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

[Thomas2]

Consider the following thought experiment: two clocks in inertial frames A and B are moving relatively to each other with speed v and both clocks are stopped and started through mechanical contacts at the end of the rod to which the clocks are mounted (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ).

What times do both clocks show after they have been stopped? A<B, A>B or A=B ? (Note: it shouldn't matter if A or B or both turn around to compare the times as the clocks are already stopped then).

 

[Janus]

Consider the following thought experiment: two clocks in inertial frames A and B are moving relatively to each other with speed v and both clocks are stopped and started through mechanical contacts at the end of the rod to which the clocks are mounted (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ).

What times do both clocks show after they have been stopped? A<B, A>B or A=B ? (Note: it shouldn't matter if A or B or both turn around to compare the times as the clocks are already stopped then).

Question: Is this diagram taken from the frame of A, B or another frame? This is important be if it is from A then From B, A will be length contracted such that the ends will not align and if it is from B then A will see B as length contracted so that the ends will not align. Also, where are the clocks located on these rods?, in the middle?

You also must realize that the clocks cannot start or stop the instant the contacts are triggered. The information that the contact has been activated cannot get from the contact to the clock at any speed greater than c.

 

[Thomas2]

Question: Is this diagram taken from the frame of A, B or another frame? This is important be if it is from A then From B, A will be length contracted such that the ends will not align and if it is from B then A will see B as length contracted so that the ends will not align.

A length contraction of either rod should be irrelevant for my question as both clocks simultaneously start at the first contact and stop at the second.

You also must realize that the clocks cannot start or stop the instant the contacts are triggered. The information that the contact has been activated cannot get from the contact to the clock at any speed greater than c.

The time it takes to relay the information about the contact to the clock merely produces a constant offset which is the same in both systems (if the clocks are in the middle of the rod) as the signal propagates in each reference frame independently.

So my question still stands: do both clocks show different or identical times when they are compared afterwards?

 

[mijoon]

A length contraction of either rod should be irrelevant for my question as both clocks simultaneously start at the first contact and stop at the second.....

The word "simultaneously" is meaningless in this context. Plotting this experiment on a Minkowski diagram will be enlightening.

 

[Thomas2]

The word "simultaneously" is meaningless in this context. Plotting this experiment on a Minkowski diagram will be enlightening.

It's only meaningless becaus for the experiment considered (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ) both clocks are started and stopped simultaneously by definition (when two point particles collide, they have identical space and time coordinates by definition).

This still doesn't answer my question though.

 

[selfAdjoint]

Consider the following thought experiment: two clocks in inertial frames A and B are moving relatively to each other with speed v and both clocks are stopped and started through mechanical contacts at the end of the rod to which the clocks are mounted (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ).

What times do both clocks show after they have been stopped? A<B, A>B or A=B ? (Note: it shouldn't matter if A or B or both turn around to compare the times as the clocks are already stopped then).

The clocks at the two ends of the rod are spacelike separated, and there is no simulaneity between them. In all prblems with different points on a rod or different locations in a spaceship you have to be aware of this important relativistic ansatz.

 

[Doc Al]

A length contraction of either rod should be irrelevant for my question as both clocks simultaneously start at the first contact and stop at the second.

Of course length contraction is relevant. As Janus points out, your diagram is ambiguous as it's not clear what viewpoint the diagram is from. My guess is that you mean for the rods to be identical--they have the same proper length.

So, I assume you intend identical clocks centered on identical rods.

The time it takes to relay the information about the contact to the clock merely produces a constant offset which is the same in both systems (if the clocks are in the middle of the rod) as the signal propagates in each reference frame independently.

Right.

So my question still stands: do both clocks show different or identical times when they are compared afterwards?

If I understand your setup correctly, the clocks will show different times. This should be no surprise, as the arrangement is not symmetric: the trigger for starting & stopping the clocks occurs at the same place in the clock A frame, but at different places in the clock B frame.

 

[Thomas2]

Of course length contraction is relevant. As Janus points out, your diagram is ambiguous as it's not clear what viewpoint the diagram is from

The http://www.physicsmyths.org.uk/imgs/timedilation.gif" is obviously drawn from the viewpoint of reference frome A, but as velocities are relative this should be immaterial. You could as well have B resting and A moving or both moving (as long as the relative speed between A and B is v)

If I understand your setup correctly, the clocks will show different times. This should be no surprise, as the arrangement is not symmetric: the trigger for starting & stopping the clocks occurs at the same place in the clock A frame, but at different places in the clock B frame.

If the clocks are in the middle of the rods this shouldn't make any difference as it takes the same time for the trigger signal to reach the clock from either side of the rod.
As the situation is consequently symmetric, it would in my opinion therefore be a logical contradiction if the clocks show different times.

 

[Zanket]

First simplify the diagram. 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.

The puzzle is analogous to the twin paradox, in which the answer depends upon how A and B accelerated to attain their velocity relative to one another.

Where A & B were initially at rest with respect to one another and acceleration is non-inertial: If they accelerated symmetrically to attain v relative to each other then A=B. If B remained inertial while A accelerated to attain v, then A<B. If A remained inertial while B accelerated to attain v, then A>B. The clock of whoever accelerated “the most” elapses less time.

Here are rough examples:

Let A & B be rockets that accelerate identically and directly toward each other from the Milky Way and Andromeda galaxies, respectively. Let both rockets attain v relative to each other at the moment of first contact. Then the situation is symmetrical, so A=B in this case.

Let the clocks on B be clocks at the Milky Way galaxy and Andromeda galaxy, respectively. Let A be a rocket launched from Earth that accelerated in a giant loop to attain v as it passes the Earth in a trip to the Andromeda galaxy. 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. So A<B in this case.

Let B be a rocket launched from rest relative to earth, accelerate in a giant loop to have attained v and have synchronized clocks as the bottom of rocket passes Earth in a trip to the Andromeda galaxy. Let the length of the rocket, as measured by us on Earth at this moment (when the rocket is length-contracted), be the distance between the Milky Way and Andromeda galaxies (so the rocket straddles the galaxies at this moment from our perspective, or from the perspective of someone in the Andromeda galaxy). Let A be a clock in the Andromeda galaxy. 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. So B<A in this case.

 

[Doc Al]

The http://www.physicsmyths.org.uk/imgs/timedilation.gif" is obviously drawn from the viewpoint of reference frome A, but as velocities are relative this should be immaterial. You could as well have B resting and A moving or both moving (as long as the relative speed between A and B is v)

OK, so the diagram is drawn from the viewpoint of reference frame A. Not only is this relevant, it is critical. Your diagram shows that the A frame measures the B rod as being equal in length to the A rod. Which obviously means that the proper length of the B rod is $\gamma L$, where L is the proper length of the A rod. The two rods are not identical. And if you drew the diagram from B's frame, it would look very different.

Your setup depends on the rods having the precise relative velocity needed to make the contracted length of the B rod equal to the proper length of the A rod. (I liked my version better!)

If I understand your setup correctly, the clocks will show different times. This should be no surprise, as the arrangement is not symmetric: the trigger for starting & stopping the clocks occurs at the same place in the clock A frame, but at different places in the clock B frame.

If the clocks are in the middle of the rods this shouldn't make any difference as it takes the same time for the trigger signal to reach the clock from either side of the rod.
As the situation is consequently symmetric, it would in my opinion therefore be a logical contradiction if the clocks show different times.

The situation is still wildly asymmetric for reasons stated previously and because the proper lengths of the rods are different. The clocks still read different times. There is no logical contradiction.

 

[Thomas2]

The puzzle is analogous to the twin paradox, in which the answer depends upon how A and B accelerated to attain their velocity relative to one another.

The whole point of the thought experiment as suggested by me (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ) 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.

 

[Doc Al]

not related to "twin paradox"

The whole point of the thought experiment as suggested by me (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ) 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.

I agree that your thought experiment has nothing to do with the "twin paradox". Nonetheless, as I've stated, your setup is not symmetric and the two clocks read different times.

 

[Zanket]

The whole point of the thought experiment as suggested by me (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ) 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 my examples the observers are moving at constant speed v. They accelerated to get to v prior to the moment that they fit your illustration. Observers don’t get to v magically; they accelerate to it at some point in the history of the universe. How they accelerated relative to each other affects whose clock elapses more time even when they subsequently move at constant speed relative to each other.

 

[Doc Al]

previous acceleration is irrelevant

How they accelerated relative to each other affects whose clock elapses more time even when they subsequently move at constant speed relative to each other.

Since the clocks in Thomas2's thought experiment are in inertial frames when they are started and stopped, I don't see how their history of acceleration can affect the times that they read. (The experiment that you analyzed in your earlier post is very different from the one proposed by Thomas2 in this thread.)

 

[Zanket]

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.

 

[Doc Al]

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.

Your examples all talk about time elapsed on the clocks during an acceleration. Who cares? The clocks aren't even on during that time. But even if they were, so what? We are only interested in the $\Delta t$ that each clock reads during the time that they are in inertial frames. Any offset due to their previous acceleration is irrelevant.

Just like with the twins. Sure, depending on their paths through spacetime, they will have different ages when they reunite. But once reunited, their clocks tick at the same rate once again.

 

[pervect]

The whole point of the thought experiment as suggested by me (see http://www.physicsmyths.org.uk/imgs/timedilation.gif" ) 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.

You need to specify a bit more about how your contacts are working. If they are electrical contacts, you might be able to get the signal transmission up near light speed. If they are actually mechanical rods transmitting a displacement, the signal will travel down the rod at the speed of sound in the rod, which will be a snails pace compared to anything relativistic.

In no case will there be any instantaneous transmission of a signal.

 

[Doc Al]

As far as I can see, the exact mechanism for transmitting the signal from the contact point to the clock doesn't matter. It won't affect the answer to the thought experiment.

 

[Zanket]

Your examples all talk about time elapsed on the clocks during an acceleration.

They do not. All time elapses in an inertial frame in the examples. For instance, the first example says “Let both rockets attain v relative to each other at the moment of first contact.” Once v is attained the frames are inertial. The clocks begin to elapse time at that moment.

Likewise, in the twin paradox, the twins’ clocks can elapse time differently while they are in inertial frames moving at constant velocity relative to each other. The twin who previously accelerated non-inertially has the slower clock when both are in inertial frames.

 

[Zanket]

In no case will there be any instantaneous transmission of a signal.

But the time it takes can be negligible. Above I suggested simplifying the experiment by putting the clocks directly at the mechanical contacts (switches). Then the transmission time between switch and clock can be infinitesimally small.

 

[pervect]

As far as I can see, the exact mechanism for transmitting the signal from the contact point to the clock doesn't matter. It won't affect the answer to the thought experiment.

Sure it will. Since the moving rod appears to be the same length as the stationary rod, it must actually be longer in its own frame. So if we consider the propagation delay effects, clock a will require an additional proper time of
$$L / v_{sound}$$ for the signal to propagate down the rod which will increase the reading of its clock by this amount, while clock b will require an additional proper time of $$\gamma \; L / v_{sound}$$ for the signal to propagate. These two factors won't be equal, and in the case where v >> vsound, they will also be the dominant factors in determining the clock's reading. Among other things, this means that control of the speed of sound would be very important for the experimental results - if the two rods did not have exactly the same speed of sound due to slight construction differences, the results would be impacted significantly.

 

[donnie]

=ab

Or

A+b=c

 

[Doc Al]

between inertial frames, SR effects are symmetric

All time elapses in an inertial frame in the examples. For instance, the first example says “Let both rockets attain v relative to each other at the moment of first contact.” Once v is attained the frames are inertial. The clocks begin to elapse time at that moment.

If the clocks start at that moment, how could any prior acceleration make any difference whatsoever? At the moment they pass each other, both clocks read zero. Since they are in inertial frames, the time dilation effect is symmetric, as usual: each sees the other clock run slow.

 

[Doc Al]

propagation delays cancel

Sure it will. Since the moving rod appears to be the same length as the stationary rod, it must actually be longer in its own frame. So if we consider the propagation delay effects, clock a will require an additional proper time of
$$L / v_{sound}$$ for the signal to propagate down the rod which will increase the reading of its clock by this amount, while clock b will require an additional proper time of $$\gamma \; L / v_{sound}$$ for the signal to propagate. These two factors won't be equal, and in the case where v >> vsound, they will also be the dominant factors in determining the clock's reading. Among other things, this means that control of the speed of sound would be very important for the experimental results - if the two rods did not have exactly the same speed of sound due to slight construction differences, the results would be impacted significantly.

Sure there is a propagation delay for the "on" signal to reach the clock, but there's an equal delay for the "off" signal to reach the clock. These propagation delays cancel.

And, yes, the propagation delays are different for each rod. But, since they cancel, it doesn't matter.

 

[Janus]

=ab

Or

A+b=c

Is there supposed to be a point to this post? If so, it eludes me.

 

[pervect]

Sure there is a propagation delay for the "on" signal to reach the clock, but there's an equal delay for the "off" signal to reach the clock. These propagation delays cancel.

And, yes, the propagation delays are different for each rod. But, since they cancel, it doesn't matter.

Hmmm, if clock B is *exactly* in the center of the rod, you're right. A small imbalance (one part in 10,000 would be way too much) error in the placement of clock B would be a problem though.

Let me work through the sequence of events if the rod were moving at velocity v , gamma = sqrt(1-v^2/c^2), and the speed of sound in the rods is Vsound.

Let 2*La be the proper length of rod A, let 2*Lb be the proper length of rod B, let both clocks be in the exact center of the rod. If the drawing is correct, Lb will be 2 * La

1) the impact occurs at the right end of both rods, starting a signal traveling at about Vsound (about 15000 ft/sec in steel) towards clock a and clock b to turn them both on. We'll call this t=0 in both frames

2) In a's frame, at Ta = 2*Lb /(gamma* v) the signal to turn off clock A will start. gamma = 1/sqrt(1-(v/c)^2). If gamma=2, v=.866c, and Lb = 2 foot, this will occur at Ta = 2.3ns, approximating c as 1 ft/ns

3) In b's frame, at Tb=2 * Lb / v = the signal to turn off clock B will start. With the same assumptions, this would occur at Tb=4.6ns

4) At time La/Vsound, the signal to start clock A will finally arrive. If we assume that La = Lb/gamma as the picture shows, and the speed of sound in steel is about 15,000 ft/second, this would occur at around 66,666 ns, when the rod has moved about 10 miles downrage.

5) 2.3ns after this, the signal to stop clock A will arive. at around 66,668 ns.

6) At Lb/vsound in B's frame, the signal to start clock B will arrive. This will occur at about 133,333 ns by clock B. At which point the rod will be 20 miles downrange using B's yardstick. I think this means around 40 miles using A's yardstick.

7) 4.6 ns later, the signal to stop clock b arrives, around 133,338 ns.

So, if no measurement errors occur (good luck), Clock a will read 2.3 ns, and clock B will read 4.6 ns

Basically, it'd be a lot better to get rid of the mechanical contacts :-)

 

[donnie]

Is there supposed to be a point to this post? If so, it eludes me.

Exactly!

 

[Thomas2]

Sure there is a propagation delay for the "on" signal to reach the clock, but there's an equal delay for the "off" signal to reach the clock. These propagation delays cancel.
And, yes, the propagation delays are different for each rod. But, since they cancel, it doesn't matter.

The delays for each rod could be different, but only if their intrinsic lengths or the positions of the clocks on each rod are different. This does however not depend on the relative velocity v of the rods (if the signal propagates with the speed of sound along the rod, the speed of sound (and hence the propagation delay that each clock will show) is referred to each rest frame separately).

 

[Doc Al]

it's a THOUGHT experiment :-)

Hmmm, if clock B is *exactly* in the center of the rod, you're right. A small imbalance (one part in 10,000 would be way too much) error in the placement of clock B would be a problem though.

Give the guy a break, pervect! It's just a highly impractical thought experiment.

Let me work through the sequence of events if the rod were moving at velocity v , gamma = sqrt(1-v^2/c^2), and the speed of sound in the rods is Vsound.

Of course, using sound to transmit the signal greatly increases the sensitivity to error. The mechanical trigger could just as well trigger a light flash that signals the clock.

If $2L$ is the proper length of rod A, then $\gamma 2L$ is the proper length of rod B. When all is done, clock A will read ${2L}/{v}$ and clock B will read ${\gamma 2L}/{v}$

---

Basically, it'd be a lot better to get rid of the mechanical contacts :-)

Agreed!

 

[Doc Al]

The delays for each rod could be different, but only if their intrinsic lengths or the positions of the clocks on each rod are different.

According to your diagram, the proper lengths of the rods are different.

This does however not depend on the relative velocity v of the rods (if the signal propagates with the speed of sound along the rod, the speed of sound (and hence the propagation delay that each clock will show) is referred to each rest frame separately).

Right, the signal delay depends on the signal propagation rate, not on the relative velocity of the rods. But the time that each clock reads does depend on the relative velocity.

 

[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.