## discuss events which are simultaneous in one frame?

JesseM

My comment was based in so much context that I don't think I can respond any better than "please see the ealier posts". I should not have tried to defend against a comment made referring to something I had written, but presented out of context. I've explained myself before. Please find the relevant posts which appear earlier in the thread.

Equally, Barney and Fred are not mine. They are kev's creations from the very start (along with the decaying lumps and disinterested robots). I will continue to try to work out if kev was claiming what I thought he was, if he had a typo or if I misread what he wrote.

cheers,

neopolitan

 Quote by kev I never said that two clocks that are co-located and syncronous in one frame would not be syncronous in another frame. What I said in post #63 was "In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned)" which is not the same thing. I also said in post #54 I said "...that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket." and in post #49 I said "Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point." I hope you can agree I never said (as far as I am aware) anything that amounts to your suggestion that I implied two clocks, which are both collocated and synchronous in one frame (Barney's) could be collocated and asychonous in another frame (Fred's). To summerise: Two clocks that are co-located and syncronised in any observers frame will be syncronised in any other observer's frame. Two clocks that are co-located but not syncronised in any observer's frame will not be syncronised in any other observer's frame. [EDIT] Can either of you show me where I said anything that contradicted that basic concept? If I did it is a typo and I will correct it.
kev,

In #54 you said
 Quote by kev [EDIT] Without doing the formal calculations we can note that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket. Since the radioctive lumps were not syncronised prior to bringing them together they will not show the same time when bringing them together.
Then in #55 I said
 Quote by neopolitan Comment one: Are we not comparing the extent to which each radiactive lump has decayed? I would have thought that this implies a synchronisation at a reference event after which the decay of each lump is measured. Comment two: I assume that, in the rest frame of the rocket, the robots move at the same speed from the nose and tail towards the middle. This means that in the rocket frame there will be a very small amount of time dilation due to the relative motion of the robots (very small because you stipulated "gradually" as the magnitude of their velocities) and that time dilation would affect each robot and lump equally. The lumps will be simultaneous at the midpoint and will have undergone the same amount of time dilation, and, if my comment one is right, still synchonised (in the rocket's frame). Since the lumps will now be collocated, and collocation applies to all frames, I believe that they will then be synchonised in all frames.
Then in #56 you said
 Quote by kev The two robots will have undergone the same amount of time dilation in the rest frame of the rocket relative to clocks on the rocket that remain at rest with rocket. However from the point of view of our single observer not onboard the rocket the two robots will not have experienced the same amount of time dilation.
Then in #62 (responding to Dr Greg) I said
 Quote by neopolitan What I was trying to say is that, within the rocket's frame, the clocks that are at first synchronous, at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation - within rocket's frame - and therefore be synchronous, at rest and collocated. Once collocated, their being synchonrous should be frame independent. I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame should explain how the clocks end up being synchronous in that other frame when they weren't initially.
As far as I can tell it was not until #63 that you introduced tamper proof clocks which you claim would lose synchronicity after acceleration - specifically, you claim that the clocks lose synchronicity in the rest frame due to acceleration:

 Quote by kev We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket. In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned). The tamper proof clock that was at the nose will be "in the future" (to use your expression) of the tamper proof clock that was at the rear.
If, for any reason, the clocks are not synchronised in their shared rest frame (the rocket's frame, or Barney's frame), then moving them to the centre of the rocket in the manner described will not make them synchronised. However, the extent to which they are not synchronised will not be affected either.

The question I must have then is, does acceleration of a rest frame - relative to an outside observer - cause synchronised clocks in that rest frame to lose synchronisation?

I doubt that it does, since each clock will be accelerated equally (since otherwise the clocks don't share a rest frame). All effects will be equal and synchronisation in that rest frame will be maintained.

Yet again, I stand ready to be corrected.

cheers,

neopolitan

Blog Entries: 6
 Quote by neopolitan ............. If, for any reason, the clocks are not synchronised in their shared rest frame (the rocket's frame, or Barney's frame), then moving them to the centre of the rocket in the manner described will not make them synchronised.
Correct

 Quote by neopolitan However, the extent to which they are not synchronised will not be affected either.
Correct

 Quote by neopolitan The question I must have then is, does acceleration of a rest frame - relative to an outside observer - cause synchronised clocks in that rest frame to lose synchronisation? I doubt that it does, since each clock will be accelerated equally (since otherwise the clocks don't share a rest frame). All effects will be equal and synchronisation in that rest frame will be maintained. Yet again, I stand ready to be corrected. cheers, neopolitan
This is a bit more tricky. While the rocket is accelerating it is length contracting according to Fred. The nose and tail therefore can not have the same velocity at all times according to Fred because the tail is catching up to the nose.

The tail is going faster at all times until the rocket starts cruising. The tail clock is time dilated more than the nose clock. When the clocks are syncronised in Barney's frame the nose clock should be showing a lesser time according to Fred using the formula $-L_o v/c^2$. The time dilation that occurs during the acceleration phase is doing exactly the opposite and is not self syncronising.

When the clocks are brought together in the centre, the tail clock is time dilated even more making the situation worse.

Blog Entries: 6
 Quote by neopolitan kev, . . As far as I can tell it was not until #63 that you introduced tamper proof clocks which you claim would lose synchronicity after acceleration - specifically, you claim that the clocks lose synchronicity in the rest frame due to acceleration:
In post #47 I introduced the decaying isotope clocks:

 Quote by kev Now say I get 3 lumps of radioactive material that decay in a consistant and predictable manner at the same rate. The half life of the material is defined as the time it takes for half the remaining atoms of isotope1 to convert to isotope2. These lumps provide a fundamental clock that cannot be arbitarily advanced, retarded or syncronised with respect to each other when they are at rest wrt each other.
"Tamper proof clocks" is my new shorthand name for the decaying isotope clocks. (Sorry, I should have made that clearer) Although if you tried really hard you could tamper with them, the rules for the thought experiment that I introduced is that we do not syncronise, advance or retard the isotope clocks. (Sorry, if I did not make that clear either). For that purpose we can imagine that the decaying isotope clocks are sealed to prevent tampering and all that is visible is a digital display of the time they record (or a display of the ratio of decayed atoms to undecayed atoms). They provide a reference clock to the conventional clocks that are syncronised when the rocket reaches its cruising speed. Each conventional clock has a tamper proof isotope clock next to it at all times for comparison.

 Quote by kev When the radiactive lumps are being brought together by the inpartial robots, the rear lump is moving even faster (relative to the one and only sentient observer) than the nose lump that is transported "backwards" from the nose to the centre and so the rear lump experiences even more time dilation relative to the front lump from the point of view of the only observer. .
You asked to point out where you made the typo (it is more like an error). See above.
The difference in the speed of lumps in the frame external to the roicket is irrelevant. You are using the differences in speeds in the observer frame in order to justify a difference in time dilation. This is incorrect: if the lumps were synchronized prior to slow transport, they will remain synchronized after slow transport, the difference in speeds between the fore and aft lump in the external observer frame is irrelevant as long as the rules of slow transport were obeyed in the rocket frame (slow and equal speeds in the rocket frame).

Blog Entries: 6
 Quote by 1effect You asked to point out where you made the typo (it is more like an error). See above. The difference in the speed of lumps in the frame external to the roicket is irrelevant. You are using the differences in speeds in the observer frame in order to justify a difference in time dilation. This is incorrect: if the lumps were synchronized prior to slow transport, they will remain synchronized after slow transport, the difference in speeds between the fore and aft lump in the external observer frame is irrelevant as long as the rules of slow transport were obeyed in the rocket frame (slow and equal speeds in the rocket frame).
In post #63 I said this:
 Quote by kev We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket.
How does that differ from the underlined part of your post?

I also said this:
 Quote by kev In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre ....
I'll say it again. "The radioactive lumps are not syncronised prior to the slow transport."

I can only assume that you also missed or skipped over the prior 2 posts (#71 and #72).

"if the lumps were synchronized prior to slow transport, they will remain synchronized after slow transport"

The point is that the lumps are not syncronised prior to the slow transport. Post #72 makes it clear that lumps are not syncronised after the rocket has accelerated from the Fred's rest frame and according to the rules of the thought experiment (post #72) the radioactive lumps are sealed or "tamper proof" and we are not allowed to to syncronise the radioactive lumps. THAT is the point of the radioctive lumps. There are conventional clock paired with each radiactive lump that can be syncronised.

So to sum up the last dozen posts where we have been repeating ourselves and making out that we are disagreeing when we are not:

1) Two clocks that are syncronised prior to slow transport will remain syncronised after slow transport.

2) Two clocks that are not syncronised prior to slow transport will not be syncronised after slow transport.

3) The radioactive lumps are not syncronised prior to slow transport (So point 2 applies)

4) The radioactive lumps are sealed in tamper proof containers and according to the rules no one is allowed to adjust, advance, retard or syncronise the radioactive lumps once the rocket has launched.

This is where me and neopolitan were at about half a dozen posts ago. The only thing neopolitan is not sure about is whether or not the the radioactive lumps will remain self syncronised during the acceleration of the rocket and after the rocket has stabilised to its cruising speed. I can assure you that the radioactive lumps will not remain syncronised after the acceleration phase.

I thought about one thing a little late last night, when I should have been asleep. I think the replies which have appeared since then answer the question, but I want to make sure.

I have assumed that the rocket is rigid and that we are not talking about a real world situation here where there would be a lag between the commencement of acceleration along the length of the rocket. If we were assuming a (semi) real world rocket, then the motor would be in the tail and the whole length would contract physically, within its own frame. This is not how I have thought about the mind experiment. The whole frame accelerates together, if it accelerates at all.

I think that kev has considered the rocket to be rigid as well, but I am not totally sure. If the rocket according to kev is not rigid, then yes there will be a loss of synchronicity in the nominal rest frame (since motions would be created by the frame's acceleration, it would no longer really be a rest frame).

 Quote by kev This is a bit more tricky. While the rocket is accelerating it is length contracting according to Fred. The nose and tail therefore can not have the same velocity at all times according to Fred because the tail is catching up to the nose.
I think this is wrong. I think I can understand what you are saying - if it relates to length contraction - but I am pretty sure there is something wrong with the conceptualisation.

I will have to ponder it a bit more though and read the comments of other contributors.

cheers,

neopolitan

U asked me to point the exact error in your post, so I did. I can only go by what you wrote in the respective post. If you intended to write something else, go ahead and correct it. :-)

Mentor
 Quote by neopolitan I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame should explain how the clocks end up being synchronous in that other frame when they weren't initially.
This is correct, although the easiest way to show it is not directly through time dilation (because there are also synchronization issues), but through the spacetime interval as follows.

The rocket has proper length L and is moving at v in the (unprimed) observer frame. Without loss of generality the origin of both frames is taken to be the event, A, when the rear clock starts moving at u in the rocket frame. The event B is when the front clock starts moving at -u in the rocket frame. The event C is when the clocks meet.
A' = (0,0) --> A = (0,0)
B' = (0,L) --> B = (γLv/c, γL)
C' = (c L/(2u),L/2) --> C = (γL(c²+uv)/(2cu), γL(u+v)/(2u))

If the clocks are initially synchronized in the rocket frame then at A/A' and B/B' the clocks read 0. The spacetime interval |C-A|=L²(c²-u²)/(4u²), so at C the rear clock reads sqrt(L²(c²-u²)/(4u²))/c. The spacetime interval |C-B|=L²(c²-u²)/(4u²), so at C the front clock also reads sqrt(L²(c²-u²)/(4u²))/c.

So, basically, in the observer's frame, the rear clock moves for longer at a higher velocity (more time dilation) which balance out to have them synchronized at their meeting.

Note that, if the clocks are initially de-synchronized in the rocket frame by an amount dt, then they will be desynchronized by dt in all frames at their meeting.
 DaleSpam, Can you confirm that acceleration does not cause a loss of synchonisation between two synchronised but non-collocated clocks in shared rest frame? This is kev's claim, not mine, my rebuttal is at post #75 (admittedly not with any hard data to back it up). cheers, neopolitan

Mentor
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 Quote by neopolitan Can you confirm that acceleration does not cause a loss of synchonisation between two synchronised but non-collocated clocks in shared rest frame? This is kev's claim, not mine, my rebuttal is at post #75 (admittedly not with any hard data to back it up).
Kev is correct. Whether clocks maintain synchronization during acceleration depends on how they are accelerated.

Blog Entries: 6
Quote by 1effect

 Quote by kev When the radiactive lumps are being brought together by the inpartial robots, the rear lump is moving even faster (relative to the one and only sentient observer) than the nose lump that is transported "backwards" from the nose to the centre and so the rear lump experiences even more time dilation relative to the front lump from the point of view of the only observer.
You asked to point out where you made the typo (it is more like an error). See above.
The difference in the speed of lumps in the frame external to the roicket is irrelevant. You are using the differences in speeds in the observer frame in order to justify a difference in time dilation. This is incorrect: if the lumps were synchronized prior to slow transport, they will remain synchronized after slow transport, the difference in speeds between the fore and aft lump in the external observer frame is irrelevant as long as the rules of slow transport were obeyed in the rocket frame (slow and equal speeds in the rocket frame).
There is no error in my original statement. It is simply the point of view of the unaccelerated observer external to the accelerated rocket.

In the case of two clocks that are syncronised from the point of view of an observer on the rocket the clocks will read the same time when slow transported to the centre. The two clocks at the nose and tail of the rocket that appear syncronised to the onboard observer are not syncronised according to the unaccelerated external observer and the difference in time dilation during the slow transport accounts for why the external observer see the two clocks as syncronised by the time they are co-located at the centre of the rocket. Both internal and external observers agree that clocks are syncronised when they are are co-located. If the clocks were not syncronised according to the onboard observer prior to the slow transport, then both observers will agree they are not syncronised when they meet at the centre.

 Quote by Doc Al Kev is correct. Whether clocks maintain synchronization during acceleration depends on how they are accelerated.
Can you check the conditions stated in the thread then, and see whether they qualify.

 Quote by kev In the case of two clocks that are syncronised from the point of view of an observer on the rocket the clocks will read the same time when slow transported to the centre. The two clocks at the nose and tail of the rocket that appear syncronised to the onboard observer are not syncronised according to the unaccelerated external observer and the difference in time dilation during the slow transport accounts for why the external observer see the two clocks as syncronised by the time they are co-located at the centre of the rocket. Both internal and external observers agree that clocks are syncronised when they are are co-located. If the clocks were not syncronised according to the onboard observer prior to the slow transport, then both observers will agree they are not syncronised when they meet at the centre.
I agree with this. This is either not what you originally wrote, or not what I initially read.

In any event, this I can agree with.

I no longer understand why you need acceleration though, since the in-frame, non-collocated synchronised clocks will not be sychronised according to an external observer (not at rest relative to the clocks) irrespective of whether they undergo acceleration or not.

You might want to review post #71

 Quote by kev The tail is going faster at all times until the rocket starts cruising. The tail clock is time dilated more than the nose clock. When the clocks are syncronised in Barney's frame the nose clock should be showing a lesser time according to Fred using the formula $-L_o v/c^2$. The time dilation that occurs during the acceleration phase is doing exactly the opposite and is not self syncronising. When the clocks are brought together in the centre, the tail clock is time dilated even more making the situation worse.
cheers,

neopolitan

Blog Entries: 6
 Quote by neopolitan I no longer understand why you need acceleration though, since the in-frame, non-collocated synchronised clocks will not be sychronised according to an external observer (not at rest relative to the clocks) irrespective of whether they undergo acceleration or not.
It is true that "the in-frame, non-collocated synchronised clocks will not be sychronised according to an external observer (not at rest relative to the clocks) irrespective of whether they undergo acceleration or not." However they will be out of sync in different ways according to the external observer. If the clocks are syncronised by someone onboard the rocket then the nose clock will show less elapsed time than the tail clock according to the external observer. If the clocks are not syncronised after the acceleration then the nose clock will show more elapsed time than the tail clock.

Basically I introduced acceleration to support your view that "the nose clock is in the future of the tail clock".

When the clocks on the rocket are syncronised according to an oboard observer (A) then to an external unaccelerated observer (B) the nose clock reads less than than the tail clock. If the rocket is 16 light seconds long and going at 0.5c relative to observer A then A will see the nose clock reading 8 seconds less than the tail clock. This is not supportive of your view that the "the nose clock is in the future of the tail clock".

When the clocks on the rocket are syncronised according to an onboard observer (A then obviously the nose clock and the tail clock will be showing the same time simultaneously as far as observer A is concerned. This is not supportive of your view that "the nose clock is in the future of the tail clock" either.

However if we place syncronised clocks on the rocket and then accelerate it, an unaccerated external observer will see that the nose clock IS in the future of the tail clock, if the onboard observer does not re-syncronise the clocks after the acceleration. The onboard observer and the external unaccelerated observer will both agree that the nose clock has aged more than the tail clock.

To give an extreme example. We place identical twin babies on a very long rocket. One twin baby is at the nose and the other at the tail of the rocket. The rocket is accelerated very hard for a very long time and then allowed to cruise for long enough to allow stresses and strains to stabilise. We bring the twins together at the centre of the rocket and we see that the nose twin is an old guy with a long white beard while the tail twin is still a baby. Since they are both co-located no observer can disagree that the nose twin is in the future of the tail twin.

By the way I am talking about a traditional rocket with a single rocket at the rear and the rocket is allowed to undergo natural length contraction.

Now if instead of accelerating the rocket that has the twins onboard, we get another rocket and accelerate away in that. We turn around and fly past the the rocket with the twins onboard and they appear to be ageing differently by our observations. Finally we land next to the twin's rocket and the twins come to meet at at the centre of their rocket. In this case we note the twins did not really age differentially. They are both older and it is us that have aged less. I am trying to show that Lorentz transformations result in real physical changes and that what may appear to be symetrical situations are not really symetrical when you take acceleration into account.

Mentor
 Quote by neopolitan Can you confirm that acceleration does not cause a loss of synchonisation between two synchronised but non-collocated clocks in shared rest frame? This is kev's claim, not mine, my rebuttal is at post #75 (admittedly not with any hard data to back it up).
As Doc Al mentioned the answer depends on the details of the motion, which I don't think have been unambiguously defined here. So instead of directly answering I will refer you to

http://www.mathpages.com/home/kmath422/kmath422.htm

Which gives a more detailed treatment of Born-rigid acceleration than I could. I am sure that you could find other pages describing other acceleration schemes.

 Quote by DaleSpam As Doc Al mentioned the answer depends on the details of the motion, which I don't think have been unambiguously defined here. So instead of directly answering I will refer you to http://www.mathpages.com/home/kmath422/kmath422.htm Which gives a more detailed treatment of Born-rigid acceleration than I could. I am sure that you could find other pages describing other acceleration schemes.
If I have this right, which is by no means a given, this theoretical effect is due to a rigidity which is not actually physically possible. One consequence is that the tail of the rocket will have to travel faster than the nose to mainain this impossible rigidity - according to an external observer - within the "rest frame" the tail will now not be at rest relative to the nose, but will have a velocity in the direction of motion as perceived by the the external observer.

That, to me, would explain why a tail observer would be younger than the nose observer.

I suspect that in reality other effects prevent this from happening.

Standing by to be corrected :)

cheers,

neopolitan