Time Dilation Question: A vs. B Clocks in Motion Experiment

[Zanket]

At the moment they pass each other, both clocks read zero.

Agreed.

Since they are in inertial frames, the time dilation effect is symmetric, as usual: each sees the other clock run slow.

It is true that when inertial observers pass right by each other, each observes the other’s clock to run slow and by the same factor. However, their clocks do not necessarily elapse the same time as one traverses a given distance in the other’s frame, which is what is being measured in Thomas2’s thought experiment.

If the clocks start at that moment, how could any prior acceleration make any difference whatsoever?

Because acceleration length-contracts space, which stays contracted when the acceleration stops, affecting measurements (like elapsed time) in the inertial frame. I gave examples of this above, but let’s do another and I’ll elaborate:

Suppose you travel to Proxima Centauri (hereafter PC), 4 light years away. You’ll need to accelerate to get there, but you want to traverse the distance between Earth and PC inertially. So you launch from Earth and accelerate in a big loop so that you have the desired final velocity v just as you pass Earth toward PC, at which moment you shut down your engine and start your clock to elapse your time during your inertial trip to PC. At this moment clocks in the Earth-PC system also start ticking.

Let v = 0.5c. Then the time dilation or length contraction multiplier is sqrt(1 - 0.5^2) = 86.6%. You have velocity v relative to the whole Earth-PC system, which includes the space between Earth and PC, so the distance between Earth and PC as you measure it is length-contracted. The distance that you measure between yourself/Earth and PC is (4 * 86.6%) = 3.46 light years. At 0.5c you’ll cross a gulf of 3.46 light years in (3.46 ly / 0.5 ly/yr) = 6.93 years. That’s the time that your clock will elapse during the trip.

Clocks at rest with respect to the Earth-PC system are in the same inertial frame. Those beside these clocks observe no length contraction of the space between Earth and PC. Hence an observer next to either clock will find it to elapse (4 ly / 0.5 ly/yr) = 8 years during your trip.

When you pass right by an Earth-PC system observer you find that their clock runs at an 86.6% rate and they find that your clock runs at an 86.6% rate, yet during your trip your clock elapses 6.93 years while Earth-PC clocks elapse 8 years. The situation is not symmetrical even though both frames are inertial.

Now let’s relate the example above to Thomas2’s thought experiment. First simplify the diagram as I gave above:

Put clocks directly at the mechanical contacts. Then there is no time delay to stop or start the clocks. In this arrangement B has 2 clocks, but we can let them be synchronized since B’s frame is inertial. Next get rid of A’s rod, which is irrelevant. A can be just a clock with a mechanical contact.

Make B’s 2 clocks a clock on Earth and a clock at PC respectively. Make A your clock. A elapses less time than B.

Put B’s 2 clocks at either end of your rocket. Make A an Earth clock. Now B elapses less time than A.

 

[Doc Al]

I suspect we are merely arguing semantics.

Since they are in inertial frames, the time dilation effect is symmetric, as usual: each sees the other clock run slow.

It is true that when inertial observers pass right by each other, each observes the other’s clock to run slow and by the same factor. However, their clocks do not necessarily elapse the same time as one traverses a given distance in the other’s frame, which is what is being measured in Thomas2’s thought experiment.

When I said that the SR effects are symmetric, I was making the trivial point that both inertial frames see the same synchronization, dilation, and contraction effects: each has the same right to apply the Lorentz transformations. Neither frame is privileged because it was the one that "really accelerated". Of course, if the situation is not symmetric they get different results. This is the case in Thomas2's thought experiment.

If the clocks start at that moment, how could any prior acceleration make any difference whatsoever?

Because acceleration length-contracts space, which stays contracted when the acceleration stops, affecting measurements (like elapsed time) in the inertial frame. I gave examples of this above, but let’s do another and I’ll elaborate:

If you are saying that in order for two things that once were in the same frame to move with respect to each other, they must accelerate, then I of course agree. But once they have attained their relative velocity, the two frames are equally inertial. How they attained their speed is irrelevant: how they accelerated, or which one accelerated, makes no difference.

Your replacing Thomas2's single clock B with two clocks is perfectly OK, but I like the original version better: Clocks A and B start and stop, each reading a single unambiguous time interval. No discussion or debate: Which clock reads the greater time, A or B?

The answer does not depend on knowing which clock "really accelerated" in attaining their relative velocity.

In Thomas2's gedanken, both frames are measuring the time it takes Clock A to traverse the length of rod B. In frame A, this is a simple time measurement made by single stationary clock. Of course, frame B sees clock A as moving and thus measures a greater time by a factor of \gamma. Clock B always reads greater than clock A.

(Now I understand why Thomas2 didn't care about the relative lengths of the rods in his diagram. It's irrelevant to answering his question.)

 

[Janus]

In Thomas2's gedanken, both frames are measuring the time it takes Clock A to traverse the length of rod B. In frame A, this is a simple time measurement made by single stationary clock. Of course, frame B sees clock A as moving and thus measures a greater time by a factor of \gamma. Clock B always reads greater than clock A.

(Now I understand why Thomas2 didn't care about the relative lengths of the rods in his diagram. It's irrelevant to answering his question.)

Which brings up the interesting question: Why even include rod A? Just place Clock A directly at the trigger point.

If the reason is so that clock B and A are opposite each other in A's frame when the start triggers are activated, what purpose will that serve? It does not mean that the clocks will both start while opposite each other; signal travel issues between triggers and clocks prevent that.

 

[Doc Al]

Which brings up the interesting question: Why even include rod A? Just place Clock A directly at the trigger point.

Right. It serves no purpose. (I think that was why Zanket got rid of it in his version.)

 

[Zanket]

Of course, if the situation is not symmetric they get different results. This is the case in Thomas2's thought experiment.

How they attained their speed is irrelevant: how they accelerated, or which one accelerated, makes no difference.

Don’t these 2 quotes conflict? I showed above that how they attained their speed does make a difference; that is, whether or not the situation is symmetric depends upon how they attained their speed. The first quote seems to agree. The second quote seems to say the opposite.

 

[Doc Al]

Don’t these 2 quotes conflict? I showed above that how they attained their speed does make a difference; that is, whether or not the situation is symmetric depends upon how they attained their speed. The first quote seems to agree. The second quote seems to say the opposite.

They don't conflict at all. What I call the "lack of symmetry" between what frame A measures and what frame B measures has nothing whatsoever to do with how they attained their relative speed.

I still don't understand why you think that the answer to Thomas2's thought experiment somehow depends on the particulars of how the clocks were accelerated. The difference in the times measured by clocks A and B is easily calculated--without any knowledge of how the two clocks may have been accelerated.

When I say that the situations in the two frames "lack symmetry", here's the kind of thing that I mean. As an example of a symmetric situation, say two rods of equal proper length are passing each other. Each frame measures the length of the other's rod: the situation is symmetric--frame A measures the length of the B rod; frame B measures the length of the A rod. It would be quite problematic if the measurements turned out to be different.

But in Thomas2's thought experiment the situation is not symmetric (in the sense that I am using the term): frame A measures the time it takes for clock A to traverse the length of rod B; frame B also measures the time it takes for clock A to traverse the length of rod B. They of course get different answers.

But symmetric or not, the answer has nothing whatsoever to do with how the two clocks were accelerated. Don't know, don't care.

 

[Zanket]

I still don't understand why you think that the answer to Thomas2's thought experiment somehow depends on the particulars of how the clocks were accelerated.

Well, I showed by example that the answer depends on how they accelerated. What part of the example do you disagree with?

 

[Doc Al]

Well, I showed by example that the answer depends on how they accelerated. What part of the example do you disagree with?

Show me where your example depends on how they accelerated. I don't see it.

In your version, you replace B's single clock with two clocks. No problem. But the frame with 2 clocks will always measure a greater time. If you disagree, show me the step in your analysis where you input the acceleration history.

You had stated:

When you pass right by an Earth-PC system observer you find that their clock runs at an 86.6% rate and they find that your clock runs at an 86.6% rate, yet during your trip your clock elapses 6.93 years while Earth-PC clocks elapse 8 years. The situation is not symmetrical even though both frames are inertial.

This I agree with completely.

And you go on to state:

Make B’s 2 clocks a clock on Earth and a clock at PC respectively. Make A your clock. A elapses less time than B.

So far, so good.

And then:

Put B’s 2 clocks at either end of your rocket. Make A an Earth clock. Now B elapses less time than A.

You state this without proof. This is the statement I challenge.

 

[Zanket]

Put B’s 2 clocks at either end of your rocket. Make A an Earth clock. Your clock elapses time at (6.93 yr / 8.00 yr) = 86.6% of the rate of Earth-PC clocks. Let your rocket be such length that it take 0.866 proper nanoseconds for your rocket to pass A, which is started when the mechanical contact at the top of your rocket hits it, and is stopped when the mechanical contact at the bottom of your rocket hits it. Then A elapses 1 nanosecond. Now B elapses less time than A.

 

[Doc Al]

Would you agree that the rocket has some proper length L? And thus, from the rocket's viewpoint it takes the rocket a time T_B = L/v to pass (front to rear) clock A. And what time does Clock A measure for this passing? From the A frame, the rocket only has a length of L/gamma, thus Clock A measures a time of only T_A = (L/gamma)/v = T_B/gamma. Clock A always measures a smaller time that clock B.

A rocket can in principle traverse between these galaxies in an arbitrarily short proper time, while clocks in the galaxies elapse at least 1 million years.

Right, but only if you reverse the situation that we are discussing! Put synchronized clocks at each galaxy. Then have the rocket (with its single clock) move from one galaxy to another. The time for that rocket to pass between those galaxies will always be measured as greater according to the frame of the galaxies! Just like the time for Clock A to traverse rod B will always be greater according to frame B.

It doesn't matter whether the two clocks are at the ends of a huge rocket, or on different galaxies: the time measured for another clock (in another frame) to traverse from one end to the other will always be greater according to the two-clock frame. (And it certainly doesn't matter how the frames were accelerated. )

 

[Zanket]

Shoot, you jumped online just as I updated my post to have a much simpler proof.

 

[Doc Al]

Shoot, you jumped online just as I updated my post to have a much simpler proof.

I'll look at it in a few minutes.

 

[Doc Al]

Put B’s 2 clocks at either end of your rocket. Make A an Earth clock. Your clock elapses time at (6.93 yr / 8.00 yr) = 86.6% of the rate of Earth-PC clocks.

I take it you want gamma = 1/.866 = 1.1544. Realize that clock dilation is completely symmetric: If we see their clocks slow by a factor of gamma, they see ours slow by a factor of gamma.

Let your rocket be such length that it take 0.866 proper nanoseconds for your rocket to pass A, which is started when the mechanical contact at the top of your rocket hits it, and is stopped when the mechanical contact at the bottom of your rocket hits it.

OK. The B clocks measure L/v = 0.866 nanoseconds.

Then A elapses 1 nanosecond. Now B elapses less time than A.

Nope. The A clock measures 0.866/gamma = 0.75 nanoseconds. (For details, see my previous post.)

 

[Zanket]

Nope. The A clock measures 0.866/gamma = 0.75 nanoseconds. (For details, see my previous post.)

You are right. Your math helped me to see this, and to see where I got off track. I agree, A<B always and prior acceleration doesn’t matter. I learned something new today. Thanks!

 

[Doc Al]

My pleasure. Now I wonder what happened to Thomas2?

 

[Thomas2]

Graphics modified

Please note that I modified the original graphic so that the design is now perfectly symmetric between A and B (see http://www.physicsmyths.org.uk/imgs/timedilation.gif ). This should make some of the replies in this thread irrelevant. (the old graphic is now under http://www.physicsmyths.org.uk/imgs/timedilation0.gif ).

There should be now not only an absence of any acceleration or change of reference frame (as with the old design already) but the complete symmetry should make it impossible to single out any of the frames as the preferred one, i.e. the twin paradox can not be resolved and time dilation should hence be logically impossible.

 

[Doc Al]

Please note that I modified the original graphic so that the design is now perfectly symmetric between A and B (see http://www.physicsmyths.org.uk/imgs/timedilation.gif ). This should make some of the replies in this thread irrelevant. (the old graphic is now under http://www.physicsmyths.org.uk/imgs/timedilation0.gif ).

Do the two rods have the same proper length? (If so, then your diagram doesn't show the length contraction.) If not, then the situation is not symmetric. The clock reading depends on the measured length of the other rod: For example: \Delta t_B = (l_A/\gamma + l_B)/v; where \Delta t_B is the time recorded on clock B, l_A is the proper length of rod A, and l_B is the proper length of rod B.

Just for argument's sake I will assume that your diagram is inaccurate and that you meant for l_A = l_B. In that case, yes, both clocks read the same. So?

There should be now not only an absence of any acceleration or change of reference frame (as with the old design already) but the complete symmetry should make it impossible to single out any of the frames as the preferred one, i.e. the twin paradox can not be resolved and time dilation should hence be logically impossible.

I have no idea why you think that this has something to do with the "twin paradox" or why you think that time dilation would not apply, as usual. While the two clocks end up recording the same time, the two frames disagree that the two clocks started and stopped at the same time. For example, frame B observers say that clock A started ticking before clock B and that clock A was still ticking after clock B was stopped. (Of course, frame A observers say the same thing about clock B.)

 

[Thomas2]

Do the two rods have the same proper length? (If so, then your diagram doesn't show the length contraction.)

Which rod would then be length contracted?

While the two clocks end up recording the same time, the two frames disagree that the two clocks started and stopped at the same time.

How could that possibly be? How could two colliding cars for instance disagree about having collided at the same time? The clocks are effectively started here here through a collision and stopped through a collision.

For example, frame B observers say that clock A started ticking before clock B and that clock A was still ticking after clock B was stopped. (Of course, frame A observers say the same thing about clock B.)

i.e. we would have a twin paradox.

 

[Doc Al]

Do the two rods have the same proper length? (If so, then your diagram doesn't show the length contraction.)

Which rod would then be length contracted?

That depends on whose viewpoint you wish to illustrate. Since your diagram shows an arrow (labeled v) illustrating the speed of clock B, I assume the diagram is from the viewpoint of the clock A frame (clock A is at rest). In which case rod B must show length contraction.

While the two clocks end up recording the same time, the two frames disagree that the two clocks started and stopped at the same time.

How could that possibly be? How could two colliding cars for instance disagree about having collided at the same time? The clocks are effectively started here here through a collision and stopped through a collision.

When two objects collide, everyone agrees that they collide "at the same time". (Otherwise, how could they collide? ) But in your example, the clocks do not start or stop at the moment of the mechanical collision. Since the clocks are not collocated with the collisions, it takes time for the signals to reach them.

For example, frame B observers say that clock A started ticking before clock B and that clock A was still ticking after clock B was stopped. (Of course, frame A observers say the same thing about clock B.)

i.e. we would have a twin paradox.

Where's the paradox? Not everything is a paradox. For example: Frame A says that rod B is contracted... But frame B says that rod A is contracted! Paradox? No... just standard relativity describing the relationship between observations made in two inertial frames.

You need to find out what the term "twin paradox" refers to.

 

[HallsofIvy]

The whole point of the thought experiment as suggested by me (see illustration) is that no accelerations occur at all. Both observers move with constant speed v relatively to each other and a mutual mechanical contact starts and stops the clocks.

In the examples in my post, the clocks are in inertial frames when they are started and stopped and during; in this way the experiments match that proposed by Thomas2. The examples show that the history of acceleration (prior to a clock starting) does affect the elapsed times on the clocks. Take a close look at the examples and see if you can find anything wrong with the conclusions.

Why do you keep repeating that? No one has disagreed with that. They have asserted that the situation is NOT symmetric because exactly what happens depends upon whether you diagram is drawn from the point of view of frame A or B and because the turning off and on of the clocks is NOT simultaneous in both frames. That has nothing to do with acceleration.

 

[Thomas2]

When two objects collide, everyone agrees that they collide "at the same time". (Otherwise, how could they collide? ) But in your example, the clocks do not start or stop at the moment of the mechanical collision. Since the clocks are not collocated with the collisions, it takes time for the signals to reach them.

Yes but this is merely a constant offset (if any) depending on the proper length of the rods and the location of the clocks and the signal propagation speed (e.g. the speed of sound within each rod).

Where's the paradox? Not everything is a paradox. For example: Frame A says that rod B is contracted... But frame B says that rod A is contracted! Paradox? No... just standard relativity describing the relationship between observations made in two inertial frames.

Likewise you couldn't agree about the frame in which the time dilation is supposed to occur, i.e. we have a ambiguous (paradoxical) situation.

 

[Doc Al]

time dilation works both ways---as always

When two objects collide, everyone agrees that they collide "at the same time". (Otherwise, how could they collide? ) But in your example, the clocks do not start or stop at the moment of the mechanical collision. Since the clocks are not collocated with the collisions, it takes time for the signals to reach them.

Yes but this is merely a constant offset (if any) depending on the proper length of the rods and the location of the clocks and the signal propagation speed (e.g. the speed of sound within each rod).

It's only a constant offset for the clock within a given frame. Frame A observers will agree that the travel time for the signal to start and stop clock A is a constant offset and can be ignored. But frame A observers will not agree that the travel time (measured in frame A) for the signals to start and stop clock B is a constant offset.

Where's the paradox? Not everything is a paradox. For example: Frame A says that rod B is contracted... But frame B says that rod A is contracted! Paradox? No... just standard relativity describing the relationship between observations made in two inertial frames.

Likewise you couldn't agree about the frame in which the time dilation is supposed to occur, i.e. we have a ambiguous (paradoxical) situation.

What are you talking about? There is nothing ambiguous or paradoxical here. Time dilation would be observed by both frames, of course, as always. Frame A sees clock B running slow, and vice-versa.

You seem to think that the clock times recorded in your setup somehow should reflect "time dilation" in a simple way. Not so! To illustrate time dilation you would need to have frame A measure the start and stop time (according to frame A clocks) of clock B and then compare that time interval to the elapsed time on clock B. If your setup did this, then you would find that if clock B says that \Delta t_B has elapsed, then frame A will say that the elapsed time according to frame A clocks is \gamma \Delta t_B because frame A sees clock B as running slow.

Of course, the same argument works for frame B observing clock A.