# B Space Like Separated Events

1. Sep 5, 2016

### morrobay

http://arxiv.org/pdf/0803.2425v1.pdf

Referring to experiment diagram how do the following statements apply ?
1 There is no particular order between events A and B ( in time) if they are spacelike separated.
2. If two events are simultaneous in any reference frame they must be spacelike separated.

2. Sep 6, 2016

### Ibix

A more precise question would help. Your statements are just statements about the properties of pairs of spacelike separated events. Which figure, and what is it you don't understand?

I haven't read the entire article. The thing that jumps out of the first page is that I've usually seen spacelike separation referring to events, not extended processes. In that terminology, what the authors call "spacelike separated" could be stated as a requirement that the start of both detection processes must be spacelike separated from the end of the other.

Last edited: Sep 6, 2016
3. Sep 6, 2016

### morrobay

Question is not related to paper. Only refer to spacelike separated measurements at detectors A and B in diagram:
Ie just read paragragh below diagram.
Re 1. Why would an observer at source in frame of experiment where A and B are 20 km separated not see detector events as simultaneous if they are ?
Re 2. Have also seen statement that spacelike events as observed from any number of moving frames have no particular order.
Are not simultaneous.
If detector events are simultaneous in rest experiment frame S, 0 .
Then i am only familiar with a moving frame S' having events non simultaneous: with Δt' = γ ( Δt - vx/c2)
The question started in CFD current topic in QM section, Post # 85 : "If measurement events are spacelike separated then there exist frames of reference for which the events occur in the reverse order." ( without involving faster than light )

Last edited: Sep 6, 2016
4. Sep 6, 2016

### Staff: Mentor

These statements are both true. The paper is pointing out that "particle arrives at detector" isn't the same thing as "detector has recorded result".

5. Sep 6, 2016

### vanhees71

Ad 1: These are very basic properties of Minkowski space. If you have two events, represented by four-vectors $x$ and $y$ space-like separated you have (using the signature (1,3) for the Minkowski product).
$$(x-y) \cdot (x-y)<0.$$
Then you can choose the direction $(x-y)$ as one of the spatial basis vectors of a Minkowksi-orthonormal basis, and for an observer in this system the events are simultaneous. Now you can Lorentz boost from this frame such that in the new frame $x$ is before $y$ or the other way around. To see this take $z=x-y$. In the just constructed frame the components are
$$(z^{\mu})=(0,x,0,0).$$
Now a Lorentz boost in $x$ direction is given by the Matrix
$$({\Lambda^{\mu}}_{\nu})=\begin{pmatrix} \cosh \eta & -\sinh \eta \\ -\sinh \eta & \cosh \eta \end{pmatrix}.$$
The boost velocity is given by $\beta=v/c=\tanh \eta$ and you can have any real value for $\eta$. Now you get
$$\tilde{z}^{\mu} = {\Lambda^{\mu}}_{\nu} z^{\nu} = \begin{pmatrix} -x \sinh \eta \\ x \cosh \eta \\0 \\0 \end{pmatrix}.$$
Depending on the sign of $\eta$ you have either $\tilde{z}^0>0$ or $\tilde{z}^0<0$, i.e., either $x$ is after $y$ or the other way around for the corresponding inertial observers in the boosted frame.

2. If in a frame $x^0=y^0$ (i.e., if the events are simultaneous in this frame), then you have obviously $(x-y) \cdot (x-y)<0$ (except if $x=y$), i.e., the events are space-like separated.

6. Sep 6, 2016

### Ibix

Simultaneity is a frame-dependent concept in relativity. So if the events are simultaneous in the frame of the observer then the observer will see them as simultaneous (possibly after correcting for light travel time). But an observer at rest in another frame will not agree that the events were simultaneous.
For spacelike separated events, there always exists a frame in which they are simultaneous. Seen from that frame, frames moving in one direction will all agree that event 1 happened after event 2, although they will not agree on the time difference. Frames moving in the other direction will all agree that event 2 happened after event 1, although they will not agree on the time difference.
Then I'm confused. Apart from a missing Δ before the x, this is exactly the formula that describes what you are talking about. There is a frame in which Δt=0. In general, Δx≠0, so Δt'≠0 in any frame moving relative to the "simultaneous" frame, and the sign of Δt' depends on the sign of v as well as the sign of Δx.

7. Sep 6, 2016

### morrobay

To clarify:(Δt): Δt' = γ ( Δt - vΔx/c2)
Is Δt the total elapsed time for experiment in rest frame of observer Δt ≠ 0
Or is Δt in rest frame of an observer at source the time difference in A and B measurement events ? Δt = 0 , simultaneous.
To clarify (Δx) in above: This is the displacement between S and S' ? From vt.
Or is Δx just any chosen distance from S to S' ?
Of course v is for S'

Last edited: Sep 6, 2016
8. Sep 7, 2016

### Ibix

x and t are coordinates. In a spacetime diagram like the one you took from the paper, x is the horizontal coordinate and t is the vertical. Δx and Δt are differences in coordinates - the horizontal and vertical separations between a pair of events (points on the graph). Δx' and Δt' are the horizontal and vertical separations of the same pair of events as measured on the spacetime diagram drawn by an observer moving at speed v relative to the one who drew this one.

Spacelike separated events are ones for which Δx>cΔt - that is, not even light can cross the distance between them in the time between them. This fact is frame invariant - the Lorentz transforms guarantee that if Δx>cΔt then Δx'>cΔt'. This means that spacelike separated events are causally disconnected. One cannot be a result of the other because there is no time even for light from one event to reach the other.

Note what I said in #2 above. The authors are using spacelike separation with respect to detection processes that are extrnded in time, rather than events which are instants. They therefore mean "every event on the vertical black line representing detection at A is spacelike separated from every event on the vertical black line representing detection at B".

9. Sep 7, 2016

### morrobay

With Δt' = γ( Δt - vΔx/c2) It was my understanding that Δx is the distance from S (rest frame of experiment) to the moving S' frame. And that could be any distance. From what you say above, that distance, Δx, would be distance between detectors A and B,
about 20 km in this case.. Since events A and B are simultaneous in frame S, Δt = 0. So is this correct: for Δt' = γ( Δt - v Δx/c2)
Δt = 0
Δx = 20 km
v = .0001c
γ = 1/√ 1- (3*104)2/ (3*108)2
Or can v2/c2 just be .00012

10. Sep 7, 2016

### Ibix

That wouldn't make any sense. Frames are just coordinate systems covering the whole of spacetime. Asking how far apart they are is like having two different scale maps of London and asking how far the Tower Bridge on this map is from the Tower Bridge on the other.

Looks fine. Replacing v/c with 0.0001 is fine (v/c is usually denoted β), but you'll have to express v and Δx in the same unit system if you want an answer without nasty conversion factors.