# Understanding Time and Space as One entity.

• dswam
It is possible, but it's not just a rotation as typically thought of in 3d space. There are other factors and considerations at play, such as the spacelike/timelike nature of the vectors involved, the choice of coordinates, and the specific transformation used.But in general, no, not all "rotations" are about the time axis.
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.

dswam said:
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.

pam said:
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

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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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.

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Observers have time-like trajectories

Antenna Guy said:
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

Antenna Guy said:
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.

Antenna Guy said:
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.

Pile on Bill day

DaleSpam said:
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

Antenna Guy said:
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.

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

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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.

dswam said:
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.

cyberdyno said:
The speed of light is constant only in relation to events happening in the same frame.
That statement doesn't make sense.

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".)

Fredrik said:
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.

cyberdyno said:
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.

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.

CaptainQuasar said:
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.

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

tiny-tim said:
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.

Antenna Guy said:
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.

cyberdyno said:
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.

DaleSpam said:
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

[deleted]

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

Antenna Guy said:
Please re-read Swaminathan's post and consider ...
I liked the OP because it seemed very clearly described to me.

dswam said:
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)

dswam said:
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.

DaleSpam said:
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

Antenna Guy said:
The difference between the two observers is what drives the lack of perceived simultineity.
What do you mean by "perceived simultaneity"?

DaleSpam said:
What do you mean by "perceived simultaneity"?

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

Regards,

Bill

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?

CaptainQuasar said:
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

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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).

Antenna Guy said:
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.)

Antenna Guy said:
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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petm1 said:
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.

petm1 said:
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.

Antenna Guy said:
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".

Antenna Guy said:
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.

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