Simultaneity of Events: Rest vs Motion

[rajeshmarndi]

I understand when two observer one at rest wrt Earth and the other in motion wrt earth, two events cannot happen at the same time to both the observer. So, is it, the moment, the observer which was in motion comes to rest wrt earth, would two events now, at once happen at the same time, to both the observer.

Or it takes some time ,after the observer that was in motion comes to rest, that two events to both the observer, will happen at the same time.

Thank you.

 

[Vitro]

I understand when two observer one at rest wrt Earth and the other in motion wrt earth, two events cannot happen at the same time to both the observer. So, is it, the moment, the observer which was in motion comes to rest wrt earth, would two events now, at once happen at the same time, to both the observer.

Or it takes some time ,after the observer that was in motion comes to rest, that two events to both the observer, will happen at the same time.

Thank you.

Earth has nothing to do with it, you can leave it out. If two observers are in relative motion then two events (separated along the direction of relative motion) that happen at the same time for one observer happen at different times for the other. If the two observers are mutually at rest then simultaneity is the same for both.

 

[phinds]

I understand when two observer one at rest wrt Earth and the other in motion wrt earth, two events cannot happen at the same time to both the observer.

Vitro's statement is correct but does not cover all cases. You can arrange the geometry so that there ARE cases where two events are simultaneous for one observer and also simultaneous for an observer in motion wrt the first observer, but they will not agree on WHEN the events occurred.

 

[rajeshmarndi]

If the two observers are mutually at rest then simultaneity is the same for both.

This is what I wanted to know, does the simultaneity for both of them come into effect, the moment the observer in motion comes to rest or it takes a while.

 

[pervect]

This is what I wanted to know, does the simultaneity for both of them come into effect, the moment the observer in motion comes to rest or it takes a while.

"Simultaneity for both of them" doesn't really ever exist in any physical sense. The question underlying your question is what do you mean by "the moment the observer's motion comes to rest"? Assuming that "the moment" exists requires that you have a simultaneity convention in place. But that's what your question is about! So one can't answer your question unless one knows what "a moment" is, but your question is about what "a moment" means.

As a result, it's difficult to see how to answer your question. The impression that one is left with is that you assume that there is some sort of universal definition of "a moment" or "a now" that is absolute, as in the Newtonian notion of time. But the relativistic notion of time says that there isn't any such universal definition. Your question appears to assume that there is such a notion.

 

[Vitro]

This is what I wanted to know, does the simultaneity for both of them come into effect, the moment the observer in motion comes to rest or it takes a while.

Simultaneity is not a physical phenomenon that has to "come into effect", it's an abstract notion, a convention. A physical procedure for synchronizing distant clocks, like Einstein synchronization, obviously takes time to accomplish due to signal propagation speed limit, if that's what you mean.

 

[Mister T]

This is what I wanted to know, does the simultaneity for both of them come into effect, the moment the observer in motion comes to rest or it takes a while.

Different observers will have different notions of simultaneity, but only if they're in motion relative to each other. If that motion ceases they will then share the same notion of simultaneity. There is no delay. But there is of course a transition from a state of relative motion to a state of relative rest, and that transition takes time, it cannot happen instantaneously.

 

[pervect]

I was just thinking that it could be helpfu to talk abstractly about what "a moment" is. As I write this I have the feeling it may appear too involved, but I don't know how to make it any simpler. The first concept one needs is the space-time diagram. A space-time diagram is a set of events. Events are things that occur at a specific time, and a specific place, they are typically distinguished from one another by giving the events unique coordinates.

Then space-time is a set of events. It's a 4 dimensional set of events, because we need four coordinates to specify when and where and event occurs. I'm assuming the concept of "a dimension" is at least somewhat familiar.

Four dimensional space-time is rather hard to visualize, though, so we can throw out 2 of the dimensions of space to create a simpler problem, something that's easier to visualize and draw diagrams of. It's still enough to illustrate the concept of the relativity of simultaneity though. This is a two dimensional space-time diagram, with one space dimension and one time dimension, that can be and usually is drawn on a plane in any reference you care to look up. This diagram represents motion along some specific line. You can regard it simply as a graph of position versus time, something that should hopefully be a familiar concept.

Note that in drawing the space-time diagram, we take the liberty here of representing the time dimension as if it were a spatial dimension in our drawings. This seems to be one of the points that people occasionally get confused by (for reasons I've never understood), but it's no more complicated than saying that we can represent events that happen in different times in history on a "timeline".

With all this background, we are then in a position to represent the concept of "now". Now is some specific set of points on the space-time diagram that happen "at the same time". Visually, it's a line we draw on the space-time diagram, generally a straight line.

And the point we are making that the set of events that comprise the "now" of an observer at rest are a different set of events that comprise the "now" of a moving observer. They are two different lines on the diagram, typically one will be horizontal, the other will be drawn at a slant.

Physically, if we label an event by a flash of light, then the "now" of an observer at rest is a set of such flashes that all occur at the same time according to that observer. We can have the observer at rest emit flashes of red light, and the moving observer emit bule flashes of light. And we can say - do all the flashes occur at the same place and time? And the answer to that question is no, they do not.

 

[stoomart]

You can arrange the geometry so that there ARE cases where two events are simultaneous for one observer and also simultaneous for an observer in motion wrt the first observer, but they will not agree on WHEN the events occurred.

It seems like the notion of "now" @pervect described is one such arrangement, because non-comoving observers will not agree on the "when" (proper time) of "now".

 

[rajeshmarndi]

Different observers will have different notions of simultaneity, but only if they're in motion relative to each other. If that motion ceases they will then share the same notion of simultaneity. There is no delay. But there is of course a transition from a state of relative motion to a state of relative rest, and that transition takes time, it cannot happen instantaneously.

Yes, this is what I mean.

 

[Ibix]

If that motion ceases they will then share the same notion of simultaneity. There is no delay.

Let’s say the moving observer decelerates instantaneously to rest in frame S as he passes through x=0 at t=0 and adopts the simultaneity convention of S at that time. But t=0 implies $t'=-vx'/c^2$. Some of that is before he decelerated, at least by the simultaneity convention of S', and some is after. So "there is no delay" is misleading at best, basically because "at the same time as I change my definition of what at the same time means" is inherently contradictory.

Much better to use something like radar coordinates. They assign S' simultaneity in the past light cone of the deceleration event, S simultaneity in the future light cone, and a consistent halfway house outside.

 

[Mister T]

Let’s say the moving observer decelerates instantaneously to rest in frame S as he passes through x=0 at t=0 and adopts the simultaneity convention of S at that time. But t=0 implies $t'=-vx'/c^2$.

Isn't that an event for which $v$ is undefined? You have things set up so that the value of $v$ is both zero and nonzero at that event. Such a thing is not possible. $v$ changes from some nonzero value to zero, and you need two events to describe that transition. What you're calling the adoption of S's simultaneity convention is not an event, it's a process that starts when $v$ starts to change from whatever nonzero value you choose for $v$ to a value of zero. When $v=0$ any spatially separated clocks that were previously synchronized in the the rest frame of S' will now be seen as not synchronized.

And isn't it correct to say that both S and S' are using the same simultaneity convention all along? It's just that when an observer at rest in S uses it he gets a different result for a pair of events than an observer at rest in S', provided they are in relative motion and those events are spatially separated along the line of that relative motion?

 

[Ibix]

What you're calling the adoption of S's simultaneity convention is not an event, it's a process that starts when vv starts to change from whatever nonzero value you choose for vv to a value of zero.

Ok. In that case there is a chunk of the accelerated observer's worldline where his definition of simultaneity is changing. But if he wants to use S' time before that chunk and S time after then he has to deal with the fact that the S' plane of simultaneity at the beginning of that chunk is not parallel to the S plane of simultaneity at the end of the chunk, and they meet somewhere. And then there is a region where the past in S' overlaps the future in S. You have to choose which frame to use in this region, and whichever you use the boundary is not a plane of simultaneity in one frame or another.

And isn't it correct to say that both S and S' are using the same simultaneity convention all along?

Hm. They are using the same clock synchronisation procedure, but they come up with different notions of simultaneity. I tend to regard that as the same clock synchronisation procedure but different simultaneity conventions. It's possible I'm using terminology incorrectly - I don't know.

 

[PeterDonis]

They are using the same clock synchronisation procedure, but they come up with different notions of simultaneity. I tend to regard that as the same clock synchronisation procedure but different simultaneity conventions.

I agree, and I think it's also good to emphasize that the same clock synchronization procedure, when conducted by different sets of inertial observers (i.e., observers who are at rest relative to the other observers in their own set, but moving relative to inertial observers in other sets), gives you the different simultaneity conventions.

 

[Mister T]

So "there is no delay" is misleading at best, basically because "at the same time as I change my definition of what at the same time means" is inherently contradictory.

So, is this the issue? That to determine the simultaneity of two events, you have to wait for the light that brings news of the event(s)?

 

[Mister T]

They are using the same clock synchronisation procedure, but they come up with different notions of simultaneity. I tend to regard that as the same clock synchronisation procedure but different simultaneity conventions. It's possible I'm using terminology incorrectly - I don't know.

I agree, and I think it's also good to emphasize that the same clock synchronization procedure, when conducted by different sets of inertial observers (i.e., observers who are at rest relative to the other observers in their own set, but moving relative to inertial observers in other sets), gives you the different simultaneity conventions.

I'm thinking that if both observers are using, say the Einstein convention, then they're using the same convention. It's all semantics, of course, but I can't reckon how you can call it the Einstein convention, it seems it should instead be called the Einstein procedure.

 

[PeterDonis]

I can't reckon how you can call it the Einstein convention

The usual term is "Einstein clock synchronization", which is a procedure. I don't think "the Einstein convention" is a standard term; that term doesn't appear in what you quoted from either Ibix or me.

 

[vanhees71]

Let's make things clear with a little math. This always helps. If we have two events $\boldsymbol{x}$ and $\boldsymbol{y}$ in Minkowski space and if there is an inertial basis $\boldsymbol{e}_{\mu}$, for which these are simultaneous, by definition you have
$$x^0=y^0 \qquad (1)$$
and thus
$$(\boldsymbol{x}-\boldsymbol{y})^2<0.$$
Then let $\boldsymbol{e}_{\mu}'$ denote a frame, which moves with constant speed $v$ along the $x^1$ axis. Then you have
$$x^{\prime 0}=\gamma(x^0-v x^1), \quad x^{\prime 1}=\gamma (x^1-v x^0), \quad x^{\prime 2}=x^{2}, \quad x^{\prime 3}=x^{3},$$
and analogous for $\boldsymbol{y}$. In this frame you get, using (1)
$$x^{\prime 0}-y^{\prime 0}=-\gamma v (x^1-y^1),$$
which is (in general) not $0$, and thus the events are not simultaneous for an observer at rest in the primed reference frame, while they are for an observer at rest in the original reference frame.

 

[Ibix]

So, is this the issue? That to determine the simultaneity of two events, you have to wait for the light that brings news of the event(s)?

Draw a Minkowski diagram of the accelerating object. Include the coordinate grid for both S and S'. The lines of equal t are not parallel to the lines of equal t', so a transition from one set of coordinates to the other cannot happen at a single time for more than one of the frames. That's the issue; it isn't consistent to say "the change of simultaneity happens immediately" because any boundary must be non-parallel with lines of now for at least one of the frames.

What you do about it is a separate thing.

There's nothing wrong with using S or S' throughout. It's not terribly natural for the observer since it implies a non-isotropic speed of light before or after the acceleration. But it's only coordinate speed, so that's fine.

There's nothing wrong with selecting some space-like 3-plane as a boundary and using S' on and before that plane and S after. This is what we've been discussing; as noted the only problem is that the boundary doesn't represent a unique time for both frames.

Neither of these is particularly natural. The beauty of Einstein's frames is that they represent your intuitive feeling of what coordinates ought to do in a flat space and do it from straightforward observations. But an infinite rigid system of rods falls apart (literally) if you accelerate. A more flexible approach is just to use radar in a completely naive manner - Dolby and Gull's paper on radar coordinates that I linked upthread has the maths and nice diagrams of the resulting simultaneity planes.

There are infinitely many ways of doing this - it's just a convention and you can use whatever works for you.

I'm thinking that if both observers are using, say the Einstein convention, then they're using the same convention. It's all semantics, of course, but I can't reckon how you can call it the Einstein convention, it seems it should instead be called the Einstein procedure.

An inertial frame is really a choice of an infinite flock of inertial clocks at rest with respect to each other and a choice of a way to set them all to zero. Given the former, Einstein proposed a way to do the latter, and that is the Einstein clock synchronisation convention (others are available, consult your physicist before use). But the resulting definition of simultaneity depends on your choice of the state of motion of the clocks. So the clock synchronisation convention does not, on its own, define a simultaneity convention and "the Einstein simultaneity convention" can't be a thing without some further qualification (I'm a bit more confident saying that now that @PeterDonis seems to be agreeing).

In other threads I have, somewhat self-referentially, referred to the "Einstein simultaneity convention for frame S". What I ought to say is more like "the result of applying Einstein's clock synchronisation convention to clocks in the state of motion I defined to be rest for S". But it's a bit of a mouthful.

I think there is a lot of room for confusion here. But there is an important distinction between a clock synchronisation convention and the resulting simultaneity convention.