discuss events which are simultaneous in one frame?

neopolitan

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

neopolitan

kev,

In #54 you said
Then in #55 I said
Then in #56 you said
Then in #62 (responding to Dr Greg) I said
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:

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

yuiop

Correct

Correct

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 [itex]-L_o v/c^2[/itex]. 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.

yuiop

In post #47 I introduced the decaying isotope clocks:

"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.

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).

yuiop

In post #63 I said this:
How does that differ from the underlined part of your post?

I also said this:
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.

neopolitan

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).

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

1effect

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. :-)

DaleSpam

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.

neopolitan

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

Doc Al

Kev is correct. Whether clocks maintain synchronization during acceleration depends on how they are accelerated.

yuiop

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.

neopolitan

Can you check the conditions stated in the thread then, and see whether they qualify.

neopolitan

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

cheers,

neopolitan

yuiop

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.

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.

neopolitan

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