concludes that A's clocks are also running slow.

The key to resolving this paradox is to note specifically that one *does* have to make the extra assumption about what defintion of simultaneity one choses to use. There are as many different defintions of "simultaneous" as there are reference frames (which you call reference states). A's defintion of simultaneity is not the same as B's defintion. This is equivalent to my remark that the results of any clock comparision will depend on the details of the path by which the clock is transported. Until a reference frame is chosen (or a clock transport method is specified), the results of a direct comparison between the spatially separated clocks is not defined.

The recognition of the fact that there *are* different defintions of "simultaneous" is what prevents the twin paradox from actually being a paradox, i.e. it is necessary to recognize the relativity of simultaneity to understand how special relativity is self-consistent.

 

[Sam Woole]

No, it is "symmetrical" because the second clock runs slow in the first clock's frame, while the first clock runs slow in the second clock's reference frame.

You have been saying this a few times. Let me show you what was my understanding. The second clock shows 3 seconds to the observer in the first frame whose clock shows 4 seconds, meaning the second clock is 1 second slower than the first clock. Symmetrically, the first clock shows 3 seconds to the observer in the second frame, whose clock shows 4 second, meaning the first clock is 1 second slower than the second clock. In short, there is always this sequence: 3-4, 4-5, 5-6 ...in each frame. This sequence means that the two clocks constantly have 1 second difference between them. It further means to me, both clocks are measuring identical time intervals, constantly synchronized. One second in one frame is always equal to the one second in another frame. To me, this means there is no time dilation.

You don't know your history. Einstein specified each reference frame must be in a state of uniform linear motion (ie no acceleration) in his original 1905 paper, and the twin paradox was certainly not thought up until after that. There is no distinction between "acceleration" and "deceleration" in physics--whether an accelerating object is increasing in velocity or decreasing in velocity depends on your reference frame, this is true in Newtonian physics as well..

Your words above mean to me, both acceleration and deceleration will cause differential aging, uniform relative motion included. But you also said earlier that the space traveling twin must accelerate "significantly"? Why significantly?

Look, whether or not you believe relativity is physically correct (and all the experimental evidence says it is), if you're trying to argue that it doesn't make mathematical sense you're just being foolish. The Lorentz transformation is a perfectly consistent way to transform between different coordinate systems, and if a clock moving at velocity v is ticking at \sqrt{1 - v^2/c^2} the normal rate in a given frame, then if you know some basic calculus it's clear that given the clock's velocity as a function of time v(t), the total time elapsed between two times t_0 and t_1 must be given by the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can prove mathematically that the answer to the total time elapsed between two points on the clock's worldline (a 'worldline' is a path through spacetime) in one frame will be equal to the answer in another reference frame if you use the Lorentz transform to describe the worldline in the two different frames. And you can also prove that the path between two points with the greatest elapsed time will always be the straight one, so any non-straight worldline between those same points will have elapsed less time, and straight worldline = no acceleration while non-straight worldline = at least some acceleration. This is analogous to the fact that the shortest distance between two points in 2D space is always a straight line, any non-straight path between the two points will have a greater length.

I myself cannot prove that Einstein's theory is correct or not. But I know there are plenty of challenges to his math. One was a Canadian Dr. Paul Marmet,
http://www.Newtonphysics.on.ca/, who said:
"Conventional Wisdom, Conventional Logic, Newton's Physics and Galilean coordinates, classical physics can explain all the observed phenomena attributed to relativity. Einstein's Relativity is completely useless." Marmet has a book to support his words.

And there were others who showed, mathematically, that Einstein's math is false as it results in 2=0, etc.

 

[JesseM]

You have been saying this a few times. Let me show you what was my understanding. The second clock shows 3 seconds to the observer in the first frame whose clock shows 4 seconds, meaning the second clock is 1 second slower than the first clock. Symmetrically, the first clock shows 3 seconds to the observer in the second frame, whose clock shows 4 second, meaning the first clock is 1 second slower than the second clock. In short, there is always this sequence: 3-4, 4-5, 5-6 ...in each frame.

No, the difference is not constant. Each frame observes the other clock as ticking slower than its own, not just being behind its own by a constant amount but ticking at the same rate. For example, suppose two clocks are moving at 0.8c relative to each other, and they start out at the same position at which point they both read "0 seconds". Then each clock will observe the other run at 0.6 times its own rate--so in the first clock's frame, when the first clock reads 4 seconds the second clock reads 2.4 seconds, when the first clock reads 5 seconds the second reads 3 seconds, when the first clock reads 6 seconds the second clock reads 3.6 seconds, and so forth. Likewise, in the second clock's frame, when the second clock reads 4 seconds the first reads 2.4 seconds, when the second clock reads 5 seconds the first reads 3 seconds, and so forth. That is what is meant by "symmetrical" here. Again, if you're having trouble understanding how this is possible, I suggest you take a careful look at the diagrams from the example I provided in An illustration of relativity with rulers and clocks.

Your words above mean to me, both acceleration and deceleration will cause differential aging, uniform relative motion included.

Relative to a particular frame, differential aging depends only on the velocity. But the point is that different frames disagree about who's aging slower when two clocks are moving apart at constant velocity, in order for there to be a single objective answer at least one of the clocks has to turn around and return to meet the other clock, and "turning around" means accelerating. As I said, it can be shown that all frames will agree on their prediction of what the clocks will read when they meet up again (each frame makes this prediction by doing the integral I posted earlier, which depends only on the velocity function v(t) in that frame), and they will all agree that the clock that moved at constant velocity elapsed more time than the clock that accelerated.

But you also said earlier that the space traveling twin must accelerate "significantly"? Why significantly?

I don't remember the context, but I'd guess I probably just said that because if the two clocks are moving apart at relativistic speeds, one will have to accelerate significantly in order for its path to change enough so that it's now moving towards the other clock at relativistic speeds (here 'significantly' can mean either a large burst of acceleration over a short time or a lower rate of acceleration but extended over a long time period). If you just had two clocks moving apart at speeds which were very small compared to light speed, then the acceleration to get one clock to turn around needn't be large. As long as two clocks depart from each other at a single point in space and then meet again later at a single point in space, it will always be the clock that accelerated that has elapsed less time, regardless of the size of the acceleration.

I myself cannot prove that Einstein's theory is correct or not. But I know there are plenty of challenges to his math. One was a Canadian Dr. Paul Marmet,
http://www.Newtonphysics.on.ca/, who said:
"Conventional Wisdom, Conventional Logic, Newton's Physics and Galilean coordinates, classical physics can explain all the observed phenomena attributed to relativity. Einstein's Relativity is completely useless." Marmet has a book to support his words.

It doesn't sound like he's challenging the math in that quote, just the theory. Can you provide a quote from his webpage or his book saying that there is some self-contradiction in the math itself?

And there were others who showed, mathematically, that Einstein's math is false as it results in 2=0, etc.

No there weren't. And if you think there were, why do you think the mathematics community wasn't convinced? Do you think they're all idiots who can't follow a simple proof showing a contradiction? Remember, once something is proved in mathematics then its completely cut-and-dried, there's no room for differences of opinion--if it's been proved, any mathematician competent to evaluate the proof should agree.

 

[Sherlock]

Just because A and B start out in a single rest frame, why should that mean that this frame's answer to which clock is later ticking faster is more correct than any other?

The selection of reference (or rest) frame is, as you note, arbitrary. But, for practical purposes we pick a location on Earth and position the clocks right next to each other. This is a state of motion of the earth-clock A-clock B system. Wrt this state, as clock A travels around the solar system, we know that clock A is the moving clock. Clock A's motion is the anomalous motion wrt the reference state. So, whether we reunite the clocks (reassemble the reference state) or not, we can make some statements about which clock's timekeeping rate has 'really' slowed wrt the rate (which clock B is still keeping time at) observed when both clocks were next to each other on earth. (We're assuming of course that the Earth's regular motion won't be appreciably changed due to A's acceleration away from it, so that if we expanded the observational context,the reference state, to include, say, the entire solar system, then we would see that clock A's motion is the anomalous motion wrt the regular motion of the solar system, and so on.) We observe that clock A's rate has changed due to changing it's state of motion.

Why does the "original state" matter, especially since the choice of what moment to pick as the beginning is totally arbitrary?

The reference state defines a specific state of motion of the clocks. We change clock A's state of motion wrt this state, and note the differences between the rate at which it's oscillator cycles and B's. They're different, so we conclude that clock A's oscillator has undergone some physical changes that clock B's oscillator hasn't -- and, therefore, that clock A's oscillator has undergone some physical changes that it would not have undergone had it remained on Earth next to clock B.

Acceleration is change in velocity per unit time--if an object isn't changing velocity, its acceleration is zero. Anyway, what's your point?

Just that when an object is accelerating, then the period of its oscillator is dilating. During intervals of uniform velocity, the dilation (the oscillator's period) is constant.

 

[Sherlock]

... my difficulty with time dilation was on the symmetrical aspect. As long as it is symmetrical, time dilation cannot be justified because both clocks are producing equal time intervals.

Your statement here suggests that you're using time dilation to mean the same thing as differential timekeeping. They're not the same thing. The dilation in timekeeping rates of two clocks, A and B, is symmetrical as long as we're only considering the two clocks and nothing else -- A wrt B, and B wrt A is the same dilation as they move wrt each other. But if we bring in another reference frame wrt which one, B, is stationary, then wrt this frame we can say that A's time is dilated while B's isn't. Also, when A and B start out next to each other on, say, a laboratory table, and then we take A on an airplane ride, then we know which clock moved and which clock didn't.
As long as we can find at least one frame of reference wrt which A and B are moving differently, then we have a basis for saying that they are keeping time differently, and, as long as we can continue to track them both, what that difference is for a given interval.

Even if we accept the change of rules, I believe time dilation still cannot be justified, especially when spaceships were meant to return to earth. For the spaceship to return, it must undergo deceleration, such as NASA's spacecraft re-entering the earth. Theoretically the spaceship's acceleration should equal its deceleration. Hence your "oscillator dilated" would be cancelled. Don't you think so? Are the twins of the same age?

Any change in speed or direction is an acceleration. As the traveling twin slows down to land back on Earth his time is still dilated wrt the earthtwin's time, but the dilation decreases as he approaches Earth to land.
As long as the traveling twin is moving wrt the earthbound twin, whether this motion is away from or toward the earthbound twin, then his timekeeping rate will be slower (the period of his oscillators dilated) wrt the earthbound twin. This is because the traveling twin's motion is the anomalous motion. We know that the traveling twin is the one who accelerated.

The twins age differently.

It took me a while to understand this. But, I think I have an intuitive grasp of it now. I know the feeling of being stuck on the idea that the twins should age the same -- even though the equations say differently. Then it sunk in that every measurement depends on time and distance units, and that these units are physically different wrt different states of motion.
I just focused on the idea that the twins are moving differently wrt the natural order of things. So their oscillators are experiencing things, interacting with things a bit differently, and as a result the periods of their oscillators are different, and the values of time and distance units (and even the biological processes) that depend on these periods are also different.

 

[Sherlock]

*IF* you make the additional assumption to use a specific reference frame's defintion of simultaneity, your logic is sound. But this additional assumption is required, and leads to a seeming paradox.

The seeming paradox is commonly called the "twin paradox", where A concludes that B's clocks are running slow, and B concludes that A's clocks are also running slow.

The key to resolving this paradox is to note specifically that one *does* have to make the extra assumption about what defintion of simultaneity one choses to use. There are as many different defintions of "simultaneous" as there are reference frames (which you call reference states).

A's defintion of simultaneity is not the same as B's defintion. This is equivalent to my remark that the results of any clock comparision will depend on the details of the path by which the clock is transported. Until a reference frame is chosen (or a clock transport method is specified), the results of a direct comparison between the spatially separated clocks is not defined.

The recognition of the fact that there *are* different defintions of "simultaneous" is what prevents the twin paradox from actually being a paradox, i.e. it is necessary to recognize the relativity of simultaneity to understand how special relativity is self-consistent.

I'd put it this way:
The resolution of paradoxes due to symmetrical time dilation is based on the observation of anomalous motion wrt some defined state or natural process or order. It's based on the observation that clock A's motion is the anomalous motion (comparing A and B) wrt the Earth reference frame. The alternative is that the Earth and clock B accelerated away from clock A -- but I think we can rule that out (anyway, we can determine what is really happening by observing the rest of the solar system).

So, yes, we make assumptions about which states to use as reference states. It's the recognition that some alternatives make more sense than others wrt our observations of the natural order of things that prevents the twin paradox from actually being a paradox.

I don't think this contradicts anything you said.

 

[JesseM]

I'd put it this way:
The resolution of paradoxes due to symmetrical time dilation is based on the observation of anomalous motion wrt some defined state or natural process or order. It's based on the observation that clock A's motion is the anomalous motion (comparing A and B) wrt the Earth reference frame. The alternative is that the Earth and clock B accelerated away from clock A -- but I think we can rule that out (anyway, we can determine what is really happening by observing the rest of the solar system).

It doesn't make any difference who accelerated away from who initially, it's only acceleration after the two clocks depart that determines which one will be behind when they meet again. The perspective of a frame where the Earth is initially moving at 0.99c, and the clock that accelerates away from it is actually slowing down initially, is every bit as valid as the perspective of the Earth's frame.

So, yes, we make assumptions about which states to use as reference states.

No, the way you are using "reference state", basically to mean using the initial conditions to determine which frame's answer is more "correct" than the others, is not an idea that any physicists would find useful, and you haven't given any justification for why the initial conditions should cause us to prefer one frame over another. It's totally arbitrary, as far as I can see (and note my comments earlier about the choice of which state to label the 'initial' one as being equally arbitrary).

I don't think this contradicts anything you said.

I don't think pervect understood what you meant by "reference state", since he seemed to think it just meant the same thing as "reference frame", when actually you are using it to mean an initial state which you use to determine a preferred frame. I'm pretty sure he wouldn't agree with this notion of yours that the initial state tells us that one frame's perspective is more valid than others.

 

[JesseM]

Any change in speed or direction is an acceleration. As the traveling twin slows down to land back on Earth his time is still dilated wrt the earthtwin's time, but the dilation decreases as he approaches Earth to land.

Only if he slows down as he approaches earth, but we don't need to make that assumption, he could equally well just pass by the Earth at constant velocity and they could compare clock readings the moment they passed.

As long as the traveling twin is moving wrt the earthbound twin, whether this motion is away from or toward the earthbound twin, then his timekeeping rate will be slower (the period of his oscillators dilated) wrt the earthbound twin. This is because the traveling twin's motion is the anomalous motion. We know that the traveling twin is the one who accelerated.

Again, it makes no difference who accelerated initially. If the Earth and the ship were originally at rest with respect to each other and in the same location, then the Earth was accelerated briefly and began to move away from the ship at constant velocity, then after moving apart for a while the ship accelerated and turned around so it was approaching the Earth rather than moving away from it, then when they reunited it would be the twin on the ship who aged less, in spite of the fact that they initially began to move apart because the Earth was accelerated, rather than the ship accelerating away from the earth. All that matters is which clock takes a straight path through spacetime after they begin to move apart.

It took me a while to understand this. But, I think I have an intuitive grasp of it now. I know the feeling of being stuck on the idea that the twins should age the same -- even though the equations say differently. Then it sunk in that every measurement depends on time and distance units, and that these units are physically different wrt different states of motion.
I just focused on the idea that the twins are moving differently wrt the natural order of things.

If by "natural order of things" you're talking about one motion being "anomolous" so that that frame is less valid then a more "natural" one, then you're still badly misunderstanding relativity.

 

[Sherlock]

It doesn't make any difference who accelerated away from who initially, it's only acceleration after the two clocks depart that determines which one will be behind when they meet again.

The departure acceleration puts clock A into different states of motion, while clock B's state of motion hasn't changed. We're assuming of course that the Earth hasn't suddenly accelerated out of it's usual motion.

The perspective of a frame where the Earth is initially moving at 0.99c, and the clock that accelerates away from it is actually slowing down initially, is every bit as valid as the perspective of the Earth's frame.

So? All I'm saying is that any change in motion (any acceleration) produces a changes in the accelerating body. If the clock accelerates away from the earth, then no matter what frame is viewing this, then the accelerated clock's rate will be altered compared to its rate when it was on the earth.

... the way you are using "reference state", basically to mean using the initial conditions to determine which frame's answer is more "correct" than the others, is not an idea that any physicists would find useful

Sure they would. They do it all the time. In experiments they always know which 'clocks' are really accelerating. There's always some way to tell. And anyway, that's how experiments are set up isn't it -- so that you know which clock has accelerated wrt some previous state and which hasn't?

... and you haven't given any justification for why the initial conditions should cause us to prefer one frame over another. It's totally arbitrary, as far as I can see (and note my comments earlier about the choice of which state to label the 'initial' one as being equally arbitrary).

Of course it's arbitrary. But we need a baseline (so to speak) state of motion for the purpose of comparison. Then we accelerate a clock wrt this state and see what happens.

I don't think pervect understood what you meant by "reference state", since he seemed to think it just meant the same thing as "reference frame", when actually you are using it to mean an initial state which you use to determine a preferred frame. I'm pretty sure he wouldn't agree with this notion of yours that the initial state tells us that one frame's perspective is more valid than others.

Valid wrt what? The state we choose as a reference state is valid as a reference state. Since we know which clock has accelerated wrt that state, then we can make some inferences about the affects of acceleration and changes in oscillators whose states of motion have been altered.

You're not saying that we don't know which clock has accelerated, are you?

 

[Sherlock]

Any change in speed or direction is an acceleration. As the traveling twin slows down to land back on Earth his time is still dilated wrt the earthtwin's time, but the dilation decreases as he approaches Earth to land.

Only if he slows down as he approaches earth, but we don't need to make that assumption, he could equally well just pass by the Earth at constant velocity and they could compare clock readings the moment they passed.

Or the traveller could keep heading away from Earth or whatever. But the question I was responding to involved the round trip scenario. 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 -- and because the traveller's clocks have been keeping time with physically altered oscillators all during his trip, they will all accumulate different total times than they would have had they remained on earth.

... it makes no difference who accelerated initially. If the Earth and the ship were originally at rest with respect to each other and in the same location, then the Earth was accelerated briefly and began to move away from the ship at constant velocity, then after moving apart for a while the ship accelerated and turned around so it was approaching the Earth rather than moving away from it, then when they reunited it would be the twin on the ship who aged less, in spite of the fact that they initially began to move apart because the Earth was accelerated, rather than the ship accelerating away from the earth. All that matters is which clock takes a straight path through spacetime after they begin to move apart.

Paths through spacetime is one way of talking about it. But, I don't think the geometrical 'explanation' for differential timekeeping is necessarily the best approach to understanding it. As I've read here on PF, "the map is not the territory."

You say that it makes no difference who accelerated initially. The point is, we *know* who accelerated and who didn't. There's a physical explanation (or at least the beginnings of one) for why the timekeeping rates of accelerated clocks are observed to be altered. The periods of their oscillators are altered.

If by "natural order of things" you're talking about one motion being "anomolous" so that that frame is less valid then a more "natural" one, then you're still badly misunderstanding relativity.

I'm not sure what you mean by "valid". We know that when a rock from space hits the earth, it's the rock that was accelerated and not the earth, etc., etc. We know that the twin who took off in the spaceship is the one that accelerated. We know these things because we can identify anomalies.

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

 

[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.

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.

You say that it makes no difference who accelerated initially. The point is, we *know* who accelerated and who didn't. There's a physical explanation (or at least the beginnings of one) for why the timekeeping rates of accelerated clocks are observed to be altered. The periods of their oscillators are altered.

What is this "physical explanation"? Do you agree that the question of who accelerates initially has no effect on the outcome of whose clock is behind when they reunite? 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?

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.

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 know that when a rock from space hits the earth, it's the rock that was accelerated and not the earth, etc., etc. We know that the twin who took off in the spaceship is the one that accelerated. We know these things because we can identify anomalies.

What do you even mean by "anomaly" in this context? If the space station is the one that fires its rockets initially in my above example, would you say that the space station's motion is "anomalous" rather than the ship's? If so, why should we care whose motion is anomalous, when it makes no difference to the answer to whose clock is behind when they reunite?

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. And again, you've given no justification for why we should say the frame of the object that accelerates initially is less objectively correct than the frame of the object that didn't. It just seems like an arbitrary rule you've made up, I could equally well say "the frame of a particle moving at 0.5c relative to the rest frames of the objects in the initial state shall hereby be declared the 'natural' frame to use, and all other frames will be declared 'anomalous'". You also ignored my point earlier about the choice of which state to label as the "initial" one being equally arbitrary--for example, if you treat the initial state as the one where the rocket and the station are at rest with respect to each other then you'd apparently say if the rocket is the one that accelerates, the rocket's motion is anomalous so its frame is less objectively correct...but what if I then told you that at an even earlier time, both the rocket and the station were moving at a different constant velocity, then they accelerated together to the frame you were previously labelling the "initial" one? Would you then say that this even earlier rest frame was the "natural" one to use, and that after they both accelerated they both had anomalous motion?

 

[DrGreg]

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 shows an earlier time than 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.

 

[Sam Woole]

No, the difference is not constant. Each frame observes the other clock as ticking slower than its own, not just being behind its own by a constant amount but ticking at the same rate. For example, suppose two clocks are moving at 0.8c relative to each other, and they start out at the same position at which point they both read "0 seconds". Then each clock will observe the other run at 0.6 times its own rate--so in the first clock's frame, when the first clock reads 4 seconds the second clock reads 2.4 seconds, when the first clock reads 5 seconds the second reads 3 seconds, when the first clock reads 6 seconds the second clock reads 3.6 seconds, and so forth. Likewise, in the second clock's frame, when the second clock reads 4 seconds the first reads 2.4 seconds, when the second clock reads 5 seconds the first reads 3 seconds, and so forth. That is what is meant by "symmetrical" here. .

JesseM, though you said the difference is not constant, yet your demonstrating numbers indicated that it is constant, constantly 0.6 seconds behind, both clocks ticking at the same rate.

I think it must be constant. If it is not, it would mean that the 0.6 seconds would become smaller and smaller and eventually to 0 and even to negative numbers. 0 and negative numbers do exist in math, but what do they mean to clocks? Applying this scenario to the twin paradox, the traveling twin might become younger than 0 years if he does not return to earth.

Again, if you're having trouble understanding how this is possible, I suggest you take a careful look at the diagrams from the example I provided in An illustration of relativity with rulers and clocks.

I did try but I am sorry I could not understand it at all.
I think we should solve the time dilation problem first.

It doesn't sound like he's challenging the math in that quote, just the theory. Can you provide a quote from his webpage or his book saying that there is some self-contradiction in the math itself?

For contradiction in Einstein's math, you can go to: <http://members.aol.com/crebigsol/section1A.html>.

I did try to move its text over to you but somehow my computer did not work on me. I must declare to you that I am not fluent in mathematics and therefore I do not know which math is correct, Einstein's or his opponents'.

 

[jtbell]

For contradiction in Einstein's math, you can go to:

http://members.aol.com/crebigsol/section1A.html

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.

 

[DrGreg]

I must declare to you that I am not fluent in mathematics.

Here is an example of the twins “paradox”, derived from first principles, direct from Einstein’s postulates and without quoting any relativistic formulas. In fact, no algebra at all!

Alice and Bob are both on Earth and synchronize their clocks to each read 12:00. Alice immediately travels away from Bob at constant velocity. As she travels, she looks through a telescope at Bob’s clock and compares it to her own. As it takes time for light to travel from Bob to Alice, Alice sees Bob’s time as being earlier than her own time.

Suppose at 12:15, by her clock, she sees Bob’s clock to read 12:10. From Alice’s point of view, it appears as though it has taken 5 minutes for the light to travel from Bob to Alice. At 12:30, by Alice’s clock, the distance will have doubled, so the delay will have doubled, too, and she sees Bob’s clock read 12:20.

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

From Alice’s point of view, it looks as though Bob’s clock is ticking at ²/₃ of the rate of her own clock. Of course, this is an illusion – it is nothing more or less than the Doppler effect, caused by the delay of the light signals.

Now consider the point of view of Ted who is stationary relative to Bob and has been waiting for Alice at the point where she arrives at 13:00 by her clock.

Bob------------------Ted
        Alice->

Ted’s clock is not synchronized to Bob’s – he is in his own time zone. When Alice arrives at 13:00 by her clock, Ted’s clock reads, say, 18:00. You already know that, at this time, Alice and Ted both see Bob’s clock to read 12:40. So Bob’s clock looks to be 5 hours 20 minutes slow from Ted’s point of view. As Bob and Ted are a constant distance apart, this delay will be constant.

You can now see what things look like from Ted’s point of view.

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

From Ted’s point of view, it looks as though Alice’s clock is ticking at ³/₂ of the rate of his own clock.

Now Alice turns round and returns to Bob at the same speed that she traveled away from earlier. As it took an hour of her time to make the outward journey, it must also take an hour of her time to cover the same distance back at the same speed.

This is where I apply the principle of Relativity and say that Alice’s view of Bob during the return journey must be like Ted’s view of Alice during the outward journey. In each case, the person being observed is moving at the same speed towards the person making the observation.

So I conclude that, from Alice’s point of view, it looks as though Bob’s clock is ticking at ³/₂ of the rate of her own clock, as follows:

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”.

You can also look at this from Bob’s point of view. Again, applying the principle of Relativity, during the outward journey, Bob sees Alice’s clock is ticking at ²/₃ of the rate of his own clock. During the return journey, Bob sees Alice’s clock is ticking at ³/₂ of the rate of his own clock.

Bob’s   Bob’s
own     view of
clock   Alice’s clock
-----   -----
12:00   12:00
12:15   12:10
12:30   12:20
12:45   12:30
13:00   12:40
13:15   12:50
13:30   13:00
-----   -----
13:30   13:00
13:40   13:15
13:50   13:30
14:00   13:45
14:10   14:00
-----   -----

Same result!

Note that I haven’t mentioned time dilation. The only assumptions I have made, I think, are that Alice’s speed relative to Bob is the same as Bob’s speed relative to Alice, and that Doppler shifts obey Einstein’s postulates of Relativity.



References: For those who don’t know, the method above derives from Sir Hermann Bondi’s http://www.geocities.com/ResearchTriangle/System/8956/Bondi/".)

 

[JesseM]

JesseM, though you said the difference is not constant, yet your demonstrating numbers indicated that it is constant, constantly 0.6 seconds behind, both clocks ticking at the same rate.

"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.

I did try but I am sorry I could not understand it at all.
I think we should solve the time dilation problem first.

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?

For contradiction in Einstein's math, you can go to: <http://members.aol.com/crebigsol/section1A.html>.

I did try to move its text over to you but somehow my computer did not work on me. I must declare to you that I am not fluent in mathematics and therefore I do not know which math is correct, Einstein's or his opponents'.

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:

And if you think there were, why do you think the mathematics community wasn't convinced? Do you think they're all idiots who can't follow a simple proof showing a contradiction? Remember, once something is proved in mathematics then its completely cut-and-dried, there's no room for differences of opinion--if it's been proved, any mathematician competent to evaluate the proof should agree.

 

[Sherlock]

Quote:
Originally Posted by Sherlock
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.
Quote:
Originally Posted by Sherlock
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. But all observers (as you note above) agree that that isn't what happens. What they do agree on is that Clock A (the accelerated clock) has recorded less time for the trip. (If follows that clock A's period of oscillation has undergone an average dilation during the trip.)
Quote:
Originally Posted by Sherlock
You say that it makes no difference who accelerated initially. The point is, we *know* who accelerated and who didn't. There's a physical explanation (or at least the beginnings of one) for why the timekeeping rates of accelerated clocks are observed to be altered. The periods of their oscillators are altered.

What is this "physical explanation"? Do you agree that the question of who accelerates initially has no effect on the outcome of whose clock is behind when they reunite? 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. 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, 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.

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. Bodies that have accelerated away from and back to some reference state of motion have undergone physical changes proportional to the accelerations. These physical changes (eg. to the periods of oscillators) entail that the values of time and length units are different for bodies in different states of motion. The Lorentz transformations allow for an effective translation between statements made using one set of unit values and statements made using another set of unit values. The fact that these transformations don't require referencing the specific interaction(s) that produce the changes indicates that (in empty space at least) the changes are due to interaction with some more or less uniform physical phenomenon. And nobody knows exactly what that is yet. But to say that it doesn't exist, or that Relativity says that it doesn't exist would be incorrect. Relativity simply doesn't *require* it, and *because* Relativity doesn't require it but still makes correct predictions, then we might make some, perhaps interesting, inferences-conjectures about it's general properties.

I've been told that this forum is not the place to speculate about the properties or contents of space. I'm not going to do that. I'm just suggesting that some relativistic considerations seem to lead to certain conclusions. I've been trying to lay out what one of those considerations is and what conclusion it seems to lead to. The net effect of this, for me at least, is that I'm learning a bit more of the nuts and bolts of Relativity at a fairly steady, if slow, pace.
Quote:
Originally Posted by Sherlock
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. 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. 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. And we *can* say with a pretty high degree of certainty that it's clock A that has accelerated and not clock B.

... why should we care whose motion is anomalous, when it makes no difference to the answer to whose clock is behind when they reunite?

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.
Quote:
Originally Posted by Sherlock
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. 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?

And again, you've given no justification for why we should say the frame of the object that accelerates initially is less objectively correct than the frame of the object that didn't.

It just seems like an arbitrary rule you've made up, I could equally well say "the frame of a particle moving at 0.5c relative to the rest frames of the objects in the initial state shall hereby be declared the 'natural' frame to use, and all other frames will be declared 'anomalous'". You also ignored my point earlier about the choice of which state to label as the "initial" one being equally arbitrary--for example, if you treat the initial state as the one where the rocket and the station are at rest with respect to each other then you'd apparently say if the rocket is the one that accelerates, the rocket's motion is anomalous so its frame is less objectively correct...but what if I then told you that at an even earlier time, both the rocket and the station were moving at a different constant velocity, then they accelerated together to the frame you were previously labelling the "initial" one? Would you then say that this even earlier rest frame was the "natural" one to use, and that after they both accelerated they both had anomalous motion?

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.

 

[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?