Must an observer be equidistant from two events for them to appear simultaneous?

[pivoxa15]

Spacelike intervals are when time separation is 0 or the observer measures a single time for the start of two events. Hence the two events are simultaneous to this observer. My question is for this to occur, must the observer be at an equal distance from the two events (or right in the middle of two events)?

I think yes.

 

[michael879]

no, you can measure two events to be simultaneous without being equidistance from them. Let's say your 5m from A and 10m from B. You know how long light takes to reach you from A and B (5/c and 10/c) so you just subtract that from whatever time you detect the light from both.

 

[Rach3]

Spacelike intervals are when time separation is 0 or the observer measures a single time for the start of two events. Hence the two events are simultaneous to this observer.

No, look at a light cone, in an arbitrary intertial frame you have spacelike intervals with all sorts of time seperation. Of course for any two spacelike intervals you can Lorentz-transform into a frame in which they are simulatneous. The definition is much more useful: ds^2<0 (or ds^2>0, depending on the signature convention... )

My question is for this to occur, must the observer be at an equal distance from the two events (or right in the middle of two events)?

I think yes.

No. Go back to the definition of the coordinate system, defined usually on page 1, with the rigid lattice of rulers and clocks. How are the clocks "synchronized"? Note in particular that if two events are simultaneous, one will NOT see the other occurring at the same time as itself! The observer is smart, and subtracts off the travel time of a light beam - this is explicit in the definition of the clocks-and-rulers coordinates.

This is necessary for consistency. Consider A, sending a red laser pulse to C. B is halfway intermediate. B sends a blue laser pulse to C upon receiving A's signal. From C's point of view, A and B appear to send their signals at the same time! There is no SR frame like this, null separated events (A,B) are NOT SR-simultaneous.The consistent coordinate system of SR is different; in C's reference frame, he measures the distances involved, substracts the travel time of light, and concludes that the event at A precedes that in B in his reference frame (despite their "apperent" simultaneity".)

In short: "apparent" simultaneity is NOT the "simultaneity" of SR.

 

[robphy]

What "spacelike" means... in a coordinate-free way.. in terms of events, "light cones", and radar experiments:

The interval between two events is spacelike when neither event is in the light cone of the other.

A related description: "A is spacelike-related to B" if the spacetime-vector from A to B does not point inside or tangent-to the light cone of A.

If A and B are spacelike-related,...
...All observers will agree on this.
...For a special subset of inertial observers, A and B are simultaneous.
...For a smaller subset of these observers, A and B are also equidistant.

Consider the following radar-experiment [it'll help to draw a spacetime diagram]:
Let an inertial observer send at event S [for send] a light ray that reaches the distant event B, and whose reflected-light echo is received by that inertial observer at event R [for receive]. Note that B is one event of the intersection of the future-light-cone of S and the past-light-cone of R. The intersection is a circle [generally, a sphere], which defines a [hyper]plane... all of whose events are [according to that inertial observer] simultaneous with B. (In fancier talk, this [hyper]plane is Minkowski-perpendicular to that inertial observer's worldline.)

If A also lies on that [hyper]plane, then, "according to this inertial observer, A is simultaneous with B".
[There are many inertial observers who will determine [with their own corresponding events S' and R'] that same hyperplane,
and will agree that "to them, A is simultaneous with B". There will, of course, be many observers that determine a different hyperplane containing B but not A... and thus will not regard A and B as simultaneous.]

If, more specifically, A also lies on the intersecting circle [sphere], then "according to this inertial observer, A is also equidistant from B".
[That is, some of those inertial observers who determine the same hyperplane containing A and B with their corresponding events S' and R' actually find that A and B are on the circle [sphere] of intersections of the future-light-cone of their S' and the past-light-cone of their R'.]

 

[franznietzsche]

no, you can measure two events to be simultaneous without being equidistance from them. Let's say your 5m from A and 10m from B. You know how long light takes to reach you from A and B (5/c and 10/c) so you just subtract that from whatever time you detect the light from both.

This is only true if you, A, and B are all in the same inertial frame.

 

[michael879]

This is only true if you, A, and B are all in the same inertial frame.

he was talking about a single observer which means there's only 1 frame..

 

[pivoxa15]

no, you can measure two events to be simultaneous without being equidistance from them. Let's say your 5m from A and 10m from B. You know how long light takes to reach you from A and B (5/c and 10/c) so you just subtract that from whatever time you detect the light from both.

Everything is in one ertial frame.

I was thinking of observing the events 'raw'. But the way you suggest is the proper way of determining whether an event is simultaneous in a single frame? Then in that case, one dosen't have to be in the centre of two events for them to be simultaneous. It seems a better way of doing things because more events can be simultaneous when we allow calculations to be made.

 

[pivoxa15]

What about the classic situation of the observer standing in the middle of a moving train, with two lightning strikes on each end of the train but the flash from one end reaches the observer earlier. Can calculations be made with this example, with the observer concluding the two strikes were simultaneous based on the speed of the train. Or is this not possible because the speed of the train cannot be known for the observer inside the train so in this case the "apparent" recording of signals must be treated as the real signal so the signals are not simultaneous.

 

[pivoxa15]

I think I have figured out an answer to my lastest question above.

The observer thinks he is stationary and if he look outside, he will think that the outside is moving. If he is in the middle of the train and receives two flashes, one after the other, he must conclude that they were not simultaneous.

If he is not in the middle of the train and in a position so that he receives the two flashes 'simultaneously' than he must subtract off the extra time, the far end light has traveled and also conclude that the two signals were not simultaneous.