Explaining Relative Simultaneity

[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

 

[cfrogue]

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

I'm sorry Grimble. I cannot figure out what you are asking.

Perhaps, someone else can answer this.

 

[JesseM]

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.

So then do you understand that when the lightning strikes at B, B and B' are not adjacent? (Remember that all frames must agree on whether two events occur at the same position and time, so all frames will agree that the second strike did not occur at B')

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

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

Why do you say that? If the second lightning strike does not occur at B', then the distance between the two strikes in the train frame is not equal to the distance A'B'. For this reason, the fact that A'B' is equal to AB does not imply that the distance between the two strikes is equal in both frames, in fact the distance would be different in each frame. Do you disagree?

Just to pick an example, suppose the train is moving relative to the embankment at 0.6c, and that the distance between the two strikes in the embankment frame is 10 light-seconds, and that this is the rest length of the train as well. Suppose the left strike occurs at x=0 light-seconds, t=0 seconds in the embankment frame, and the right strike occurs at x=10 l.s., t=0 s. If the left end of the train is at x=0 at t=0, then since the train's length is contracted to $$\sqrt{1 - 0.6^2}*10$$ = 0.8*10 = 8 light seconds, the right end of the train must be at x=8 l.s. at t=0 s in the embankment frame...so in this frame we predict that the right end of the train is not struck by lightning, and since all frames must agree in their predictions about which events locally coincide, this must be true in the train frame as well. To figure out what position the right strike occurs in the train frame, we can take the coordinates in the embankment frame and plug them into the Lorentz transformation:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2)
with gamma = 1/sqrt(1 - v^2/c^2)

In this case we have gamma=1.25, and since the right strike occurred at x=10 l.s., t=0 s in the embankment frame, we can plug this into find the x' and t' coordinates of the same event in the train frame:

x' = 1.25*(10 - 0.6*0) = 12.5 light seconds
t' = 1.25*(0 - 0.6*10/1) = -7.5 seconds

So, these are the coordinates of the right strike in the train frame. The Lorentz transformation is also based on the assumption that the origins of the two frames coincide (x=0, t=0 coincides with x'=0, t'=0), so since the left strike occurred at x=0, t=0 in the embankment frame, the left strike must have occurred at x'=0, t'=0 in the train frame. Thus in the train frame the distance between the two strikes was 12.5 light seconds, and the time between them was 7.5 seconds. Plug this into the formula for the invariant interval and you get $$\sqrt{12.5^2 - (-7.5)^2}$$ = $$\sqrt{156.25 - 56.25}$$ = $$\sqrt{100}$$ = 10, which is the same as what you get when you plug the distance and time in the embankment frame (10 light-seconds and 0 seconds) into the same formula.

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.

What cannot occur? If the strike at B does not coincide with B', then the two strikes do not occur at A' and B', so the distance A'B' does not appear in the calculation of the spacetime interval between the two strikes. Are you suggesting we add a third lightning strike at B' that is simultaneous with the strike at A and A'? If so, this would be fine, I don't see what you think "cannot occur". Using the above coordinate systems, if we have a third strike at x'=10, t'=0 in the train frame (which is the position of the end of the train, and which is simultaneous with the first strike at x'=0, t'=0 in this frame), then we can just use the inverse Lorentz transformation to find the coordinates of this event in the embankment frame:

x = gamma*(x' + vt')
t' = gamma*(t' + vx'/c^2)

So, the third strike's coordinates in the embankment frame would be:

x = 1.25 * (10 + 0.6*0) = 12.5 light seconds
t = 1.25 * (0 + 0.6*10/1) = 7.5 seconds

So, in the embankment frame this third strike is not simultaneous with the first strike, which occurred at x=0 and t=0 in the embankment frame. But the spacetime interval between the third strike and the first strike is 10, just like the spacetime interval between the second strike and the first strike.

 

[Grimble]

So then do you understand that when the lightning strikes at B, B and B' are not adjacent? (Remember that all frames must agree on whether two events occur at the same position and time, so all frames will agree that the second strike did not occur at B')

Yes, I do understand what you are saying but please consider the following diagrams:

http://img697.imageshack.us/img697/8710/train15.jpg

Taking the situation described by Einstein where he states quite clearly

But the events A and B also correspond to positions A and B on the train.

Figures 1 and 2 show this and the corresponding spacetime coordinates. They have to be the same, therefore the event A/A' and B/B' must take place at the same time, they have identical spacetime coordinates and as it is is given that A and B are simultaneous, how can A' and B' not be simultaneous in their own reference frames?

Einstein is quite right as shown in figure 4 that the lightning strikes will not be simultaneous to the train, as observed with reference to the embankment.

However, as seen in figures 3 and 5 they will be seen from the train just as they are seen from the embankment, after all that is relativity.

So what concerns me is that there is something here that doesn't add up and I can't see what it is.

Grimble.

 

[Janus]

I don't know if this will help much, but here are two animations showing events from both the frames of the embankment and the train.

From the the embankment:

from the train:

Note that the position of the train observer relative to the embankment is the same when he sees each flash in both frames and that in both frames the embankment observer sees the flashes at the same time.

 

[matheinste]

Hello Grimble,

I am in a hurry so the reasoning below is a bit rambling and not very ordered but I think everything relevant is in there

The lightning strikes in your diagrams 2 and 5, shown in the train rest frame are 4 light seconds apart. In diagram 4, the embankment rest frame you have also shown them 4 light seconds apart. The spatial separation cannot be the same in both frames. The spacetime coordinates are not the same in both frames. Remember the definition of length is taken as a measurement between two points at the same time, or simultaneouly. So if the distance between strikes AB (or A'B') is 4 light seconds when shown in the embankment frame where the strikes are simultaneous it is not 4 light seconds in the train frame.

The train's length appears contracted as viewed form the embankment frame and vice versa; that is symmetrical. BUT the scenario is not symmetrical in all respects because the lightning strikes are defined to be simultaneous in the frame of the embankment.

In this scenario the rest length, or proper length, of the train would be greater than that of the rest length, or proper length, of the distance between the strikes.

The scenario is set up to show that the strikes cannot be simultaneous in both frames by setting the strikes to be simultaneous in the embankment frame and showing that they cannot also be simultaneous in the train frame. Of course you could, if you wished, do the opposite and set the strikes to be simultaneous in the train frame and so show that they cannot also be simultaneous in the embankemnt frame. However in this second case the rest length of the distance between strikes in the embankment frame would be longer than the rest length of the train.

Of course these lengths are not normally taken into consideration because the thought experiment is only designed to show the frame dependence of simultaneity.

I hope this is of some help

Matheinste.

 

[JesseM]

Yes, I do understand what you are saying but please consider the following diagrams:

http://img697.imageshack.us/img697/8710/train15.jpg

You say you understand what I am saying, but the diagrams show you clearly don't! These diagrams are incompatible according to relativity--in diagram 1 you show the distance between A and B as 4 in the embankment frame, in diagram 2 you show the distance between A' and B' as also being 4 in the train frame, but then in diagram #4, drawn from the perspective of the embankment frame, you show A' and B' lining up with A and B at t=0! This is just wrong, if the distance between A' and B' is 4 in their own rest frame, then in the embankment frame the distance between A' and B' should be shorter than 4 due to length contraction, so if A lines up with A' at t=0 in the embankment frame, then B' should be somewhere short of 4 at t=0 (for example, if v=0.6c, then at t'=0, B' would be at position 3.2). You have to either change the rest length of the train (changing diagram 2 so that the train's rest length is greater than 4, so its contracted length in the embankment frame can be 4), or else redraw diagram #4 to show that the worldline of B' starts at some position short of 4 on the x-axis of the embankment frame.

 

[Janus]

Yes, I do understand what you are saying but please consider the following diagrams:

http://img697.imageshack.us/img697/8710/train15.jpg

Taking the situation described by Einstein where he states quite clearly

Figures 1 and 2 show this and the corresponding spacetime coordinates. They have to be the same, therefore the event A/A' and B/B' must take place at the same time, they have identical spacetime coordinates and as it is is given that A and B are simultaneous, how can A' and B' not be simultaneous in their own reference frames?

Einstein is quite right as shown in figure 4 that the lightning strikes will not be simultaneous to the train, as observed with reference to the embankment.

However, as seen in figures 3 and 5 they will be seen from the train just as they are seen from the embankment, after all that is relativity.

So what concerns me is that there is something here that doesn't add up and I can't see what it is.

Grimble.

Here's the actual S-T diagrams from each frame:

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/simul.gif

The point is that both frames have to agree that M' does not see both flashes at the same time. Anything else would result in physical contradictions. And since only way that this can be true in the Train's frame, (Since M is an equal distance from A' and B') is for the strikes not to have taken place simultaneously in that frame.

 

[Grimble]

Thank you, good people for your comments, (and I do accept that in diagram 4 I neglected the length contraction - sorry)

I have therefore redrawn it:

http://img4.imageshack.us/img4/4831/train2r.jpg

But, back to the point I was raising, if we consider the spacetime coordinates at the time of the lightning strikes in the embankment frame we have:
A = 0,0,0,0
M = 0,2,0,0
B = 0,4,0,0

While at the same time, when A and A' are adjacent in the Train's frame we have:
A' = 0,0,0,0
M' = 0,2,0,0
B' = 0,4,0,0

Now the two frames with which we are concerned are inertial frames, their units are proper units.
They have a common origin the event of kightning striking A, when A' is adjacent to A.

This has NOTHING to do with what is seen of one frame by any observer in the other frame.

This is solely about the spactime coordinates of six points in two separate frames of refrence that have a common origin.

If they are both inertial frames, if therefore they have common units, and if two points have the same coordinates, then they must represent the same event.

Therefore the lightning must hit B' at the same time as B and the lightning strikes must be simultaneous in both frames of reference, but only in those frames of reference, not in one frame viewed from the other.

So where am I going wrong?

Grimble

 

[matheinste]

Hello Grimble.

But, back to the point I was raising, if we consider the spacetime coordinates at the time of the lightning strikes in the embankment frame we have:
A = 0,0,0,0
M = 0,2,0,0
B = 0,4,0,0

While at the same time, when A and A' are adjacent in the Train's frame we have:
A' = 0,0,0,0
M' = 0,2,0,0
B' = 0,4,0,0

This is incorrect. The coordinates are not the same in both frames.

This is obvious for the time coordinate of B' as it icannot have the same time coordinate of A' as the strikes are not simultaneous in the train frame, as has already been established.

Matheinste.

 

[Janus]

But, back to the point I was raising, if we consider the spacetime coordinates at the time of the lightning strikes in the embankment frame we have:
A = 0,0,0,0
M = 0,2,0,0
B = 0,4,0,0

While at the same time, when A and A' are adjacent in the Train's frame we have:
A' = 0,0,0,0
M' = 0,2,0,0
B' = 0,4,0,0

No, for example if the relative speed of the train to the embankment is 0.5c:

Then in the embankment frame, the distance between A and B is 4 light sec, and the distance between A' and B' is 4 light sec. This is the proper distance between A and B and the length contracted distance between A'and B'

Therefore, in the train frame, the proper distance between A' and B' is 4.62 light sec, and the length contracted distance between A and B is 3.46 light sec.

Therefore, in the embankment frame, when A is next to A'
we get

A = 0,0,0,0
M = 0,2,0,0
B = 0,4,0,0

And
A' = 0,0,0,0
M' = 0,2,0,0
B' = 0,4,0,0

but in the train frame we would get:

A = 0,0,0,0
M = 0,1.73,0,0
B = 0,3.46,0,0

And
A' = 0,0,0,0
M' = 0,2.32,0,0
B' = 0,4.62,0,0

Assuming that both frames assign the zero time coordinate to the moment that A' and A meet, then while both point B and B' are the same distance from AA' in the embankment frame at t=0, B' is further from AA' than B is in the train frame at t=0. B' has already past B and therefore B and B' met before t=0