Clocks Within Each Ship in Bell Spaceship Paradox

[stevendaryl]

I have no idea what two effects you talk about. Both clocks are subject to time dilation in the initial inertial rest frame. But during the acceleration phase their time dilation is different.

The two effects (if I'm thinking of the same two) for a Born-rigid accelerating rocket are:

1. Because of the relativity of simultaneity, a comoving inertial frame will find the front clock ahead of the rear clock, even if they show the same time in the "launch frame".
2. Because of length contraction of the rocket itself (as viewed in the "launch frame"), the front clock is traveling at a slightly slower speed than the rear clock. So the time dilation factor effects the rear clock more.

The first effect is easy to get exactly backwards (as I think 1977ub does).

Here's what I hope is the correct analysis: Suppose that, according to the "launch frame", the two clocks are synchronized. Let e_1 be the event at which the rear clock shows time T. Let e_2 be the event at which the front clock shows time T. If these events are simultaneous in the "launch frame", then in the comoving frame, e_2 takes place BEFORE e_1. That means that the front clock shows time T before the rear clock shows time T. That means that the front clock is running ahead of the rear clock.

Both effects 1 and 2 above have the same sign; they both tend to make the front clock ahead of the rear clock, as seen in the comoving frame. Early on, the first effect dominates. Much later, the second effect dominates.

 

[1977ub]

I'm referring to observer in the initial frame S, not a comoving frame.
1) for S, as rigid body accelerates and the clock at the front ticks faster and gets ahead as determined by/in S. This is not due to RoS, and would be confirmed in all frames, right?

2) then body stops accelerating - ticks are now at the same rate to all inertial observers including S and in the body - (if observers on the body synch the clocks, observer in S will find that the back clock hits any given clock time first. This is due to RoS.)

If the clocks were not synched, then to find the combined difference in clock times/ages - for S -, the RoS #2 effect combines with the acceleration effect #1 - negatively - so that it seems to me that if everything were arranged properly, the 2 effects might cancel out - for observer in S. The front has more ticks, but ticks in a=0 phase arrive later. The back has had fewer ticks, but ticks in a=0 phase arrive sooner. (Actually effect #2 should exist as soon as body starts moving, but is more ambiguous to define during acceleration as I understand it since they are no longer in the same frame.)

 

[stevendaryl]

Let us make clear that I refer to a specific frame - let's say S, where the rigid object begins its acceleration.
There is a fixed quantity of difference which builds up between the clocks over the period of the acceleration, as measured by S, and makes the back clock appear to be younger. Ok so far?
variant A: after reaching target velocity V, observers in the object synchronize their clocks. Thereafter, everyone in S (or any other frame moving wrt the object) finds that the clock at the back reads a later time than the clock at the front.
variant B: but what if the synchronization never takes place. Since the acceleration difference puts the back clock ahead and the velocity difference puts the back clock behind, there might in fact be a perfect arrangement whereby the front and back clocks could start out reading the same time as one another before the acceleration begins, and then also end up reading the same time as one another during the velocity phase according to observers in S - right?

I think you're getting something confused. Let's make our assumptions explicit:

1. Initially, in frame S, the two clocks are synchronized.
2. The two clocks accelerate (with the front clock accelerating slightly less) until they are both at rest in a new frame S'.

Since the front clock accelerates slightly less than the rear clock, then in frame S, the front clock is ahead. In frame S', the front clock is even FURTHER ahead.

 

[1977ub]

I think you're getting something confused. Let's make our assumptions explicit:

1. Initially, in frame S, the two clocks are synchronized.
2. The two clocks accelerate (with the front clock accelerating slightly less) until they are both at rest in a new frame S'.

Since the front clock accelerates slightly less than the rear clock, then in frame S, the front clock is ahead. In frame S', the front clock is even FURTHER ahead.

Agreed. All of this relates to my #1.

#2 relates to clocks being/appearing out of synch in a moving inertial frame due to RoS.

On a moving (relative to S) train, if 2 clocks are created at the center, and one is moved to the back and the other is moved to the front (I presume we can neglect acceleration here for moving the clocks), then observer in S finds the back clock to hit any given time before the front clock - it is "older", has ticked more since it was created. This is RoS not acceleration.

https://www.google.com/search?q=lig...a.edu%2Flectures%2Fsynchronizing.html;544;184
http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html

 

[stevendaryl]

I'm referring to observer in the initial frame S, not a comoving frame.
1) for S, as rigid body accelerates and the clock at the front ticks faster and gets ahead as determined by/in S. This is not due to RoS, and would be confirmed in all frames, right?

2) then body stops accelerating - ticks are now at the same rate to all inertial observers including S and in the body - (if observers on the body synch the clocks, observer in S will find that the back clock hits any given clock time first. This is due to RoS.)

If the clocks were not synched, then to find the combined difference in clock times/ages - for S -, the RoS #2 effect combines with the acceleration effect #1 - negatively

I don't understand your scenario. If you don't resynchronize the clocks, then RoS isn't relevant for frame S. You just need to consider velocity-dependent time dilation. If the two clocks start off synchronized in frame S, and the rear clock travels faster than the front clock, then the front clock will continually get ahead of the rear clock, as viewed in both frame S and the comoving frame. There is no compensating RoS effect.

 

[stevendaryl]

Agreed. All of this relates to my #1.

#2 relates to clocks being/appearing out of synch in a moving inertial frame due to RoS

Yes, but those aren't compensating effects; they work in the same direction: to make the front clock ahead of the rear clock.

• Let e_1 be the moment when the rear clock shows time T_1
• Let e_2 be an event at the front clock that is simultaneous with e_1, according to frame S.
• Let T_2 be the time on the front clock at event e_2
• Let e_3 be an event at the front clock that is simultaneous with e_1, according to frame S' (the new rest frame).
• Let T_3 be the time on the front clock at event e_3

Event e_3 takes place AFTER event e_2, so T_3 > T_2. So RoS makes the time discrepancy bigger, not smaller.

 

[stevendaryl]

Agreed. All of this relates to my #1.

#2 relates to clocks being/appearing out of synch in a moving inertial frame due to RoS.

On a moving (relative to S) train, if 2 clocks are created at the center, and one is moved to the back and the other is moved to the front (I presume we can neglect acceleration here for moving the clocks), then observer in S finds the back clock to hit any given time before the front clock - it is "older", has ticked more since it was created. This is RoS not acceleration.

That's talking about a case in which the two clocks are synchronized in the comoving frame. In that case, the rear clock will be ahead, according to frame S. But if you don't make an effort to synchronize them--you just start with them synchronized in frame S, and accelerate both clocks--then the clocks won't be synchronized in the comoving frame--the front clock will be ahead.

 

[1977ub]

you just start with them synchronized in frame S, and accelerate both clocks--then the clocks won't be synchronized in the comoving frame--the front clock will be ahead.

"the front clock will be ahead" for comoving observer. But not necessarily for observer in S. For him, the two effects do not add, they interfere. As soon as the body measures contracted to observer in S, this is an effect which also sets the rear clock ahead - and the faster the body is to S, the more contracted, and the more ahead the rear clock is. Effect #1 can be made arbitrarily small - is that right? The acceleration can be made to last a very small amount of time, Meanwhile effect #2 can be made as large as one likes - approaching c. If I missed anything I don't know what it is.

 

[stevendaryl]

"the front clock will be ahead" for comoving observer. But not necessarily for observer in S.

Are you talking about the case where there is no effort to resynchronize the clocks? In that case, the clocks start out synchronized, and the front clock has a smaller acceleration than the rear clock, so has less time dilation. So the front clock gets ahead of the rear clock.

For him, the two effects do not add, they interfere.

If you don't resynchronize, then there is no "two effects" for frame S. There's only time dilation.

As soon as the body measures contracted to observer in S, this is an effect which also sets the rear clock ahead

That's backwards. Length contraction means that the rear clock travels faster than the front clock. The clock that travels fastest has the most time dilation, and so shows the smallest elapsed time.

- and the faster the body is to S, the more contracted, and the more ahead the rear clock is. Effect #1 can be made arbitrarily small - is that right? The acceleration can be made to last a very small amount of time, Meanwhile effect #2 can be made as large as one likes - approaching c. If I missed anything I don't know what it is.

From the point of view of frame S, there is only one effect: time dilation. The clock that travels the fastest shows the most time dilation, and so shows the smallest elapsed time. Since the rear clock travels fastest, that's the one that shows the smallest elapsed time.

The two effects that I was talking about were:

1. Time dilation, which causes the rear clock to tick slower than the front clock. This applies to both the initial frame and the comoving frame.
   
2. Relativity of simultaneity, which causes the front clock to be even further ahead in the comoving frame than it is in the initial frame.

They aren't compensating effects, they work in the same direction. Both tend to make the front clock tick faster than the rear clock.

 

[1977ub]

Are you talking about the case where there is no effort to resynchronize the clocks? In that case, the clocks start out synchronized, and the front clock has a smaller acceleration than the rear clock, so has less time dilation. So the front clock gets ahead of the rear clock.
If you don't resynchronize, then there is no "two effects" for frame S. There's only time dilation.
That's backwards. Length contraction means that the rear clock travels faster than the front clock. The clock that travels fastest has the most time dilation, and so shows the smallest elapsed time.
From the point of view of frame S, there is only one effect: time dilation. The clock that travels the fastest shows the most time dilation, and so shows the smallest elapsed time. Since the rear clock travels fastest, that's the one that shows the smallest elapsed time.

The two effects that I was talking about were:

1. Time dilation, which causes the rear clock to tick slower than the front clock. This applies to both the initial frame and the comoving frame.
   
2. Relativity of simultaneity, which causes the front clock to be even further ahead in the comoving frame than it is in the initial frame.

They aren't compensating effects, they work in the same direction. Both tend to make the front clock tick faster than the rear clock.

I don't have any math to show you, but you don't appear to be describing the situation I am. Maybe we can try this.

Phase 1) A train is moving with constant speed V relative to frame S. Two identical synchronized clocks are created by train denizens in the center of the train and moved to the ends. They appear to read same time and same rate to train observers. To S, they tick at same (slow) rate and the back clock is set later than the front clock once they are separated, and the effect increases as they are separated.

Phase 2) The train is decelerated to rest wrt frame S. I take it that the rear clock will encounter more ticks during deceleration than the front clock (the inverse of the accelerating case) - and this holds for both train observers and S observers.

For observers in the center of the train, this is the only effect they encounter which desynchronizes the clocks, and when the train comes to rest, they find that the clocks now are ticking at the same rate again, though the rear clock is set later, having experienced more ticks during deceleration.

For observers in S, who always thought that the rear clock was set later than the front (once they were separated), they now find that the rear clock is set later than the front but by a different amount, since caused by a different effect.

If this is all agreed, I would state that in the acceleration case, the two different affects creating observed clock difference to S observer run in counter directions, just as here in the deceleration case they run in the same direction.

 

[stevendaryl]

I don't have any math to show you, but you don't appear to be describing the situation I am.

That's possible, but my scenario is pretty simple:

1. Initially, both clocks (one in the front of a rocket, one in the rear) are synchronized in frame S.
2. Initially, the rocket is at rest in frame S.
3. Initially the rocket has length L in frame S.
   
4. The rocket undergoes "Born rigid" acceleration until it reaches some cruising speed, and then stops accelerating.
5. Afterward, the rocket has length L/\gamma.
6. Afterward, the front clock shows more elapsed time than the rear clock.

I thought that 1-5 was your scenario. Conclusion 6 just follows from time dilation: the front clock travels slowest, and so has the smallest time dilation, so has the largest elapsed time.

Phase 1) A train is moving with constant speed V relative to frame S. Two identical synchronized clocks are created by train denizens in the center of the train and moved to the ends. They appear to read same time and same rate to train observers. To S, they tick at same (slow) rate and the back clock is set later than the front clock once they are separated, and the effect increases as they are separated.

Agreed. If you make sure that the clocks are synchronized according to the comoving frame, then the rear clock will be ahead, according to frame S.

Phase 2) The train is decelerated to rest wrt frame S. I take it that the rear clock will encounter more ticks during deceleration than the front clock (the inverse of the accelerating case) - and this holds for both train observers and S observers.

Yes, during deceleration, the rear clock gains time, relative to the front clock.

For observers in the center of the train, this is the only effect they encounter which desynchronizes the clocks, and when the train comes to rest, they find that the clocks now are ticking at the same rate again, though the rear clock is set later, having experienced more ticks during deceleration.

Right.

For observers in S, who always thought that the rear clock was set later than the front (once they were separated), they now find that the rear clock is set later than the front but by a different amount, since caused by a different effect.

Right.

If this is all agreed, I would state that in the acceleration case, the two different affects creating observed clock difference to S observer run in counter directions, just as here in the deceleration case they run in the same direction.

I don't understand why you would go through so much detail about the deceleration case, if what you're interested in is the acceleration case. The same reasoning does not apply.

If you want to account for the time discrepancy (whether for acceleration or deceleration), we can break it up into three parts:

1. \delta T_1: the initial discrepancy (time difference between front and rear clocks; positive means the front clock is ahead), as measured in the initial rest frame of the train.
2. \delta T_2: the additional discrepancy, as viewed in frame S, due to relativity of simultaneity.
3. \delta T_3: the additional discrepancy, due to time dilation, as measured in frame S.

So the final discrepany, as viewed in frame S, is \delta T = \delta T_1 + \delta T_2 + \delta T_3

Now, we have two different scenarios: (1) Acceleration from rest in frame S. (2) Deceleration to rest in frame S.

Acceleration case:

1. \delta T_1 = 0. Initially, the clocks are synchronized in their own rest frame.
2. \delta T_2 = 0. Initially, the clocks are at rest in frame S, so there is no RoS correction.
3. \delta T_3 > 0. The front experiences less time dilation, and so has more elapsed time.

Conclusion: \delta T > 0. The front clock is ahead. Time dilation contributes positively, and there is no contribution due to RoS.

Deceleration case:

1. \delta T_1 = 0: Again, the clocks are synchronized in their own frame.
2. \delta T_2 < 0: Initially, because of RoS, the rear clock is ahead of the front clock, as viewed in frame S.
3. \delta T_3 < 0: The front clock experiences more time dilation, and so has less elapsed time.

Conclusion: \delta T < 0. The rear clock is ahead. Time dilation contributes negatively, and there is a negative contribution due to RoS.

 

[1977ub]

I don't understand why you would go through so much detail about the deceleration case, if what you're interested in is the acceleration case. The same reasoning does not apply.

I have the clocks to be created during fixed-velocity phase, in order to eliminate the objection in the acceleration case that the moving frame denizens synchronize the clocks, thus breaking each clock's measure of it's own proper age. I will go through the rest of it later.

You've gone out of your way to omit the one thing I'm going out of my way to highlight - the RoS which goes hand in hand with the Lorentz contraction (as determined in S) , and which contributes some effect of the rear clock being set later for S observers.

Also: how about this: In a modified Bell scenario, 2 rockets with clocks of the same age in S accelerate together wrt S and then hit a level velocity. At no time is there any difference between the 2 clocks for S - during acceleration phase nor during velocity phase. They have slowed wrt his own but are still in lockstep between each other. Note there is no Lorentz contraction in this example. The distance between the clocks remains the same over time for S. *Correspondingly* there is no RoS which makes the clocks tick out of synch from each other as far as S obs is concerned.

A train goes by at relativistic velocity. On the platform, you hit a button on the outside of the train at the middle, it flashes a lightbulb in the middle of the train which sends light toward the front and back ends. From your perspective, the light hits the back end first, triggering the clock or the bomb or whatever is back there - at some small or large time before the light hits the front. This is not an effect which only matters "if the clocks have been synchronized" - there is a real "before"-ness to the back of a vehicle moving at relativistic velocity.

 

[harrylin]

I have the clocks to be created during fixed-velocity phase, in order to eliminate the objection in the acceleration case that the moving frame denizens synchronize the clocks, thus breaking each clock's measure of it's own proper age. I will go through the rest of it later.

You've gone out of your way to omit the one thing I'm going out of my way to highlight - the RoS which goes hand in hand with the Lorentz contraction (as determined in S) , and which contributes some effect of the rear clock being set later for S observers. [..]

I addressed much or perhaps all of your clock issues in posts #3, 5, 7 etc., and next others did the same. But here I see that, very many posts later, the situation appears to be unchanged!

As mere words and equations could not solve this, probably the only way to clarify this matter is by means of a calculation example with numbers. Please present yours, and next others will correct it. Surely that will clarify the issues.

 

[stevendaryl]

You've gone out of your way to omit the one thing I'm going out of my way to highlight - the RoS which goes hand in hand with the Lorentz contraction (as determined in S) , and which contributes some effect of the rear clock being set later for S observers.

I took into account both RoS and Lorentz contraction. You're right, that Lorentz contraction itself can be understood in terms of RoS and time dilation.

Also: how about this: In a modified Bell scenario, 2 rockets with clocks of the same age in S accelerate together wrt S and then hit a level velocity. At no time is there any difference between the 2 clocks for S - during acceleration phase nor during velocity phase. They have slowed wrt his own but are still in lockstep between each other. Note there is no Lorentz contraction in this example. The distance between the clocks remains the same over time for S. *Correspondingly* there is no RoS which makes the clocks tick out of synch from each other as far as S obs is concerned.

Right. But what's your point?

A train goes by at relativistic velocity. On the platform, you hit a button on the outside of the train at the middle, it flashes a lightbulb in the middle of the train which sends light toward the front and back ends. From your perspective, the light hits the back end first, triggering the clock or the bomb or whatever is back there - at some small or large time before the light hits the front. This is not an effect which only matters "if the clocks have been synchronized" - there is a real "before"-ness to the back of a vehicle moving at relativistic velocity.

Right. If the flashes are simultaneous in the train frame, then the rear flash arrives before the front flash, in the "stationary" frame. That's RoS.

 

[1977ub]

I took into account both RoS and Lorentz contraction. You're right, that Lorentz contraction itself can be understood in terms of RoS and time dilation.
Right. But what's your point?
Right. If the flashes are simultaneous in the train frame, then the rear flash arrives before the front flash, in the "stationary" frame. That's RoS.

And is an effect which sets the rear clock ahead of the front clock in the 'platform' frame, just as the acceleration period is an effect which puts the front clock ahead of the rear clock for the 'platform' frame. Either way, I'm done here. No more need to chew up the scenery. Thanks everyone.

 

[stevendaryl]

And is an effect which sets the rear clock ahead of the front clock in the 'platform' frame, just as the acceleration period is an effect which puts the front clock ahead of the rear clock for the 'platform' frame. Either way, I'm done here. No more need to chew up the scenery. Thanks everyone.

The flash is a synchronization mechanism for the train frame. That's why I said that IF there is an attempt to synchronize the clocks in the moving frame, then that would result in setting the rear clock ahead of the front clock, as viewed in the stationary frame. If there is no effort to synchronize the clocks in the moving frame, then there is no such effect.

 

[1977ub]

The flash is a synchronization mechanism for the train frame. That's why I said that IF there is an attempt to synchronize the clocks in the moving frame, then that would result in setting the rear clock ahead of the front clock, as viewed in the stationary frame. If there is no effort to synchronize the clocks in the moving frame, then there is no such effect.

I think my example of the light-triggered bombs at each end of the train, stranger on platform activates light in middle of the car, causing bomb at the rear to go off before the front bomb goes off - as viewed in platform - without anybody going to any effort to synchronize any clocks - makes my point quite viscerally.

 

[stevendaryl]

I think my example of the light-triggered bombs at each end of the train, stranger on platform activates light in middle of the car, causing bomb at the rear to go off before the front bomb goes off - as viewed in platform - without anybody going to any effort to synchronize any clocks

Okay, in this case, you're synchronizing bomb explosions instead of clocks.

 

[1977ub]

If the sphere was linearly accelerated from rest, then the back has aged less, not more, than the front in that process.

I totally get that now, thank you.

 

[1977ub]

Okay, in this case, you're synchronizing bomb explosions instead of clocks.

Ok thank you.

 

[1977ub]

On a moving (relative to S) train, if 2 clocks are created at the center, and one is moved to the back and the other is moved to the front (I presume we can neglect acceleration here for moving the clocks), then observer in S finds the back clock to hit any given time before the front clock - it is "older", has ticked more since it was created. This is RoS not acceleration.

I think this is where I went wrong. I wasn't really getting that all the effects seen in orig frame S can be made sense of without caring about what train frame regards as simultaneous. So - the clocks are created together on the moving train and then moved to the two ends of the train. The one at the rear now appears to be older than the front for S, and indeed it is the velocity differential (for S) related to moving the clocks in the opposite directions, which becomes what S finds to be the clocks reading different times.

There's no "effect" other than what is accounted for by following the individual paths of the clocks.

 

[harrylin]

I think this is where I went wrong. I wasn't really getting that all the effects seen in orig frame S can be made sense of without caring about what train frame regards as simultaneous.

Great - that's very important insight! In many cases, jumping reference frames is more confusing than useful. It was however very useful for developing the theory.