Understanding Time and Space as One entity.

[dswam]

I ( Swaminathan) started studying theory of relativity. I need some help in understanding the following:

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).

I assume that Two events are happening. An observer views these two events happening at same space but at different times. Another observer views these two events happening at same time but at different space.

Is my assumption is correct. If so, can someone give real world example for the above situation. I searched for examples in textbooks and it is not available.

- Swaminathan.

 

[pam]

I assume that Two events are happening. An observer views these two events happening at same space but at different times. Another observer views these two events happening at same time but at different space.

That can't happen. Your first example describes a "space-like" interval (dx>dt).
Your second example is a "time-like" interval (dt>dx).
A Lorentz transformation (a rotation in space-time) cannot transform from a space-like to a time-like interval.

 

[Antenna Guy]

That can't happen. Your first example describes a "space-like" interval (dx>dt).
Your second example is a "time-like" interval (dt>dx).
A Lorentz transformation (a rotation in space-time) cannot transform from a space-like to a time-like interval.

I don't think that's correct.

Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis?

Regards,

Bill

 

[CaptainQuasar]

I think what pam is saying is true strictly within within the “http://en.wikipedia.org/wiki/Minkowski_spacetime" . Minkowski coordinates are “flat spacetime”, an approximation not taking into account the GR effects of gravity. The Lorentz transformation is sort of embedded in the coordinate system which is the reason for the special property pam is talking about.

I think what Bill is saying is sort of true generally within any 4-space, since it isn't usually as straightforward to classify intervals as spacelike or timelike in that case.

dswam you may be interested in [THREAD=215019]this thread[/THREAD], in which I tried to discuss the effects of relativity in non-technical, mostly non-mathematical terms with a film student.

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[dx]

dswam, the idea of spacetime as one entity is embodied in what is called the spacetime interval. say (t,x) are the coordinates of an event E according to observer A. Then the coordinates of the same event according to another observer B (t',x') must satisfy

t^2 - x^2 = t'^2 - x'^2

This spacetime interval is invariant. It is more fundamental than any particular set of coordinates assigned to it by an observer.

 

[tiny-tim]

Observers have time-like trajectories

I don't think that's correct.

Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis?

I'm with Pam on this.

dswam asked about the view of the same things by two different observers.

Observers (who move slower than light) have time-like trajectories.

A "rotation in space-time" which rotates one time-like vector onto another (one observer's trajectory onto another's) cannot rotate any time-like vector onto any space-like vector (including the t-axis).

Different observers (who move slower than light) will always agree on what is time-like and what is space-like

 

[Fredrik]

Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis?

A space-time rotation (i.e. a homogeneous Lorentz transformation) around the time axis in Minkowski space is just an ordinary rotation around an arbitrary axis in space. So no, they (homogeneous Lorentz transformations) are not all rotations around the time axis.

I don't know what you mean by "change the projection of a 4-space vector onto the time axis". If you mean "why can't a space-time rotation take the t axis to the x axis?", the short answer is that a function that does that doesn't preserve the "length" of all vectors as a "rotation" must do by definition. It would take 4-vectors with positive "length" to 4-vectors with negative "length".

Note that the "length" in this case is defined in a different way than what (I guess) you're used to. For example, the "distance" between the origin and (t,x,y,z) isn't defined as t²+x²+y²+z². It's defined as -t²+x²+y²+z².

The fact that "length" is something different than we're used to implies that rotations are different too. (Because a rotation is defined as a linear map that preserves the length of vectors). In particular, a homogeneous Lorentz transformation in 1+1 dimensions (1 time and 1 space dimension) that rotates (in the usual sense of that word) the t axis down a bit towards the x-axis also rotates the x-axis in the opposite direction by the same amount.

 

[Dale]

I don't think that's correct.

Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis?

pam is correct

It doesn't matter what axis the rotation is about, all rotations preserve the length of a vector. Spacelike and timelike intervals have different lengths, so they cannot be exchanged through a rotation.

 

[Antenna Guy]

Pile on Bill day

pam is correct

Sorry, but I still disagree.

It doesn't matter what axis the rotation is about, all rotations preserve the length of a vector.

I never said that length wasn't preserved - what I tried to say was: To assume that a 4-space vector never changes its' projection along the time axis implies that all rotations of that vector are about the time axis.

If I were to suggest that a 3-vector rotated about a vector in 3-space never changes its' projection along the z-axis, would I not imply that all rotations are about z?

Spacelike and timelike intervals have different lengths, so they cannot be exchanged through a rotation.

I'm afraid we're talking apples and oranges. A space-time interval in Minkowski space is not a 4-space vector - it's a scalar.

Regards,

Bill

 

[George Jones]

Why can't a rotation in space-time change the projection of a 4-space vector onto the time axis? Are all rotations about the time axis?

While it is true that a spacetime "rotation" can change the projection onto the time axis (of a particular observer), it is also true that a spacetime rotation, i.e., a (restricted) Lorentz transformation, cannot, by definition, transform a spacelike interval into a timelike interval.

[edit]I somehow missed seeing StatusX's post before I posted.[/edit]

 

[belliott4488]

AntennaGuy (and anyone else who's interested): this page: http://casa.colorado.edu/~ajsh/sr/sr.shtml is the start of one of my favorite expositions of Relativity and Minkowski Space. There are some very nice illustrations and animations of Lorentz transformations, i.e. the "rotations" of Minkowski space-time, that really make it clear how to think about these things.

If I understand you correctly, your question about projections of vectors onto axes still retains the usual Euclidean ideas about vector transformations. If you read through the pages I linked above, I think you'll find your answer.

 

[cyberdyno]

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).

The speed of light is constant only in relation to events happening in the same frame.

 

[Fredrik]

The speed of light is constant only in relation to events happening in the same frame.

That statement doesn't make sense.

 

[robphy]

Note: "Rotations" are about "an axis" only in a 3-dimensional space.
(In a similar vein, the cross-product of two vectors is more about the parallelogram determined by the two vectors rather than a "vector perpendicular to the parallelogram".)

 

[cyberdyno]

That statement doesn't make sense.

It means that for there to be an observed, there needs to be an observer, and both need to be in the same frame. The speed of light is the same for all observers but not to all frames. In other words, its constancy is local.

 

[CaptainQuasar]

It means that for there to be an observed, there needs to be an observer, and both need to be in the same frame. The speed of light is the same for all observers but not to all frames. In other words, its constancy is local.

That is not consistent with my understanding. I'm pretty sure that the speed of light is constant in all frames and it's time and space that vary between frames, via time dilation and length contraction.

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[tiny-tim]

observer = frame

Still doesn't make sense.

An observer has one frame - like an astronomer has one telescope.

But an observed can be seen in every frame, just as a star can be seen in every telescope.

A frame is simply the choice of x y z and t which the observer makes for himself.

Basically, observer = frame.

"The speed of light is the same for all observers" is the same as "The speed of light is the same for all frames."

Only observers have frames, not the observed.

 

[cyberdyno]

That is not consistent with my understanding. I'm pretty sure that the speed of light is constant in all frames and it's time and space that vary between frames, via time dilation and length contraction.

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Right, c is constant in all frames... but not to all frames at once.

 

[cyberdyno]

Still doesn't make sense.

For there to be an observer there needs to be an observed... in any frame.

But whatever is happening in one frame does not affect the speed of light on another. The speed of light does not need to be constant among frames, only within them.

 

[Dale]

Sorry, but I still disagree.

I never said that length wasn't preserved - what I tried to say was: To assume that a 4-space vector never changes its' projection along the time axis implies that all rotations of that vector are about the time axis.

I think you miss pam's point. The OP wanted to find a case where two co-located events would transform into two simultaneous events. In other words, for coordinates s = (ct,x) he wanted to find some Lorentz transform, L, where Δs = (c dt,0) transforms to Δs' = L.Δs = (0,dx').

This is not possible for any L since any L preserves the interval c²t² - x² and c²dt² ≠ -dx'² for all real dt and dx' (except zero).

The "projection along the time axis" does change. That is time dilation. But it does not change in the way requested by the OP, as pam correctly stated. In fact, by looking at the interval you can see that the projection must always lengthen along the time axis (relative to the "proper" frame). This is because the rotation is hyperbolic along that axis rather than circular.

 

[CaptainQuasar]

Right, c is constant in all frames... but not to all frames at once.

At this point I have to agree with the others that you simply aren't making any sense. If you can give an example of a reference frame where light moves at a speed other than 299,792,458 meters per second, or a way in which it would appear to differ between two different reference frames, go ahead and enlighten us.

The speed of light is constant in all frames at once. There isn't any incongruity or illogic to that - time dilation and length contraction for each observer / reference frame “correct” everything so that the speed of light is the thing that is not relative, it's absolute to all reference frames. I think you may be misunderstanding this and imagining that the speed of light is relative too.

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[Antenna Guy]

I think you miss pam's point. The OP wanted to find a case where two co-located events would transform into two simultaneous events.

Please re-read Swaminathan's post and consider whether or not changing a 4-space vector's projection along the time axis (between observer frames) satisfies the original request. Changing a spatial component in one frame, to a time component in another frame seems like a more appropriate intepretation of what was asked.

Regards,

Bill

 

[kenewbie]

[deleted]

 

[pam]

I am (not) amused that a trivial question and an obvious answer generates a thread of
(now) 24 posts.

 

[Dale]

Please re-read Swaminathan's post and consider ...

I liked the OP because it seemed very clearly described to me.

I assume that Two events are happening. An observer views these two events happening at same space but at different times.

Δs = (c dt,0)

Another observer views these two events happening at same time but at different space.

Δs' = L.Δs = (0,dx')

I really don't know how you could come to another interpretation.

 

[Antenna Guy]

I liked the OP because it seemed very clearly described to me.

I really don't know how you could come to another interpretation.

Now that we've beat the assumption to a pulp, can we look at what the interpretation is supposed to relate to?

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).

Does the above not equate to changing a vector's projection along the time axis of two different observers?

The difference between the two observers is what drives the lack of perceived simultineity.

I think the train struck by two lightning bolts might apply here.

Regards,

Bill

 

[Dale]

The difference between the two observers is what drives the lack of perceived simultineity.

What do you mean by "perceived simultaneity"?

 

[Antenna Guy]

What do you mean by "perceived simultaneity"?

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

Regards,

Bill

 

[CaptainQuasar]

I know that there are general simultaneity problems within GR, but that's not a matter of perception, is it? Don't length contraction and time dilation leave events just as simultaneous in a particular inertial reference frame as they would be from an omniscient / Lorentz-corrected viewpoint?

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[Antenna Guy]

I know that there are general simultaneity problems within GR, but that's not a matter of perception, is it?

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).

Don't length contraction and time dilation leave events just as simultaneous in a particular inertial reference frame as they would be from an omniscient / Lorentz-corrected viewpoint?

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

Regards,

Bill

 

[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.