Discuss events which are simultaneous in one frame?

[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

 

[Dale]

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.

There are several meaningful differences. One very important difference is that absolute past events are timelike separated and relative past events are spacelike separated. Another very important difference is that the absolute past can cause the event and the relative past cannot. A third very important difference is that the absolute past is frame invariant while the relative past is frame variant.

PS If you had instead said that there was no "meaningful distinction between relative future and relative past" I would not have been able to use any of these distinctions as counter-examples.

 

[neopolitan]

There are several meaningful differences. One very important difference is that absolute past events are timelike separated and relative past events are spacelike separated. Another very important difference is that the absolute past can cause the event and the relative past cannot. A third very important difference is that the absolute past is frame invariant while the relative past is frame variant.

PS If you had instead said that there was no "meaningful distinction between relative future and relative past" I would not have been able to use any of these distinctions as counter-examples.

Ok, we seem to be looking at the diagram from different perspectives which leads to different interpretations.

The intersection of the two cones is an Event E. I personally was looking at Event E as being me at a particular, but unspecified, point in time. I agree that all the events in the cone are events which include the past version of me, and therefore only events within the cone could have caused me. This is not what I meant though.

I also agree about the distinction between spacelike and timelike separations (between Event E and events in the relevant zones) but this is just saying the same thing using fancier words.

That leaves "the absolute past is frame invariant while the relative past is frame variant". Actually, I am not so sure about that. I am not so sure about the blue surface being skewed for observers in other frames, either.

Indulge me, if you could be so kind.

Think about the meaning of the lorentz transformations. What are they describing?

Consider an event F, at a distance (x=x) from event E (t=0, x=0) such that the information about that event is received at a third event E2 (t=t, x=0) which means that x=ct. At event E, there are two observers collocated, but not in the same frame. All the details given so far relate to an observer who is nominally at rest. The other observer is nominally in motion, with a velocity of v, relative to the "rest" observer, in the direction of event F.

The observer who is nominally in motion observes event F at t'=(t-vx/c^2).gamma and considers its location to be x'=(x-vt).gamma.

Since x=ct, then t=x/c and
t'=(t-vx/c^2).gamma=(x/c-v.t/c).gamma=(x-vt).gamma/c=x'/c
so x'=ct'.

Do not both of them therefore restrospectively consider event F to have been an event which was simultaneous with event E (since x=ct and x'=ct'). Since x is not bounded, then this applies to all values of x (and by extension all values of x'), so I really do not think that blue surface is frame variant at all.

If the blue surface is not frame variant, then neither is the "relative past".

Standing by to have my error pointed out to me.

cheers,

neopolitan

 

[Dale]

Since x=ct

Only for light. This is not a general proof. You are actually only proving here that the light cone is frame invariant, which we already agreed on.

Here is a prooof by counter example. Let E=(0,0) and let F=(0,1), so F is simultaneous with E in the unprimed reference frame. Let the primed reference frame be moving at v=-0.6c wrt the unprimed frame, then E'=(0,0) and F'=(0.75,1.25). So E and F are simultaneous in the unprimed frame, but F is in the relative future of E in the primed frame.

 

[neopolitan]

I think you need to read what I wrote again. Your counter proof doesn't follow since you are not considering the same events that I considered.

Please take another look, try to work out what I was saying. Note that your comment "only for light" indicates that you have either misread or misinterpreted what I wrote. I was discussing observations and the transmission of information, light is involved.

cheers,

neopolitan

 

[Dale]

I think you need to read what I wrote again. Your counter proof doesn't follow since you are not considering the same events that I considered.

Huh? Yes I did. You said E=(0,0) and F=(0,x). If your arguments fail for x=1 then they are not valid in general.

Actually, why don't you start over. Perhaps I am misunderstanding your point. I certainly have no idea why you put E2=(x/c,0) in the problem.

But I guarantee, choose any event in the absolute past and I will not be able to find a reference frame where it is not still in the absolute past, and specify any event in the relative past or relative present and I will be able to find a reference frame where it is in the relative future.

 

[Dale]

Here is a proof that the absolute past is frame variant:

Without loss of generality choose units where c=1, let E be the origin of both reference frames, set the x, y, z axes parallel to the x', y' ,z' axes respectively, and consider only a boost of speed v along the x-axis.

The absolute past of E is bounded by the hyper-surface:
x² + y² + z² = t²
and the event
E = (0,0,0,0)

by the Lorentz transform
t = γ(t'+vx')
x = γ(x'+vt')
y = y'
z = z'

by substitution
γ²(x'+vt')² + y'² + z'² = γ²(t'+vx')²
and
E'=(0,0,0,0)

which simplifies to
x'² + y'² + z'² = t'²
and
E'=(0,0,0,0)

So the absolute past of E has the same form in the primed and unprimed frames. Therefore the absolute past of E is frame invariant. QED

 

[neopolitan]

It seems I have not made myself sufficiently clear. I don't disagree that the "absolute past" of DrGreg's diagram is frame invariant.

Note the post in which I discussed that perhaps this region could be better described "observed past" - since photons from that region have already reached (and maybe gone by) us. The frame invariance of this region is not in question.

You may have misinterpreted what I said in that post:

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.

This was not intended to say that DrGreg's "absolute past" is not absolute, nor that it is not frame invariant. What I was saying is that this region is a subset of a greater region of events which have happened prior to Event E - all of which could therefore be considered absolute past. The only difference between the two regions below the blue surface relates to whether events in the past are observed or unobserved.

DaleSpam, you have still not properly read the post which you are responding to, this is made clear when you state:

Huh? Yes I did. You said E=(0,0) and F=(0,x). If your arguments fail for x=1 then they are not valid in general.

Actually, why don't you start over. Perhaps I am misunderstanding your point. I certainly have no idea why you put E2=(x/c,0) in the problem.

If you don't understand the inclusion of E2, your best option would be to seek clarification first, rather than firing off a response when even you admit that you don't understand what you are responding to.

Here is what I wrote, I will make bold the section which it seems you didn't read or didn't understand. I will also colour code two statements which are linked and have added a comment which may also help. Please let me know if it is still not clear once the important sections are highlighted.

Consider an event F, at a distance (x=x) from event E (t=0, x=0) such that the information about that event is received at a third event E2 (t=t, x=0) which means that x=ct. At event E, there are two observers collocated, but not in the same frame. All the details given so far relate to an observer who is nominally at rest. The other observer is nominally in motion, with a velocity of v, relative to the "rest" observer, in the direction of event F.

The observer who is nominally in motion observes event F at t'=(t-vx/c^2).gamma and considers its location to be x'=(x-vt).gamma. (EDIT: this observation, where the second observer observes Event F, is actually a fourth event, and could be labelled Event E3 - the location where photons from Event F and the observer nominally in motion are collocated.)

Since x=ct, then t=x/c and
t'=(t-vx/c^2).gamma=(x/c-v.t/c).gamma=(x-vt).gamma/c=x'/c
so x'=ct'.

Remember our working definition of simultaneity:

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.

I assume that if I released a photon, I would receive it instantaneously. This is of course not absolutely correct, but compared to a photon I am gargantuan - rather than being a nice point-event. So, I hope we can be sensible and either allow a slight fudging or make the assumption that when I talk about "me" and "my location" we choose one suitable reference point in the general area of where I am. In any case, the slight delay while one of my components emits a photon and another absorbs it is very slight indeed, when compared to a sufficiently large value of x associated with Event F.

cheers,

neopolitan

 

[Dale]

Look Neopolitan, it isn't that I don't understand your events, I don't understand the point you are trying to make. So why don't we just start this part over.

The region labeled "absolute past" in DrGreg's diagram is frame-invariant. In other words, given an event E with an event PA in its absolute past then in any other inertial reference frame PA' is in the absolute past of E'. Similarly with the absolute future. I think we agree on this.

The region labeled "relative past" in DrGreg's diagram is frame-variant. In other words, given an event E with an event PR in its relative past then in some other inertial reference frame PR' is not in the relative past of E'. Similarly with the relative present and future. Do you agree with this?

 

[Dale]

Here is a proof that the relative past is frame variant:

Without loss of generality choose units where c=1, let E be the origin of both reference frames, set the x, y, z axes parallel to the x', y' ,z' axes respectively, and consider only a boost of speed v along the x-axis.

The relative past of E is bounded by the hyper-surface:
x² + y² + z² > t²
and the hyper-surface
t < 0

by the Lorentz transform
t = γ(t'+vx')
x = γ(x'+vt')
y = y'
z = z'

by substitution
γ²(x'+vt')² + y'² + z'² > γ²(t'+vx')²
and
γ(t'+vx') < 0

which simplifies to
x'² + y'² + z'² > t'²
and
t' < -vx'

which is only the same as the expression in the unprimed frame for the special case that v=0.

So the relative past of E does not have the same form in the primed and unprimed frames. Therefore the relative past of E is frame variant. QED

 

[neopolitan]

DaleSpam,

No, I am saying that I am not convinced that the "relative past" is frame variant. I am also not convinced agree that the reconstructed blue surface of simultaneity is frame variant - as it pertains to events (not the labelling of events). That is what I am trying to point out with what I wrote before.

If you understand that, and you understand my events, can we work with that? Or should I try again with a diagram?

cheers,

neopolitan

 

[neopolitan]

I want to edit what I wrote just before more extensively, but that would have left the poorly phrased version there longer.

I think I can see the source of the confusion. I am talking about events, DaleSpam, the actual events. You are talking about the labels for the events.

I agree that the two observers will not agree about when each event happens nor when. To that extent, the surface of simultaneity is frame variant. I agree.

However, the two observers will also disagree about when and where they are collocated - unless they both artificially zero both their clocks and the null point of their x-axes. Despite this disagreement, the two observers will, when collocated at Event E, share the same plane of simultaneity - in that any event that occurs simultaneously with event E in the unprimed frame (as retrospectively calculated) will also be retrospectively calculated as being simultaneous with event E in the primed frame.

Your calculations are not incorrect, they just aren't relevant to what I am saying. Can you see that?

Do you understand what I am saying yet? (I have a diagram in mind, I just need to find the time to devote to it.)

cheers,

neopolitan

 

[neopolitan]

Here is the diagram. The first is most important (simultaneity1.jpg). The second is just showing that I know it looks frightful when you have an event which is not in the direction of the second observer's apparent motion (relative to the first "at rest" observer). It can be because the location of Event "F minus" is -x or it can be because the relative velocity involved is negative, it makes no real difference since in both cases it works out that the event is retrospectively considered to be simultaneous with Event E by both observers.

cheers,

neopolitan

 

[JesseM]

Hi again neopolitan, back from my trip to Mexico...still at my friend's place in NY so I can't talk to much until I get home tomorrow, but I'll respond to this:

Consider an event F, at a distance (x=x) from event E (t=0, x=0) such that the information about that event is received at a third event E2 (t=t, x=0) which means that x=ct. At event E, there are two observers collocated, but not in the same frame. All the details given so far relate to an observer who is nominally at rest. The other observer is nominally in motion, with a velocity of v, relative to the "rest" observer, in the direction of event F.

The observer who is nominally in motion observes event F at t'=(t-vx/c^2).gamma and considers its location to be x'=(x-vt).gamma. (EDIT: this observation, where the second observer observes Event F, is actually a fourth event, and could be labelled Event E3 - the location where photons from Event F and the observer nominally in motion are collocated.)

If x and t in your equations above are supposed to represent specific positions and times in the unprimed frame (the position of the photon at E and the time of E2) rather than variables, then you have the coordinates wrong for E3 here. In the unprimed frame the observer "nominally in motion" has position as a function of time x(t) = vt while the photon has position as a function of time x(t) = x - ct (because you have the photon moving to the left from x=x...your notation would be less confusing if you used labels like x1 and t1 for specific positions and times in the unprimed frame, with x and t always being variables, this is what is normally done in relativity problems, but I'll use your notation for now). So, in the unprimed frame E3 happens when vt = x - ct, meaning t = x/(v+c). Since the observer "nominally in motion" has position x(t) = vt, E3 must happen at position x = vx/(v+c) in the unprimed frame. So, the observer who is nominally in motion sees the light from event F (with E3 being the event of his seeing this light) at the following time coordinate in his own system:

t' = gamma*(x/(v+c) - v^2x/(v+c)*c^2) = gamma*(x*(c^2 - v^2))/((v+c)*c^2) = gamma*(x*(c-v))/c^2

This does not work out to 0, so this is not the same time-coordinate as the event of the observer "nominally in motion" passing next to the other observer at E, which in the primed coordinate system has t' = gamma*(0 - v*0/c^2) = 0. So, your statement here is incorrect:

However, the two observers will also disagree about when and where they are collocated - unless they both artificially zero both their clocks and the null point of their x-axes. Despite this disagreement, the two observers will, when collocated at Event E, share the same plane of simultaneity - in that any event that occurs simultaneously with event E in the unprimed frame (as retrospectively calculated) will also be retrospectively calculated as being simultaneous with event E in the primed frame.

As shown above, at event E the two observers do not share the same plane of simultaneity, since the observer "nominally at rest" thinks F happened at the same time-coordinate as E (t=0), while the observer "nominally in motion" thinks E happened at t'=0 while F happened at t'=gamma*(x*(c-v))/c^2 (where again, x is meant to represent the specific position of the photon at t=0 in the unprimed frame, rather than being used as a variable).

 

[DrGreg]

I've been offline all weekend, so there's a lot to respond to. Here goes:

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?

"Where" is a relative concept, too. If Alice says two events occurred at the same distance from her (i.e. at the same place), Bob, moving towards those events, would say the second event occurred closer to him than the first (i.e. at two different places). It is meaningless to ask whether two distinct* events occur at the same place unless you specify relative to a specific frame or observer. There is no "absolute position", just as there is no absolute time.

*The one exception is that we can say if two events occur at the same place and the same time, something which everyone can agree on; but these are not two distinct events, in fact mathematically they are the same event.

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"?

But theres's something missing here. You talk of two linked events E1 and E2, and their spatial offset. But what about their temporal offset? You know where E2 is relative to E1, but when is it? If we say they are "both in the same frame", this might be interpreted as "they are both at the same t-coordinate". But as soon as you say that, you are making use of relative simultaneity. If you (static in the frame) say E1 and E2 are simultaneous, someone else (moving relative to you) will say they are not. So the problem hasn't gone away, it's just been concentrated onto the events E1 and E2.

Consider an event F, at a distance (x=x) from event E (t=0, x=0) such that the information about that event is received at a third event E2 (t=t, x=0) which means that x=ct. At event E, there are two observers collocated, but not in the same frame. All the details given so far relate to an observer who is nominally at rest. The other observer is nominally in motion, with a velocity of v, relative to the "rest" observer, in the direction of event F.

The observer who is nominally in motion observes event F at t'=(t-vx/c^2).gamma and considers its location to be x'=(x-vt).gamma.

This is a misapplication of the Lorentz transform, in which x and t must refer to the same event. In your example, x refers to event F but t refers to event E2. You need to carry out two separate transformations for F(t=0,x=X) and E2(t=X/c,x=0).

No, I am saying that I am not convinced that the "relative past" is frame variant. I am also not convinced agree that the reconstructed blue surface of simultaneity is frame variant - as it pertains to events (not the labelling of events). That is what I am trying to point out with what I wrote before.

I don't understand your point. Describing an event as "past, present or future" is labelling an event.

Despite this disagreement, the two observers will, when collocated at Event E, share the same plane of simultaneity - in that any event that occurs simultaneously with event E in the unprimed frame (as retrospectively calculated) will also be retrospectively calculated as being simultaneous with event E in the primed frame.

That isn't true (except in the special case when both observers are stationary relative to each other, in which case they are effectively a single observer).

 

[Dale]

Despite this disagreement, the two observers will, when collocated at Event E, share the same plane of simultaneity - in that any event that occurs simultaneously with event E in the unprimed frame (as retrospectively calculated) will also be retrospectively calculated as being simultaneous with event E in the primed frame.

This is precisely what I disproved with my counter-example and the second post of algebra. The hyper-plane of simultaneity is t=0 which transforms to t'=-vx', not t'=0. They do not share the same plane of simultaneity, despite being co-located at E.

I have shown the moving observer's plane of simultaneity on your previous diagram (the first one).

 

[neopolitan]

Ok, I think I am now convinced. I also think I understand where I was screwing up, by having two incompatible concepts in my head simultaneously (ouch, unintended pun).

I must admit it is a relief because I just couldn't reconcile two simultaneous events at rest in the nominally "in motion" observer's frame - which we have previously agreed will not be simultaneous according to the nominally "at rest" observer.

It is around this point where my error lay, I think. An example of the argument I was ready to present (greyed out because I know it is wrong):

The nominally "in motion" observer will consider the spatial separation between event E and event F (say rest value of L) to be foreshortened by Lorentzian contraction (to L'). So, a photon released by event F will, according to the nominally "in motion" observer, only have to travel a shorter period of time to reach the location of event E, a period of L'/c.

I can see this is a mix of frames and that the timing I refer to for the photon to reach E is the time it takes for the photon to reach the nominally "at rest" observer, according to the nominally "in motion" observer - this is not the time it takes for the photon to reach the nominally "in motion" observer according to either observer. When I look at it now, I honestly can't understand why I was messing it up before.

Anyways, DaleSpam's modification of my diagram does clear things up and I thank him for that. Sorry about my continued muddleheadedness.

Now I need try to remember where I was heading with this before I had that headcold. Catching up with work will have to take priority though.

cheers,

neopolitan

 

[Dale]

Anyways, DaleSpam's modification of my diagram does clear things up and I thank him for that. Sorry about my continued muddleheadedness.

You're welcome. Don't worry about it. Studies have shown how difficult the relativity of simultaneity is to learn. It is probably the single most difficult relativity concept.

 

[neopolitan]

Still mulling this over, I have something I want to explore but I need to think how best to express it without causing confusion. And I need the time to sit down and do it. Ten minutes here and there doesn't suffice.

cheers,

neopolitan

 

[neopolitan]

Hm,

I put a bit of effort into a post and it has disappeared. Can someone PM me if they saw it please.

It is possible that in my rush to leave work I merely previewed the post and, in shutting down the computer, then cast it all to oblivion. In that case I will redo it, when I get the time.

cheers,

neopolitan

 

[neopolitan]

I am going to have to use three observers to explain something and pose a question. This is indeed a break with (my) tradition!

My three observers are me (who is nominally at rest, or the NAR observer) and two other inertial observers who are nominally in motion (NIM) so we will call them NIM1 and NIM2. NIM1 and NIM2 do not share the same rest frame, so each is in motion relative to the other (and I am in motion relative to both).

If we look at DaleSpam's modification of my diagram which was a modification of DrGreg's, we can see that it applies to any two of the three observers. (Note that six diagrams can be made depending on whose reference frame is used to form the vertical and horizontal axes and which other observer is being considered, so: me looking at NIM1 (me-NIM1), NIM1-me, me-NIM2, NIM2-me, NIM1-NIM2 and NIM2-NIM1. Three of these diagrams will actually be mirrored versions of DaleSpam's diagram, because the relative velocity of the observer being considered will be negative.)

Event E is when all three of us are collocated.

Event F is an event which I subsequently calculate to have been simultaneous with Event E.

My question is this:

Given that I know the relative velocities of NIM1 and NIM2, will I be able to use their observations to accurately locate Event F (in my reference frame), if I happened to not notice it myself? I can work out when and where each actually is (in my reference frame) when they observe Event F and I know how their perceptions of time and space are skewed by their relative motions.

I am pretty certain that I can.


My mind-experiment question which follows from this:

Say, by staggering co-incidence, I just happen to share a frame with Chronos (who was known by the ancient Greeks to be responsible for Time, but we modern folk know that he actually looks after Space-Time). By virtue of this happy circumstance, my rest frame just happens to be the "one true rest frame".

What, if anything, makes this "one true rest frame" totally impossible? As far as I can tell, you could never detect this "one true rest frame" - or absolute at rest (AAR) frame - since the skewing of all other frames, each of which is nominally at rest in terms of itself, will make the AAR frame appear like any other NIM frame. However, this does not make the AAR frame impossible per se. Does it?

Why are there three observers? Well, this is because I wonder if we could not always have this situation even when we are considering two observers -

Both of them may be in motion relative to some indetectible AAR frame, making both of them absolutely in motion (AIM) relative to the AAR frame. This does not stop us from nominating one of them as "at rest" and the other as "in motion". Despite this arbitrary assignment of NAR and NIM frames, would we not still work out the "absolute" space-time locations of events? - we would just express them in terms of our own frames. Chronos and we would agree where all events take place, we would just express those events differently.

(An analogy which may help is two observers in two boats, adrift on an ocean but currently becalmed, who observe a rock. The observers will be able to say "According to my clock, at time t, the rock is a distance x from me, and q degrees to my left/right". Between themselves, the observers will be able to agree that they both observed the same rock. Chronos stands on a shrouded island and can observe all three, and he can say precisely where all of them are (and his watch also reads the absolute correct time). He will agree that the rock is a certain distance and angle from each of the observers (he can also reconcile any errors on their watches, but since the observers are becalmed, timestamps not really necessary) - and he can precisely locate the rock too. The inability of the observers on the boat to determing their "absolute" location does not necessarily invalidate the fact they they may have an "absolute" location.
- This is meant only to help grasp the concepts - select it to be able to read what it says. If it doesn't help, ignore it. Please address the actual question in responses, not the analogy. Thanks.)

Note that this is all speculation. I am aware that absolutes are not part of relativity and I am not saying there are any. I just would like to hear why there can't be any.

cheers,

neopolitan

 

[Mentz114]

Neopolitan:

Note that this is all speculation. I am aware that absolutes are not part of relativity and I am not saying there are any. I just would like to hear why there can't be any.

Formally in relativity, nowhere does it say there is no absolute frame. What is said is that 'it is impossible to detect inertial motion without reference to another frame'.
In other words, all motion is relative. Therefore, in practice, even if an 'absolute frame' existed, it would not make any difference to the way we percieve things.

(I'm putting on my tin-hat and waiting for the onslaught of the enraged relativists).

 

[Dale]

Given that I know the relative velocities of NIM1 and NIM2, will I be able to use their observations to accurately locate Event F (in my reference frame), if I happened to not notice it myself? I can work out when and where each actually is (in my reference frame) when they observe Event F and I know how their perceptions of time and space are skewed by their relative motions.

Certainly. In fact, you only need one NIM observer. Simply Lorentz transform his coordinates into yours.

What, if anything, makes this "one true rest frame" totally impossible? As far as I can tell, you could never detect this "one true rest frame" - or absolute at rest (AAR) frame - since the skewing of all other frames, each of which is nominally at rest in terms of itself, will make the AAR frame appear like any other NIM frame. However, this does not make the AAR frame impossible per se. Does it?

Both of them may be in motion relative to some indetectible AAR frame, making both of them absolutely in motion (AIM) relative to the AAR frame. This does not stop us from nominating one of them as "at rest" and the other as "in motion". Despite this arbitrary assignment of NAR and NIM frames, would we not still work out the "absolute" space-time locations of events? - we would just express them in terms of our own frames. Chronos and we would agree where all events take place, we would just express those events differently.

This is essentially the Lorentz ether theory, which is experimentally indistinguishable from special relativity. Analytically it goes something like this: set up all reference frames, randomly pick any one, call it the "ether frame", proceed from that frame to make all special relativity predictions. I have no problem with people who prefer the Lorentz interpretation to the Einstein interpretation. Personally, I think it is wise to learn both and use whichever fits the situation best. In my case, I like the Lorentz interpretation for explaining relativistic Doppler effects, but the Einstein interpretation is for most other things.

 

[neopolitan]

Certainly. In fact, you only need one NIM observer. Simply Lorentz transform his coordinates into yours.

Indeed, I explained why I had three later in the post.

This is essentially the Lorentz ether theory, which is experimentally indistinguishable from special relativity. Analytically it goes something like this: set up all reference frames, randomly pick any one, call it the "ether frame", proceed from that frame to make all special relativity predictions. I have no problem with people who prefer the Lorentz interpretation to the Einstein interpretation. Personally, I think it is wise to learn both and use whichever fits the situation best. In my case, I like the Lorentz interpretation for explaining relativistic Doppler effects, but the Einstein interpretation is for most other things.

This is not quite what I am asking. I know that you could select any inertial frame and use that as what could be called an "ether frame". But what I am asking is, is there anything preventing a "one true rest frame" or an "absolute (at) rest frame"?

Then going further, if there is nothing preventing such a frame (even if we cannot distinguish it), would not that frame's hypersurface of simultaneity be the boundary of the universe? It seems to me that the boundary of the universe is more of a "when" question than a "where" question.

This would mean that, in the terms that we normally use for thinking about such things, there would be no 3 dimensional edge to the universe and no 3 dimensional centre. Instead there would be a 4 dimensional edge and a 4 dimensional centre. (It may help to remove one dimension and think of a sphere. The two dimensional surface of the sphere has no centre and no edge. The three dimensional sphere itself, however, has the two dimensional surface as the boundary and the centre of the sphere is surrounded by and separated from the surface - so two dimensional people living on the sphere would never be able to reach the centre of the universe. The centre of our universe would, therefore, be in the past - in a big bang event, or some equivalent to a big bang event.)

The standard cosmological model has the universe expanding like the surface of a balloon. Is what I have expressed above just saying the same thing, perhaps in another way?

I wonder if a few people thought I was wandering away from the simultaneity topic. I wasn't after all.

cheers,

neopolitan

 

[Dale]

Then going further, if there is nothing preventing such a frame (even if we cannot distinguish it), would not that frame's hypersurface of simultaneity be the boundary of the universe?

You have to be very careful here. My GR is not strong enough to provide a lot of detail or arguments, but everything I have been describing here has been exclusively SR. If you are going to be talking about the universe as a whole then you cannot simply use SR and you would need to use GR.

My understanding is that the notion of simultaneity is much more difficult to define in GR than in SR. In other words, there are many spacetimes that you simply cannot draw a continuous surface of simultaneity that covers the whole spacetime. For example, in a rotating spacetime a surface of simultaneity unavoidably has a discontinuous jump along some radius.

However, even if the universe has a simple geometry for which a universal notion of simultaneity could be defined, your "true boundary" of the universe and "true center" of the universe would have no physical significance. Any arbitrary reference frame could construct its "center" and "boundary" which would be equally valid and experimentally indistinguishable.

 

[pervect]

I've never seen a convincing argument that simultaneity has any physical significance whatsoever.

Events with the same time coordinate in a given coordinate system can be regarded as simultaneous, but the choice of coordinate systems is a purely arbitrary human choice.

To me this implies that simultaniety is also a human choice, a matter of labels on a map, rather than having any fundamental significance.

 

[Dale]

I've never seen a convincing argument that simultaneity has any physical significance whatsoever.

Events with the same time coordinate in a given coordinate system can be regarded as simultaneous, but the choice of coordinate systems is a purely arbitrary human choice.

To me this implies that simultaniety is also a human choice, a matter of labels on a map, rather than having any fundamental significance.

I agree completely. The universe simply doesn't care about simultaneity as far as I can tell.

 

[neopolitan]

I have read this a few times, that "simultaneity has no physical significance".

I wasn't necessarily saying that simultaneity does have "physical significance" (although I am not 100% what you mean by that, so I can't say whether I agree or disagree). If you could please look at what I did say, can you then make comment on whether the boundary of the universe has any physical significance?

My understanding is that the notion of simultaneity is much more difficult to define in GR than in SR. In other words, there are many spacetimes that you simply cannot draw a continuous surface of simultaneity that covers the whole spacetime. For example, in a rotating spacetime a surface of simultaneity unavoidably has a discontinuous jump along some radius.

I note an earlier comment DaleSpam made in the post quoted, about insufficiently strong GR knowledge, so this is not an attack - I just want to work through something here.

What exactly is "a rotating spacetime"? What is meant by "many spacetimes"? Is "a spacetime" used here to replace "an inertial frame" (noting that frames in GR don't have to be inertial)?

My understanding, which may be wrong, is that the universe doesn't care what you do in it (as equally as it doesn't care about simultaneity). I can choose whatever frame of reference I like, I can consider ourselves to be at rest, and I can subsequently work out the spacetime locations of events around me, in terms of my selected frame of reference. Other observers can follow the same process, choosing whatever frame of reference they like, considering themselves to be at rest and they can subsequently work out the spacetime locations of events around them, in terms of their own selected frames of reference.

If we all then consider the same event, we can use transformations between our frames of reference and work out that we are indeed considering the same event. Is this correct?

If it is correct, then I don't see what relevance there is to "a rotating spacetime". I don't see how it relates to what I was initially pondering, the possibility of a "one true rest frame" (or "absolute at rest frame", or AAR frame) which is indistinguishable from any "nominally in motion" NIM frame. I don't think there is any reason to assume that an AAR frame would be, or should be, rotating. It could be, I guess. You are just left with the question "what is it rotating in reference to?" Since this is a conceptual "absolute at rest" frame, it is at rest. So really what you would be saying is that all other frames, which appear to be inertial, are for some reason actually rotating relative to the AAR frame (actually, they would be orbiting - the vast majority of them impossibly, since their rotation velocity would be greater than the speed of light).

cheers,

neopolitan

 

[Dale]

Any real GR expert reading this, please feel free to correct any mistakes I make. This is all according to my rather uninformed understanding.

What exactly is "a rotating spacetime"? What is meant by "many spacetimes"? Is "a spacetime" used here to replace "an inertial frame" (noting that frames in GR don't have to be inertial)?

My understanding, which may be wrong, is that the universe doesn't care what you do in it (as equally as it doesn't care about simultaneity). I can choose whatever frame of reference I like, I can consider ourselves to be at rest, and I can subsequently work out the spacetime locations of events around me, in terms of my selected frame of reference. Other observers can follow the same process, choosing whatever frame of reference they like, considering themselves to be at rest and they can subsequently work out the spacetime locations of events around them, in terms of their own selected frames of reference.

In GR it is a little more complicated than that. Sure, you can draw whatever coordinate system you want, but the underlying spacetime itself is curved. This means that certain complexities will arise in any coordinate system that you use.

Let's consider 26 inertial clocks, A-Z, arranged in a stable ring in flat spacetime (SR). Clock A synchronizes with clock B using Einstein synchronization, then B with C, ... When we get around the ring we find, as expected, Z is synchronized with A. In flat spacetime there is a well-defined global notion of simultaneity.

Consider the same inertial clocks in stable orbit around a non-rotating massive body (curved spacetime) and further consider that they are close enough to each other to consider the spacetime between any two neighbors to be flat. A synchronizes with B which synchronizes with C ... When we get around the ring we find again that Z is synchronized with A. Again, there is a well-defined global notion of simultaneity.

Now, consider the same inertial clocks in stable orbit around a rotating massive body (rotating spacetime). A synchronizes with B ... When we get around the ring, due to the frame dragging effect, we find that Z is not synchronized with A. There is no well-defined global notion of simultaneity, simultaneity can only be defined locally in a rotating spacetime.

 

[neopolitan]

Hi DaleSpam,

Any real GR expert reading this, please feel free to correct any mistakes I make. This is all according to my rather uninformed understanding.

What exactly is "a rotating spacetime"? What is meant by "many spacetimes"? Is "a spacetime" used here to replace "an inertial frame" (noting that frames in GR don't have to be inertial)?

My understanding, which may be wrong, is that the universe doesn't care what you do in it (as equally as it doesn't care about simultaneity). I can choose whatever frame of reference I like, I can consider ourselves to be at rest, and I can subsequently work out the spacetime locations of events around me, in terms of my selected frame of reference. Other observers can follow the same process, choosing whatever frame of reference they like, considering themselves to be at rest and they can subsequently work out the spacetime locations of events around them, in terms of their own selected frames of reference.

In GR it is a little more complicated than that. Sure, you can draw whatever coordinate system you want, but the underlying spacetime itself is curved. This means that certain complexities will arise in any coordinate system that you use.

Let's consider 26 inertial clocks, A-Z, arranged in a stable ring in flat spacetime (SR). Clock A synchronizes with clock B using Einstein synchronization, then B with C, ... When we get around the ring we find, as expected, Z is synchronized with A. In flat spacetime there is a well-defined global notion of simultaneity.

Consider the same inertial clocks in stable orbit around a non-rotating massive body (curved spacetime) and further consider that they are close enough to each other to consider the spacetime between any two neighbors to be flat. A synchronizes with B which synchronizes with C ... When we get around the ring we find again that Z is synchronized with A. Again, there is a well-defined global notion of simultaneity.

Now, consider the same inertial clocks in stable orbit around a rotating massive body (rotating spacetime). A synchronizes with B ... When we get around the ring, due to the frame dragging effect, we find that Z is not synchronized with A. There is no well-defined global notion of simultaneity, simultaneity can only be defined locally in a rotating spacetime.

None of this answers my questions. I understand what you mean by "rotating spacetime" and I accept that it may only be possible to define simultaneity "locally in a rotating spacetime" but I fail to see the relevance.

If anything, it gives me reason to wonder if the idea of an AAR has more relevance than I initially though, since your explanation forces me to ask this:

Accepting what you have to say about "rotating spacetimes", in reference to what is this spacetime rotating?

I know that GR is not where you are at at the moment, but it seems that you didn't address the second paragraph of mine that you quoted. Can you or can you not, given details of another frame's relative motion (inertial, "rectilinear and non-rotating" or curved/rotating), work out where events are in your own frame of reference and confirm that other frame's interpretation of events are valid? Or are you suggesting that, if you are in a rotating spacetime, you cannot perform a transformation to obtain that event's spacetime location in terms of an observer in an inertial, rectilinear and non-rotating frame?

Note, this is pretty much the same question as I asked before, viz (with slight editing)

If we (all observers in our individual frames) all ... consider the same event, we can use transformations between our frames of reference and work out that we are indeed considering the same event. Is this correct?

cheers,

neopolitan

 

[Mentz114]

Forgive me chipping in here, but you've gone a bit off track from simultaneity.

What exactly is "a rotating spacetime"? What is meant by "many spacetimes"? Is "a spacetime" used here to replace "an inertial frame" (noting that frames in GR don't have to be inertial)?

In GR, we refer to any particular metric as a 'space-time'. A metric can define an entire universe. An observer is someone inhabiting the space-time, who uses a local ( usually Minkowski) co-ordinate system to relate to things close to her.

My understanding, which may be wrong, is that the universe doesn't care what you do in it (as equally as it doesn't care about simultaneity). I can choose whatever frame of reference I like, I can consider ourselves to be at rest, and I can subsequently work out the spacetime locations of events around me, in terms of my selected frame of reference. Other observers can follow the same process, choosing whatever frame of reference they like, considering themselves to be at rest and they can subsequently work out the spacetime locations of events around them, in terms of their own selected frames of reference.

That's what happens in practice. We each have a laboratory frame for close-in work, but in astronomy we might choose a frame with the sun at the center.

If we all then consider the same event, we can use transformations between our frames of reference and work out that we are indeed considering the same event. Is this correct?

Yes.

I think you're making heavy weather of this simultaneity thing. What does it matter if two observers disagree about two events being simultaneous or not ?