Do clocks on Bell’s spaceships stay synchronized?

[Foppe Hoekstra]

TL;DR

If both of Bell’s spaceships have a clock that are synchronised when the spaceships are at rest, will they still be synchronized in the moving frame of the spaceships, after the spaceships are accelerated to a constant speed?

If both of Bell’s spaceships have a clock that are synchronised when the spaceships are at rest, will they still be synchronized in the moving frame of the spaceships, after the spaceships are accelerated to a (the same) constant speed?
(Personally I think they do, but I often seem to think wrong when it comes to RT.)

 

[stevendaryl]

If you have two rockets that are both accelerating identically, with one rocket ahead of the other in the direction they are accelerating then:

• In the "launch" frame (the inertial frame in which they are initially at rest), the clocks in the two rockets will remain synchronized. The distance between the rockets remains constant.
• In the noninertial frame of either rocket, the clock in the front rocket will be running faster than the clock in the rear rocket. The distance between the rockets grows with time.

 

[PeroK]

Summary: If both of Bell’s spaceships have a clock that are synchronised when the spaceships are at rest, will they still be synchronized in the moving frame of the spaceships, after the spaceships are accelerated to a constant speed?

If both of Bell’s spaceships have a clock that are synchronised when the spaceships are at rest, will they still be synchronized in the moving frame of the spaceships, after the spaceships are accelerated to a (the same) constant speed?
(Personally I think they do, but I often seem to think wrong when it comes to RT.)

Suppose we start with a frame in which two clocks, separated by a certain distance $d$, are at rest. If the two clocks are given equal acceleration in this frame, so that they remain the same distance apart and end up at some common velocity $v$, then by symmetry they must remain synchronised in this frame.

This means, however, that in the eventual rest frame of the clocks, they are a distance $\gamma d$ apart and out of sync by the converse of the"leading clocks lag" rule.

 

[Meir Achuz]

A Lorentz transformation shows that when the clocks are each moving at velocity v, there is a time difference
$\Delta t=vL/c^2$. Because they are separated, the clocks run a different rates during acceleration.

 

[Foppe Hoekstra]

Suppose we start with a frame in which two clocks, separated by a certain distance $d$, are at rest. If the two clocks are given equal acceleration in this frame, so that they remain the same distance apart and end up at some common velocity $v$, then by symmetry they must remain synchronised in this frame.

This means, however, that in the eventual rest frame of the clocks, they are a distance $\gamma d$ apart and out of sync by the converse of the"leading clocks lag" rule.

Glad I got that right.
So if these clocks would simultaneously in their own moving frame, print dots on the ground in the rest frame, these dots would for an observer in the rest frame be printed at a distance γ.γd (due to time dilation or "leading clocks lag" as you call it). Right?

 

[PeroK]

Glad I got that right.
So if these clocks would simultaneously in their own moving frame, print dots on the ground in the rest frame, these dots would for an observer in the rest frame be printed at a distance γ.γd (due to time dilation or "leading clocks lag" as you call it). Right?

In the instantaneous rest frame of the clocks, the dots get further apart as the clocks accelerate. But that's because the clocks are getting further apart.

 

[PeterDonis]

In the instantaneous rest frame of the clocks

There is no single "instantaneous rest frame of the clocks" while the clocks are accelerating.

 

[PeterDonis]

simultaneously in their own moving frame

What does "simultaneously" mean when the clocks are not synchronized?

 

[Meir Achuz]

"There is no single "instantaneous rest frame of the clocks" while the clocks are accelerating."

Consider an instant in the original rest frame where each ship has velocity v.
Make a Lorentz transformation with velocity -v.
Each ship will be at rest. Their Minkowski times will be different, but the clock in each ship will have the same reading.

This is what happens for passengers in an accelerating spaceship (Ask any asronaut.)
The passengers will all be at rest in their seats during the acceleration.
Their Minkowski times will be different, but they won't notice this because their clocks will all show the same time.

 

[PeterDonis]

Consider an instant in the original rest frame where each ship has velocity v.
Make a Lorentz transformation with velocity -v.
Each ship will be at rest.

You can't Lorentz transform "at an instant". You have to Lorentz transform the entire spacetime (more precisely, you have to transform from one coordinate chart to another over the entire spacetime). That includes the entire worldlines of the ships.

You can certainly pick some speed $v$ and do a Lorentz transformation by $-v$ from the original frame. What you will get is a new frame in which clock #1 is momentarily at rest at some time $t_1$, and clock #2 is momentarily at rest at some time $t_2 \neq t_1$. Whatever this is, it is certainly not a "single instantaneous rest frame of the clocks".

 

[PeterDonis]

This is what happens for passengers in an accelerating spaceship

The worldlines of the passengers in an accelerating spaceship (more precisely, sitting at different heights in an accelerating spaceship) are not the worldlines of Bell's spaceships. They are worldlines of the Rindler congruence, not the Bell congruence. Their experience is not the same as what is being discussed in this thread.

 

[pervect]

There is an instantaneous co-moving inertial reference frame for every point on the worldline of the acclerating clocks. By context, I assume that is what the other posters were talking about.

 

[PeterDonis]

There is an instantaneous co-moving inertial reference frame for every point on the worldline of the acclerating clocks.

But there is no such frame in which both clocks are at rest at the same coordinate time while the clocks are accelerating.

 

[A.T.]

So if these clocks would simultaneously in their own moving frame

I would suggest clarifying that the printing happens, after the acceleration of both clocks stopped, to avoid the problems with the above specification mentioned by others here. If they both have the same acceleration program, they will both end up moving at the same constant speed v in the launch frame. Then you can re-synch their clocks in their own rest frame, and do the printing simultaneously in their own rest frame.

Is that what you mean?

 

[Foppe Hoekstra]

I would suggest clarifying that the printing happens, after the acceleration of both clocks stopped, to avoid the problems with the above specification mentioned by others here. If they both have the same acceleration program, they will both end up moving at the same constant speed v in the launch frame. Then you can re-synch their clocks in their own rest frame, and do the printing simultaneously in their own rest frame.

Is that what you mean?

Yes indeed. (Otherwise I should have had to specify how simultaneity was obtained. Isn't it?)
But what do you mean by "re-synch"? I just learned (form PeroK) that they are sync (in their own moving frame) when the clocks are at constant speed again.

 

[Foppe Hoekstra]

while the clocks are accelerating.

The printing is suposed to take place after accelerating the clocks, so when they are both at the same constant speed.

 

[PeroK]

Yes indeed. (Otherwise I should have had to specify how simultaneity was obtained. Isn't it?)
But what do you mean by "re-synch"? I just learned (form PeroK) that they are sync (in their own moving frame) when the clocks are at constant speed again.

No. They are out of synch in their "comoving" frame. They remain in synch in the original rest frame.

 

[A.T.]

But what do you mean by "re-synch"? I just learned (form PeroK) that they are sync (in their own moving frame) when the clocks are at constant speed again.

When at constant speed again, they tick at the same rate in their common rest frame, but they have accumulated an offset.

 

[Foppe Hoekstra]

When at constant speed again, they tick at the same rate in their common rest frame, but they have accumulated an offset.

So in their co-moving frame the clocks undergo the same acceleration and thus the same deviation to the original time in the rest frame, but nevertheless they also develop an offset of time in between them (to an observer traveling along in the commoving frame). How is that possible?

 

[Nugatory]

How is that possible?

The same way that the length between them changes in any frame except the original rest frame: They are flying the same acceleration profile according to the ground-based observer, meaning that they change their speed by the same amount at the same time using that frame - but not other frames. For example, they don't both take off at the same time using the final comoving frame.

 

[PeroK]

So in their co-moving frame the clocks undergo the same acceleration and thus the same deviation to the original time in the rest frame, but nevertheless they also develop an offset of time in between them (to an observer traveling along in the commoving frame). How is that possible?

On general, acceleration introduces as asymmetry between the objects being accelerated.

The simplest manifestation is length contraction. The two ships cannot remain the same distance apart in both the initial and final reterence frames.

The key to understanding this is to look at the mathematics of accelerating reference frames. once you see what is happening mathematically gives you a better insight into this asymmetry.

 

[Ibix]

So in their co-moving frame the clocks undergo the same acceleration and thus the same deviation to the original time in the rest frame, but nevertheless they also develop an offset of time in between them (to an observer traveling along in the commoving frame). How is that possible?

It's closely analogous to the following scenario.

Two people start side by side a few meters apart and walk forward, dropping markers every pace. When they have traveled a specified distance, both turn 30° to the right and carry on walking in their new forward direction, still dropping markers.

According to a reference frame that regards their original direction as "forwards" then the walkers' markers are always level with one another. However, according to a frame that regards their new direction as "forwards" the markers are not and never have been level with one another. According to the walkers, they were initially level and stopped being level after the turn. How is this possible? Clearly the answer is that the walkers' definition of "forward", and hence their definition of "level" has changed.

Similarly, Bell's ships' definition of not moving in space, only advancing in time, has changed. Hence their definition of "now" has changed and hence whether their distance markers (clock ticks) are simultaneous has changed.

 

[PeroK]

The key to understanding this is to look at the mathematics of accelerating reference frames. once you see what is happening mathematically gives you a better insight into this asymmetry.

Let's look at the maths. We have an inertial reference frame (IRF) in which two objects an initial distance $d$ apart accelerate identically at a constant acceleration, $a$. In the original IRF, their positions are:

$x_1(t) = \frac12 at^2$ and $x_2(t) = d + \frac12 at^2$

We have, therefore, two events that represent the two objects reaching a speed $v$ relative to the original IRF. And the coordinates of those events in the original IRF are:

$E_1 = (t, x_1(t)), \ E_2 = (t, x_2(t))$

Let's take a second IRF traveling at the speed $v$ relative to the first. And, let's transform the time coordinate of the two events to this IRF. We have:

$t_1' = \gamma(t - \frac{vx_1}{c^2}), \ t_2' = \gamma(t - \frac{vx_2}{c^2})$

But, these times are different! In this IRF we have an asymmetry. The objects are at rest in this reference frame at different times. If $v$ is less than the final velocity, then this frame is never a common instantaneous rest frame for both objects. An observer using this frame would measure the objects coming to rest (relative to this IRF) at different times.

If $v$ is the final velocity, at which point the acceleration stops, then in this IRF one object reaches the final velocity first, then stops accelerating, and then the other object reaches the final velocity later, at which time this becomes the common rest frame for both objects.

If the objects are equipped with clocks, then - by symmetry of the motion in the original IRF - the clocks will remain synchronised in this frame. But, they are not synchronised in the frame defined by their final velocity.

Or, if we look at this another way. Imagine the frame traveling at the final speed $v$ at the beginning of the experiment. In this frame the clocks are not synchronised at the outset. Again, the relativity of simultaneity applies. The clocks start out out of synch. In this frame, the objects start and stop the acceleration phase at different times, but have otherwise identical acceleration profiles. So, at the end of the acceleration the original offset in their times is preserved. The same total time dilation applies to both and so in this frame they end the acceleration phase as they started it: out of synch by the same offset.

Either way you study it, the clocks remain in synch in the original IRF and remain out of synch in the final IRF.

 

[PeterDonis]

How is that possible?

Please read our Insights article on the Bell Spaceship Paradox:

https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/

 

[vanhees71]

For a comparison between the two spaceships a la Bell (i.e., starting from different positions with constant and equal proper acceleration) to two spaceships connected with a Born-rigid rod, which in my opinion resolves the apparent paradox very intuitively, see my SRT article,

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
which is a bit growing (but in the section about electromagnetism rather than mechanics) right now.

 

[stevendaryl]

So in their co-moving frame the clocks undergo the same acceleration and thus the same deviation to the original time in the rest frame, but nevertheless they also develop an offset of time in between them (to an observer traveling along in the commoving frame). How is that possible?

Whether two clocks are synchronized or not is frame-dependent. If two rockets are traveling at constant velocity, and the clocks on the rockets are synchronized in the frame of the Earth, then they will not be synchronized in the frame of the rockets. Why is that?

Well, Einstein gave an operational definition of when two clocks are synchronized in a frame in which they are at rest: Send a light signal from the rear clock to the front clock, and back again. Then there are three relevant events: (1) the signal is sent from the rear clock, (2) the signal is received by the front clock and a reply is sent, (3) the reply reaches the rear clock. The clocks are synchronized if the time of the second event, as shown on the front clock, is halfway between the times of the first and third events.

If we apply Einstein's definition of synchronization to moving rockets, then we find that in the trip from the rear clock to the front clock, the light signal takes longer (according to the Earth's frame) than in the return trip. That's because the front rocket is moving away from the light signal, so it takes longer for the light signal to catch up, and the rear rocket is moving toward the light signal, so it takes less time for the light signal to catch up. So if the clocks in the two rockets are synchronized, according to Earth's frame, then the front clock will be ahead of the rear clock, according to the rocket's frame (that is, the time shown on the front clock will be more than half-way between the first and third events, as described above). If the clocks are synchronized in the Earth's frame, then there will be a positive offset of the front clock in the rocket frame. If the rockets are traveling at constant velocity, then this offset will be constant.

If instead of the rockets traveling at constant velocity, the rockets are accelerating, then the offset will keep growing with time. So the riders in the rocket will interpret this as the front clock getting more and more ahead of the rear clock.

 

[Ibix]

I thought I'd generate some Minkowski diagrams, the rigorous version of the analogy from my earlier post. Here's one in the initial frame of the ships:

There is a red ship half a light year behind a blue ship. They are initially at rest, and they have synchronised clocks that tick every tenth of a year (marked on their worldlines by black crosses). At $t=0$ they both start accelerating with a proper acceleration of 1 ly/y², which is about 1g, until they reach 0.6c compared to their initial state, then they switch off their engines and travel inertially. I've linked the clock ticks with fine grey lines, and you can see that they are simultaneous and the spacing increases as the ships accelerate.

This is the same situation from the final rest frame of the ships:

You can see that the ticks were never simultaneous in this frame, and the ships didn't accelerate simultaneously, which is why the distance changes. The grey lines do get closer together during the acceleration, but you'll probably have to take my word for that unless you've got a ruler.

What I haven't drawn is the point of view of the ships. That isn't inertial, of course, and needs to be drawn separately for the two ships since "during" the acceleration isn't the same block of spacetime for both ships. One set of reasonable simultaneity surfaces is shown in figure 6 of Dolby and Gull's paper on Radar Time.

 

[Foppe Hoekstra]

How is that possible?

Apparently there is a lot more possible in RT then I can imagine. However, I fear, at the cost of a specific possibility I always took for granted. But for overview convenience I’ll open a new thread on that. I sincerely thank you all for the efforts you make to enlighten me and hope to see you at my next post.