Discuss events which are simultaneous in one frame?

[neopolitan]

Hi Neopolitan,

Say I place two mickey mouse clocks next to each other on a table. Say the clocks are syncronised with each and nothing is moving except for the hands of the clocks advancing at the same rate as far as I am concerned. Now if I advance the hands of the right hand clock by 2 hours, is the right hand clock 2 hours in the future of the left hand clock in the meaning of "future" that you are using? Or is it just that I advanced the hands of the right hand clock and there is nothing fundamental about the time displayed. For instance the atoms of the two clock cases would have (near enough) the same "age".

The clocks are desynchronised as soon as you move the hands. It's not what I meant at all.

cheers,

neopolitan

 

[neopolitan]

Jesse,

I will get back to you tomorrow. For the moment, I was adamant that there be only one observer so you knew which equations mattered. We have had so much trouble in other exchanges by your swapping observers.

One observer, only one observer, and we should get past that.

cheers,

neopolitan

 

[neopolitan]

I can partly justify the nose isotope lump being ahead of the rear lump in a fundamental sense by getting the onboard robots to slowly transport both lumps to the centre of the ship so that they are right next to each other. We would notice that the lump originally at the nose has decayed more than the isotope lump that was originally at the rear. Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point.

Um, where are "we"? And in any case, I think when you bring them together you should find that they have decayed equally.

I am ready to be proved wrong on this though.

cheers,

neopolitan

 

[yuiop]

Um, where are "we"? And in any case, I think when you bring them together you should find that they have decayed equally.

I am ready to be proved wrong on this though.

cheers,

neopolitan

"We" are in the original frame before the rocket accelerated away from "us". We are not onboard the rocket and have been stationary and never experienced any acceleration for the duration of the experiment. We count as one observer because we are next to each other and at rest with respect to each other. Sorry, I should have said "I" or "you" rather than "we" to stay in line with your condition of only one observer, but I get lonely sometimes :P The robots are programmed not to have an opinion and so do not count as observers ;)

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.


When I get time I will do the calculations using the equations for the proper time experienced by accelerated and transported clocks.

[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.

 

[neopolitan]

[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.

I should let someone else respond, but I make two comments anyway.

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.

Again, I may be wrong and stand ready to be corrected :)

cheers,

neopolitan

 

[yuiop]

I should let someone else respond, but I make two comments anyway.

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.

We do a crude form of syncronisation by starting with one large lump of radioactive material that consists of billions of atoms and is essentially homogenous and then divide the large lump into 3 equal portions to form 3 crude clocks.

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.

Again, I may be wrong and stand ready to be corrected :)

cheers,

neopolitan

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.

I added an edit to the end of my last post (#47) that you may have missed.

--> 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. <--

Does that additional comment help any?

 

[neopolitan]

kev,

If you check my previous post, you will see that your edit was the topic of my reply - I quoted your edit and nothing else.

I would appreciate a third opinion on this. I do still think that, when collocated, the lumps will have decayed equally - in all frames.

cheers,

neopolitan

MY EDIT - Here is an explanation for why. Perhaps it works, perhaps not. The time dilation for both lumps will be the same, relative to an outside observer - so long as the lumps are at rest in the rest frame they share with the rocket. There will just be an offset between due to their not being collocated. The reduction of that offset, to zero, can be accounted for by the different rates of time dilation as observed by the outside observer when they are not at rest in what was otherwise their shared rest frame (and, for the duration, only the rocket's rest frame).

A related effect must take place if a moving pair of clocks are slowed, so that they are and remain synchonised in their rest frame and become synchonised in a frame in which they were not at rest before deceleration but are at rest after deceleration. I am not saying it is the deceleration that does it.

 

[neopolitan]

Back to topic - Simultaneity

I would like to return to the original core topic, simultaneity.

Jesse confirmed that what we call simultaneity is "transmission simultaneity". Here is my definition:

Transmission simultaneity - photons from two events are released simultaneously, such that if the sources were equidistant (and remain equidistant - in other words the observer is at rest), the photons would reach the observer at rest together. Under most circumstances however, the photons will not reach the observer simultaneously and knowledge of where the photons were released is required to know that their release was in fact simultaneous.

If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.

Are there any conceptual problems with that?

To prevent diversions, I state explicitly that I am aware that relative to other observers who might not share my rest frame the event frame I just define is not necessarily an instant or a surface of simultaneity. I will get to that later, if there are no real conceptual problems with what I am proposing.

cheers,

neopolitan

 

[yuiop]

I would like to return to the original core topic, simultaneity.

Jesse confirmed that what we call simultaneity is "transmission simultaneity". Here is my definition:



If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.

Are there any conceptual problems with that?

To prevent diversions, I state explicitly that I am aware that relative to other observers who might not share my rest frame the event frame I just define is not necessarily an instant or a surface of simultaneity. I will get to that later, if there are no real conceptual problems with what I am proposing.

cheers,

neopolitan

What you are describing here is pretty standard. See this link http://casa.colorado.edu/~ajsh/sr/simultaneous.html. Ceruleans "hypersurface of simultaneity" is the tilted blue plane as seen by Vermilion whos own hypersurface of simultaneity is the red plane orthogonal to the time axis of his coordinate system.

 

[neopolitan]

Thanks kev,

I'll wait a bit to see if anyone thinks it is not standard.

cheers,

neopolitan

 

[DrGreg]

If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.

All of the above is correct. An observer has, for each event, a surface of simultaneity. In special relativity (i.e. ignoring gravity) the surface is a 3-dimensional "plane" in 4-dimensional spacetime. For a single observer, all the planes of simultaneity (for different events) stack up in parallel. But, as you suggest, different observers have different planes of simultaneity. This is standard, mainstream special relativity.

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.

If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

This may seem reasonable on the grounds that both clocks undergo the same time dilation. This would be true relative to the rocket, but it's not true relative to the outside observer; from the observer's point of view, one clock experiences more dilation than the rocket and the other less.

You might then argue that any change of dilation can be ignored if the clocks move slowly enough. However, this ignores the fact that the slower the clocks move (relative to the rocket), the longer it will take to bring them together. This lengthening of time taken increases the effect of the dilation. In fact the decrease in dilation-change and increase in duration tend to cancel each other out, and no matter how slow the clocks are moved, the there is a change between the clocks that will not go away, relative to the outside observer.

If you really wanted I could prove all this mathematically, but it would take a page or two of calculation.

Note that synchronising clocks by slowly moving them apart is referred in the literature as "slow clock transport" or "ultra slow clock transport". It can be proved that synchronisation by slow clock transport is exactly the same as Einstein synchronisation. For example, see this post.

 

[neopolitan]

If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

This may seem reasonable on the grounds that both clocks undergo the same time dilation. This would be true relative to the rocket, but it's not true relative to the outside observer; from the observer's point of view, one clock experiences more dilation than the rocket and the other less.

You might then argue that any change of dilation can be ignored if the clocks move slowly enough. However, this ignores the fact that the slower the clocks move (relative to the rocket), the longer it will take to bring them together. This lengthening of time taken increases the effect of the dilation. In fact the decrease in dilation-change and increase in duration tend to cancel each other out, and no matter how slow the clocks are moved, the there is a change between the clocks that will not go away, relative to the outside observer.

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.

It is not a case of the problem "going away".

cheers,

neopolitan

 

[yuiop]

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.

You started out this thread stating you wanted to stick to one observer (Let's call him Fred) to avoid confusion. Fred is an observer that is not onboard the rocket but was at one point at rest with rocket before the rocket accelerated. I have inserted red text in square brakets to make clear which measurements are Fred's observations. Consciously or unconsiously you now talking in terms of another observer (call him Barney?) who's observations I have inserted in square brakets and blue text in your statements.

"What I was trying to say is that, within the rocket's frame [Barney's frame], the clocks that are at first synchronous [According to the observer onboard the rocket], at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation [As measured by Barney] - within rocket's frame - and therefore be synchronous, at rest and collocated."

Once collocated, their being synchonrous should be frame independent.

Correct

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.

It is not a case of the problem "going away".

cheers,

neopolitan

"I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame [By Fred] should explain how the clocks end up being synchronous in that other frame [In Fred's frame] when they weren't initially [According to Fred].


Ok, you seem to ready to talk in terms of two observers.

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.

 

[1effect]

You started out this thread stating you wanted to stick to one observer (Let's call him Fred) to avoid confusion. Fred is an observer that is not onboard the rocket but was at one point at rest with rocket before the rocket accelerated. I have inserted red text in square brakets to make clear which measurements are Fred's observations. Consciously or unconsiously you now talking in terms of another observer (call him Barney?) who's observations I have inserted in square brakets and blue text in your statements.

"What I was trying to say is that, within the rocket's frame [Barney's frame], the clocks that are at first synchronous [According to the observer onboard the rocket], at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation [As measured by Barney] - within rocket's frame - and therefore be synchronous, at rest and collocated."



Correct



"I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame [By Fred] should explain how the clocks end up being synchronous in that other frame [In Fred's frame] when they weren't initially [According to Fred].


Ok, you seem to ready to talk in terms of two observers.

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.

Sorry for intervening but I think you need to pay attention to what neopolitan just told you:

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.

The fact that the two clocks move at different speeds from the point of view of an observer external to the rocket is purely irrelevant. The fact that the clocks are synchronized and collocated takes precedence.

 

[neopolitan]

My strict regime of one and only one observer was an attempt to avoid a situation as happened when I talked about the dorky physics guy (see my earlier post and a later reply from JesseM) in which I wanted to look at things from one specific perspective, where one specific set of equations applied, and JesseM wanted to look at things from another perspective, where another specific set of equations applied.

In my reply to you, it is not possible or sensible to only consider one observer, since we are inherently, as you point out, discussing two points of view.

What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.

Is this not paradoxical?

You may understand that it is difficult to accept this as objectively true.

cheers,

neopolitan

 

[1effect]

If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

.

I think (IMHO) that neopolitan talks about two clocks synchronized from the perspective of an on-board observer , i.e. inside the rocket.
This is why, he claims , correctly (IMHO) that once the two clocks are "slow transported" to the center of the rocket, they will be still synchronized (by the very definition of slow clock transport).This is regardles as to how long the transport took.

He further claims that, since the clocks are now collocated , they are synchronized in any frame of reference.

If this is what neopolitan is saying, I think (again, IMHO) that he is correct.

 

[JesseM]

I'm away from home but I've been using my friend's laptop a bit this afternoon, so I thought I'd give a quick reply to this:

My strict regime of one and only one observer was an attempt to avoid a situation as happened when I talked about the dorky physics guy (see my earlier post and a later reply from JesseM) in which I wanted to look at things from one specific perspective, where one specific set of equations applied, and JesseM wanted to look at things from another perspective, where another specific set of equations applied.

But as I said, I don't think it's possible to make sense of your claim that the nose is "more in the future" in the outside observer's frame without making some reference to the rocket's own rest frame--the idea is to pick to clock readings which are simultaneous in the rocket's frame, then switch to the outside observer's frame, and note that the reading on the nose happens at a later time in this frame than the reading on the tail. But even if we do have two frames, I suppose it might be a bit easier to have just one "observer" who measures the rocket moving, that way we can refer to "the observer's frame" and "the rocket's frame" instead of "the frame of the outside observer who sees the rocket moving" and "the frame of the observer on board the rocket".

Do you disagree with the basic point that two frames are needed? If so, how would you explain the notion that the nose clock is "further in the future" without making reference to the rocket frame, and without assuming the clocks have actually been synchronized in the rocket frame?

What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.

We don't (referring to the bolded part). When they're collocated, both frames agree they show the same time, assuming they were brought together at equal speeds in the frame where they were synchronized when located at opposite ends of the rocket (the rocket's rest frame). In this case, that means that in the frame where the rocket is moving forward, the clock that is being pushed in the same direction that the rocket is moving (from tail to middle) will have a slightly higher speed than the clock that is being pushed opposite the rocket's direction of motion (from nose to middle), so the clock that's pushed from tail to middle will tick slightly slower as it's brought together with the clock that's pushed from nose to middle, which means even though the clock at the tail was originally ahead of the clock at the nose, the time difference decreases as they're brought together until they show the same time at at the middle (and also show the same time as a clock that was fixed at the middle and previously synchronized with the other two in the rocket's frame, before they were moved).

I gave a numerical example showing that slow transport should always cause the clock that gets slowly transported to read the same time as local untransported clocks at rest and synchronized in the frame where its speed is arbitrarily close to zero (in the above example, this means if you had a row of synchronized clocks on board the rocket, then during the process of transporting two clocks at either end to the middle, they would always read the same time as whichever of the other clocks in the row they were passing next to), even if you analyze the sitation from a frame where the transported clock's speed is not close to zero and the untransported clocks are not in sync, in post #37 of this thread, and a more general proof at the end of post #41.

 

[yuiop]

I think (IMHO) that neopolitan talks about two clocks synchronized from the perspective of an on-board observer , i.e. inside the rocket.
This is why, he claims , correctly (IMHO) that once the two clocks are "slow transported" to the center of the rocket, they will be still synchronized (by the very definition of slow clock transport).This is regardles as to how long the transport took.

He further claims that, since the clocks are now collocated , they are synchronized in any frame of reference.

If this is what neopolitan is saying, I think (again, IMHO) that he is correct.

You are correct and so is neopolitan in this particular aspect and I never disagreed with this view. In the last paragraph of post #63 I showed I also agreed with this aspect in this sentence --> "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."

...

What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.

Is this not paradoxical?

You may understand that it is difficult to accept this as objectively true.

cheers,

neopolitan

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.

 

[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]

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

[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

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

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

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:

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

 

[yuiop]

....

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

However, the extent to which they are not synchronised will not be affected either.

Correct

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.

 

[yuiop]

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:

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.

 

[1effect]

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).

 

[yuiop]

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:

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:

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.

 

[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).

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

 

[1effect]

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.

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. :-)

 

[Dale]

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.

 

[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]

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.

 

[yuiop]

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.

 

[neopolitan]

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.

 

[neopolitan]

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

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

 

[yuiop]

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.

 

[Dale]

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.

 

[neopolitan]

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

 

[neopolitan]

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.

See earlier post on acceleration, I don't think that this is what I had in mind since, as DaleSpam pointed out, we have not sufficiently defined the scenario. When I wanted a rocket that didn't compress due to the acceleration, I didn't mean that I somehow wanted to do away with length contraction. The reasoning behind this is that the rocket will compress mechanically if the motor is at the tail (if the rocket is pushed) and stretch mechanically if the motor is at the nose (if the rocket is somehow pulled) and I wanted to eliminate the differences involved.

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".

First, thanks for coming out in support. I do appreciate that.

The problem, however, is that we have a disagreement about the meaning of "the nose clock is in the future of the tail clock". I explained what I meant to JesseM in post #42.

If you read that post, you can hopefully see why I think that, in the scenario you describe above, the nose clock is in the future of the tail clock.

cheers,

neopolitan

 

[yuiop]

Hi neopolitan,

Essentially the method of acceleration i was describing is born-rigid acceleration as described in the link. I was totally ignoring Newtonian compression forces under acceleration and that is why I mention allowing time for the rocket to settle to its natural length contracted time. For example a fast accelerating rear wheel drive car will get shorter (or a front wheel drive car get longer) for reasons that have nothing to do with relativity. The impossibility of ideal born-rigid motion is due to consideration of those Newtonian compression and tension effects under acceleration. Ideal born rigid acceleration would require each particle of the rocket to have its own rocket motor. However, we can ignore those petty details if we allow the rocket to cruise for a while after acceleration to settle down.

Dalespam correctly points out that how the clocks go out of sync during acceleration depends on the acceleration scheme that is employed. However, I have been looking at acceleration methods in relation to this topic and I have come to this conclusion:

There is no acceleration scheme that can be applied to two clocks (in flat space) (that are spatially separated by a non zero distance along the x axis) that will keep the clocks syncronised as viewed by an observer co-moving with the clocks when the clocks have reached a constant and equal (non zero) velocity along the x axis, relative to the initial frame.

 

[Dale]

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.

It is indeed impossible to have a perfectly rigid object, but it is theoretically possible to have perfectly rigid motion. If you read more about Born rigidity you will notice that it is always used to describe motion rather than objects.

If you look at the diagram on the page I linked to such a motion is possible, but each particle would require it's own "engine".

 

[neopolitan]

It is indeed impossible to have a perfectly rigid object, but it is theoretically possible to have perfectly rigid motion. If you read more about Born rigidity you will notice that it is always used to describe motion rather than objects.

If you look at the diagram on the page I linked to such a motion is possible, but each particle would require it's own "engine".

Would not then each particle in the engine require an engine, and so on. Which makes it sort of impossible?

cheers,

neopolitan

 

[Dale]

Would not then each particle in the engine require an engine, and so on. Which makes it sort of impossible?

Sure, it is practically impossible, but not logically or theoretically impossible. Additionally, it still makes sense to understand Born rigid motion as a close approximation to easily achievable situations, e.g. a short stiff rocket with gentle thrust.

On the other hand a perfectly rigid object is not even theoretically possible and inherently leads to logical contradictions. Such an idea is generally not useful even as an approximation.

 

[yuiop]

Dalespam is right that that gentle acceleration of a short rocket is a good aproximation to born rigid motion. We could also achieve a very close aproximation by accelerating some particles in a particle accelerator with precisely timed dynamically changing electromagnetic fields. On a larger scale we could accelerate two unconnected clocks, each with their own rocket so that the clocks maintain constant proper separation. If we ignore the rockets then the two clocks would be undergoing close to ideal born rigid motion. Gravity acts on an object so as to accelerate each particle of an object individually so that is close to the ideal of having a little hypothetical rocket for each particle. Unfortunately the tidal effects of gravity tend to stretch an object so the motion of a falling object is probably not born rigid motion.

 

[neopolitan]

Given that we seem to have come to a reasonable conclusion on the acceleration strand, can we return to the situation which I raised a while ago in which there is no acceleration, just two inertial frames.

Is there still dissent about what I mean with "the nose clock will be in the future relative to the tail clock"?

Note that post #42 applies.

cheers,

neopolitan

 

[yuiop]

Given that we seem to have come to a reasonable conclusion on the acceleration strand, can we return to the situation which I raised a while ago in which there is no acceleration, just two inertial frames.

Is there still dissent about what I mean with "the nose clock will be in the future relative to the tail clock"?

Note that post #42 applies.

cheers,

neopolitan

To be honest, I do not agree or disagree with you on the grounds that I am not clear on what you are getting at. Could you try explaining again?

Do you mean a situation where a rocket is going past you and you observe the nose clock when it is level with you and later the tail clock as it passes your location?

 

[Dale]

Is there still dissent about what I mean with "the nose clock will be in the future relative to the tail clock"?

Note that post #42 applies.

Why would we go back to that discussion? Was there something unclear or wrong about the diagrams in post 39? If not then all further discussion is purely semantics, which I am not interested in arguing.

 

[yuiop]

Why would we go back to that discussion? Was there something unclear or wrong about the diagrams in post 39? If not then all further discussion is purely semantics, which I am not interested in arguing.

There is nothing wrong or unclear with your arguments or diagrams in post #39. Your viewpoint is clear. I am just not clear what neopolitan's viewpoint is, but you are probably right that is just semantics that is causing the problem.

 

[neopolitan]

This is the relevant section from #42.

To try to clarify again, in a now moment in the observer's frame (all now moments are relative, since "now" changes all the time), the observer may observe the tail clock reading 10s and the nose clock reading 2s. IF the clocks are synchonised relative to their rest frame - noting that the observer can work this out from the relative velocity of the clocks and their apparent separation from each other - THEN the observer can further deduce that the nose clock he sees "now" is a younger version of the nose clock and an older version of the tail clock (the observed nose clock manifests earlier in the clocks's rest frame than the observed tail clock - in our example 8s earlier). The nose clock, if you like, has reached the observer's "now" before the tail clock has.

I am sorry to have to do this, but I hope I can justify it. Let's introduce a third clock - on the rocket, in the midpoint between the nose and the tail. That clock will read a midpoint value. Without thinking too deeply about the specifics, I suspect it is 6s (midway between 10s and 2s) but the acutal reading is immaterial - what is important is that it is more than 2s and less than 10s.

If the observer not at rest relative to the clocks observes a reading of 6s on the midpoint clock, 2s on the nose clock and 10s on the tail clock - and knows from his deductions that in their own rest frame the clocks are synchonised then he can say, taking the midpoint clock as his reference, that the nose clock he "should" (see since the clocks are synchronised) is in the future and the tail clock he "should" see is in the past. Whose past and whose future? the past and future of the observer.

What that observer sees, as you point out (I think), is a past version of the nose clock, relative to the midpoint clock, and a future version of the tail clock, relative to the midpoint clock.

You don't really need the third clock, since the same logic applies with only two points, but hopefully the temporary introduction of a third clock makes it easier to understand.

DaleSpam is most probably right, we are probably arguing over semantics.

The situation described is equivalent to that illustrated in the first diagram in #39. I doubt the validity of the second diagram to be honest, since the perspectives are a bit screwy. In diagram one there is the perspective of the external observer observing the rocket and the perspective of the rocket. In diagram two there seem to be "corrected" perspecitives and I wonder if that is valid (since if there were a rocket in each frame, pointed in opposite directions, they would both, in my scenario, consider that other's nose clock is in the future. It is not so that one frame's observer would think that the second frame's nose clock is in the future and the second frame's observer would think that the first frame's nose clock is in the past).

I do want to go further, but I would like to get over this question of whether a clock that reads less is in the future or in the past, from an external observer's frame. Is it at all possible to go from what I have written in the quoted text above?

If necessary I can try to rephrase it, if it is not sufficiently clear.

cheers,

neopolitan

 

[Dale]

I doubt the validity of the second diagram to be honest, since the perspectives are a bit screwy. In diagram one there is the perspective of the external observer observing the rocket and the perspective of the rocket. In diagram two there seem to be "corrected" perspecitives and I wonder if that is valid (since if there were a rocket in each frame, pointed in opposite directions, they would both, in my scenario, consider that other's nose clock is in the future. It is not so that one frame's observer would think that the second frame's nose clock is in the future and the second frame's observer would think that the first frame's nose clock is in the past).

The second diagram is the rocket's rest frame where v=0. The solid outline in the second diagram shows the position of the rocket at a given instant in the rest frame.

The first diagram is the observer's frame where v=.6c. The solid outline in the first diagram shows the position of the rocket at a given instant in the observer's frame.

The dashed outline in each diagram is the solid outline from the other diagram, so the dashed outline in the first diagram shows the rocket's perspective and the dashed outline in the second diagram shows the external observer's perspective. The two diagrams show the same two things in different frames.

What specifically do you think is wrong with the second diagram?

 

[yuiop]

If a rocket is moving relative to us and an explosion occurs at the nose and tail of the rocket simultaneously in the rocket frame, then we would see the tail explode first and then at a later time we would see the nose explode. From the time we saw the tail explode to the time we see the nose explode are we not seeing a past version of the nose that has not yet exploded?

 

[neopolitan]

If a rocket is moving relative to us and an explosion occurs at the nose and tail of the rocket simultaneously in the rocket frame, then we would see the tail explode first and then at a later time we would see the nose explode. From the time we saw the tail explode to the time we see the nose explode are we not seeing a past version of the nose that has not yet exploded?

We specified earlier that we took into account travel times for photons from each clock.

cheers,

neopolitan

 

[neopolitan]

The second diagram is the rocket's rest frame where v=0. The solid outline in the second diagram shows the position of the rocket at a given instant in the rest frame.

The first diagram is the observer's frame where v=.6c. The solid outline in the first diagram shows the position of the rocket at a given instant in the observer's frame.

The dashed outline in each diagram is the solid outline from the other diagram, so the dashed outline in the first diagram shows the rocket's perspective and the dashed outline in the second diagram shows the external observer's perspective. The two diagrams show the same two things in different frames.

What specifically do you think is wrong with the second diagram?

I may have it wrong but in the second diagram it seems you are saying "if the nose and the tail of the rocket are simultaneous in the external observer's frame, this is what happens in the rocket's frame" - the clocks in the rocket's frame will not be synchronous, and being synchronous in the rocket's frame was inherent in the scenario.

cheers,

neopolitan