Explaining Relative Simultaneity

[Max™]

 

[Saw]

it would be stretching the anthropomorphism to the point of absurdity to talk of frames making "judgments" about what is true in other frames besides themselves, or of two frames A and B "agreeing" that two events are simultaneous in A (even though they are not simultaneous in B). Trust me, physicists never talk this way.

I agree with everything you are saying to help Grimble understand the train thought experiment. Only… I would not share this particular statement. It’s true that you do not need a human observer sitting in a physical vehicle to construct a reference frame. A reference frame can be an idealized region, with origin at an idealized point (which may not be occupied and, what is more, cannot be occupied by any physical object) and despite that be useful for calculation purposes in your attempt at predicting what may happen and what may not. However, this idealized frame can make “judgments about what is judged in other frames”, namely the embankment frame may agree that the lightning strikes will not be labelled as simultaneous in the train frame. For this purpose, the embankment frame relies on two elements:

- One is factual: the light flashes coming from the two sides do not arrive simultaneously at the center of the train. This is an event, so it is true for any reference frame. The embankment frame may perfectly take it as basis for making some judgment.

- The other is conceptual: the embankment frame knows that the train frame will label the lightning strikes as simultaneous, taking into account that they happened at points equidistant from the center of the train, if after reflecting towards the latter, they meet at it simultaneously. In other words, the embankment frame is aware that the train frame constructs its concept of simultaneity on the basis of the assumption that the speed of light is c in both directions, in the train frame. So it “agrees” that the strikes are not deemed to be simultaneous in the train frame.

In fact, the point of Lorentz transformations is that: every frame “agrees” that the measurements of time and distances or lengths made from other frames are what the latter should have obtained from their respective physical perspectives and conceptual assumptions. That’s why, after mixing them in the corresponding formula, you get the measurements that you would have obtained if you had clocks and rulers at the relevant place. And all sets of values for x and t of different frames lead to the same predictions in terms of events.

Well, this is little more than semantics. But it may help Grimble. Grimble, I think that you are having the same misconception I had when I first read Einstein’s account. Maybe you think that he is saying that the lightning strikes meet the center of the platform simultaneously because they “happen in the platform”, but flashes projected (at the same time as the lightning strikes) from sources on the train would also meet at the center, in this case, of the train, because they “happen in the train”. It would not be so: the flashes projected from the train will always keep in parallel with those projected from the platform, so that (i) the four flashes will meet at the center of the platform simultaneously, whereas (ii) the two from the front will reach the center of the train before the two from the back. Once that we have agreement between frames on events, we can talk about concepts, like simultaneity. Both frames agree that the judgment on simultaneity must be made this way: two flashes are simultaneous in a given frame if, happening at points equidistant from another point of that frame, their light reaches the latter simultaneously. Wrt the platform center, that happens, so the embankment frames labels the flashes as simultaneous and the train frame agrees that the embankment frame should make that judgment, although it doesn’t make it for its own purposes. Wrt the train center, that doesn’t happen, so the train frame does not label the flashes as simultaneous and the embankment frame agrees that the train frame should make that judgment, although it doesn’t make it for its own purposes.

How can that be? Agreement on disagreement? Well, the paradox quickly dissolves if you take into account that the “disagreement” only projects on an instrumental concept. Just with simultaneity, none of them does anything. If they want to predict events (i.e. what will happen), they need something more. With simultaneity, you set the clocks running but then you need that they run, you need a time lapse. You also need length. All observers disagree on each of these particular labels. But when they combine them in their formulas, they all get the same predictions about events or happenings. Thus you can understand what I meant when I said “for its own purposes”. “Purpose” here is the combination with the rest of their own concepts.

 

[JesseM]

I agree with everything you are saying to help Grimble understand the train thought experiment. Only… I would not share this particular statement. It’s true that you do not need a human observer sitting in a physical vehicle to construct a reference frame. A reference frame can be an idealized region, with origin at an idealized point (which may not be occupied and, what is more, cannot be occupied by any physical object) and despite that be useful for calculation purposes in your attempt at predicting what may happen and what may not. However, this idealized frame can make “judgments about what is judged in other frames”, namely the embankment frame may agree that the lightning strikes will not be labelled as simultaneous in the train frame.

A person at rest in the embankment frame can judge that, but they do so by making calculations in the train frame rather than the embankment frame. It simply confuses things to say the embankment frame is making that judgment when you are making calculations that have nothing to do with the embankment frame, and even if you think it would make sense to talk this way if you were inventing the terminology from scratch, the fact is that physicists don't in fact talk about frames this way.

- The other is conceptual: the embankment frame knows that the train frame will label the lightning strikes as simultaneous,

But a frame is just a coordinate system, it's purely an anthropomorphic figure of speech to talk about it "knowing" anything. And it's misleading to equate a frame with observers at rest in that frame, since after all a given human observer is free to use any frame they like for the purpose of making calculations, it's not as if they live in one coordinate system but not others...it's purely a matter of linguistic convention that we refer to the frame where an observer is at rest as "their frame".

Well, this is little more than semantics. But it may help Grimble.

I would say the opposite, it seems to me from previous posts (on this thread and the other thread) that the idea of frames having "opinions" about other frames besides themselves is precisely one of the main things that has led Grimble into confusion (see the comments on the other thread about 'inertial units' vs. 'transformed units', which I think may have something to do with this confusion though I'm not sure)

 

[Rasalhague]

Suppose we have a value correct with respect to one frame, call it frame A. And suppose we figuratively say that the value is such in frame A's "opinion" or "judgement". If this value is different in another frame, call it frame B, we could likewise figuratively call the value according to frame B the "opinion" of frame B. But I agree with Jesse that it seems needlessly confusing to talk about frame B having an opinion about (or an awareness of) frame A's opinion since "frame B's (C's, D's, E's...) opinion about frame A's opinion" will always mean exactly the same thing as simply "frame A's opinion". So the words "frame B's opinion about..." are meaningless, and give the misleading impression that the value according to frame A depends on something other than how the events are located in frame A.

 

[Grimble]

Thank you, Good People, I do understand the point that Jesse is making and appreciate your comments.


May I say that when doing that I was being a little lazy and should have been saying "an observer in frame x would think; would agree; would conclude; would judge..."

Jesse, I can see, in reflection, that I was allowing excessive hyperbole to obscure what I was saying. I realize that this can colour the way that such statements are received and make it uncomfortable for the reader; and that that is not good in scientific discussion.

Thank you for pointing that out to me.

Grimble, I think that you are having the same misconception I had when I first read Einstein’s account. Maybe you think that he is saying that the lightning strikes meet the center of the platform simultaneously because they “happen in the platform”, but flashes projected (at the same time as the lightning strikes) from sources on the train would also meet at the center, in this case, of the train, because they “happen in the train”. It would not be so: the flashes projected from the train will always keep in parallel with those projected from the platform, so that (i) the four flashes will meet at the center of the platform simultaneously, whereas (ii) the two from the front will reach the center of the train before the two from the back. Once that we have agreement between frames on events, we can talk about concepts, like simultaneity. Both frames agree that the judgment on simultaneity must be made this way: two flashes are simultaneous in a given frame if, happening at points equidistant from another point of that frame, their light reaches the latter simultaneously. Wrt the platform center, that happens, so the embankment frames labels the flashes as simultaneous and the train frame agrees that the embankment frame should make that judgment, although it doesn’t make it for its own purposes. Wrt the train center, that doesn’t happen, so the train frame does not label the flashes as simultaneous and the embankment frame agrees that the train frame should make that judgment, although it doesn’t make it for its own purposes.

Thank you Saw, what you say is true!

The mist is clearing...!

But let me test my understanding...

Points A & A' are adjacent in time and space, as are points B and B'.

We know that the light will meet at point M because that is a given, in the problem's description.

Because the light meets at M we know that A and B are simultaneous to the embankment.

Because A & B are simultaneous to M, they cannot be simultaneous to M'.

M & M' will both agree that they are simultaneous to M but not to M'.

Right so far?

I was thinking "but what if we were not told that the light met at M? How could we determine to which of them it would be simultaneous?

Then I realized the stupidity in that line of argument, for unless we are told that the strikes at A & B are simultaneous to one frame, we have no indication that they were simultaneous in any frame!

But please let me suggest one more variation:
The embankment is solid and rigid.
If we, not unreasonably, stipulate that the same is true of the train, and say that two lights are placed alongside the track such that they shine their lights upwards where mirrors reflect the light towards our observer M.
Now if part of the train obscures the lights except at two points A' and B' which coincide with A & B as the train passes, such that the lights both reach their mirrors, then will the resulting flashes of light be simultaneous at A & B or A' & B', for we have agreed that they cannot be simultaneous at both?

And thinking about the above scenario raises another little question to my fevered brain:
A & B, and A' & B' must be equidistant for the above to work.
But observer's in either frame would know that that distance A - B, or A' - B', observed in the other frame is length contracted and therefore would not coincide with their non contracted distance.
Yet at the same time, taking into account what I have learned here, both these distances exist within whichever frame they are measuring in...?
So let me restate my question:
Observer M', sitting on the train knows where A' and B' are, how far they are from her.
And she also knows that in in the same frame of reference A & B have the same separation as A' and B'.
And the same is true for M on the embankment (or Platform).
Yet if either regards the moving system those moving distances would be length contracted and not meet up with the stationary (within that frame of reference) points.

Help!

I am confusing myself again!

Grimble

 

[Rasalhague]

But let me test my understanding...

Points A & A' are adjacent in time and space, as are points B and B'.

We know that the light will meet at point M because that is a given, in the problem's description.

Because the light meets at M we know that A and B are simultaneous to the embankment.

Because A & B are simultaneous to M, they cannot be simultaneous to M'.

M & M' will both agree that they are simultaneous to M but not to M'.

Right so far?

Yes.

I was thinking "but what if we were not told that the light met at M? How could we determine to which of them it would be simultaneous?

Then I realized the stupidity in that line of argument, for unless we are told that the strikes at A & B are simultaneous to one frame, we have no indication that they were simultaneous in any frame!

That's right, although we know that, so long as there's a spacelike separation between the events (meaning that it's impossible for a signal to pass from one event to the other without traveling faster than c), then there will always be some frame we could chose in which they'd be simultaneous (not necessarly M or M').

But please let me suggest one more variation:
The embankment is solid and rigid.
If we, not unreasonably, stipulate that the same is true of the train, and say that two lights are placed alongside the track such that they shine their lights upwards where mirrors reflect the light towards our observer M.
Now if part of the train obscures the lights except at two points A' and B' which coincide with A & B as the train passes, such that the lights both reach their mirrors, then will the resulting flashes of light be simultaneous at A & B or A' & B', for we have agreed that they cannot be simultaneous at both?

I'm not sure if I'm visualising your scenario correctly, but if I understand what you're saying, then the situation is exactly the same as in the case of the lightning strikes example. It doesn't matter whether the light is reflected off the train or off something at rest in the embankment frame. If the light from each side reaches M at the same time, then the light will have left the two points simultaneously in the embankment frame (and not in the train frame). But if the light from each side reaches M' at the same time, then it will have left the two points simultaneously in the train frame (and not in the embankment frame).

And thinking about the above scenario raises another little question to my fevered brain:
A & B, and A' & B' must be equidistant for the above to work.

Yes. Otherwise, the same principles about simultaneity apply, except that we'd have to take into account the difference in the distances the light would have to travel from A = A' and B = B'. So if they're not equidistant, we can still determine whether the events were simultaneous in a particular frame; it just makes the calculation a little bit more complicated.

 

[Rasalhague]

And thinking about the above scenario raises another little question to my fevered brain:
A & B, and A' & B' must be equidistant for the above to work.
But observer's in either frame would know that that distance A - B, or A' - B', observed in the other frame is length contracted and therefore would not coincide with their non contracted distance.
Yet at the same time, taking into account what I have learned here, both these distances exist within whichever frame they are measuring in...?
So let me restate my question:
Observer M', sitting on the train knows where A' and B' are, how far they are from her.
And she also knows that in in the same frame of reference A & B have the same separation as A' and B'.
And the same is true for M on the embankment (or Platform).
Yet if either regards the moving system those moving distances would be length contracted and not meet up with the stationary (within that frame of reference) points.

Let's suppose the lightning strikes are simultaneous in the rest frame of the embankment. We can define points in space A and B, at rest with respect to the embankment, and points in space A' and B' at rest with respect to the train, such that the lightning strikes occur (in all frames) one at the intersection of the world lines of A and A', the other at the intersection of the world lines of B and B'. (Note that points in space are curves in spacetime, specifically straight lines if they're at rest in some inertial frame, as is the case here. The intersection of such curves defines a point in spacetime, which we call an "event".)

A is adjacent to A' when, in the rest frame of the embankment, B is adjacent to B', and so |B - A| = |B' - A'| in that frame. That's to say, the distance between A and B is the same as the distance between A' and B', by our definition of these points, in the rest frame of the embankment. Let's call this distance \Delta x.

The "separation" between two events (points in spacetime) is a spacetime vector, the four dimensional analogue of a displacement vector in three dimensional space. The length of this vector (called the spacetime "interval") is the same in all frames:

|| \mathbf{s} || = \sqrt[]{\left| (c \; \Delta t)^{2} - (\Delta x)^{2} \right|},

where \Delta x is the distance in space between the events, and (\Delta t)^{2} the time between them. For this equation to hold, and for the length of this vector to be the same in all frames, if a given pair of events are further apart in time in one frame than another, they must also be further apart in space. As you can see, the spacetime interval between the two events is equal to the spatial distance between them in the embankment's rest frame where there's no difference in time between them (and only in that frame).

In the rest frame of the train, the lightning strikes are further apart in time than they were in the rest frame of the embankment, since there was no time at all between them in the rest frame of the embankment, so they must also be further apart in space, by a factor of

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

times as far apart in space. In the train's rest frame, the stationary points A' and B' are this much further apart than the moving points A and B. But there's no contradiction with the fact that the lightning strikes when A = A' and B = B' in all frames. That's because these events aren't simultaneous in the rest frame of the train; the lightning bolts don't strike at the same time, and so A' doesn't line up with A at the same time as B' lines up with B.

If we change the scenario and think about a pair of lightning strikes that are simultaneous at A = A' and B = B' in the rest frame of the train, then the distance between A and B would equal the distance between A' and B' in the rest frame of the train, and it would be the distance between A and B that would be greater by

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

in the rest frame of the embankment.

 

[oldman]

For what it's worth here are some thoughts that might help Grimble feel more comfortable with relativity.

The concept of simultaneity, in the context of this thread, was refined by physicists from
evolution-conditioned prejudice. People are sighted creatures, like many other animals that have evolved along with us over the millennia. Sight is a survival advantage in a world full of dangerous happenings. So we, and no doubt many other animals, are conditioned by evolution to accept the world exactly as we see it -- we prosper as wysiwyg critters. That's why we invented the word ‘simultaneous’ to label events we see 'now', why two lightning flashes are judged to be ‘simultaneous’ when their light arrives at our eyes at the same instant and why we feel in our bones that there is a universal ‘now’ that is the same for everyone, everywhere in spacetime.

If you doubt this, imagine how humanity would define ‘simultaneous’ if it could only hear
thunder, and not see the lightning flashes that caused it. Imagine further the complications a hearing-based concept of simultaneity would cause in formulating an understanding of the way things work around us. Physics would be very hard indeed to invent, especially for understanding faraway happenings, things traveling at speeds comparable with Mach 1 and happenings in space. We’d then land up with a rather different set of evolution-conditioned prejudices -- if we survived, that is.

Although light is fast, it is not infinitely fast, and this causes similar (but certainly not identical) complications in using ordinary laboratory physics to understand the fast life and exporting it to remote locations in spacetime. It turns out that to match observation, preserve logical consistency and make verifiable predictions, some of our evolution-conditioned prejudices have to go.

In formulating relativity Einstein (remember --- he was a genius) most ingeniously effected a compromise by preserving an evolution-conditioned prejudice (our very local concept of simultaneity); by introducing a sensible way of gauging times and distances (the light-ranging method of special relativity); by discarding other prejudices (as far as time is concerned those of a universal ‘now’ and a universal measure of duration) and so created a universal foundation of local physics.

I sometimes wonder whether in doing so he uncovered eternal Platonic truths, or if he ‘only!’
devised an esoterically clever but anthro’centric description of the way things work verywhere and everywhen.

 

[Grimble]

Let's suppose the lightning strikes are simultaneous in the rest frame of the embankment. We can define points in space A and B, at rest with respect to the embankment, and points in space A' and B' at rest with respect to the train, such that the lightning strikes occur (in all frames) one at the intersection of the world lines of A and A', the other at the intersection of the world lines of B and B'. (Note that points in space are curves in spacetime, specifically straight lines if they're at rest in some inertial frame, as is the case here. The intersection of such curves defines a point in spacetime, which we call an "event".)

A is adjacent to A' when, in the rest frame of the embankment, B is adjacent to B', and so |B - A| = |B' - A'| in that frame. That's to say, the distance between A and B is the same as the distance between A' and B', by our definition of these points, in the rest frame of the embankment. Let's call this distance \Delta x.

The "separation" between two events (points in spacetime) is a spacetime vector, the four dimensional analogue of a displacement vector in three dimensional space. The length of this vector (called the spacetime "interval") is the same in all frames:

|| \mathbf{s} || = \sqrt[]{\left| (c \; \Delta t)^{2} - (\Delta x)^{2} \right|},

where \Delta x is the distance in space between the events, and (\Delta t)^{2} the time between them. For this equation to hold, and for the length of this vector to be the same in all frames, if a given pair of events are further apart in time in one frame than another, they must also be further apart in space. As you can see, the spacetime interval between the two events is equal to the spatial distance between them in the embankment's rest frame where there's no difference in time between them (and only in that frame).

In the rest frame of the train, the lightning strikes are further apart in time than they were in the rest frame of the embankment, since there was no time at all between them in the rest frame of the embankment, so they must also be further apart in space, by a factor of

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

times as far apart in space. In the train's rest frame, the stationary points A' and B' are this much further apart than the moving points A and B. But there's no contradiction with the fact that the lightning strikes when A = A' and B = B' in all frames. That's because these events aren't simultaneous in the rest frame of the train; the lightning bolts don't strike at the same time, and so A' doesn't line up with A at the same time as B' lines up with B.

If we change the scenario and think about a pair of lightning strikes that are simultaneous at A = A' and B = B' in the rest frame of the train, then the distance between A and B would equal the distance between A' and B' in the rest frame of the train, and it would be the distance between A and B that would be greater by

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

in the rest frame of the embankment.

THANK YOU, THANK YOU, THANK YOU

That really does make beautiful sense :

Grimble

 

[Grimble]

Let's suppose the lightning strikes are simultaneous in the rest frame of the embankment. We can define points in space A and B, at rest with respect to the embankment, and points in space A' and B' at rest with respect to the train, such that the lightning strikes occur (in all frames) one at the intersection of the world lines of A and A', the other at the intersection of the world lines of B and B'. (Note that points in space are curves in spacetime, specifically straight lines if they're at rest in some inertial frame, as is the case here. The intersection of such curves defines a point in spacetime, which we call an "event".)

A is adjacent to A' when, in the rest frame of the embankment, B is adjacent to B', and so |B - A| = |B' - A'| in that frame. That's to say, the distance between A and B is the same as the distance between A' and B', by our definition of these points, in the rest frame of the embankment. Let's call this distance \Delta x.

The "separation" between two events (points in spacetime) is a spacetime vector, the four dimensional analogue of a displacement vector in three dimensional space. The length of this vector (called the spacetime "interval") is the same in all frames:

|| \mathbf{s} || = \sqrt[]{\left| (c \; \Delta t)^{2} - (\Delta x)^{2} \right|},

where \Delta x is the distance in space between the events, and (\Delta t)^{2} the time between them. For this equation to hold, and for the length of this vector to be the same in all frames, if a given pair of events are further apart in time in one frame than another, they must also be further apart in space. As you can see, the spacetime interval between the two events is equal to the spatial distance between them in the embankment's rest frame where there's no difference in time between them (and only in that frame).

In the rest frame of the train, the lightning strikes are further apart in time than they were in the rest frame of the embankment, since there was no time at all between them in the rest frame of the embankment, so they must also be further apart in space, by a factor of

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

times as far apart in space. In the train's rest frame, the stationary points A' and B' are this much further apart than the moving points A and B. But there's no contradiction with the fact that the lightning strikes when A = A' and B = B' in all frames. That's because these events aren't simultaneous in the rest frame of the train; the lightning bolts don't strike at the same time, and so A' doesn't line up with A at the same time as B' lines up with B.

If we change the scenario and think about a pair of lightning strikes that are simultaneous at A = A' and B = B' in the rest frame of the train, then the distance between A and B would equal the distance between A' and B' in the rest frame of the train, and it would be the distance between A and B that would be greater by

\frac{1}{\sqrt[]{1 - \left( \frac{v}{c}\right)^{2}}}

in the rest frame of the embankment.

THANK YOU, THANK YOU, THANK YOU

That really does make beautiful sense :

Grimble

BUT, having pondered this at some length I have another question to ask.

If AB on the embankment is the same distance as A'B' in proper units, in their respective frames of reference, such that they would line up as equidistant if the train stopped, then the space time interval, which has to be the same in all frames, includes the term \Delta{x^2} which is the spatial distance between the two events A/A' and B/B', being AB in the frame of the embankment; but is this not also the same distance as A'B' in the frame of the train and would this not mean that the two space time intervals would have to be equal, with the two time intervals consequently being equal?

So I suppose that what I am asking is: if A and A' are adjacent and if the distance AB in the embankment's frame is equal to the distance A'B' in the train's frame then would this not be a definition of simultaneity between two events in two frames rather than simultaneity between two events in one frame?

And would this not give the expected reciprocality of SR? with each observer saying that the events were simultaneous to them but not to the other?

Or do I need my thoughts straightening out again?

Grimble

 

[JesseM]

If AB on the embankment is the same distance as A'B' in proper units, in their respective frames of reference, such that they would line up as equidistant if the train stopped,

This would be different than Einstein's scenario--if AB and A'B' had the same proper length, and the train was in motion relative to the embankment, then in both the embankment frame and the train frame AB and A'B' would be different lengths, so if the two strikes happened simultaneously at AB in the embankment frame, the strikes could not coincide with both A' and B' as well.

then the space time interval, which has to be the same in all frames,

The space time interval between what pair of events? The lightning strikes? Again, if you make AB and A'B' the same proper length, and still have the strikes be simultaneous in the embankment frame, then the strikes won't be next to both A' and B' on the train. So, the delta-x' for the two strikes in the train frame would not be equal to the distance A'B' in this case.

 

[cfrogue]

BUT, having pondered this at some length I have another question to ask.

If AB on the embankment is the same distance as A'B' in proper units, in their respective frames of reference, such that they would line up as equidistant if the train stopped, then the space time interval, which has to be the same in all frames, includes the term \Delta{x^2} which is the spatial distance between the two events A/A' and B/B', being AB in the frame of the embankment; but is this not also the same distance as A'B' in the frame of the train and would this not mean that the two space time intervals would have to be equal, with the two time intervals consequently being equal?

So I suppose that what I am asking is: if A and A' are adjacent and if the distance AB in the embankment's frame is equal to the distance A'B' in the train's frame then would this not be a definition of simultaneity between two events in two frames rather than simultaneity between two events in one frame?

And would this not give the expected reciprocality of SR? with each observer saying that the events were simultaneous to them but not to the other?

Or do I need my thoughts straightening out again?

Grimble

There are two different ideas here.
The space time interval for all observers will be invariant with respect to two light sources is decided by SR.


But, that has nothing to do with the ordinality of the two light sources in terms of which occurred in which order. That is determined by the position and relative velocity.
This is not decidable from SR as there is no absolute simultaneity.

Thus, you are mixing space time intervals which are invariant with the ordinality of events which are not.

 

[matheinste]

There are two different ideas here.

But, that has nothing to do with the ordinality of the two light sources in terms of which occurred in which order. That is determined by the position and relative velocity.
This is not decidable from SR as there is no absolute simultaneity.

The order of events is frame dependent and thus dependent on relative velocity. When light travel times are taken into account the order of events is not dependent on postion. But of course the order in which we actually see events is position dependent.

Matheinste

 

[cfrogue]

The order of events is frame dependent and thus dependent on relative velocity. When light travel times are taken into account the order of events is not dependent on postion. But of course the order in which we actually see events is position dependent.

Matheinste

Assume we have two light sources A and B.

Let O be located at A and O' be located at B.

Assume all observers are in the same frame.

Now, assume we sync the clocks according to Einstein's clock synchronization method.
At an agreed upon time, both A and B emit.

Obviously, position is important to determining the relative ordinality of these light sources.

Thus, you cannot discount position.

 

[JesseM]

Assume we have two light sources A and B.

Let O be located at A and O' be located at B.

Assume all observers are in the same frame.

Now, assume we sync the clocks according to Einstein's clock synchronization method.
At an agreed upon time, both A and B emit.

Obviously, position is important to determining the relative ordinality of these light sources.

Thus, you cannot discount position.

All observers have to use position to figure out the travel time between the emission events and their seeing the light from these events, but when they subtract out the travel time to find the time the events "really" occurred in their own frame, all observers who are at rest relative to one another will agree on whether the events were simultaneous or not...their different positions won't cause them to have different judgments about the answer to this question, as long as they all share the same inertial rest frame.

 

[cfrogue]

All observers have to use position to figure out the travel time between the emission events and their seeing the light from these events, but when they subtract out the travel time to find the time the events "really" occurred in their own frame, all observers who are at rest relative to one another will agree on whether the events were simultaneous or not...their different positions won't cause them to have different judgments about the answer to this question, as long as they all share the same inertial rest frame.

No problem with the above.

My issue was position and relative velocity are both a part of how a frame will determine the order of two light events. matheinste seemed to disagree with this, perhaps not.

I put them in one frame to emphasize this fact, that is all.

Both position and relative speed will have an impact on how an event will be seen by a frame.

I could have just made 3 at rest above and O' moving at .0000000000000000000000000001 relative to O.

As long as the two light sources were far enough apart, the relative motion would not be the major determining factor.

It would be the position.

 

[matheinste]

Cfroque,

Consider two light emmiting sources in the same inertial frame in which clocks have been synchronize using the Einstein procedure. Let them both emit a short light pulse simultaneously in that same inertial frame, using the usual definition of simultaneity. The order in which an observer in that frame SEES the flashes (events) will depend upon the observer's position, however, they are sinultaneous because that is how we have set up the scenario. If the sources are set so as not to emit simultaneously the order in which an observer SEES them is position dependent, but their time coordinates or "real" times or the order in which they occur is not position dependent. When you SEE an event is not when it happened. Alll observers can, if they know the distances or positions involved can calculate light transmission times and determine when the emissions "really" happened. They will all agree on the result.

Matheinste

 

[cfrogue]

Cfroque,

Consider two light emmiting sources in the same inertial frame in which clocks have been synchronize using the Einstein procedure. Let them both emit a short light pulse simultaneously in that same inertial frame, using the usual definition of simultaneity. The order in which an observer in that frame SEES the flashes (events) will depend upon the observer's position, however, they are sinultaneous because that is how we have set up the scenario. If the sources are set so as not to emit simultaneously the order in which an observer SEES them is position dependent, but their time coordinates or "real" times or the order in which they occur is not position dependent. When you SEE an event is not when it happened. Alll observers can, if they know the distances or positions involved can calculate light transmission times and determine when the emissions "really" happened. They will all agree on the result.

Matheinste

If the sources are set so as not to emit simultaneously the order in which an observer SEES them is position dependent, but their time coordinates or "real" times or the order in which they occur is not position dependent.

what?

 

[matheinste]

If the sources are set so as not to emit simultaneously the order in which an observer SEES them is position dependent, but their time coordinates or "real" times or the order in which they occur is not position dependent.

what?

If two observers at rest relative to each other, at different positions, SEE events in a certain order that is not necessarily the order in which the events occurred. But they can, from their relative positions, calculate and agree on when and in what order they occurred.

These are basic facts. I cannot add any more to them or put them more simply.

Matheinste.

 

[cfrogue]

If two observers at rest relative to each other, at different positions, SEE events in a certain order that is not necessarily the order in which the events occurred. But they can, from their relative positions, calculate and agree on when and in what order they occurred.

These are basic facts. I cannot add any more to them or put them more simply.

Matheinste.

Yea, I was not saying anyting about the order in which events occured. I was talking about the order in which they were seen.

I cannot put it any more simply that both relative motion and position are key factors in determining when an event is seen.

You already agreed relative motion is a factor.

Now, assume an event is located at postive x-axis 4.

Now assume a and b are in the same frame relative to the event toward the negative xaxis and both are in the same frame but a is one unit more negative than b.

Clearly, whenever the event strikes at 4, b will see it first and a second because of position and not relative motion because they are in the same frame and not in the same frame as the event.

Thus, one cannot discount initial distance to the event as a factor.

So, I simply stated both as possible conditions, are you saying this is not true?

 

[matheinste]

Yea, I was not saying anyting about the order in which events occured. I was talking about the order in which they were seen.

I cannot put it any more simply that both relative motion and position are key factors in determining when an event is seen.

You already agreed relative motion is a factor.

Now, assume an event is located at postive x-axis 4.

Now assume a and b are in the same frame relative to the event toward the negative xaxis and both are in the same frame but a is one unit more negative than b.

Clearly, whenever the event strikes at 4, b will see it first and a second because of position and not relative motion because they are in the same frame and not in the same frame as the event.

Thus, one cannot discount initial distance to the event as a factor.

So, I simply stated both as possible conditions, are you saying this is not true?

If you are talking about when an event or events are seen then I agree that it is position dependent. And also that the order in which they are seen is also thus dependent. So if you are talking about what is seen then I agree.

However, it is your statement below that I disagree with.

----the ordinality of the two light sources in terms of which occurred in which order. That is determined by the position and relative velocity.----

You use the word occurred and this is not normally taken to mean seen. For observers at rest relative to each other, when events occur is not position dependent. When thay are seen is position dependent.

Matheinste.

 

[Grimble]

This would be different than Einstein's scenario--if AB and A'B' had the same proper length, and the train was in motion relative to the embankment, then in both the embankment frame and the train frame AB and A'B' would be different lengths, so if the two strikes happened simultaneously at AB in the embankment frame, the strikes could not coincide with both A' and B' as well.

Yes, indeed, Jesse, this is a different circumstance

In this case I am saying that we stipulate that AB equals A'B';
that the lightning strikes at A and B are simultaneous (in the frame of the embankment);
and that when the lightning strikes, A and A' are adjacent.

Now if the spacetime interval between two events is the same in any frame of reference,

The "separation" between two events (points in spacetime) is a spacetime vector, the four dimensional analogue of a displacement vector in three dimensional space. The length of this vector (called the spacetime "interval") is the same in all frames:

|| \mathbf{s} || = \sqrt[]{\left| (c \; \Delta t)^{2} - (\Delta x)^{2} \right|},

where \Delta x is the distance in space between the events, and (\Delta t)^{2} the time between them. For this equation to hold, and for the length of this vector to be the same in all frames, if a given pair of events are further apart in time in one frame than another, they must also be further apart in space. As you can see, the spacetime interval between the two events is equal to the spatial distance between them in the embankment's rest frame where there's no difference in time between them (and only in that frame).

And we know that A and A', being adjacent are two references for the same event and that
A'B' is equal to AB, then by the above definition they not only have the same spacetime interval, it comprises the same terms. I.E. if the lengths are equal, then the times must be equal too

So we have AB simultaneous in the embankment's frame and A'B' simultaneous in the train's frame, yet as seen in either frame this cannot occur as the other's distances will be length contracted.

As you say:

The space time interval between what pair of events? The lightning strikes? Again, if you make AB and A'B' the same proper length, and still have the strikes be simultaneous in the embankment frame, then the strikes won't be next to both A' and B' on the train. So, the delta-x' for the two strikes in the train frame would not be equal to the distance A'B' in this case.

So what is wrong here?

P.S. as another little musing, if we are talking about relativity, then surely relativity is reciprocal - we can switch the participants and the relationship remains the same?

Something doesn't fit here : :

 

[cfrogue]

If you are talking about when an event or events are seen then I agree that it is position dependent. And also that the order in which they are seen is also thus dependent. So if you are talking about what is seen then I agree.

However, it is your statement below that I disagree with.

----the ordinality of the two light sources in terms of which occurred in which order. That is determined by the position and relative velocity.----

You use the word occurred and this is not normally taken to mean seen. For observers at rest relative to each other, when events occur is not position dependent. When thay are seen is position dependent.

Matheinste.

Oh, yes, then we are in agreement. It was poor wording on my part. Here is another statement to the effect.

both relative motion and position are key factors in determining when an event is seen.

Occurred to me in this case means when it was seen.

I was talking about the observers and the ordinality in which the strikes occurred to them.

I clearly indicated, this ordinality can differ among the observers based on relative velocity and initial position though that is most likely not known, it is a factor.

The poster was suggesting an invariant space time interval implied an invariant ordinality for the strikes. I was showing that implication is not valid as each observer in collinear relative motion will see the strikes occur at different times and further at possibly a different ordinality, meaning the order of the strikes.

 

[matheinste]

Hello cfroque,

Just for clarification, I think that in normal usage the time in an inertial frame at which an event is seen is the time coordinate of the observer, whereas the time at which it occurred is the the time coordinate of the event.

Matheinste.

 

[Grimble]

The poster was suggesting an invariant space time interval implied an invariant ordinality for the strikes. I was showing that implication is not valid as each observer in collinear relative motion will see the strikes occur at different times and further at possibly a different ordinality, meaning the order of the strikes.

Hello cfrogue, thanks for your help with this question but I am a little perplexed by your reference to an invariqant ordinality for the strikes.

As far as I am concerned I made no reference to how this situation was seen.
My point is concerned with spacetime intervals.

We have A adjacent to A', an event.
We have two lightning strikes at A and B with A occurring when A and A' are adjacent.
A and B are simultaneous in the embankment frame, that is a stipulation (a 'given').
The spacetime interval that separates events A and B has two terms: LaTeX Code: {(c \\Delta t)^2} and LaTeX Code: {\\Delta x^2} .
Now, as A and B are simultaneous the time element is equal to their spatial separation;
and as A'B' is the same distance in the trains frame as AB in the embankment's frame (when both are in proper units); these two terms are equal across the frames.
Therefore as the spacetime interval has to be the same in both frames and the distances are identical then the times also have to be identical. Therefore the lightning strikes also occur simultaneously in the train's frame.

This argument has nothing to do with ordinality, or where anything is seen from; it is all to do with a common event, a common distance between points in each of two frames and the fact that a spacetime interval is the same in all frames.

The outcome as far as I can work it out, is that Mand M' will each see the lightning strikes as simultaneous - in their own frames but that neither will see the other as simultaneous.

 

[cfrogue]

Hello cfroque,

Just for clarification, I think that in normal usage the time in an inertial frame at which an event is seen is the time coordinate of the observer, whereas the time at which it occurred is the the time coordinate of the event.

Matheinste.

OK, but with the train/embankment experiment,
Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:

http://www.bartleby.com/173/9.html

Since this is a R of S thread, took place earlier and occured earliermean the same thing. The observation is that of the observer which is what I said.

But, it is just semantics.

You and I agree conceptually.

 

[matheinste]

My point is concerned with spacetime intervals.

We have A adjacent to A', an event.
We have two lightning strikes at A and B with A occurring when A and A' are adjacent.
A and B are simultaneous in the embankment frame, that is a stipulation (a 'given').
The spacetime interval that separates events A and B has two terms: {(c \Delta t)^2} and {\Delta x^2} and that these two terms are equal.

Now, as A and B are simultaneous the time element is equal to their spatial separation.
But A'B' is the same distance in the trains frame as AB in the embankment's frame when both are in proper units.
Therefore as the spacetime interval has to be the same in both frames and the distances are identical then the times also have to be identical. Therefore the lightning strikes also occur simultaneously in the train's frame.

If two events are simultaneous in an inertial frame their time separation is zero in that frame.

The spatial and time components are equal only if a photon can be present at both events, whch is impossible non colocated simultaneous events.

Matheinste

Matheinste.

 

[cfrogue]

Hello cfrogue, thanks for your help with this question but I am a little perplexed by your reference to an invariqant ordinality for the strikes.

As far as I am concerned I made no reference to how this situation was seen.
My point is concerned with spacetime intervals.

We have A adjacent to A', an event.
We have two lightning strikes at A and B with A occurring when A and A' are adjacent.
A and B are simultaneous in the embankment frame, that is a stipulation (a 'given').
The spacetime interval that separates events A and B has two terms: {(c \Delta t)^2} and {\Delta x^2} and that these two terms are equal.

Now, as A and B are simultaneous the time element is equal to their spatial separation.
But A'B' is the same distance in the trains frame as AB in the embankment's frame when both are in proper units.
Therefore as the spacetime interval has to be the same in both frames and the distances are identical then the times also have to be identical. Therefore the lightning strikes also occur simultaneously in the train's frame.

This argument has nothing to do with ordinality, or where anything is seen from; it is all to do with a common event, a common distance between points in each of two frames and the fact that a spacetime interval is the same in all frames.

The outcome as far as I can work it out, is that Mand M' will each see the lightning strikes as simultaneous - in their own frames but that neither will see the other as simultaneous.

Yes, but you forgot to mention you said also the below in the same post.

So I suppose that what I am asking is: if A and A' are adjacent and if the distance AB in the embankment's frame is equal to the distance A'B' in the train's frame then would this not be a definition of simultaneity between two events in two frames rather than simultaneity between two events in one frame?

So, from my POV you are arguing that invariant space time intervals implies invariance with respect to two events as well since you said a definition of simultaneity between two events in two frames .

Perhaps, I am misreading what you are saying.

 

[Grimble]

If two events are simultaneous in an inertial frame their time separation is zero in that frame.

The spatial and time components are equal only if a photon can be present at both events, whch is impossible non colocated simultaneous events.

Matheinste

Matheinste.

Many apologies, you are quite right, it was badly worded - editing error! I should preview more carefully before posting

Let me rephrase:
The spacetime interval that separates events A and B has two terms: {(c \Delta t)^2} and {\Delta x^2}.
Now, as A and B are simultaneous the time element is equal to their spatial separation;
and as A'B' is the same distance in the trains frame as AB in the embankment's frame (when both are in proper units); these two terms are equal across the frames.

Hopefully that clarifies what I am asking?

Grimble

 

[Grimble]

Yes, but you forgot to mention you said also the below in the same post.




So, from my POV you are arguing that invariant space time intervals implies invariance with respect to two events as well since you said a definition of simultaneity between two events in two frames .

Perhaps, I am misreading what you are saying.

Sorry again

It was all a question, not a statement.

I am unsure how to refer to it as it doesn't seem to fit any labels.

We have:

1) Due to the spacetime interval being the same in all frames;
2) comprising only two terms;
3) one of them being the same in two separate frames,
4) but ONLY in those two frames:

we arrive at the other term, the temporal separation (zero = simultanaity?) being the same but again ONLY in those two frames.

Now is the time being common across two frames, but only two frames, simultaneity or invariance?

Grimble