Discuss events which are simultaneous in one frame?

[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

 

[Dale]

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"

Yes, that is the dashed outline in the rocket frame diagram.

the clocks in the rocket's frame will not be synchronous, and being synchronous in the rocket's frame was inherent in the scenario.

The rocket clocks are synchronous in the rocket frame, that is the solid outline in the rocket frame diagram. The clocks on the rocket read the corresponding t' value. So you can see that in the rocket frame diagram the clocks are synchronized (solid outline) at t'=2, while in the observer's frame the rocket clocks are not syncronized (solid outline) with the nose reading approximately t'=0.5 and the tail reading approximately t'=3

 

[neopolitan]

Ok. Happy with that, DaleSpam. Putting that aside, since there is no dissention, I will restate what I had a couple of posts ago, with the hope that we can go further.

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.

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 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.

When you talk about future or past it is a comparison between two events. So which two events are you referring to?

My guess is that you are referring to two spacelike separated events in which case the distinction between future or past is frame-variant.

 

[neopolitan]

When you talk about future or past it is a comparison between two events. So which two events are you referring to?

My guess is that you are referring to two spacelike separated events in which case the distinction between future or past is frame-variant.

Any two simultaneous events at the nose and the tail of the rocket, where the clocks are, which are described as simultaneous in the rocket's frame - in which case two simultaneous events are neither in the past nor the future relative to each other. The point of the synchronised clocks is that, when a simultaneous reading is taken in the rocket's frame, they read the same.

When two simultaneous events are observed from the external observer's frame, the clocks do not read the same. However, with knowledge of the frame's relative velocity, the external observer can figure out that the clocks are indeed synchronous, in their own rest frame.

Then, thinking a little more, the external observer can figure out that he simultaneously sees one clock which is - figuratively - transported through time, by the effects of relativity, into the future relative to the other clock.

Let's think about two time travellers, just for the purposes of nutting out which clock is in the future. Both hold clocks which are initially synchronised. Both of them travel through time at different rates and compare clocks when they meet again. Which one gets to the future "first"? (relative to an external observer - remember I really wanted to only have one observer to get around problems with people wanting to swap observation points around?) Specifically, which could be said to be in the future - the time traveller who ends up with 2s on his clock, or the time traveller with 10s on his clock?

Clearly the one who has 2s on his clock is in the future, if the other clock reads 10s.

This is what I mean. On the rocket, the nose clock reads less than the tail clock. The nose clock is in the future relative to the tail clock.

Is that any clearer?

cheers,

neopolitan

PS Since this seems to just go round and round, I may move on to my next point anyway. Not tonight though.

 

[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?

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

cheers,

neopolitan

I always take light travel times into account except on accasions where we are expliicitly talking about optical illusions as for example in Penrose-Terrell rotation.

In the example I gave you have to imagine you have a whole grid of clones all at rest with respect to each other and all with clocks syncronised to each other. When the explosion occurs at the tail and nose of rocket (simultaneously as far as observers on the rocket are concerned) one of your clones is standing right next to the tail when it explodes and another is standing right next to the nose when it explodes. Light travel times are not required. The clone that was standing next to the tail of the passing rocket when it exploded says the tail exploded at 3 PM. The clone that was standing next to the nose of the passing rocket when it exploded says it exploded at 3.30 PM.

 

[Dale]

Any two simultaneous events at the nose and the tail of the rocket, where the clocks are, which are described as simultaneous in the rocket's frame - in which case two simultaneous events are neither in the past nor the future relative to each other.

OK, I modified my previous drawings to identify two such events. In the second drawing, the rocket's rest frame, you can see that the tail event (yellow dot) and the nose event (green dot) are simultaneous (t'=2) in the rocket frame. In the first drawing, the outside observer frame, you can see that the nose event is unambiguously in the future of the tail event in the observer frame.

Of course, the events are spacelike separated so other frames will disagree about the ordering.

Let's think about two time travellers, ...

You have got to be kidding. Do you just enjoy adding unnecessary confusion?

 

[neopolitan]

You have got to be kidding. Do you just enjoy adding unnecessary confusion?

No, but I do know that people think in different ways. It seems you don't need to to understand what I mean.

cheers,

neopolitan

 

[neopolitan]

I always take light travel times into account except on accasions where we are expliicitly talking about optical illusions as for example in Penrose-Terrell rotation.

In the example I gave you have to imagine you have a whole grid of clones all at rest with respect to each other and all with clocks syncronised to each other. When the explosion occurs at the tail and nose of rocket (simultaneously as far as observers on the rocket are concerned) one of your clones is standing right next to the tail when it explodes and another is standing right next to the nose when it explodes. Light travel times are not required. The clone that was standing next to the tail of the passing rocket when it exploded says the tail exploded at 3 PM. The clone that was standing next to the nose of the passing rocket when it exploded says it exploded at 3.30 PM.

Ah, kev, I have reread what you had written and see what you mean.

Yes, the tail explodes first. Then the nose. This is consistent with what I said since the explosion of the nose happens in the future relative to the tail (, ieafter the explosion of the tail).

Sorry about that, it was my misreading.

cheers,

neopolitan

 

[neopolitan]

Ok then, time to move on.

With reference to #58 (mine), #59 (kev's) and #61 (DrGreg's), in which I asked about the validity of the concept of an event-space which constitutes an "instant" or a "surface of simultaneity" and I was told that the concept of "hyper-surfaces" of simultaneity "is standard, mainstream special relativity":

is it not so that every event that is bounded by the hyper-surface of simultaneity has already happened and is therefore immutable? (I think it applies to whoever's hyper-surface, but let's take it one observer at a time, first any single observer in an inertial frame - it might as well be me.)

I think this also applies to every event on the hyper-surface also. The distinction between "now" and "a fraction of a picosecond ago" is reasonably minor. I don't think that even quantum uncertainty really implies that outcomes are undefined until they are observed. (If so this would imply instantaneous communication at a distance, since information about an outcome would radiate outwards, undefined, until the first "observer" is encountered and then the defined outcome would somehow have to be communicated to all packets of information being dispersed.)

So, restating: has every event bounded by my hyper-surface of simultaneity already happened? If so, is this past of mine immutable?

cheers,

neopolitan

 

[Dale]

I am not sure I understand your question. In classical physics, including relativity, the future is as immutable as the past. It follows inevitably and deterministically from the initial-condition of the past according to all the laws of physics.

What do hyper-surfaces of simultaneity have to do with immutability?

 

[DrGreg]

The attached diagrams might just clarify the notions of "past", "present" and "future".

Everything is relative to one specific event (a place and a time) E.

The diagram on the left shows two dimensions of space horizontally and one dimension of time vertically. The diagram on the right is a vertical cross-section through the first diagram, and shows one dimension of space horizontally and one dimension of time vertically.

The red conical region marked "absolute past" contains all events that occurred indisputably before E. All observers agree on this. A signal can be sent, no faster than the speed of light, from any event in the absolute past to reach event E.

The green conical region marked "absolute future" contains all events that will occur indisputably after E. All observers agree on this. A signal can be sent, no faster than the speed of light, from event E to any event in the absolute future.

The region in between, which some people call "elsewhen", can be further subdivided, but each observer makes their own divisions and they do not agree with each other. Each observer can decide the "relative past", "relative future", and separating them the blue "hyper-surface of relative simultaneity", the "relative present", or "now". The distinction between past, present and future depends on the conventions and procedures adopted by each observer. Each observer has their own definition and those definitions are incompatible. Each observer draws the blue plane at a different angle through E (but never within the red or green cones).

(The two cones and the blue disk continue outwards to infinity.)

So, if two observers pass by each other, meeting at event E, one oberver might say that some other event F had already happened before E while the other would say it had yet to happen after E. However they would each only be able say this retrospectively, some time after event E. At the moment of E, neither would be aware of event F, as it would not then be in the absolute past, and so no signal could ever travel from F to E.

The only events that an observer can ever be aware of are events in the absolute past. All other events are yet to be detected. You can only ever retrospectively decide that two past events must have occurred at the same time according to your definition of simultaneity. You have no way of knowing what is happening "now".

In a sense, you can regard anything that is not in the absolute past as being "in the future", in that we can only try to predict it and cannot measure it.

Simultaneity is really an artificial man-made concept; it is whatever we define it to be, and doesn't have much physical significance.

In classical physics, including relativity, the future is as immutable as the past. It follows inevitably and deterministically from the initial-condition of the past according to all the laws of physics.

That's one way of looking at classical (non-quantum) physics, although it goes against our intuitive notions of "free will". In quantum theory, the future is less certain.

 

[Dale]

Simultaneity is really an artificial man-made concept; it is whatever we define it to be, and doesn't have much physical significance.

I agree 100% with this, in fact I cannot think of any physical significance at all. IMO, the universe cares about causality, and since two simultaneous events by definition cannot be causally connected the universe simply doesn't care about the ordering.

 

[neopolitan]

Sorry about the long reply time, I have been and am still a bit sick with a head cold.

DrGreg,

I have like most here seen that diagram before but usually to explain the concepts "spacelike", "timelike" and "lightlike". If I have it right, in my currently delicate state, labelling an event "spacelike" indicates if I had the will and resources available, I could change my inertia in such a way as to reach that event as it happened. "Lightlight" events have such separation that I would need to attain lightspeed to reach an event as it happened and "timelike" events have such separation that I cannot reach them, even if I could reach lightspeed. The "timelike" events are your absolute past and absolute future, if I have it correct.

If so, then the definition of simultaneous (earlier in the strand) is such that only what you call absolute past is in the past, the lightlike cone bordering the absolute past is "now" and everything else is in the future - for me.

The diagram only applies for one inertial ("rest") frame, every inertial frame will have a similar diagram that applies with the only difference being the distribution of events within the "relative past" and "relative future" sections. What doesn't change is the "absolute past" and the "absolute future", if I have it correct, of course.

If that is the case, then the cone which bounds the "absolute past" and the "relative past" will be the same for all inertial frames, which in turn means "now" is the same - as we defined it earlier. I would have thought that this had some sort of physical significance.

I have in mind another conceptualisation using three events (an unprimed observer nominally at rest, a primed observer in a frame which is not at rest relative to the unprimed observer and an observed "now" event such that the the unprimed observer calculates where the primed observer thinks he is at time "now" and where he thinks the observed event takes place, and whether the primed observer also considers the event to be a "now" event). Just at the moment, however, I am too foggy to be reliable - so it will have to wait.

I await your comments,

cheers,

neopolitan

 

[DrGreg]

Sorry about the long reply time, I have been and am still a bit sick with a head cold.

As I log on only for a short while once a day, any response within 24 hours is fast for me! Hope you are well soon.

I have like most here seen that diagram before but usually to explain the concepts "spacelike", "timelike" and "lightlike". If I have it right, in my currently delicate state, labelling an event "spacelike" indicates if I had the will and resources available, I could change my inertia in such a way as to reach that event as it happened. "Lightlight" events have such separation that I would need to attain lightspeed to reach an event as it happened and "timelike" events have such separation that I cannot reach them, even if I could reach lightspeed.

The wrong way round. If F is the event we are measuring relative to E then the vector EF is timelike if F is in the absolute past or absolute future of E, lightlike = null if it lies on the surface of one of the cones, or spacelike otherwise.

For any forward-timelike direction in spacetime there is an inertial observer who could travel along it. For such an observer the direction would lie along his or her t-axis. Anyone could, with enough energy, accelerate from E to arrive at F.

For any spacelike direction in spacetime there are no inertial observers who could travel along it. Nobody could ever accelerate from E to arrive at F. But there is an inertial observer for whom the events E and F occur simultaneously. For such an observer the direction would lie within his or her xyz-hyperplane, i.e. their "hypersurface of simultaneity", their "relative present".

The "timelike" events are your absolute past and absolute future, if I have it correct.

Yes! To be more accurate, it is the vector (or displacement, or offset, or difference) between two events that is "timelike" (or spacelike or null). Where only one event is mentioned, the other is taken to be the observer "at time zero".

If so, then the definition of simultaneous (earlier in the strand) is such that only what you call absolute past is in the past, the lightlike cone bordering the absolute past is "now" and everything else is in the future - for me.

Sorry, I haven't read in detail every post in this thread. Could you point out which post you mean? The standard definition of simultaneity in special relativity is what I labelled the "relative present" in my diagram, the blue hyperplane. It is not the backwards light cone surface. Maybe this is why you are still having some difficulties.

It is possible to define "simultaneous" in some other, non-standard way, and as long as everyone understands which definition is being used, that is not a problem, in theory. But in practice most people are so used to the standard definition, they will find it extremely difficult to think in terms of a different definition. In all discussions of relativity, the standard definition is assumed unless explicitly stated otherwise (and that is very rare).

If you choose to define "simultaneous" as lying on the surface of the backward light cone, that is actually a form of absolute simultaneity amongst observers passing through event E, because they all agree what the light cone is. Under that definition, if you see, with your eyes, two events at the same time, those events are deemed "simultaneous". However, it is a peculiar definition because it is not "transitive": if an observer at E says that F occurs "simultaneously", an observer at F does not say that E occurs "simultaneously"! Also, under your definition, if you think about it, the "speed" of light towards you would be infinite -- the emission and reception occur "simultaneously". (It would be c/2 away from you.)

 

[neopolitan]

I need to think about this while my head is clearer. I will look more closely at the implications of the standard definition of simultaneity.

The one we were using is described at post #1 and confirmed as the right one of two options by JesseM in post #2.

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.

cheers,

neopolitan

 

[DrGreg]

The one we were using is described at post #1 and confirmed as the right one of two options by JesseM in post #2.

Hmm. I think you are misunderstanding what JesseM was trying to say then. Even you admit, post #1:

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.

That is entirely correct. If you ignored where the photons were released you would get the red surface, what you called "reception simultaneity" in post #1. As soon as you apply your knowledge of where the photons were released you get the blue surface instead (on the assumption that the speed of light is a constant c in all directions).

It should be emphasised that you can only construct the blue surface retropectively some time after E has occurred, when you, the observer, are further up the diagram.

 

[neopolitan]

Actually I don't think I was misunderstanding JesseM (certainly not with post #2). I may have been confused over the past week, more so than usual :) and I feel like it could be the case right now.

The lower cone, which bounds the absolute past, represents all the events for which information reaches me now - assuming that I am at the origin. That makes those events absolute past, since I observe them now (event E being "now"). Yes?

Any other observer at event E, irrespective of inertial frame will also observe all those events making them absolute past. This, I think, gives double meaning to the tag "absolute past", although I am not sure that it was intended so. You can't get more "past" than "I saw it happen" and this past is not relative since any frame will consider it the past, with the precondition of collocated observers.

Then, I think, we get to a point that DaleSpam was making elsewhere. If I have it right, he says there can be no physical meaning to "when". The blue surface is a recreation, after the fact, not something that actually existed. All that matters really are time intervals between events.

Does the same apply to spatial intervals? Is the concept "where" somehow more physical than "when" and, if so, in what way?

cheers,

neopolitan

BTW - I want to come back the reconstructing the blue surface concept later. There is something about the diagram and the speed of light that strikes me as odd, but I want to be clearer headed before I take it on.

 

[neopolitan]

I believe that I am past the worst of my headcold, so here goes.

Dr Greg, you pointed out that the blue surface must be reconstructed later, I assume this is using one's own value for the speed of light (which in one's own frame will always be c).

The red surface represents a set of events from which photons reach one at event E, where the cones meet. Since photons from those events have reached the observer at E, those events and all those events inside the cone are irrefutably the past, the "absolute past" as you label it. The red surface photons reached the observer after traveling at "observer speed of light".

As I understand it, an observer in another inertial frame who is collocated at E, would create another diagram which looks entirely the same using "second observer speed of light". This will, from the point of view of the first observer, skew the cone, so that while the second observer has a blue surface which is not horizontal, the second observer's red cone will fit neatly over the first observer's red cone. They will agree that the events are all in the past, but not over "when" or "where" all those events took place. The diagram is rough but shows what the second observers diagram looks like in terms of the first observer, as far as I can work out. Is that right?

Now, another thing about the diagram is that it is not just frame dependent but also location dependent. Say I shared a frame with someone else, so that we were both at rest with respect to each other but not collocated (I don't mind two observers in one frame, since we don't get tempted to swap perspectives between frames without mentioning it). We would have linked events, E1 and E2, such that the only difference betwee E1 and E2 is the spatial offset. Both of us would have cones of "absolute past" representing all the photons which reach us at events E1 and E2. Now, since we are in the same frame, both at rest, is it not reasonable to say that what is in my colleagues absolute past is also in my absolute past? If all the photons from his "absolute past" reach him together at the same time that all the photons from my own personal "absolute past" reach me, and we are at rest, then the events from which those photons were emitted are irrefutably in the past.

If we take that a step further and conceptually have an infinitite number of observers at rest with me (on my blue plane), each with their own cone of "absolute past", do we not build up to a situation where everything below my blue surface is -at least in effect- a composite "absolute past"?

You might think that that means that an observer in motion relative to me will build up a similar diagram and come to the conclusion that what is in my "relative future" is in his composite "absolute past", but the skewing of the cone and reconciliation of the disagreements about simultaneity will result in overall agreement about which events are in a composite "absolute past", leaving just disagreement about "where" and "when" they actually happened.

Diagram two shows, very roughly, three observers in a shared frame. You hopefully can see how if you have enough observers (and real observers are not required, of course), you end up with a composite "absolute past" which completely fills the area under the blue surface.

cheers,

neopolitan

 

[Dale]

As I understand it, an observer in another inertial frame who is collocated at E, would create another diagram which looks entirely the same using "second observer speed of light". This will, from the point of view of the first observer, skew the cone,

The second observer speed of light = the first observer speed of light = c. So from the point of view of the first observer the light cone is unchanged.

so that while the second observer has a blue surface which is not horizontal, the second observer's red cone will fit neatly over the first observer's red cone. They will agree that the events are all in the past, but not over "when" or "where" all those events took place.

This is correct.

The diagram is rough but shows what the second observers diagram looks like in terms of the first observer, as far as I can work out. Is that right?

In the diagram the red should be unskewed, and the blue should be skewed.

Now, another thing about the diagram is that it is not just frame dependent but also location dependent.

Well, this is a little complicated. Different features of the diagram behave differently. The light cones are indeed event-specific (not just location but time), however they are not frame dependent. Because the speed of light is the same in all frames all frames agree on the light cones. The plane of simultaneity is frame dependent and also dependent on the coordinate time of the event. So the diagram as a whole is indeed frame dependent and event dependent, but you have to be careful about generalizing that overall statement when referring to different parts of the diagram.

Now, since we are in the same frame, both at rest, is it not reasonable to say that what is in my colleagues absolute past is also in my absolute past?

No, this is incorrect, as is all that followed. At a given event E, events which are not inside or on the past light cone are not absolute past. However, if they are in your colleague's absolute past then they are in your relative past, not your future or relative present.

 

[neopolitan]

No, this is incorrect, as is all that followed. At a given event E, events which are not inside or on the past light cone are not absolute past. However, if they are in your colleague's absolute past then they are in your relative past, not your future or relative present.

If a range of events are observed by an observer at rest relative to me, but with a non-zero separation, how can they meaningfully not be as equally in my past as they are in that observer's past? What am I saying is that I don't agree about there being a meaningful distinction between absolute past and relative past for any single observer.

Perhaps you could call the events inside the red cone "observed past" since the photons from them have already gone by. Everything under the blue surface would then be "unobserved past" but the reason they are not observed is not because they are in the future, but because their spatial location has made it impossible for the photons to reach us yet. Then above the blue line, not inside the upper cone, are unreachable future events, to reach which we would have to travel faster than the speed of light. Inside the cone, are reachable future events.

No matter how fast another observer is, they cannot reach events that are for us unreachable. They may consider themselves stationary whereas we think they are traveling at 0.8c, but that does not mean that they can jump in a spacecraft and zoom off at a relative 0.7c and, adding that to the velocity that we think they have, thereby reach 1.5c.

The only reason why someone can observe what is, for me, the "unobserved past" is because they are located between me and that "unobserved past". Being observed does not really make the past absolute. Having happened makes it absolute.

I think your comment here was made a little too quickly.

I am tempted to give another conceptualisation, but I would prefer a response to this first.

cheers,

neopolitan