Understanding Time and Space as One entity.

[petm1]

Special relativity says that what looks like time to one observer(in a moving frame) looks like space to another observer(in a stationary frame).

Try thinking of a skydiver (moving frame) and an observer on the ground (stationary frame). The observer on the ground sees the sky diver as falling through space (space like motion) while the skydiver sees the Earth dilating out to meet him (time like motion).

 

[CaptainQuasar]

We're not talking about GR. In SR, simultaneity is a matter of perception; specifically, the distance between the observer and an event determines when an event can be perceived by the observer (t=r/c).

Okay, but the rules of simultaneity are going to be the same between SR and GR, right? Events which are simultaneous in SR are still going to be simultaneous in GR, right? Or is relativity relative to which relativity you're using? (Though GR opens up more possibilities for how events can be simultaneous - it just doesn't violate simultaneity from SR.)

How would your "omniscient/Lorentz-corrected viewpoint" be any different than your "particular inertial reference frame"?

Well, since you're talking about perception, I assumed that you're talking about what an observer experiences / observes without “knowing” that they need to Lorentz-correct things. My “omniscient” viewpoint is referring to an observer who is somehow capable of directly perceiving events in something like Minkowski spacetime without any need for relative corrections (which doesn't imply an absolute reference frame, it would be a viewpoint that is independent of inertia, basically).

Let's call it a “Minkowski observer” rather than what I said before. I'm saying that what the Minkowski observer would regard as simultaneity is the “real” simultaneity, as well as any ambiguity about simultaneity that the Minkowski observer would perceive, rather than any confusions based upon how light reaches an inertially-relative observer. I think there's a single, unified reference for whether two things are simultaneous or not - two events aren't simultaneous for one observer, but not simultaneous for another, as long as all observers are correctly making adjustments per the known physics.

(I know this isn't a standard physics term - I'm not pretending to perform original research or anything, I'm just attaching labels to existing mathematical formulations of relativity. Minkowski spacetime is more than a hundred years old now and is a mathematical formulation that actually predates GR.)

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[CaptainQuasar]

Try thinking of a skydiver (moving frame) and an observer on the ground (stationary frame). The observer on the ground sees the sky diver as falling through space (space like motion) while the skydiver sees the Earth dilating out to meet him (time like motion).

You seem to be suggesting that from the skydiver's point of view the Earth would increase in length. But that doesn't happen, does it? I'm pretty sure only length contraction occurs, the same way that time dilation only slows down the apparent passage of time in another reference frame, it never speeds it up.

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[tiny-tim]

Try thinking of a skydiver (moving frame) and an observer on the ground (stationary frame). The observer on the ground sees the sky diver as falling through space (space like motion) while the skydiver sees the Earth dilating out to meet him (time like motion).

No, each regards the other's motion (strictly, worldine) as time-like.

Two events have a space-like relationship if an observer can be found who regards them as simultaneous (because then they are separated only by space).

Two events have a time-like relationship if an observer can be found who regards them as in the same place (because then they are separated only by time). So any observer, and any material object, always follows a time-like course.

 

[Dale]

Simultaneous in one frame does not (necessarily) equate to simultaneous in another frame.

OK, that is usually called "relativity of simultaneity" rather than "percieved simultaneity".

In SR, simultaneity is a matter of perception; specifically, the distance between the observer and an event determines when an event can be perceived by the observer (t=r/c).

No it is not. In SR the actual time that the event occurred is determined by correcting for perception. In other words, in SR all observers are intelligent and account for the delay in signal reception. Therefore, an observer who, on his 50th birthday, receives a photon from an event 1 light year away determines that the emission event was simultaneous with his 49th birthday. The point of the relativity of simultaneity is that even correcting for those perceptual delays you still find that intelligent observers disagree about the actual simultaneity of events.

That said, you are correct that the relativity of simultaneity can be thought of as a rotation from the space axis into the time axis. Because this is a hyperbolic rotation if you increase the time separation then you will increase the space separation also. This is opposed to the spatial circular rotations that you are used to where increasing one axis reduces the other. This is the fact that prevents you from rotating a spacelike interval into a timelike interval.

 

[Antenna Guy]

In SR, simultaneity is a matter of perception; specifically, the distance between the observer and an event determines when an event can be perceived by the observer (t=r/c).

No it is not.

Yes, it is.

The point of the relativity of simultaneity is that even correcting for those perceptual delays you still find that intelligent observers disagree about the actual simultaneity of events.

And that is why.

Lorentz correction does not compensate for \delta r between observers - only \delta v within a particular observer's frame. Displaced observers disagree on the velocity of objects common among their frames - even if they record those velocities within their respective frames at a "synchronized" time.

Each observer's perception of velocity at a given time is relative to their position in space.

Regards,

Bill

 

[CaptainQuasar]

Each observer's perception of velocity at a given time is relative to their position in space.

Okay, but can it not be said of a given pair of events either "yes, these events are simultaneous" or "no, these events are not simultaneous" regardless of any observer's perceptions? From the discussion above it seems that for any given pair of events that are spatially separated from each other it can definitely be said "no, these events are not simultaneous." Unless I'm misinterpreting something.

And from what I've been reading in various web searches related to this discussion, this principle was known even before SR - Lorentz had derived it (separately from the Lorentz transformation) and referred to it as "local time."

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[tiny-tim]

From the discussion above it seems that for any given pair of events that are spatially separated from each other it can definitely be said "no, these events are not simultaneous."

No. What do you mean by "spatially separated"?

If you mean separated in space but not in time for a particular observer, then that observer must regard them as simultaneous.

Or, if you mean space-like, then there will always be some observers who regard them as simultaneous.

Okay, but can it not be said of a given pair of events either "yes, these events are simultaneous" or "no, these events are not simultaneous" regardless of any observer's perceptions? Unless I'm misinterpreting something.

A time-like or light-like pair are never simultaneous for any observer.

But a space-like pair will always be simultaneous for some observers.

 

[Dale]

Lorentz correction does not compensate for \delta r between observers - only \delta v within a particular observer's frame. Displaced observers disagree on the velocity of objects common among their frames - even if they record those velocities within their respective frames at a "synchronized" time.

What are you talking about here? It sounds like you are saying that SR claims that two observers, stationary wrt each other but separated by some distance, will disagree on the velocity of an object.

That is simply false. Two such observers will agree on all time, distance, velocity, and other measurements. They will each account for any perceptual delays in order to arrive at a common conclusion about any event.

 

[CaptainQuasar]

A time-like or light-like pair are never simultaneous for any observer.

But a space-like pair will always be simultaneous for some observers.

Yeah, I'm just trying to understand what you were saying there, thank you for responding to that. The variety of different terms beginning with space- and time- are tripping me up, I think.

In the second case, of a space-like pair that is simultaneous for some observers (but not for others, is the corollary?) what does the fact that an event is "simultaneous to a given observer" mean? If the simultaneity in this case is a relative phenomenon, it isn't establishing any sort of objective relationship between the two events, right? Just trying to integrate this into my existing understanding and/or learn the right words to describe these things.

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[tiny-tim]

Yeah, I'm just trying to understand what you were saying there, thank you for responding to that. The variety of different terms beginning with space- and time- are tripping me up, I think.

Yes, I always think "space-like" and "time-like" are the wrong way round!

When I see one, I think "well, it means the opposite …"

In the second case, of a space-like pair that is simultaneous for some observers (but not for others, is the corollary?)

Yes!

what does the fact that an event is "simultaneous to a given observer" mean?

A pair of events is "simultaneous to a given observer" simply means that that observer's t co-ordinate for each is the same.

If the simultaneity in this case is a relative phenomenon, it isn't establishing any sort of objective relationship between the two events, right?

You got it! That's the difference between relative and absolute.

Separation (time-squared minus distance-squared, of course) is absolute. Time and length aren't.

Only two events at the same time and place are objectively simultaneous.

™​

 

[Antenna Guy]

It sounds like you are saying that SR claims that two observers, stationary wrt each other but separated by some distance, will disagree on the velocity of an object.

That is simply false. Two such observers will agree on all time, distance, velocity, and other measurements. They will each account for any perceptual delays in order to arrive at a common conclusion about any event.

How can these two observers agree (generally) on velocity while they have conflicting perceptions of \frac {\delta r}{\delta t} (radial velocity) for an event common to both frames? The only regions I figure these two observers will agree on velocity are where an event is traveling along a ray that passes through the (spatial) origin of both frames - or on a plane perpendicular to that ray that is equidistant from each observer.

That said, let's assume that both observers agree on a correct velocity, and project an event to its' "corrected" position. The only region where the two observers will agree on a common time ordinate for the "corrected" event (with respect to the observer frames) is on the plane I described earlier. You will note, however, that even in this case (common time) the two observers cannot agree on a spatial location of the event (due to sign discrepancies arising from the displaced frames). Averaging the spatial locations of the two frames (as means of removing the spatial discrepancy between the two) leads to a time offset that is positive in one observer's 4-space, and negative in the other (for a common 3-space direction among the frames, the time components must differ in sign to arrive at the same 4-space location relative to the "average" frame [edit: not "to each"].

Regards,

Bill

 

[Antenna Guy]

I always think "space-like" and "time-like" are the wrong way round!

When I see one, I think "well, it means the opposite …"

I agree.

Regards,

Bill

 

[Dale]

... two observers, stationary wrt each other but separated by some distance, ...

... an event common to both frames? ... a ray that passes through the (spatial) origin of both frames ... spatial locations of the two frames

Two observers that are at rest wrt each other have the same reference frame. Do you have some idea that a reference frame is some sort of bubble that surrounds each observer to a short distance? If so, don't feel bad, that is a common misconception.

In SR a reference frame is simply a coordinate system that extends infinitely in all four dimensions of spacetime. Clocks at rest in the reference frame are synchronized and different observers at rest in a reference frame agree about all temporal and spatial coordinates of any event.

 

[tiny-tim]

Perception time

In SR, simultaneity is a matter of perception; specifically, the distance between the observer and an event determines when an event can be perceived by the observer (t=r/c).

Each observer's perception of velocity at a given time is relative to their position in space.

Hi Bill!

By "perception", you're meaning the time at which light from the event reaches the observer.

And by "simultaneous" you're meaning "having the same perception times" - so two events are simultaneous if the light from them reaches the observer at the same time.

Am I right?

If so, I think the confusion is that everyone else is using "simultaneous" to mean having the same perception-time-minus-r/c. In other words: having the same time coordinate.

Displaced observers disagree on the velocity of objects common among their frames - even if they record those velocities within their respective frames at a "synchronized" time.

No - displaced observers (with the same velocity) always agree on the velocity of objects.

How can these two observers agree (generally) on velocity while they have conflicting perceptions of \frac {\delta r}{\delta t} (radial velocity) for an event common to both frames?

Because an observer does not use perceived time to calculate velocity - he uses coordinate time.

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[Antenna Guy]

Two observers that are at rest wrt each other have the same reference frame.

How do you figure that?

Do you have some idea that a reference frame is some sort of bubble that surrounds each observer to a short distance?

Drop the "to a short distance" part, and that would be about it. Each "bubble" corresponds to a constant time (t=r/c) with respect to the origin.

If so, don't feel bad, that is a common misconception.

I don't feel bad. How is it amiss?

In SR a reference frame is simply a coordinate system that extends infinitely in all four dimensions of spacetime.

OK.

Clocks at rest in the reference frame are synchronized and different observers at rest in a reference frame agree about all temporal and spatial coordinates of any event.

Consider Einstein's clock synchronization procedure for any pair of clocks. Are the two clocks not on a sphere of constant radius about the flash when they set their times? If I add a third clock at the midpoint of the other two and repeat the synchronization procedure, I what frame do the three clocks simultaneously show the same time?

Answer: in the frame of the third clock.

Regards,

Bill

 

[matheinste]

Quote:-

""""""Consider Einstein's clock synchronization procedure for any pair of clocks. Are the two clocks not on a sphere of constant radius about the flash when they set their times? If I add a third clock at the midpoint of the other two and repeat the synchronization procedure, I what frame do the three clocks simultaneously show the same time?

Answer: in the frame of the third clock."""""""

Not if you carry out the procedure properly.

By the way all three clocks, if not moving relative to each other, are in the same frame however far apart they may be.

Matheinste.

 

[Dale]

In SR a reference frame is simply a coordinate system that extends infinitely in all four dimensions of spacetime.

OK.

So, starting from this point of agreement I can demonstrate the rest clearly and conclusively.

Two observers that are at rest wrt each other have the same reference frame.

How do you figure that?

So let's assume that two observers at rest wrt each other do not share a common coordinate system (reference frame). So one observer will describe things in terms of his coordinates (t, x, y, z) and the other will describe things in her coordinates (t', x', y', z').

Now, as I am sure you know, in special relativity different reference frames are related to each other by the http://en.wikipedia.org/wiki/Lorentz_transformation" :
t' = γ(t-vx/c²)
x' = γ(x-vt)
y' = y
z' = z
where γ = 1/sqrt(1-v²/c²)

Because the two observers are at rest wrt each other we evaluate the above equations for v=0 and obtain:
t' = t
x' = x
y' = y
z' = z

So since the two coordinate systems are identical and since a reference frame is a coordinate system, we say that two observers that are at rest wrt each other share a common reference frame.

 

[Antenna Guy]

Because the two observers are at rest wrt each other we evaluate the above equations for v=0 and obtain:
t' = t
x' = x
y' = y
z' = z

So since the two coordinate systems are identical and since a reference frame is a coordinate system, we say that two observers that are at rest wrt each other share a common reference frame.

Then I suppose it will be easy for you to show that an event occurring directly between two stationary observers will exhibit identical 4-space positions in either frame.

Please do.

Regards,

Bill

 

[Dale]

Then I suppose it will be easy for you to show that an event occurring directly between two stationary observers will exhibit identical 4-space positions in either frame.

I just showed that it isn't "either frame" it is the same frame.

But if you insist on being pedantic then let the first observer follow the worldline (t, 1, 2, 3) in "the first" reference frame and let the second observer follow the worldline (t', 3, 6, 9) in "the second" reference frame and let the event of interest be (0, 2, 4, 6) in the coordinates of "the first" reference frame.

t' = t = 0
x' = x = 2
y' = y = 4
z' = z = 6

so the coordinates are also (0, 2, 4, 6) in "the second" reference frame.

This should come as no surprise to anyone since as I already proved in the previous post "the first" and "the second" frames are the same frame. It should come as no surprise that a general result that holds for any and all events holds for some particular event.

 

[matheinste]

Hello Antenna Guy.

I think perhaps where you are going wrong is by considering each observer as presiding over his own exclusive reference frame with himself at the origin of any coordinate system which he ascribes to this frame. An observer is free to choose a coordinate system for the reference frame and to consider himself as being at the origin of such a coordinate system, but no other observer in the same frame can also at the origin unless he is colocated with the frst observer.

Matheinste.

 

[Dale]

If you are correct that is fine, we simply use the http://en.wikipedia.org/wiki/Poincar%C3%A9_group" instead of the Lorentz transform. The Poincare group includes translations and spatial rotations in addition to boosts. Although they will then disagree about the coordinates assigned to any given event two observers at rest wrt each other will agree on the time between any two events as well as the spatial distance between them. They will therefore also agree on any speed measurement.

 

[matheinste]

Hello DaleSpam.

Looks good to me but perhaps a bit technical. I think Antenna Guy needs to be clear about what constitutes an inertial reference frame.

Matheinste.

 

[CaptainQuasar]

Two observers that are at rest wrt each other have the same reference frame.

How do you figure that?

Isn't that the exact definition of an inertial reference frame? The set of things that are at inertial rest with respect to each other?

Consider Einstein's clock synchronization procedure for any pair of clocks. Are the two clocks not on a sphere of constant radius about the flash when they set their times? If I add a third clock at the midpoint of the other two and repeat the synchronization procedure, I what frame do the three clocks simultaneously show the same time?

Answer: in the frame of the third clock.

If I understand things properly, distance doesn't have anything to do with SR, just inertial reference frames. Two observers in the same inertial reference frame could be a galaxy's width apart¹ in intergalactic space and they'd be able to see each other (yes, with a delay for the light to travel) but they would not perceive any relativistic effects.

Light cones for talking about causality involve distance and then that subject also discusses SR and GR too… is that what you're thinking of, Antenna Guy? Hey the rest of you guys… when you're looking at the 2D or 3D diagram of a light cone, can't you measure interval vectors within the diagram and based upon whether the interval vector has a steeper or more shallow slope than the surface of the light cone decide some property related to whether it's a spacelike or timelike interval?

¹Not that galaxies have standardized widths.

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[Dale]

can't you measure interval vectors within the diagram and based upon whether the interval vector has a steeper or more shallow slope than the surface of the light cone decide some property related to whether it's a spacelike or timelike interval?

Yes, if it is inside the light cone it is timelike, if it is outside the light cone it is spacelike, and if it is on the light cone it is lightlike or null. Since another way of restating the second postulate is that the light cone is preserved then no change of inertial reference frame will change the spacelike, timelike, or lightlike designation.

 

[belliott4488]

Yes, if it is inside the light cone it is timelike, if it is outside the light cone it is spacelike, and if it is on the light cone it is lightlike or null. Since another way of restating the second postulate is that the light cone is preserved then no change of inertial reference frame will change the spacelike, timelike, or lightlike designation.

To clarify, this is true for a vector between an event at the origin, or apex of the light cone, and a second event that is located as DaleSpam described. You could have two events with a space-like separation that both lay with in a given light cone, provided that the slope of line between them was less than c, i.e. the angle of the cone. Just wanted to be what what "it" referred to in DaleSpam's post ...

 

[Antenna Guy]

let the first observer follow the worldline (t, 1, 2, 3) in "the first" reference frame and let the second observer follow the worldline (t', 3, 6, 9) in "the second" reference frame and let the event of interest be (0, 2, 4, 6) in the coordinates of "the first" reference frame.

t' = t = 0
x' = x = 2
y' = y = 4
z' = z = 6

so the coordinates are also (0, 2, 4, 6) in "the second" reference frame.

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity). If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

Thus, information is traveling from our event in opposite spatial directions along a new world-line (that of the event) to get to our two observers. In order to express this transferrence of information in terms of a single world-line, we have t'=-t (n.b. this is a very specific case).

Now, why might these two observers disagree regarding the velocity of some new event that occurs elsewhere in space-time...

Let's arbitrarily choose the location of our new event (denoted event_new) to be (0,1,2,6), and have it traveling along a path parallel to the ray passing through our two observers. Expressed in observer 1's frame, I think (1,1/t,2/t,3/t) will do for the velocity.

As before, let's look at the 4-space displacements first:

observer_1-event_new=(t,1,2,3)-(0,1,2,6)=(t,0,0,-3)

observer_2-event_new=(t',3,6,9)-(0,1,2,6)=(t',2,4,3)

Now, let's look at the displacements after some differential time has passed:

observer_1-event_new=(t+dt,1,2,3)-{(0,1,2,6)+dt*(1,1/t,2/t,3/t)}
=(t,-dt/t,-2dt/t,-3-3dt/t)

observer_2-event_new=(t'+dt',3,6,9)-{(0,1,2,6)+dt'*(1,1/t',2/t',3/t')}
=(t',2-dt'/t',4-2dt'/t',3-3dt'/t')

During this differential time interval, 4-space deltas are then:

observer_1->(event_new'-event_new): (0,-dt/t,-2dt/t,-3dt/t)

observer_2->(event_new'-event_new): (0,-dt'/t',-2dt'/t',-3dt'/t')

This is somewhat problematic from a 4-velocity standpoint (spatial displacement in zero time in either frame).

Near as I can tell, the two spatially displaced observers will agree on the spatial direction of the event, but that's about it (generally).

It appears implicit that some differential amount of time must pass between the event and each observer for the velocity vectors to become meaningful. I think the Lorentz transform accomplishes this by mapping dt from the observer frame to dt in the event frame - leaving a net time displacement in the above intervals.

BTW - since both our observers agree on space, time, and the speed of light, is it not obvious that the two observers will record the 4-space properties of our new event at different positions on the event world-line wherever the observers' spatial displacements from the event differ?

Regards,

Bill

 

[belliott4488]

Bill, you're not going to like this, so I hope you'll take it in the spirit in which it's offered.

You do not understand Special Relativity. You can continue to argue with people until they go away and leave in frustration, or you can accept the possibility that you've got some misconceptions and maybe try to learn something. I hope you'll try the latter, since you seem genuinely interested and have a basic grasp of some of the concepts. You're simply making too many incorrect statements, though - you need to get clearer on much of this.

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

[Edited] I'm not sure what you mean by "simultaneous positions". A "position" in Minkowsky space has four components, which makes it an event. That's what any point in Minkowsky space is. So, to talk about 3 positions, only one of which is an event, makes no sense. Three events can certainly be simultaneous in one reference frame, in which case they will all have the same time coordinate, so by setting t = t' = 0, perhaps you meant simultaneous events?

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity).

I'm not sure where you get this, but it's not strictly true. There is nothing wrong with spatial translations within a given reference system; they're translations that happen in the 3-d subspace spanned by the spatial axes. As long as you don't translate along the time direction, the translation is spatial. Since the space-like nature of a displacement 4-vector is an invariant, however, this translation will be space-like in all reference frames. It will be simultaneous in only one reference frame, however; in others it will combine a shift in time as well.

In any case, translations are not "limited in rate." You can mathematically translate any point to another by a suitable translation. Whether or not this corresponds to physical motion with a velocity relative to some coordinate frame depends on the translation. A translation consisting of time-like displacements of space-time events could indeed represent motion with velocity, the magnitude of which would depend on the chosen coordinate frame. For one frame, the translation would be a pure time translation, i.e. no spatial motion at all.

If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

This is true in Euclidean space, but not in Minkowsky space. Consider two events with a light-like separation - the "distance" between them, i.e. the space-time interval, is zero, regardless of how much time has elapsed between them in any particular reference frame. Also, as has been pointed out, time-like intervals have negative magnitudes, so talking about the "shortest distance" doesn't have the same meaning it has in Euclidean space.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

You have three points in M. space, i.e. three events. You can certainly talk about their vector components in a particular reference frame, as you have, and then you can calculate the interval between any two of them, which you have not. The interval you get in each case is invariant, i.e. it will be the same regardless of which reference frame you choose to express the vectors in - this is the analog of the vector magnitude in Euclidean space.

There is no meaning to the "velocity of information" in this case, since you have not spoken of anything propagating from one event to any other. If you do, however, then by examining the interval between the events, you can say whether it is space-like, time-like, or light-light, which will be invariant, i.e. the same in all reference frames. Only time-like propagation is physical.

I'm going to drop it here, since you can see where this is going - I could continue picking at your statements, but I suspect we'd both lose interest.

Please don't take offense; I don't mean to be insulting. In fact, there is nothing insulting about the suggestion that someone doesn't understand SR, since most people don't. It's just that based on my experience teaching SR, right now I'd say you have not yet mastered it.

 

[Dale]

Since you seem to have chosen your world-lines arbitrarily within some third coordinate system, is there any reason not to use (t,-1,-2,-3) and (t',1,2,3) as the observer world-lines, and (0,0,0,0) as the event?

Not an important reason. It is just that, after proving that it applied to all events in general I didn't want to do an example using a "special" event like the origin.

anyhoo, each world line [allegedy] describes a direction in 4-space. If I were to substitute 0 for t and t' in your observer world lines, I would have what amounts to three simultaneous positions in 4-space (two observers + one event).

Yes.

SR implies that translations are not strictly spatial - they are spatio-temporal; and as such, limited in rate (velocity). If "information" (I like that descriptor) is to travel from an event to an observer, it starts at a point in 4-space, and ends on an observer's "world-line". Since the shortest distance between two points is a straight line (please don't bring up spheres, etc. ), let's look at the delta between the event and each observer as you've characterized them.

observer_1-event=(t,1,2,3)-(0,2,4,6)=(t,-1,-2,-3)

observer_2-event=(t',3,6,9)-(0,2,4,6)=(t',1,2,3)

OK, I follow you so far.

The (average) velocity of information in the respective cases is:

event->observer_1: (1,-1/t,-2/t,-3/t)

event->observer_2: (1,1/t',2/t',3/t')

I don't know how you arrived at this, but it looks wrong to me. If (the usual case) information is sent via an EM pulse then the speed will always be c. If the information is sent via some material object (e.g. via FedEx) then the speed will be the speed of the material object, and if that object is traveling inertially then that speed will be constant. Here, since your velocity is a function of time and not c it seems that you are transfering information via some accelerating material object. Also, the velocity function you are using here is undefined for t=t'=0, and for small values of t it is faster than light.

Thus, information is traveling from our event in opposite spatial directions along a new world-line (that of the event) to get to our two observers.

First, events do not have world-lines. Events are the spacetime equivalent of points. Second, the information will travel (usually at c) along two generally different worldlines to reach each observer. The intersection of the information's worldlines with each observer's worldline are other events (the "signal reception" events). The locations and times of the signal reception events are different from the location and time of the signal transmission event. None of this is problematic since intelligent observers always work backwards to determine when and where the transmission event occurred. This is the "correcting for perception" that we discussed earlier.

In order to express this transferrence of information in terms of a single world-line, we have t'=-t (n.b. this is a very specific case).

Yes, this is a special case. Since in general t'=t then the condition t'=-t implies t'=-t=-t'=0. I don't know why you want to restrict your analysis to this specific case, particularly since your velocity is undefined at t=0, but OK.

Now, why might these two observers disagree regarding the velocity of some new event that occurs elsewhere in space-time...

Let's arbitrarily choose the location of our new event (denoted event_new) to be (0,1,2,6), and have it traveling along a path parallel to the ray passing through our two observers. Expressed in observer 1's frame, I think (1,1/t,2/t,3/t) will do for the velocity.

Again, events don't travel, and again observer 1 and observer 2 are stationary wrt each other so they share the same reference frame. But, OK, I follow.

As before, let's look at the 4-space displacements first:

observer_1-event_new=(t,1,2,3)-(0,1,2,6)=(t,0,0,-3)

observer_2-event_new=(t',3,6,9)-(0,1,2,6)=(t',2,4,3)

OK

Now, let's look at the displacements after some differential time has passed:

observer_1-event_new=(t+dt,1,2,3)-{(0,1,2,6)+dt*(1,1/t,2/t,3/t)}
=(t,-dt/t,-2dt/t,-3-3dt/t)

observer_2-event_new=(t'+dt',3,6,9)-{(0,1,2,6)+dt'*(1,1/t',2/t',3/t')}
=(t',2-dt'/t',4-2dt'/t',3-3dt'/t')

During this differential time interval, 4-space deltas are then:

observer_1->(event_new'-event_new): (0,-dt/t,-2dt/t,-3dt/t)

observer_2->(event_new'-event_new): (0,-dt'/t',-2dt'/t',-3dt'/t')

OK, both observers agree on your displacement measurements.

This is somewhat problematic from a 4-velocity standpoint (spatial displacement in zero time in either frame).

Of course you get a 0 time difference, you added the dt to the worldline expression (essentially an event on the future of the worldline of the observer). Also, none of your velocities in this post are 4-velocities, they are not Minkowski 4-vectors of any kind. That is not a big problem since you are working in a single frame, but just FYI your "delta displacements" will not Lorentz transform properly. Given that we are working in a single frame and doing no transforms I won't worry about that.

Near as I can tell, the two spatially displaced observers will agree on the spatial direction of the event, but that's about it (generally).

How do you figure that? You haven't shown anything that they disagree on. Even using some strange concepts like your "delta displacement" and your accelerating information.

It appears implicit that some differential amount of time must pass between the event and each observer for the velocity vectors to become meaningful. I think the Lorentz transform accomplishes this by mapping dt from the observer frame to dt in the event frame - leaving a net time displacement in the above intervals.

Events don't have frames. Also, I am not sure what your point about a differential amount of time and "meaningful" is about. The concept of the velocity at an instant is well defined. It is the limit as your differential time goes to zero.

BTW - since both our observers agree on space, time, and the speed of light, is it not obvious that the two observers will record the 4-space properties of our new event at different positions on the event world-line wherever the observers' spatial displacements from the event differ?

Again, events do not have worldlines. The bottom line is that if two observers agree on t, x, y, and z then they must agree on any f(t,x,y,z). This includes any 4-velocities, 4-accelerations, and even your displacements and delta-displacements. This isn't any complicated physics, just very basic math. Your conclusion is completely backwards and has not been substantiated by anything you have done in this post. Actually, everything you have done in this post contradicts this conclusion.

It is honestly difficult for me to know where to go from here in this conversation. You do a bunch of strange work that confirms my claims and then claim that the opposite is obvious. I don't think there is anything I can do to help you out here.

 

[Antenna Guy]

Bill, you're not going to like this, so I hope you'll take it in the spirit in which it's offered.

You do not understand Special Relativity.

Well there's a resounding endorsement!

Seriously - I'm not offended. I've spent the last 20yrs working with antennas, not relativity. I just see a lot of parallels regarding interpretations (even if it is awkward trying to relate how I see them as such). If I learn something new or correct something that is wrong with my understanding while trying to describe these parallels, it's for the better. However, as I'm sure any other "old dog" on this forum will attest, it can be difficult (though not impossible) to teach me new tricks.

BTW - Have you ever thought of a near-field antenna measurement as analogous to an array of observers about a single event? Never mind...

Keep up the good work.

Regards,

Bill