Relative simultaneity...

[Grimble]

Ref:Chapter IX. The Relativity of Simultaneity (http://www.bartleby.com/173/9.html)

I am having a little bother with this and hope that someone will be able to explain it for me:doh:

If lightening strikes A & B simultaneously then, as those strikes are space-time 'events' they have no motion, only a time and a place. OK?

Then, as they also occur adjacent to points A' & B' on the train, and, if the light were reflected by mirrors attatched to those points, it would travel at 'c' relative to the train, in which co-ordinate system the observer at M' is not moving!:(

Therefore the two lightening strikes at A & B, A' & B', are also simultaneous to the observer on the train, as perceived by that observer:confused:

Einstein wrote that is the observer on the train
Then was he not saying that the lightening strikes were not simultaneous with respect to the train, as perceived from the embankment?:confused:

[ZikZak]

Ref:Chapter IX. The Relativity of Simultaneity (http://www.bartleby.com/173/9.html)
If lightening strikes A & B simultaneously then, as those strikes are space-time 'events' they have no motion, only a time and a place. OK?

No, you're equivocating a specific "time and place" with "spacetime event." A spacetime event is a specific geometrical location in spacetime, which to any given observer will be at a specific time and place. But different observers will *disagree* on the particular time and place of the event.

In other words, yes, spacetime events are specific locations in spacetime, but the actual x and t coordinates assigned to them by different observers will differ.

The example is meant to illustrate that the strikes cannot be simultaneous in both frames, because if the light flashes reach M simultaneously (a single event, since it happens at a specific time and place), then they cannot have reached M' simultaneously (since M' is not at M at the event when both signals arrive).

There is only one event where the light flashes meet, and in the example, this is given to be at observer M. Since observer M' is not at that event, the light flashes cannot have reached him simultaneously.

Since the light flashes did not reach M' simultaneously, even though he is equidistant from A' and B', where the lightning struck, he can only conclude that the flash at B' (=B) happened first. Thus in his reference frame the lightning flashes are not simultaneous. i.e., even though observer M assigns the same value of the time coordinate t to the two lightning flashes, observer M' does not.

[HallsofIvy]

I might add that you go off the track when you start with "If lightening strikes A & B simultaneously". There is no such thing without stating a frame of reference in which the two events are simultaneous.

[Rasalhague]

In Einstein's example, the lightning strikes are simultaneous in the rest frame of the embankment, and therefore not simultaneous in the rest frame of the train:

"Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative."

In fact they can't be simultaneous in the rest frame of the train if they're simultaneous in the rest frame of the embankment. Likewise, if another pair of lightning bolts struck A and B simultaneously in the rest frame of the train, then this second pair of lightning strikes would not be simultaneous in the rest frame of the embankment.

If the train stopped relative to the embankment, then the rest frames of train and embankment would coincide and be the same frame; only then could lightning strike A and B simultaneously in the rest frame of the train and the embankment. But the lightning strikes could only be simultaneous in all frames moving at some nonzero velocity relative to each other if the two bolts of lightning struck the same place as each other and at the same time as each other. This is what "relativity of simultaneity" means; in general, whether two events are simultaneous depends on which frame of reference they're described with respect to.

Here are three animations illustrating the relativity of simultaneity.

http://video.google.co.uk/videoplay?docid=4820883104387202016&ei=QnjiSteyONOg-Aai6LyoAg#

(Animation begins at 08:00 minutes.)
http://video.google.co.uk/videoplay?docid=-6328514962912264988&ei=EXziSoGiI9yf-AbftsDzBw#

(Animation begins 01:16.)
http://video.google.co.uk/videoplay?docid=6322511432077219124&ei=OoHiSvT3Msnm-AbmtOTRDA

I found it very confusing when I first saw these. I remember watching the animation in The Mechanical Universe again and again as I tried to get some intuition for it, but I think they're well worth a study, even if it's not obvious at first. Also worth scrutinising spacetime diagrams.

A personal way I have of imagining it that I have is to think of a line (or plane) of simultaneity in a Minkowski spacetime diagram as being like a speedboat that tilts upwards in the direction that the speedboat is going (and down at the back). But that's just my personal way of remembering which way round it goes. So if that makes no sense, just ignore this paragraph!

[Rasalhague]

I might add that you go off the track when you start with "If lightening strikes A & B simultaneously". There is no such thing without stating a frame of reference in which the two events are simultaneous.

Yes, this is the source of a lot of apparent paradoxes. For the statement "A and B are simultaneous" to be meaningful, you have to say in which frame of reference they're simultaneous - because if they're simultaneous in one frame of reference, they won't be in others.

[Grimble]

OK, sorry, take out If lightening strikes A & B simultaneously

No, let's start again.

Einstein wrote in chapter IX (http://www.bartleby.com/173/9.html) UP to now our considerations have been referred to a particular body of reference, which we have styled a “railway embankment.” We suppose a very long train travelling along the rails with the constant velocity v and in the direction indicated in Fig. 1. People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.


But then he writes,
Now in reality (considered with reference to the railway embankment) he that is the observer on the train is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.

Was he not saying that the lightening strikes were not simultaneous with respect to the train, as perceived from the embankment?:confused:

[Grimble]

And again, if we were to consider the observer in the train, what does he perceive?

Again I quote People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Then every event which takes place along the line also takes place at a particular point of the train. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.


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

And it is surely not unreasonable to suppose that at night, for instance, the traveller is completely unaware of the embankment. No all he perceives is himself and his reference-body, the train.

Now when the lightning strikes he will see the light from A and the light from B, each travelling at c relative to the train reach him at the same time. For him, only his own reference-body exists and to him, this reference-body is not moving.

I do understand what Einstein was saying, the relativity of simultaneity; but it is, for one observer, between his own reference-body and his perception of another reference body; where he will of course also perceive length contraction and time dilation, not that they are relevant to this discussion.

[Rasalhague]

Was he not saying that the lightening strikes were not simultaneous with respect to the train, as perceived from the embankment?:confused: [...] And again, if we were to consider the observer in the train, what does he perceive?

In the first example, lightning strikes A and B simultaneously with respect to a coordinate system in which the embankment is still and the train moving, but these strikes are not simultaneous with respect to a coordinate system in which the train is still and the embankment moving. An observer sitting on the embankment at M will see the flashes at the same time, whereas an observer sitting on the train at M' will see the flash from B before the flash from A. This is no illusion. In the latter observer's rest frame (the rest frame of the train), the lightning will really have struck B before it struck A, even though in the rest frame of the observer at M (the rest frame of the embankment), both strikes happened at the same time as each other.

It's completely cointerintuitive. I'm sure everyone finds this a confusing idea when they first encounter it.

In the second example, Einstein imagines a different pair of lightning strikes which happen to be simultaneous with respect to the train's rest frame, in which case they're not simultaneous in the rest frame of the embankment. In this example, it's the train passenger at M' who sees the flashes at the same time, while the person sitting on the embankment at M will see the lightning strike A before it strikes B. In this case, in the rest frame of the embankment, the lightning really does strike A first, even though the strikes were simultaneous in the train's rest frame.

And it is surely not unreasonable to suppose that at night, for instance, the traveller is completely unaware of the embankment. No all he perceives is himself and his reference-body, the train.

This is a good idea to think about.

Now when the lightning strikes he will see the light from A and the light from B, each travelling at c relative to the train reach him at the same time. For him, only his own reference-body exists and to him, this reference-body is not moving.

This is not what will happen if the lightning strikes are simultaneous in the rest frame of the (invisible) embankment. Even at night and even if there is no embankment there at all, so long as the lightning strikes A and B simultaneously according to a frame in which the train is defined as moving, these events will not be simultaneous in a reference frame defined such that the train is at rest in it.

The choice of reference frame is arbitrary, a matter of convenience. We can define a frame in terms of some object being at rest in that frame, but we don't have to. We could just as well define a frame moving at some velocity relative to the train even if we didn't know of any object which was at rest in such a frame, or of any literal observer at rest with respect to it. Or if there was only an embankment and no train, we could just as easily define a frame moving at some velocity relative to the embankment and calculate time, distances and the order of events in such a frame, regardless of whether there is any physical object at rest with respect to that frame.

[Grimble]

Yes but...

Let me simplify my understanding to the very basics as I see them:

We have two entities, the embankment and the train which are moving relative to one another.

Each is stationary wth respect to its own reference-body and each is moving with respect to the other's reference body.

Lightning strikes points A & B which exist in both reference bodies.

Observers in each reference-body will surely see the same effects with respect to their own reference-body; which, at the expense of repeating myself, is stationary with respect to that observer.

The light will travel from A & B at the speed of light c relative to the observer's reference-body and so will meet at either M or M' depending on which reference-body the observer is in.

It is all a completely self contained, symmetrical, set of circumstances.

From the above, please tell me how one can determine that one case is simultaneous and the other is not:confused:

As far as I can see each would say that the lightning strikes were simultaneous for him but not for the other. What is it I am missing here:frown:

[Doc Al]

Lightning strikes points A & B which exist in both reference bodies.
Think of A and B as fixed points on the embankment. Both reference frames agree that the lightning strikes A and B, but the train frame will measure the positions differently. Think of a position A' fixed in the train that coincides with A at the instant the lightning strikes A; similarly, think of train location B' coinciding with B at the instant the lightning strikes B.

Observers in each reference-body will surely see the same effects with respect to their own reference-body; which, at the expense of repeating myself, is stationary with respect to that observer.
Not sure what that means.

The light will travel from A & B at the speed of light c relative to the observer's reference-body and so will meet at either M or M' depending on which reference-body the observer is in.
No. M' coincides with M at the moment the lightning strikes according to the embankment frame. Light traveling from A & B will meet at M, but not at M'. By the time that the light reaches M, M' has moved along.

It is all a completely self contained, symmetrical, set of circumstances.
Not symmetric at all.

From the above, please tell me how one can determine that one case is simultaneous and the other is not:confused:
The embankment frame sees the light go from A and B and meet in the middle at M. So they surely think that the lightning strikes were simultaneous. The train, on the other hand, sees the lightning striking at point A' and B' (equidistant from M'), but the light does not meet in the middle of the train at M'. So they must conclude that the lightning strikes did not occur simultaneously.

[Bob_for_short]

If in one reference frame you have a solution Acos(ωt) (a long horizontal rod oscillating up and down as a whole), in a moving RF (with velocity V) you will see it as a wave: A'cos(ω't'-kx') with k being dependent on V.

[Doc Al]

If in one reference frame you have a solution Acos(ωt) (a long horizontal rod oscillating up and down as a whole), in a moving RF (with velocity V) you will see it as a wave: A'cos(ω't'-kx') with k being dependent on V.
Well, that's true. But I'm not sure how helpful it will be in sorting out the specific example being discussed in this thread.

[ZikZak]

Lightning strikes points A & B which exist in both reference bodies.

But not simultaneously in both. The example demonstrates how assuming that the strikes occur simultaneously in both reference frames leads to an inescapable contradiction.

Observers in each reference-body will surely see the same effects with respect to their own reference-body; which, at the expense of repeating myself, is stationary with respect to that observer.

The light will travel from A & B at the speed of light c relative to the observer's reference-body and so will meet at either M or M' depending on which reference-body the observer is in.

The light does indeed travel at c relative to either reference body. Therefore, the fact that the flash from B' reaches M' before the flash from A' does requires that M' conclude that the strikes were not simultaneous.

It is all a completely self contained, symmetrical, set of circumstances.

No it isn't. You are forgetting the givens. The given is that the flashes reach M simultaneously. There is no such given for M'. The assumption of complete symmetry you are trying to make leads to a logical contradiction. That is the point of the exercise.

From the above, please tell me how one can determine that one case is simultaneous and the other is not:confused:

As far as I can see each would say that the lightning strikes were simultaneous for him but not for the other. What is it I am missing here:frown:

In the example, it is given that the flashes are received simultaneously by M. There is one event at which the world-lines of both flashes and M himself meet. Because all the worldlines meet at one event, all observers must agree that this event occurred. This is the basic assumption of reality in science: that observers might disagree on when and where an event happened, but they should always agree that it did.

Therefore, M' also observes that the light rays meet with observer M at a single event. If an external reality exists, this is it. There is no escaping this. Things that happen, happen. M' may label the "both light rays hit M" event with different coordinate values than M does, but it is one single event with one single set of coordinates in any particular reference frame.

Working in the reference frame of M, M observes the flash from B hit M' at a different time (different event) from the flash from A. Two separate events. Changing reference frames cannot possibly combine two different events into one. In other words, M observes that when flash A reaches M', flash B is not there at that event. When flash B reaches M', flash A is not there. When the two flashes meet each other, M' is not there. Merely changing reference frames cannot magically stitch all these different events together. M' therefore observes the flashes at two different times (two different events). He must conclude that the bolts were not simultaneous.

Summary: assuming that the bolts were simultaneous in M *forbids* them from being simultaneous in M'. Both observers agree on the events that occur: light flashes A and B reach M at a single event, and M' at two different events. All observers in the universe must agree that both light flashes and M meet at some event, while M' meets the light flashes separately at two different events.

You could do the same argument starting from the assumption that the flashes reach M' simultaneously, which would forbid M from seeing them simultaneously, but that would be a different universe. The two situations would be two completely different sets of events. Only one of them can occur, not both.

EDIT: And I should add, that since the given is that the light rays reach M simultaneously, and not that the bolts were simultaneous in all frames, that the correct calculation to make in the M' frame would be to ask when the bolts must occur in order for their flashes to meet at M, NOT at M'. Since M is traveling backwards, bolt B must occur before bolt A in order for the flashes to meet at M as given.

[Rasalhague]

We have two entities, the embankment and the train which are moving relative to one another.

Yes.

Each is stationary wth respect to its own reference-body and each is moving with respect to the other's reference body.

I'm more familiar with the synonyms "reference frame" and "coordinate system", but this is what the book says "reference body" means, a coordinate system. Bear in mind that although the names train and embankment can be given to these two coordinate systems, the coordinate systems themselves are abstract entities. There doesn't have to be a physical embankment present for us to define a coordinate system moving relative to the train.

Lightning strikes points A & B which exist in both reference bodies.

Yes.

Observers in each reference-body will surely see the same effects with respect to their own reference-body; which, at the expense of repeating myself, is stationary with respect to that observer.

They won't see the same effect if lightning strikes A and B simultaneously in one frame. If this was the case, there would be no need for a theory of relativity. It's because the effects referred to one coordinate system differ from the effects referred to another, travelling at some velocity relative to it, that we need to take care to specify which coordinate system events are simultaneous in. If they're simultaneous in one frame, they can't be in the other (except in the trivial cases where A = B or the frames are the same).

The light will travel from A & B at the speed of light c relative to the observer's reference-body and so will meet at either M or M' depending on which reference-body the observer is in.

The light travels at c in both frames. If the strikes were simultaneous in the rest frame of the embankment, the light from each strike will reach M at the same time. According to this frame, the light from B will reach M' first. In this frame, the light reaches B first because M' is moving to meet the light from B and away from the light from A, so the light from A has further to go to get to M' and so takes longer to reach M'. Likewise in the rest frame of the train, the light from B will reach M' first. This couldn't happen if the strikes were simultaneous in this frame too unless the light from B was approaching M' at a different speed than the light from A. But we know the light always travels at speed c, so that can't be the reason. Instead we conclude that B must have been struck first, since the light from this strike arrived first, and the only way it could arrive first is if it set off first.

It is all a completely self contained, symmetrical, set of circumstances.

The symmetry becomes apparent when you compare what happens in the case of a pair of lightning strikes at A and B which are simultaneous in one frame with the case of a different pair of lightning strikes, this latter pair being simultaneous in the other frame. The amount of time that M' will have to wait in the first case after seeing the light from B till the light from A arrives will be the same as the amount of time that M will have to wait in the second case after seeing the light from A till the light from B arrives.

Did you look at the videos I linked to? They might not make it immediately clear. They certainly didn't for me. But I think that poring over them did eventually help get me a little closer to understanding. But it was really when I learnt about spacetime diagrams that the concept of relative simultaneity began to sink in.

EDIT: The book talks about "two strokes [sic] of lightning A and B", which we could think of as points in spacetime (events). Alternatively, we can think of a space coordinate A (constant in the embankment's rest frame) and distinguish it from a space coordinate A' (constant in the train's rest frame), so A makes a line in spacetime, as does A', such that their intersection is the point in spacetime where and when the lightning strikes on the left. And we can define B and B' likewise.

[Grimble]

Observers in each reference-body will surely see the same effects with respect to their own reference-body; which, at the expense of repeating myself, is stationary with respect to that observer.

Not sure what that means.

What that means is that as each observer is stationary within his own reference-body and the points A&B are stationary with respect to each body of reference they will experience similar phenomena. Light from A&B will meet at the mid-point between A&B in each reference frame. i.e. at M or M' depending on which frame the observer is in.

No. M' coincides with M at the moment the lightning strikes according to the embankment frame. Light traveling from A & B will meet at M, but not at M'. By the time that the light reaches M, M' has moved along. But not according to the train the train's frame where it is permanently midway between A&B

Look, I am not saying that M and M' remain adjacent, They are in different reference-bodies.
Light is travelling at c within each reference body, relative to each reference body, so it will meet at two separate points in two seperate reference bodies. Surely that is fundamental to Relativity.

The embankment frame sees the light go from A and B and meet in the middle at M. So they surely think that the lightning strikes were simultaneous. The train, on the other hand, sees the lightning striking at point A' and B' (equidistant from M'), but the light does not meet in the middle of the train at M'.

But why not? It is travelling at c from A' and from B' to M', MIDWAY, and permanently midway, between A' & B', from the train's reference body.

[Doc Al]

What that means is that as each observer is stationary within his own reference-body and the points A&B are stationary with respect to each body of reference they will experience similar phenomena. Light from A&B will meet at the mid-point between A&B in each reference frame. i.e. at M or M' depending on which frame the observer is in.
Note that A & B are stationary with respect to the embankment; while A' & B' are stationary with respect to the train. The light meets at M, not at M'!

But not according to the train the train's frame where it is permanently midway between A&B
Wrong: M is a fixed point in the embankment that is permanently midway between A & B. M' is a fixed point in the train that is permanently midway between A' and B'. M is not fixed from the train's viewpoint, just like M' (the middle of the train) is not fixed from the embankment's view.

Look, I am not saying that M and M' remain adjacent, They are in different reference-bodies.
Light is travelling at c within each reference body, relative to each reference body, so it will meet at two separate points in two seperate reference bodies. Surely that is fundamental to Relativity.
The light meets at point M--not M'. Everyone agrees on that. M is midway between the A and B (as it always is). M is not midway between A' and B' at the moment the light reaches M.

But why not? It is travelling at c from A' and from B' to M', MIDWAY, and permanently midway, between A' & B', from the train's reference body.
No. Light does not travel from A' and B' and meet at M'--the light meets at M. Which is the entire point. Since everyone agrees that the light meets at M, and since M is much closer to the rear of the train when the light reaches M, the train observers conclude that the lightning strikes could not have been simultaneous. (If the lightning strikes were simultaneous according to train observers, then the light would meet in the middle of the train--at M', not M--since the lightning struck at points A' and B' which were equidistant from the middle of the train. But that doesn't happen.)

[ZikZak]

Look, I am not saying that M and M' remain adjacent, They are in different reference-bodies.
Light is travelling at c within each reference body, relative to each reference body, so it will meet at two separate points in two seperate reference bodies. Surely that is fundamental to Relativity.

No, a universe which would permit real paradoxes (M is not equal to M' and the light beams meet at M and the light beams meet at M') is anathema to Relativity, and all of science.

[Rasalhague]

If there's a spacelike separation between two events (i.e. the sum of the squares of each space components of the separation vector is greater than the square of the time component (when space and time are measured in the same units)), then--given that nothing can go faster than c--it's never possible for a single particle to be present at both events, and so neither event can have an effect on the other. The lightning strikes are two such events. And when there is a spacelike separation between two events, there's no absolute way of ordering them in time; we can only say which happens first and how far apart they are in time with respect to a particular coordinate system, knowing that there are infinitely many other coordinate systems we could chose to describe them, in some of which the other event will happen first. Although bizarre and counterintuitive to us humans, these rules are consistent with causality because events which have no absolute order can't influence each other, so the contradiction of cause preceding effect never arises.

But if there's a timelike separation between two events (i.e. the square of the time component of the separation vector is greater than the sum of the squares of the space components), or a lightlike separation (i.e. the square of the time component equals the sum of the squares of the space components), then it is possible for a particle to be present at both events. There is a causal connection between the two events: one can have an effect on the other. Events which a single particle is present at are said to lie on the particle's worldline. Worldline means a trajectory through spacetime. Consider the light from the lightning strike at B = B' reaching M' as one event, and the light from the lightning strike at A = A' reaching M' as another event. Both events lie on the worldline of M' because M' is present at both events. It doesn't matter which coordinate system we describe this pair of events in, this pair of events will always happen in the same order, otherwise there would be a genuine contradiction: no way of ordering cause and effect. If there wasn't an absolute order to events on a worldline, then we'd lose the causal structure of Minkowski spacetime. So if this pair of events (the light from each strike reaching M') happens in one order in the rest frame of the embankment, it must happen in the same order in all frames (coordinate systems), including the rest frame of the train.

Likewise the simultaneous arrival of the light from A and B at M. When things happen simultaneously at the same location in one frame, they must happen simultaneously in all frames otherwise there'd be a contradiction about real physical events.

[Grimble]

No. Light does not travel from A' and B' and meet at M'--the light meets at M. Which is the entire point. Since everyone agrees that the light meets at M, and since M is much closer to the rear of the train when the light reaches M, the train observers conclude that the lightning strikes could not have been simultaneous. (If the lightning strikes were simultaneous according to train observers, then the light would meet in the middle of the train--at M', not M--since the lightning struck at points A' and B' which were equidistant from the middle of the train. But that doesn't happen.)

OK OK, let us examine what you are saying here:

Everyone agrees (which somehow gives it authority?) that the lightning strikes at A & B are simultaneous as observed from the embankment. . . . . . (1)

At the instant that the lightning strikes hit, points A' & B' are adjacent to points A & B, so the lightning strikes them too. . . . . . (2)

In order to establish whether this appears to be simultaneous as observed from the train, we need to establish when the light from those lightning strikes reaches an observer at M', midway between A' and B'. . . . . . (3)

If the distance between the points A' & B' is 2L the light has to travel the distance L to reach M'. . . . . . (4)

Then the time it will take from A' to M' t = \frac{L}{c} where c is the speed of light in the train's reference-body. . . . . . (5)

And the time it will take from B' to M' is also t = \frac{L}{c}.. . . . . (6)

So the transit times for the light to travel from A' to M' and from B' to M' are equal - (5),(6) above. . . . . . (7)

And as we saw in (2) above the lightning struck A' and B' at the same time as it struck A & B which strikes we have established was simultaneously - (1)

So if the light started from A' and B' at the same instant and the travel times were equal - (7)

The light must meet at point M' in the Train's reference-body! - in the same way that it meets at M in the embankment's reference-body.

That surely is relativity - What happens is relative to where it is viewed from.

Or is there something wrong with my maths???

It has to be straightforward and logical, there is nothing mysterious about relativity, it is not some sort of dark magic only known to a few initiates

Please note that my last comment is intended to be humerous and is in no way intended to be sarcastic.o:)o:)o:)

Grimble:smile:

[Rasalhague]

If the distance between the points A' & B' is 2L the light has to travel the distance L to reach M'. . . . . . (4)

Here's where the problem is. It's not an algebraic mistake, but a logical one. A' and B' are indeed equidistant from M', but when you use (4) in your argument for simultaneity, you assume the conclusion you expect, namely that the lightning strikes are simultaneous in the rest frame of M' (the rest frame of the train). But if this was the case, we'd have a contradiction because, according to M (in the rest frame of the embankment), the light from the right reaches M' before the light from the left, not at the same time. In the rest frame of the embankment this is because M' has moved closer to B (while B' has moved further to the right beyond B). But M' hasn't moved in the train's rest frame. The light from B can't both arrive first at M' and not arrive first at M', nor (according to observation and Einstein's second postulate) can its speed be different. The only remaining alternative is that in the train's rest frame, the light struck B' first, before it struck A', and that's why it arrives first at M'.

This conclusion offends our intuition, and seems superficially like a contradiction too. But the fact that information can't travel faster than c ensures that causality is respected, i.e. that cause always precedes effect. The only events whose order isn't frame invariant are events which can't be affected by each other because there isn't time in any frame for information to travel from one to the other at speed c or less.

Similarly, if the light from both strikes does reach an observer simultaneously in one frame, it must reach that observer simultaneously in all frames. When two things happen at the same time and place, we call them a "spacetime coincidence". Anything that's a spacetime coincidence in one frame is a spacetime coincidence in all frames, and would have to be for causality to be preserved.

It has to be straightforward and logical, there is nothing mysterious about relativity, it is not some sort of dark magic only known to a few initiates

It is logical, but I think it's only natural that most people find it far from obvious at first (I certainly did, and there's lots of concepts I still struggle with). That's because it's so counterintuitive. In our everyday experience, at the human scales we're used to, simultaneity is not relative, and there's nothing directly comparable to this effect. But of course what one person can learn, another can, and with determination you'll get there in the end!

[Doc Al]

Everyone agrees (which somehow gives it authority?) that the lightning strikes at A & B are simultaneous as observed from the embankment. . . . . . (1)
That's a basic stipulation of the setup. (The lightning strikes must be simultaneous according to somebody, otherwise we don't have much of a thought experiment.)

At the instant that the lightning strikes hit, points A' & B' are adjacent to points A & B, so the lightning strikes them too. . . . . . (2)
Careful here. More accurate to make these two statements: (2a) When lightning hits A, A' is adjacent to A. (2b) When lightning hits B, B' is adjacent to B. Of course--by stipulation--the embankment observers claim that those strikes were simultaneous, thus they say A & A' and B & B' were adjacent at the same time.

In order to establish whether this appears to be simultaneous as observed from the train, we need to establish when the light from those lightning strikes reaches an observer at M', midway between A' and B'. . . . . . (3)

If the distance between the points A' & B' is 2L the light has to travel the distance L to reach M'. . . . . . (4)

Then the time it will take from A' to M' t = \frac{L}{c} where c is the speed of light in the train's reference-body. . . . . . (5)

And the time it will take from B' to M' is also t = \frac{L}{c}.. . . . . (6)

So the transit times for the light to travel from A' to M' and from B' to M' are equal - (5),(6) above. . . . . . (7)
All good.

And as we saw in (2) above the lightning struck A' and B' at the same time as it struck A & B which strikes we have established was simultaneously - (1)
Careful here: Those strikes were simultaneous according to the embankment frame. You cannot assume that they were simultaneous from the train frame.


So if the light started from A' and B' at the same instant and the travel times were equal - (7)

The light must meet at point M' in the Train's reference-body! - in the same way that it meets at M in the embankment's reference-body.
Your reasoning is correct, but that's a big if! IF the lightning strikes were simultaneous according to the train observers, then the light would meet at M'. Absolutely!

Note that this reasoning works the other way around: If the light does not meet at M', then the lightning strikes could not have been simultaneous according to the train observers. (7a)


That surely is relativity - What happens is relative to where it is viewed from.
Actually, no. "What happens" generally does not depend on where it is viewed from.

Consider these questions:
Do you agree that the light meets at M? (You must agree, because it was stipulated that according to the embankment frame, the lightning strikes were simultaneous. Using the same math you used above, that means the light must meet at M.) This is a physical fact agreed to by everyone. For example, I can place a bomb at M with a light detector on each side and set it to explode if a signal is detected on both sides within some arbitrarily small time interval. Either it explodes or it doesn't--everyone will observe or not observe the explosion.

When the light reaches M, is M' still adjacent? Of course not. M' & M were adjacent (according to embankment observers) when the lightning struck. So by the time the light reaches M, M' is long gone. Note that the light from B/B' reaches M' before it reaches M. (And if we put a similar bomb on the train at M', it would not explode.)

Thus we must conclude that the light does not meet at M'. And then, using the reasoning of 7a above, we conclude that according to train observers the lightning strikes could not have been simultaneous.

Or is there something wrong with my maths???
Your math is fine; it's your premise that is incorrect.

It has to be straightforward and logical, there is nothing mysterious about relativity, it is not some sort of dark magic only known to a few initiates
Logical? Absolutely. Straightforward? I'd say it's pretty tricky stuff since it goes against the intuitions we've built from dealing with things at non-relativistic speeds.

If it were easy and obvious, where would be the fun? :tongue:

[ZikZak]

Everyone agrees (which somehow gives it authority?)

What authority is there in science, besides observation?

[matheinste]

Originally Posted by Grimble
Everyone agrees (which somehow gives it authority?) hat the lightning strikes at A & B are simultaneous as observed from the embankment.

In the text Einstein asks---
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? ----

So they are defined as being simultaneous in the embankment frame. No more authority required.

He then continues----
We shall show directly that the answer must be in the negative. -----

The statement, by Einstein himself, is pretty unambiguous. I used to have problems with it as well but realised that, given the scenario and its unambiguous answer, it might be a sensible option to try to understand it rather than disagree with it. Many things which follow, although a little odd and seemingly complicated at first, will eventually make sense and you will wonder why you ever had a problem with it.

Matheinste.

[Grimble]

Originally Posted by Grimble
Everyone agrees (which somehow gives it authority?) hat the lightning strikes at A & B are simultaneous as observed from the embankment.

In the text Einstein asks---
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? ----

So they are defined as being simultaneous in the embankment frame. No more authority required.

He then continues----
We shall show directly that the answer must be in the negative. -----

The statement, by Einstein himself, is pretty unambiguous. I used to have problems with it as well but realised that, given the scenario and its unambiguous answer, it might be a sensible option to try to understand it rather than disagree with it. Many things which follow, although a little odd and seemingly complicated at first, will eventually make sense and you will wonder why you ever had a problem with it.

Matheinste.

But I have absolutely no problem with what Einstein says!

1)Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.

2)Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

3)Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.

4)Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).

It all makes sense to me:smile:

[Grimble]

Please accept my apologies for the state of this thread, good people, it is so easy to be diverted by arguments of minutiae.

Let me say that I have come to SR on my own using the Einstein paper that I quote from.

I found that to be clear, concise and easy to understand.

I then looked further, on the web, principally in Wikipedia etc. and was intersted to find things that did not match what I had learned from Einstein.

I do not pretend (honestly) to know the answers, but for me, just being told 'thats the way it is' doesn't satisfy, I like to know WHY and HOW.

One of the major problems I find is the constant pulling apart every statement I make and telling me to rephrase it or disecting what the words mean!

I have been told (not just in this thread) that proper time is the term to use and that there is no such thing as proper time...

that there is no need to use A' and B', but that A & B will do: then I am told that I should use A' and B'...

having experienced so much criticism for using the wron terms, I tried defining my own - inertial and transformed units, and earning immediate criticism despite attempting to define exactly what I meant by the use of those terms.

No matter how I try and ask questions or address the points that don't seem to add up for me all I get is constant critical disection of the language I am trying to use.

And it is not just that I am unused to the particular terms you use, but you can't even agree amongst yourselves about the use of the terms.

Top this with a tendency to read into what I am saying, what you expect me to be saying, without bothering, it seems, to actually reading it, and the whole exercise becomes frustrating.

One thing which I find particularly annoying (and which I am sure will annoy any one who experiences it) is to be told what I am thinking, when what I am told is not, and sometimes is the very opposite, of what I am thinking.

Moaning over!

My background is scientific, I studied physics at university, many years ago, followed by 25 years in computing, where I spent many years solving problems, designing systems and in support work, where the prime skill was to be able to take written documents, designs, and complete software systems and find the bugs in them.

I have come to you for assistance in understanding SR and answering questions that arise where the modern understanding seems to fit uneasily with what Einstein wrote.

So I ask for your patience and your help

Thank you, Grimble:smile::smile::smile::smile::smile:

[JesseM]

OK OK, let us examine what you are saying here:

Everyone agrees (which somehow gives it authority?) that the lightning strikes at A & B are simultaneous as observed from the embankment. . . . . . (1)

At the instant that the lightning strikes hit, points A' & B' are adjacent to points A & B, so the lightning strikes them too. . . . . . (2)

In order to establish whether this appears to be simultaneous as observed from the train, we need to establish when the light from those lightning strikes reaches an observer at M', midway between A' and B'. . . . . . (3)

If the distance between the points A' & B' is 2L the light has to travel the distance L to reach M'. . . . . . (4)
Only in the frame where M' is at rest, i.e. the train's frame. In the embankment frame, M' is moving towards the point where one strike occurred (let's say this is B) and away from the point where the other strike occurred (say this is A), so naturally the light from B will hit the observer on the train at position M' before the light from A hits him.
Then the time it will take from A' to M' t = \frac{L}{c} where c is the speed of light in the train's reference-body. . . . . . (5)

And the time it will take from B' to M' is also t = \frac{L}{c}.. . . . . (6)
Again, these results only hold in the train's rest frame.
So the transit times for the light to travel from A' to M' and from B' to M' are equal - (5),(6) above. . . . . . (7)
Yes, in the train's frame the transit times were equal, but you can't assume the strikes were simultaneous in this frame, and thus can't assume that equal transit times implies the light from each strike reaches M' at the same moment.
And as we saw in (2) above the lightning struck A' and B' at the same time as it struck A & B which strikes we have established was simultaneously - (1)
They were only simultaneous in the embankment frame, again you can't assume they were simultaneous in the train frame.
So if the light started from A' and B' at the same instant
But you have no reason to assume that it "started from A' and B' at the same instant" in the train's frame, and in fact Einstein's whole point was to show that it must not have. The idea is that different frames can never disagree on coincidences of local events (this is a very crucial idea in relativity), like what two clocks read when they pass next to each other, or whether two light beams coming from different directions strike a single observer at the same moment or at different moments. If they did, it would be easy to settle which frame's predictions were correct by experiment (thus defining a preferred frame and violating the first postulate), since it's impossible to have different truths about local events without invoking parallel universes or something of the sort. To see this, suppose one frame predicted a given clock would read 30 seconds when it passed an observer, and another frame predicted it would read 60 seconds when it passed that observer. Well, we could easily see which frame's prediction was objectively correct by attaching a bomb to the clock which was set to go off at 60 seconds--one frame could then predict the clock would already have passed the observer and gotten far enough away that the explosion wouldn't harm him, while the other would predict the observer would be right next to the bomb when it went off and would be killed. To see which frame's prediction is correct, you need only check whether the guy is alive or dead--they can't both be right! Similarly with the train thought experiment, it would be easy enough to put a small device at the center of the train which would sound an alarm (or set off a bomb) only if light detectors on opposite sides of the device detected bright light simultaneously (the detectors could convert light above a certain threshold into electrical signals which would be channeled through an AND gate, for example)--if different frames made different predictions about whether this device goes off, it would be easy enough to check which prediction was correct and thus establish a preferred frame, since again they couldn't both be right.

Einstein didn't explicitly state this point about different frames needing to agree on local events, but it's understood as a basic point in physics that different predictions about local events are mutually exclusive, and thus different frames cannot disagree in these predictions. We already know that the embankment frame predicts that the light from the flashes will reach the observer at the center of the train at different moments (a purely local prediction), so Einstein's point was that the only way this can also be true in the train's rest frame, without violating the assumption that the light from each flash should travel at c in this frame too (so the travel times for the light to get from the strike to the center must be equal in this frame), is if the strikes were not simultaneous in this frame. Hence the "relativity of simultaneity".

Please read the above carefully and then reread the text by Einstein to see if there is anything he says that is incompatible with this interpretation of his meaning (this is how every physicists since has interpreted him, and the fact that events which are simultaneous in one frame are non-simultaneous in others is built into the Lorentz transformation, so it seems a priori pretty unlikely that everyone has been misunderstanding him). And if after doing so you still think that somehow Einstein was saying that the strikes at A' and B' were simultaneous in the train frame just like they were in the embankment frame, how can you possibly make sense of this statement by Einstein?
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.

[Grimble]

But you have no reason to assume that it "started from A' and B' at the same instant" in the train's frame, and in fact Einstein's whole point was to show that it must not have. The idea is that different frames can never disagree on coincidences of local events (this is a very crucial idea in relativity), like what two clocks read when they pass next to each other, or whether two light beams coming from different directions strike a single observer at the same moment or at different moments. If they did, it would be easy to settle which frame's predictions were correct by experiment (thus defining a preferred frame and violating the first postulate), since it's impossible to have different truths about local events without invoking parallel universes or something of the sort.

Hello Jesse, I have been re-reading and considering this as you suggested and it seems to me that you are saying on the one hand that different frames can never disagree on coincidences of local events (this is a very crucial idea in relativity),

And then, on the other hand, that the two strikes of lightening that coincide (are simultaneous) in the embankment frame do not coincide in the train's frame.

Which could imply that the embankment is a preferred frame:confused:

And I agree that Einstein said: Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative

But he also implies that this is only from the point of view of the embankment: Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A

That which frame we are viewing from is an essential piece of information: Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.

And he says near the start of this chapter that: Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.

And I think you can see how confused I am for when I studied this, the understanding that I formed, was that Einstein was saying that there was one event (no, wrong use of that word); one coincidence - the simultaneous strikes of lightning, that could only be simultaneous from the frame it is being viewed from, which ever that was, and that from any other frame it would be asynchronous.

That reading seems to me to answer all your criteria:
The two frames would both agree that it was simultaneous, but only to the viewing frame;
Neither frame would be in any way a preferred frame;
That from the embankment frame that the observer on the train is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A;
That the two frames are agreeing about what happened locally within their two frames, whilst agreeing that the other would see it differently, as would be expected; For as Einstein said Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.

I am not trying to be difficult here, Jesse, I am only trying to shew how I have become confused, as I like to dot every I and cross every T!

I am going over and over all that you and the other kind contributers to my threads, say and looking at all the references; it does take a little time, but it is fascinating.

I may seem a bit of a recalcitrant old fuddy-duddy, but, wanting to be clear in my understanding, I like everything to fit neatly into place.

Thank you for your patience, your humble student, Grimble :smile:o:):smile:

[JesseM]

Hello Jesse, I have been re-reading and considering this as you suggested and it seems to me that you are saying on the one hand
But you have no reason to assume that it "started from A' and B' at the same instant" in the train's frame, and in fact Einstein's whole point was to show that it must not have. The idea is that different frames can never disagree on coincidences of local events (this is a very crucial idea in relativity), like what two clocks read when they pass next to each other, or whether two light beams coming from different directions strike a single observer at the same moment or at different moments. If they did, it would be easy to settle which frame's predictions were correct by experiment (thus defining a preferred frame and violating the first postulate), since it's impossible to have different truths about local events without invoking parallel universes or something of the sort.
And then, on the other hand, that the two strikes of lightening that coincide (are simultaneous) in the embankment frame do not coincide in the train's frame.
You misunderstand, when I said "coincidences of local events" I was referring to events that coincide right next to each other in both space and time--all my examples were of that sort (that's why I keep talking about bombs going off, only if you're right next to the bomb in space at the time it explodes are you going to be killed, and all frames should agree about whether or not you get killed!) That's also what "locally" always means in relativity, "in the same local neighborhood of spacetime" (for the purposes of idealized problems, this means that the separation in space and time is treated as zero). The two lightning strikes happen at the same time in one frame, but they do not happen at the same position in space in any frame, so this is not what I meant by a "coincidence of local events". On the other hand, the event of the lightning hitting point A on the tracks and point A' on the end of the train is treated as happening at the same point in both time and space, so these events would coincide locally. Likewise, if two light beams coming from different directions strike a single observer at the same moment, the events of each beam striking him would coincide locally, so all frames would have to agree on this; and the converse of this is that if one frame predicts they hit an observer at different times, then all frames must agree these events do not locally coincide. So, Einstein was using the fact that in the embankment frame, we predict that the light from the strikes will hit the observer at the center of the train (at position M') at different times to show that in order for the train frame to agree that the light hits the observer at the center of the train at different moments, it must judge that the strikes happened non-simultaneously in this frame.

Does this help?

And I agree that Einstein said:
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative
But he also implies that this is only from the point of view of the embankment
How could it only be from the point of view of the embankment? He's explicitly comparing simultaneity in two frames here, the embankment frame and the train frame, and saying that if they are simultaneous in the first frame ('simultaneous with reference to the railway embankment') then they are not "simultaneous relatively to the train". It doesn't make any sense that a statement about what is true "relatively to the train" could be from the perspective of the embankment frame, the phrase "relatively to the train" is synonymous with "in the train's frame".

In the section you quote next, "Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A", he was talking about what is true in the embankment frame, but the point of doing this was to show that the embankment frame predicts the light strikes the observer at the center at different moments, which is an objective fact that must be true in all frames (that's why he next says 'Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A', without bothering to specify a frame). This implies (by the principle that all frames must agree on coincidences of local events, as understood above) that the train frame must also predict the light strikes the observer at the center of the train at different moments, which is why he goes on to say "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".
And he says near the start of this chapter that:
Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.
He is not saying that all frames agree on simultaneity, just that they all define simultaneity in their own frame using the physical procedure given in the previous section (http://www.bartleby.com/173/8.html) of the book, i.e. they say two events are simultaneous if an observer at the exact midpoint between the position of those events recieves light-signals from each event at the same moment. The train thought experiment shows that if observers in both frames use this definition, then a pair of events which one observer defines to be simultaneous must not be judged to be simultaneous by the other; if the light from each flash reaches an embankment observer at M at the midpoint of A and B at the same moment, then the light from each flash will reach the train observer at M' at the midpoint of A' and B' at different moments.
And I think you can see how confused I am for when I studied this, the understanding that I formed, was that Einstein was saying that there was one event (no, wrong use of that word); one coincidence - the simultaneous strikes of lightning, that could only be simultaneous from the frame it is being viewed from, which ever that was, and that from any other frame it would be asynchronous.

That reading seems to me to answer all your criteria:
The two frames would both agree that it was simultaneous, but only to the viewing frame;
This phrase doesn't make any sense to me. How can they both "agree that it was simultaneous" if it was only simultaneous in the viewing frame? If the events aren't simultaneous in the train frame, then by definition both frames do not "both agree it was simultaneous". When physicists use anthropomorphic language about frames, like talking of frames making judgments or agreeing about things, they do not mean to imply that frames can "think" about any other frame besides themselves and form opinions about what is true in that other frame; a "judgment" made by a frame is just a fact which you arrive at by making an analysis that involves the coordinates of that frame alone, like the judgment that two events are simultaneous in that frame (they happen at the same t-coordinate), and for two frames to "agree" on something just means that a thing which is true when you analyze it from one frame's perspective is also true when you analyze it from another frame's perspective. So 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.
Neither frame would be in any way a preferred frame;
If different frames made different predictions about whether the light from each strike would reach the observer at the center of the train at the same moment or at different moments (a disagreement about whether two events happen at the same local point in both space and time), then surely you agree only one frame's prediction can be empirically correct, and whichever one had made the correct prediction would be preferred over the one that made the incorrect prediction? Again consider the apparatus I imagined in the earlier post:
Similarly with the train thought experiment, it would be easy enough to put a small device at the center of the train which would sound an alarm (or set off a bomb) only if light detectors on opposite sides of the device detected bright light simultaneously (the detectors could convert light above a certain threshold into electrical signals which would be channeled through an AND gate, for example)--if different frames made different predictions about whether this device goes off, it would be easy enough to check which prediction was correct and thus establish a preferred frame, since again they couldn't both be right.

[matheinste]

Hello Grimble,

This may be of some use.

Consider the scenario where the strikes are simultaneous in the embankment frame. We know they are simultaneous in the embankment frame because the light from the strikes meets at the location of an observer, on the embankment, midway between the points where the lightning struck. This is the requirement for simultaneity.

Now consider an observer at rest in the train frame at the mid point of the train, which is also midway between the points at which the lightning struck. He will not consider the strikes to be simultaneous in the train frame because the light from the strikes did not meet at his location as would be necessary for the requirements of the definition of simultaneity to be met.. HOWEVER, with the information available to him, and with an understanding of the postulates of relativity, he can calculate that the events WERE simultaneous in the embankment frame. So we could say that that although the strikes were not simultaneous in both frames, observers in both frames can conclude that they were simultaneous in the embankment frame. The opposite applies if the strikes were simultaneous in the train frame.

As a little extra comment, which can be ignored for the present purposes if it in any way adds to the confusion, any observer at any point at rest in the embankment frame can, by knowledge of data which is available to him, also conclude that the strikes were simultaneous in the embankment frame but not in the train frame. The observer does not have to be at the mid point between the strikes to be able to decide upon simultaneity. The same reasoning applies with regards to the train frame.

Matheinste

[Grimble]

Thank you, both, that is a great help to understand what you are saying, I must re-think and understand it now.

Grimble:smile:

[Max™]

http://www.mcasco.com/Now/Relativity_of_Simultaneity_Animation.gif

[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 (http://www.physicsforums.com/showthread.php?t=333112&page=10)) 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 realise 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 realised 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! :doh:

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:doh::doh::doh:

[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 realised 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! :doh:

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 travelling 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:surprised:

Grimble :approve:

[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:surprised:

Grimble :approve:

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?

:confused::confused::confused:Grimble:confused::co nfused::confused:

[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?

:confused::confused::confused:Grimble:confused::co nfused::confused:


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:smile:

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:eek:

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???:confused::confused::confused:

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:surprised::surprised:

[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 spacial 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 spacial 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 spacial 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:redface:

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 spacial 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?:smile:

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:redface:

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:confused:

[cfrogue]

Sorry again:redface:

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:confused:


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:smile:

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:eek:
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 in to 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 (http://img697.imageshack.us/i/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:

http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul1.gif

from the train:

http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul2.gif

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 (http://img697.imageshack.us/i/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 (http://img697.imageshack.us/i/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:redface:)

I have therefore redrawn it:

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

[JesseM]

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

I have therefore redrawn it:

http://img4.imageshack.us/img4/4831/train2r.jpg (http://img4.imageshack.us/i/train2r.jpg/)
The altered diagram still shows the worldline of B' intersecting with B at x=4 in the embankment frame. Are the new red lines supposed to be the light rays as seen in the train frame? You can't combine what's seen in the embankment frame and what's seen in the train frame in a single diagram, a spacetime diagram in SR is supposed to represent the perspective of one frame. There should only be a single pair of lines representing the light rays, to fix diagram #4 you need to erase the current B' worldline and draw a new one (still parallel to the A' worldline) that intersects the x-axis at a different position than x=4, showing that at t=0 in the embankment frame, the distance between A' and B' is less than 4 due to length contraction.
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.
You're not making any sense to me. What do you mean "if therefore they have common units"? By definition different frames have different ways of assigning coordinates to the same event. If the second lightning strike occurs at x=4 and t=0 in the embankment frame, its coordinates of the second strike are different in the train frame. Do you understand what the "Lorentz transformation" is? Its whole purpose is to take the coordinates assigned to a single physical event in one frame, and tell you the different set of coordinates assigned to that same physical event in another frame. If an event has coordinates x,t in one inertial frame, and we want to know the coordinates x',t' assigned to that same event in a different inertial frame whose spacetime origin coincides with the first frame (i.e. an event at x=0 and t=0 in the first frame has coordinates x'=0 and t'=0 in the second frame) and which is moving at speed v relative to the first frame, then the Lorentz transformation gives the answer:

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

Try picking a value for v (v=0.6c is a nice one, it leads to a gamma factor of 1.25) and then plugging in the event (x=4, t=0), you'll see that the x' and t' coordinates of this event are different.

[Grimble]

Hello Grimble.

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

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

Matheinste.

Then tell me what the coordinates are.
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

There is no reference to LT, because all I am doing is looking at the spacetime coordinates of the two separate frames.
LT is only relevant when observing one from the other.
In each frame all the units are proper units.
You are describing how the train will look from the embankment, which involves their relative velocity and the length contraction that is dependent thereon. If that velocity changed so would your values for the train's coordinates, which has nothing to do with the train's spacetime coordinates, in the train's frame of reference, considered purely on it's own.
I am only comparing the spacetime coordinates after calculating what they are; considering that we have one common event and then seeing how they compare.


The altered diagram still shows the worldline of B' intersecting with B at x=4 in the embankment frame. Are the new red lines supposed to be the light rays as seen in the train frame? You can't combine what's seen in the embankment frame and what's seen in the train frame in a single diagram, a spacetime diagram in SR is supposed to represent the perspective of one frame. There should only be a single pair of lines representing the light rays, to fix diagram #4 you need to erase the current B' worldline and draw a new one (still parallel to the A' worldline) that intersects the x-axis at a different position than x=4, showing that at t=0 in the embankment frame, the distance between A' and B' is less than 4 due to length contraction.

Yes, very good! but the red lines were to shew that that diagram was wrong and the replacement is just below it and agrees with your comments -- I think!:rofl:

You're not making any sense to me. What do you mean "if therefore they have common units"? By definition different frames have different ways of assigning coordinates to the same event.

OK let me restate this so you can understand what I am saying.

We have two inertial frames of reference.
They have identical physical laws.
Therefore identical clocks and measuring devices must keep identical time and measurements in each frame.
We do therefore have common units

In the first frame, the embankment, we describe the space time coordinates of three events; A,M,B.

AM = 2 light-seconds and MB = 2 lightseconds. All three have a time = 0

In the second frame, the train, we describe another three points, A',M',B'.

A'M' = 2 light-seconds and M'B' = 2 lightseconds. All three have a time = 0

Then if event A and event A', coincide, i.e. are adjacent, then M and M', and B and B' must also coincide.

Now as lightning strikes A and B simultaneously at the events A/A' and B/B' they must be simultaneous in each frame.

None of this has anything to do with Lorentz Transformations.
We are considering the spacetime coordinates of each frame, from that same frame, from which point of view that frame is at rest.

When we expand this to view how this appears relative to an observer in each reference frame then LT is of prime imortance and length contraction would be evident.

If you consider that this reasoning is mistaken please tell me what the spacetime coordinates would be. For the spacetime coordinates of one frame must be made from the point of view that an observer within that same frame considers it to be stationary.

Grimble

[matheinste]

----------Now as lightning strikes A and B simultaneously at the events A/A' and B/B' they must be simultaneous in each frame.----------

This is wrong.

The basic Einstein thought experiment already explains this fact. The strikes are simultaneous in the embankment frame. That is given. They are not simultaneous in the train frame. That is derived by logic from the postulates. So if you choose to assign the same coordinate to A snd A' when viewed from both frames ( effectively making those events occur at the origin in both frames ), then if B and B' have the same time coordinate as each other in the embankment frame, which is how the scenario is set up, they cannot have the same time coordinate as each other in the train frame. The same applies to M and M'.

Matheinste.

[Grimble]

----------Now as lightning strikes A and B simultaneously at the events A/A' and B/B' they must be simultaneous in each frame.----------

This is wrong.

The basic Einstein thought experiment already explains this fact. The strikes are simultaneous in the embankment frame. That is given. They are not simultaneous in the train frame. That is derived by logic from the postulates. So if you choose to assign the same coordinate to A snd A' when viewed from both frames ( effectively making those events occur at the origin in both frames ), then if B and B' have the same time coordinate as each other in the embankment frame, which is how the scenario is set up, they cannot have the same time coordinate as each other in the train frame. The same applies to M and M'.

Matheinste.

Not as seen/observed/referenced from the embankment as Einstein said but as Einstein said:
Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.
but this has, it seems to me been mistakenly taken as referring to the definition for simultaneity. But it could just as easily/correctly be taken as meaning "the definition of simultaneity can be applied relative to the train in exactly the same way as with respect to the embankment.

Then the argument referring to the coordinates works. Otherwise we have a conflict based on nothing more than the statement that simultaneity can only happen in one frame, and as I keep saying Einstein only applied that with reference to the embankment

Grimble

[JesseM]

Hello Grimble.

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

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

Matheinste.
Then tell me what the coordinates are.
If the velocity of the train relative to the embankment is 0.6c, and in the embankment frame the first strike occurs at x=0 light-seconds, t=0 seconds while the second strike occurs at x=4 l.s., t=0 s, then in the train frame the first strike occurs at x'=0 l.s, t'=0 s and the coordinates of the second strike can be found using the Lorentz transformation:

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

Here gamma=1.25, so we have x' = 1.25*(4) = 5 l.s., and t' = 1.25*(-0.6*4) = -3.25 s. So, in the train frame the coordinates of the second strike are x' = 5 l.s. and t' = -3.25 s.
There is no reference to LT, because all I am doing is looking at the spacetime coordinates of the two separate frames.
LT is only relevant when observing one from the other.
I've already told you this notion of frames "observing" each other doesn't make any sense to me, the Lorentz transformation simply tells you the coordinates of a single spacetime event in one frame's own coordinates if you know the coordinates of the same event in the other frame. As I said back in post #28:
If the events aren't simultaneous in the train frame, then by definition both frames do not "both agree it was simultaneous". When physicists use anthropomorphic language about frames, like talking of frames making judgments or agreeing about things, they do not mean to imply that frames can "think" about any other frame besides themselves and form opinions about what is true in that other frame; a "judgment" made by a frame is just a fact which you arrive at by making an analysis that involves the coordinates of that frame alone, like the judgment that two events are simultaneous in that frame (they happen at the same t-coordinate), and for two frames to "agree" on something just means that a thing which is true when you analyze it from one frame's perspective is also true when you analyze it from another frame's perspective. So 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.
If this point is still not clear to you, please take a look at my thread An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023), which shows the ruler/clock systems of two different frames moving alongside each other. Although the diagrams drawn from each frame's perspective look different, you can see that they always agree on which pairs of events coincide locally, as I've discussed earlier on this thread. For example, the top part of the diagram below shows what things look like at t=1 microsecond in frame A, the bottom part shows what things look like at t'=0 microseconds in frame B, even though the two diagrams look different they both agree that the event of (A's clock at position 346.2 meters on A's ruler showing a time of 1 microsecond) coincides locally with the event of (B's clock at 173.1 meters on B's ruler showing a time of 0 microseconds).

http://www.jessemazer.com/images/MatchingClocks.gif

So, if there was an event like a lightning flash that occurred next to the 346.2 meter mark on A's ruler when the clock at that mark read 1 microsecond, then naturally by the principle that there are objective frame-independent truths about which events locally coincide, it would have to be true that this same lightning flash occurred next to the 173.1 meter mark on B's ruler when the clock at that mark read 0 seconds. These are just statements about what coordinates each frame assigns to the flash using their own rulers and clocks, nothing to do with one frame "observing" the other one! And this is exactly what the Lorentz transformation is designed to do--if you plug in x=346.2 meters and t=1 microsecond into the Lorentz transformation, along with v=0.866c and gamma=2 as I assumed in the diagrams, then you will in fact get x'=173.1 meters and t=0 microseconds (I showed the math in post #134 of the Time dilation formula thread (http://www.physicsforums.com/showthread.php?p=2410200)).
Yes, very good! but the red lines were to shew that that diagram was wrong and the replacement is just below it and agrees with your comments -- I think!:rofl:
My bad! OK, then I agree with your redrawn diagram #4.
You're not making any sense to me. What do you mean "if therefore they have common units"? By definition different frames have different ways of assigning coordinates to the same event.
OK let me restate this so you can understand what I am saying.

We have two inertial frames of reference.
They have identical physical laws.
Therefore identical clocks and measuring devices must keep identical time and measurements in each frame.
We do therefore have common units
Does "common units" mean nothing more than the idea that clocks at rest in each frame behave identically with respect to the coordinates of that frame, and likewise for rulers at rest in each frame? Of course each frame's own clocks run at a rate of 1 tick per second of coordinate time in their own frame--you can see this is true if you compare the first and second diagram in the An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023) thread, the first showing things from frame A's perspective (and showing that at t=0 microseconds in A's frame, every one of A's clocks reads 0 microseconds, then at t=1 microseconds in A's frame, every one of A's clocks reads 1 microsecond, and finally at t=2 microseconds in A's frame, every one of A's clocks reads 2 microseconds) the second showing things from frame B's perspective (and showing that at t'=0 microseconds in B's frame, every one of B's clocks reads 0 microseconds, then at t'=1 microseconds in B's frame, every one of B's clocks reads 1 microsecond, and finally at t'=2 microseconds in B's frame, every one of B's clocks reads 2 microseconds). But this notion of "common units" does not mean that an event with a given set of coordinates in A's frame (like the event in the third diagram at x=346.2 meters, t=1 microsecond in A's frame) will have the same coordinates in B's frame (in this case the event would have coordinates x'=173.1 meters, t'=0 microseconds in B's frame). So why do you think that "common units" somehow shows that if an event occurs at x=4 light seconds, t=0 seconds in one frame, it should occur at x'=4 light seconds, t'=0 seconds in the other frame? There seems to be no logical connection here.
In the first frame, the embankment, we describe the space time coordinates of three events; A,M,B.

AM = 2 light-seconds and MB = 2 lightseconds. All three have a time = 0

In the second frame, the train, we describe another three points, A',M',B'.

A'M' = 2 light-seconds and M'B' = 2 lightseconds. All three have a time = 0

Then if event A and event A', coincide, i.e. are adjacent, then M and M', and B and B' must also coincide.
Why??? This is a complete non sequitur! In the embankment frame the distance A'M' is less than 2 light seconds, and in the train frame the distance AM is likewise less than 2 light seconds, so there's no reason to expect M' to coincide with M in either frame. Again, please look at the An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023) thread. Suppose I say that in frame A, event A is at (x=0 meters, t=0 microseconds) and event M is at (x=346.2 meters, t=0 microseconds). Likewise I say that in frame B, event A' is at (x'=0 meters, t'=0 microseconds) and event M' is at (x'=346.2 meters, t'=0 microseconds). By your reasoning, would you say that if events A and A' coincide, then events M and M' must coincide as well? But if you look at the diagrams, you can easily verify that events A and A' do coincide (the event of A's clock at the 0-meter mark reading 0 microseconds lines up with the event of B's clock at the 0-meter mark reading 0 microseconds) while M and M' do not coincide (the first diagram shows that the event of A's clock at the 346.2-meter mark reading 0 microseconds lines up with the event of B's clock at the 692.3-meter mark reading -2 microseconds, while the second diagram shows that the event of B's clock at the 346.2-meter mark reading 0 microseconds lines up with the event of A's clock at the 692.3-meter mark reading +2 microseconds). And yet, clearly the diagrams respect the principle that identical clocks and rulers behave identically in their own frames--in frame A, A's clocks tick forward at a rate of 1 microsecond per microsecond of coordinate time, and in frame B, B's clocks also tick forward at a rate of 1 microsecond per microsecond of coordinate time.

Please go over the diagrams and verify that what I am saying is correct, then tell me if you still think the principle that "identical clocks and rulers behave identically in their own frame" somehow implies that if A and A' coincide, M and M' coincide (if so, how do you explain the fact that the diagrams indicate otherwise, in spite of the fact that they do seem to respect the 'identical clocks and rulers behave identically in their own frame' principle?)
None of this has anything to do with Lorentz Transformations.
We are considering the spacetime coordinates of each frame, from that same frame, from which point of view that frame is at rest.
Again you appear confused about what the Lorentz transformation is for. It simply tells you which ruler/clock readings in one frame line up with which ruler/clock readings in the other--everyone should agree on which pair of readings line up, based on the principle that there must be an objective truth about which events locally coincide. And each set of ruler/clock readings represent that frame's own coordinates, nothing to do with one frame "observing" the other. Again, based on the idea that there must be an objective truth about which events locally coincide, then if a given event like a lightning flash coincides locally with the event of a clock at 346.2 meters on frame A's ruler reading a time of 1 microsecond (with both the ruler and clock at rest in frame A), and we also know that this ruler/clock reading in frame A coincides locally with the event of a clock at 173.1 meters on frame B's ruler reading a time of 0 microseconds, then obviously it must also be true that the same lightning flash coincides locally with the event of the clock at 173.1 meters on frame B's ruler reading a time of 0 microseconds, that's just the coordinates that B will assign to the same event using his own set of rulers and clocks.
If you consider that this reasoning is mistaken please tell me what the spacetime coordinates would be. For the spacetime coordinates of one frame must be made from the point of view that an observer within that same frame considers it to be stationary.
Again, the whole point of the Lorentz transform is that if you know the spacetime coordinates of an event in frame #1, it will give you the spacetime coordinates of the same event in frame #2. As I said, if the second lightning strike occurs at x=4 l.s. and t=0 s in the embankment frame, and if the train is moving at 0.6c relative to the embankment, then in the train frame's own coordinates this same lightning strike will have coordinates x'=5 l.s and t'=-3.25 s.

[Grimble]

First of all let me apologise to you; whenever I use the anthropomorphic language that offends you, would you please excuse me, and read it as saying that someone in that frame 'thinks'/'judges'/'agrees', I do understand how annoying something such as this can be and I will try not to do it again.:redface:

Secondly let me assure you that I do understand the points you raise but I do have some difficulty accepting them. (I must seem like hard work to you and I do appreciate the time you are spending to help me).

As far as the 'common' units are concerned, it seems to me that if one has identical clocks, kept under identical conditions the time that they keep will be identical, i.e. 1 second on one clock will be exactly equal to one second on another clock - this is after all only an extension of Einstein's own theory's statement viz:

It is clear that this definition can be used to give an exact meaning not only to two events, but to as many events as we care to choose, and independently of the positions of the scenes of the events with respect to the body of reference 1 (here the railway embankment). We are thus led also to a definition of “time” in physics. For this purpose we suppose that clocks of identical construction are placed at the points A, B and C of the railway line (co-ordinate system), and that they are set in such a manner that the positions of their pointers are simultaneously (in the above sense) the same. Under these conditions we understand by the “time” of an event the reading (position of the hands) of that one of these clocks which is in the immediate vicinity (in space) of the event. In this manner a time-value is associated with every event which is essentially capable of observation.
This stipulation contains a further physical hypothesis, the validity of which will hardly be doubted without empirical evidence to the contrary. It has been assumed that all these clocks go at the same rate if they are of identical construction. Stated more exactly: When two clocks arranged at rest in different places of a reference-body are set in such a manner that a particular position of the pointers of the one clock is simultaneous (in the above sense) with the same position of the pointers of the other clock, then identical “settings” are always simultaneous (in the sense of the above definition).

For surely any such clock, in any inertial frame of reference, will be subject to identical physical conditions, identical physical laws, and must therefore keep the same time. How could it not do so?

And I quite agree that the LT enable one to see (calculate) how the coordinates from one frame can be transformed to another frame and what they will be after transformation.

But the point I have been trying to put across is that I am looking at two seperate frames and how the spacetime coordinates of each frame can be arrived at, before comparing them.

The two sets of coordinates exist entirely on their own, independent of each other.
It is only when we state that A and A' (and the lightning strike are, in fact, all one event) that we compare them. And at that time the LT have not been involved.

This is just my explanation for you.

Grimble:smile:

[A.T.]

But the point I have been trying to put across is that I am looking at two seperate frames and how the spacetime coordinates of each frame can be arrived at, before comparing them.

What are "spacetime coordinates of each frame"? Events have spacetime coordinates, not frames. A frame defines spacetime coordinates of each event.

The two sets of coordinates exist entirely on their own, independent of each other.

The spacetime coordinates of the same unique events in each frame are not independent. They are coupled via the LT.

[JesseM]

First of all let me apologise to you; whenever I use the anthropomorphic language that offends you, would you please excuse me, and read it as saying that someone in that frame 'thinks'/'judges'/'agrees', I do understand how annoying something such as this can be and I will try not to do it again.:redface:
You're missing the main point, which is that the Lorentz transformation is not about what someone in one frame "judges" about another frame, it's about which ruler/clock readings in one frame objectively coincide with ruler/clock readings in another frame. Did you look over my example in the the "illustration of relativity with rulers and clocks" thread like I asked?
Secondly let me assure you that I do understand the points you raise but I do have some difficulty accepting them. (I must seem like hard work to you and I do appreciate the time you are spending to help me).
Then please actually address the questions and examples I bring up, which are intended to either convince you that you are wrong, or clarify where you disagree with me.
As far as the 'common' units are concerned, it seems to me that if one has identical clocks, kept under identical conditions the time that they keep will be identical, i.e. 1 second on one clock will be exactly equal to one second on another clock
The meaning of this statement is ambiguous. 1 second of time on clock A in clock A's rest frame is the same as 1 second of time on clock B in clock B's rest frame (they both last 1 second of coordinate time in their respective frames), but if we pick a single frame to analyze both clocks, and the clocks have different speeds in this frame, then in this frame 1 second of time on clock A is not equal to clock B. Agreed?
For surely any such clock, in any inertial frame of reference, will be subject to identical physical conditions, identical physical laws, and must therefore keep the same time. How could it not do so?
"Same time" is again ambiguous. It keeps the same time in its own rest frame as any other clock does in its own rest frame. But from the perspective of any one frame, clocks with different speeds keep different times.
And I quite agree that the LT enable one to see (calculate) how the coordinates from one frame can be transformed to another frame and what they will be after transformation.

But the point I have been trying to put across is that I am looking at two seperate frames and how the spacetime coordinates of each frame can be arrived at, before comparing them.
But what you don't seem to understand is that when we talk about "transforming into another frame", all we are talking about is finding out what x',t' coordinates that other frame assigned to the event on its own, totally independent of any use of the Lorentz transformation! That's why I am trying to steer you to discuss the specifics of my example with rulers and clocks, so that you will actually see how this works instead of just talking in generalities. Please, look at this diagram and actually think about it for just a minute!

http://www.jessemazer.com/images/MatchingClocks.gif

Again, suppose an event like a lightning flash happens in the same local vicinity as the circled region. Does observer A have to use the Lorentz transformation, or to think about observer B's coordinates, in order to figure out the coordinates of the event in his frame? No, he can just take a picture of the lightning flash as it happened, and see that as the flash occurred it was right next to the 346.2 meter mark on his ruler, and that at that moment the clock attached to that mark read 1 microsecond. Likewise, does observer B have to think about observer A's coordinates or use the Lorentz transformation to figure out the coordinates of the event in his frame? No, he too can take a picture of the lightning flash as it happened, and see that as the flash occurred it was right next to the 173.1 meter mark on his ruler, and that at that moment the clock attached to that mark read 0 microsecond. The Lorentz transformation simply relates the readings that one observer will get using his own ruler/clock system to the readings another observer will get using his own ruler/clock system, it has nothing to do with introducing some new notion called "transformed coordinates" which represent one observer's view of another frame or which are fundamentally different from the coordinates each observer gets using only his own measurements.
The two sets of coordinates exist entirely on their own, independent of each other.
It is only when we state that A and A' (and the lightning strike are, in fact, all one event) that we compare them. And at that time the LT have not been involved.
Well, hopefully you agree that it's not just a matter of us arbitrarily "stating" which events coincide in spacetime, there must be an objective physical truth about this. And you're right, we do not need to ever use the Lorentz transformation if different frames actually have their own physical measurement apparatuses (like the ruler/clock systems at rest in each frame) to assign coordinates to events. It will nevertheless be true that when they compare the coordinates each has found using their own inertial measurement apparatuses, they will find after-the-fact that the coordinates will be coupled in exactly the manner predicted by the Lorentz transformation, assuming that the two postulates of relativity actually hold true. Do you disagree with this? If so, why?

[Grimble]

You're missing the main point, which is that the Lorentz transformation is not about what someone in one frame "judges" about another frame, it's about which ruler/clock readings in one frame objectively coincide with ruler/clock readings in another frame. Did you look over my example in the the "illustration of relativity with rulers and clocks" thread like I asked?

Yes, that makes sense, it is at last an easy way to think of it; and yes, I have been looking at your illustration, and I will do so again.

Then please actually address the questions and examples I bring up, which are intended to either convince you that you are wrong, or clarify where you disagree with me.

I will go over all you have been saying and try and do so.

The meaning of this statement is ambiguous. 1 second of time on clock A in clock A's rest frame is the same as 1 second of time on clock B in clock B's rest frame (they both last 1 second of coordinate time in their respective frames), but if we pick a single frame to analyze both clocks, and the clocks have different speeds in this frame, then in this frame 1 second of time on clock A is not equal to clock B. Agreed?

Oh yes, I agree absolutely with what you say here, but if my statement was ambiguous I need to refine my thinking and improve my expression of my thoughts.

"Same time" is again ambiguous. It keeps the same time in its own rest frame as any other clock does in its own rest frame. But from the perspective of any one frame, clocks with different speeds keep different times.

I agree with you again but as that was not what I was saying, I must go away and think about it again!

But what you don't seem to understand is that when we talk about "transforming into another frame", all we are talking about is finding out what x',t' coordinates that other frame assigned to the event on its own, totally independent of any use of the Lorentz transformation! That's why I am trying to steer you to discuss the specifics of my example with rulers and clocks, so that you will actually see how this works instead of just talking in generalities. Please, look at this diagram and actually think about it for just a minute!

http://www.jessemazer.com/images/MatchingClocks.gif

I can appreciate much of this but I do have a couple of observations:

But for another observer who sees the ship moving, the back of the ship is moving towards the point where the flash happened and the front is moving away from that point, so his own network of clocks will be synchronized in such a way that the clock next to the event "light hits the back of the ship" will read an earlier time than the clock next to the event "light hits the front of the ship".

Surely either frames clock's should be synchronised in it's own frame? That is how Einstein's clock synchronisation works, doesn't it?

In "time = 0 microse. in Ruler A's Frame" the clock marked in green at -519.3 m shews time 1.5 microsec, yet 1 micrsec. later in Ruler A's Frame, it shews 2 microsec but has advanced 519.3 m.? Which by my calculation is 1038.6 m/microsec. or almost 3.5c!??

Again, suppose an event like a lightning flash happens in the same local vicinity as the circled region. Does observer A have to use the Lorentz transformation, or to think about observer B's coordinates, in order to figure out the coordinates of the event in his frame? No, he can just take a picture of the lightning flash as it happened, and see that as the flash occurred it was right next to the 346.2 meter mark on his ruler, and that at that moment the clock attached to that mark read 1 microsecond. Likewise, does observer B have to think about observer A's coordinates or use the Lorentz transformation to figure out the coordinates of the event in his frame? No, he too can take a picture of the lightning flash as it happened, and see that as the flash occurred it was right next to the 173.1 meter mark on his ruler, and that at that moment the clock attached to that mark read 0 microsecond. The Lorentz transformation simply relates the readings that one observer will get using his own ruler/clock system to the readings another observer will get using his own ruler/clock system, it has nothing to do with introducing some new notion called "transformed coordinates" which represent one observer's view of another frame or which are fundamentally different from the coordinates each observer gets using only his own measurements.

OK.

Well, hopefully you agree that it's not just a matter of us arbitrarily "stating" which events coincide in spacetime, there must be an objective physical truth about this. And you're right, we do not need to ever use the Lorentz transformation if different frames actually have their own physical measurement apparatuses (like the ruler/clock systems at rest in each frame) to assign coordinates to events. It will nevertheless be true that when they compare the coordinates each has found using their own inertial measurement apparatuses, they will find after-the-fact that the coordinates will be coupled in exactly the manner predicted by the Lorentz transformation, assuming that the two postulates of relativity actually hold true. Do you disagree with this? If so, why?

No problem with this.

But one thing is still bothering me; if all things are relative does that mean that there are no standard units by which different frames can be related?
By this, I am thinking of inertial frames with no external forces acting upon them. In such a frame, a muon's half-life, 2.2 microseconds, would surely be the same as the half-life in another inertial frame, it is a physical property, a law, and one would think, therefore, the same in any inertial frame? And I don't mean just to say it's half-life would be the same number of microseconds, I mean that the microseconds would have the same duration in any inertial frame?

After all, this is something that can be reasoned/deduced from your An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023)
In which you state: This means that each ruler will observe the other one’s clocks tick exactly half as fast as their own, and will see the other ruler's distance-markings to be squashed by a factor of two.
And this could only be true if the two ruler's markings and clocks had identical (in magnitude) units.
So, do we imply from this, that the magnitude of the units of measurement in inertial frames will be constant? That inertial units can be taken as the 'real' units and that it is these units that are 'contracted' or 'dilated' as a function of two frames relative velocity?

In post #75 you wrote:
I've already told you this notion of frames "observing" each other doesn't make any sense to me, the Lorentz transformation simply tells you the coordinates of a single spacetime event in one frame's own coordinates if you know the coordinates of the same event in the other frame.

which is a nice concise description, however, it does lead to another consideration:
If the coordinates of 'a single spacetime event' are known in frame A, the Lorentz transformation tells (an observer in) frame B what those coordinates are in frame B.
Frame C, however, can use LT to determine the coordinates of that same event from frame B (as frame B now knows what those coordinates are in frame B's coordinates).
But if frame C were then to use LT to determine what the coordinates are from those in frame A would he get the same result? I think not??

This may seem a little contrived but consider, if Einstein had two railway tracks and two trains running through the embankment, such that A and B lined up for all three frames; and if the relative velocity between train1 and the embankment were the same as the relative velocity between train2 and the embankment, then the length AB on each train would be the same in the embankments coordinates but not from the coordinates of the other train.
(This does of course require that A is the front of one train and the back of the other)

Grimble:smile:

[JesseM]

Thanks for focusing on those diagrams Grimble...you ask some good questions, I'll try to address them:
I can appreciate much of this but I do have a couple of observations:
But for another observer who sees the ship moving, the back of the ship is moving towards the point where the flash happened and the front is moving away from that point, so his own network of clocks will be synchronized in such a way that the clock next to the event "light hits the back of the ship" will read an earlier time than the clock next to the event "light hits the front of the ship".
Surely either frames clock's should be synchronised in it's own frame? That is how Einstein's clock synchronisation works, doesn't it?
Yes, but that isn't contradicted by my quote above. In the ship's own frame the clocks at the front and back of the ship are synchronized, what I said was that "for another observer who sees the ship moving", the light hits the back clock on the ship later than the light hits the front clock on the ship (as measured by 'his own network of clocks', so if he looks at the time on a clock in his network which is next to the back of the ship when the light reaches it, and compares to the time on a clock in his network which is next to the front of the ship when the light reaches it, the former time will be earlier than the latter time). So he sees the two clocks on the ship being out-of-sync, since the ship observer set them both to read the same time when the light reached them.
In "time = 0 microse. in Ruler A's Frame" the clock marked in green at -519.3 m shews time 1.5 microsec, yet 1 micrsec. later in Ruler A's Frame, it shews 2 microsec but has advanced 519.3 m.? Which by my calculation is 1038.6 m/microsec. or almost 3.5c!??
But the speed of B's green clock in A's frame is defined solely by measurements of position and time on A's own rulers and clocks, not by the time on B's clock itself. If we define event #1 as "B's green clock at the -519.3 m mark on B's ruler showing a time of 1.5 microseconds", and event #2 as "B's green clock at the -519.3 m mark on B's ruler showing a time of 2 microseconds", then if you look at the diagram you can see that event #1 occurs halfway between the -173.1 meter mark and the -346.2 meter mark on A's ruler, so that'd be at position -259.6 meters on A's ruler, and you can also see that it occurs at a time of 0 microseconds in A's frame; meanwhile, event #2 occurs at the 0.0 meter mark on A's ruler, and at a time of 1 microsecond in A's frame.

http://www.jessemazer.com/images/RulerAFrame.gif

So the distance between event #1 and event #2 as measured by A is 259.6 meters, and the time between these events as measured by A is 1 microsecond, so A will conclude that the green clock had a speed of 259.6 meters/microsecond (and the speed of light is 299.792458 meters/microsecond, so that's a speed of 0.866c, which gives the correct gamma factor of 2).
But one thing is still bothering me; if all things are relative does that mean that there are no standard units by which different frames can be related?
By this, I am thinking of inertial frames with no external forces acting upon them. In such a frame, a muon's half-life, 2.2 microseconds, would surely be the same as the half-life in another inertial frame, it is a physical property, a law, and one would think, therefore, the same in any inertial frame? And I don't mean just to say it's half-life would be the same number of microseconds, I mean that the microseconds would have the same duration in any inertial frame?
All frames agree on the proper time for the muon to decay--proper time is just the time that would elapse on a clock moving along with the muon. For an object moving at constant velocity, you always can calculate the proper time between two events on its worldline by first taking the time t in your rest frame between the events and then multiplying by \sqrt{1 - v^2/c^2}. For example, if a muon is traveling at 0.6c, and I observe that it takes a time of 2.75 microseconds to decay according to clocks in my frame, then I can calculate that the proper time must have been 2.75 * sqrt(1 - 0.6^2) = 2.75 * 0.8 = 2.2 microseconds. In other frames the muon's decay time t and its velocity v will be different, but all will find that when they calculate t * sqrt(1 - v^2/c^2) they will get an answer of 2.2 microseconds.

Not sure if this answers your question, I'm still having trouble understanding what you mean by "standard units".
After all, this is something that can be reasoned/deduced from your An illustration of relativity with rulers and clocks (http://www.physicsforums.com/showthread.php?t=59023)
In which you state:
This means that each ruler will observe the other one’s clocks tick exactly half as fast as their own, and will see the other ruler's distance-markings to be squashed by a factor of two.
And this could only be true if the two ruler's markings and clocks had identical (in magnitude) units.
What do you mean by "identical" in this context? If in ruler A's rest frame, ruler B's markings are only half the size of ruler A's, doesn't this mean the markings are not identical in this frame?
So, do we imply from this, that the magnitude of the units of measurement in inertial frames will be constant? That inertial units can be taken as the 'real' units and that it is these units that are 'contracted' or 'dilated' as a function of two frames relative velocity?
Again, I don't understand what you mean by "the magnitude of the units of measurement in inertial frames will be constant"...magnitude relative to what? Constant relative to what? Would it be possible to give some sort of specific physical example of what you're talking about?
If the coordinates of 'a single spacetime event' are known in frame A, the Lorentz transformation tells (an observer in) frame B what those coordinates are in frame B.
Frame C, however, can use LT to determine the coordinates of that same event from frame B (as frame B now knows what those coordinates are in frame B's coordinates).
But if frame C were then to use LT to determine what the coordinates are from those in frame A would he get the same result? I think not??
There can be no disagreement about which ruler markings/clock readings on a given frame's ruler/clock system coincide with a given event. If in frame A a lightning flash happens at x=4 light-seconds and t=0 seconds on A's ruler/clock system, and that if frame B has a velocity of +0.6c along A's x-axis, then this same flash will happen at coordinates x'=5 l.s. and t'=-3 s in B's frame. Now suppose we have a third frame C which is moving at -0.8c along A's x-axis, in the opposite direction from frame B. From the perspective of frame C, it must be true that A is moving at +0.8c along C's x''-axis, and we can use the relativistic velocity addition formula (http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html) to show that from C's perspective, B must be moving at (0.8c + 0.6c)/(1 + 0.8*0.6) = 1.4c/1.48 = +0.945945...*c along C's x''-axis.

In C's frame, this same lightning flash happens at coordinates x''=6.666... l.s., and t''=5.333... s. So what happens if we start with these coordinates, and try to apply the Lorentz transformation to them to re-derive the coordinates in A's frame and B's frame? Well, for A we have a velocity of 0.8c and a gamma factor of 1.666..., so this calculation would give the coordinates in A's frame as:

x = gamma*(x'' - v*t'') = 1.666... * (6.666... - 0.8*5.333...) = 1.666... * (2.4) = 4
t = gamma*(t'' - v*x''/c^2) = 1.666... * (5.333... - 0.8*6.666...) = 0

And for B we have a velocity of 0.945945...*c and a gamma factor of 3.08333..., so this calculation would give the coordinates in B's frame as:

x' = gamma*(x'' - v*t'') = 3.08333... * (6.666... - 0.945945...*5.333...) = 3.08333... * (1.621621...) = 5
t' = gamma*(t'' - v*x''/c^2) = 3.08333... * (5.333... - 0.945945...*6.666...) = 3.08333... * (-0.972972...) = -3 s.

So, you can see that starting with the coordinates in frame C and then applying the Lorentz transformation to re-derive the coordinates in frames A and B yields the same result as starting from the coordinates in frame A and applying the Lorentz transformation to find the coordinates in frames B and C (likewise you'd get the same results if you started with the coordinates in frame B and applied the Lorentz transformation to find the coordinates in frames A and C).

[Grimble]

Thank you Jesse, this is great, having your help! It is clearing up a lot of points: sometimes all it needs is to see how to view or interpret what we see but I do still have a couple of queries...
Thanks for focusing on those diagrams Grimble...you ask some good questions, I'll try to address them:

Yes, but that isn't contradicted by my quote above. In the ship's own frame the clocks at the front and back of the ship are synchronized, what I said was that "for another observer who sees the ship moving", the light hits the back clock on the ship later than the light hits the front clock on the ship (as measured by 'his own network of clocks', so if he looks at the time on a clock in his network which is next to the back of the ship when the light reaches it, and compares to the time on a clock in his network which is next to the front of the ship when the light reaches it, the former time will be earlier than the latter time). So he sees the two clocks on the ship being out-of-sync, since the ship observer set them both to read the same time when the light reached them.

Yes, that certainly makes sense, I'm OK with that.

But the speed of B's green clock in A's frame is defined solely by measurements of position and time on A's own rulers and clocks, not by the time on B's clock itself. If we define event #1 as "B's green clock at the -519.3 m mark on B's ruler showing a time of 1.5 microseconds", and event #2 as "B's green clock at the -519.3 m mark on B's ruler showing a time of 2 microseconds", then if you look at the diagram you can see that event #1 occurs halfway between the -173.1 meter mark and the -346.2 meter mark on A's ruler, so that'd be at position -259.6 meters on A's ruler, and you can also see that it occurs at a time of 0 microseconds in A's frame; meanwhile, event #2 occurs at the 0.0 meter mark on A's ruler, and at a time of 1 microsecond in A's frame.
http://www.jessemazer.com/images/RulerAFrame.gif

So the distance between event #1 and event #2 as measured by A is 259.6 meters, and the time between these events as measured by A is 1 microsecond, so A will conclude that the green clock had a speed of 259.6 meters/microsecond (and the speed of light is 299.792458 meters/microsecond, so that's a speed of 0.866c, which gives the correct gamma factor of 2).

Yes, but that would work out wouldn't it? For are you not merely reversing the process by which you arrived at the figures in the first place?
It seems to me, without wishing to be confrontational, that we need a separate way to verify your calculations and that we have here Einstein's system K, represented by Ruler A, and system K', represented by Ruler B?
Now, Einstein demonstrated in section 6 (http://www.bartleby.com/173/11.html) that x = ct and x' = ct'.
Now, if I have worked this out correctly, when x' = {519.3} and t' = 0.5 c = \frac{519.3}{0.5} = {1038.6}

All frames agree on the proper time for the muon to decay--proper time is just the time that would elapse on a clock moving along with the muon. For an object moving at constant velocity, you always can calculate the proper time between two events on its worldline by first taking the time t in your rest frame between the events and then multiplying by \sqrt{1 - v^2/c^2}. For example, if a muon is traveling at 0.6c, and I observe that it takes a time of 2.75 microseconds to decay according to clocks in my frame, then I can calculate that the proper time must have been 2.75 * sqrt(1 - 0.6^2) = 2.75 * 0.8 = 2.2 microseconds. In other frames the muon's decay time t and its velocity v will be different, but all will find that when they calculate t * sqrt(1 - v^2/c^2) they will get an answer of 2.2 microseconds

The following was the post that introduced me to the muons;
Here is a very specific real-world test. BNL (Brookhaven Nat. Lab.) physicists stored muons with γ=29.4 in a circular ring. The muon's lifetime at rest is about 2.2 microseconds. In the ring, their lifetime was about 65 microsecons in the lab reference frame.
Bob S
So taking the proper time for the muon to decay to be 2.2 microseconds, the time in the lab rest frame to be approximately 65 microseconds and γ = 29.4 that fits.
This comes close to answering what bothers me; my difficulty is, I suppose, that so many contributors seem to be very reluctant to commit themselves to definite statements and this has led to much confusion.

Not sure if this answers your question, I'm still having trouble understanding what you mean by "standard units".

Let me try to simplify my question. There is a lot of talk of 'seconds', 'microseconds', 'proper time' and 'coordinate time' but there is some confusion in my mind about what is being referred to.

It seems to me that proper time is as you defined it above:
coordinate time is that same proper time but transformed, as you would measure it in your rest frame.
The tricky question is what scales are the different seconds measured upon?
Let me put it this way, if two separate inertial frames A & B are moving with the same speed relative to a third frame C, and let us say that 0.866c is that relative velocity, giving us a gamma factor of 2, as in your example above.
Then the C will measure the seconds in each of the other two frames, moving at the same relative velocity, as being half as long as the proper seconds in A and B.
Now my question is, in frame C will the transformed seconds from A equal those transformed from B? i.e. are the proper seconds in A equal in magnitude to the proper seconds in B?
One would assume so from the way your Rulers are drawn in your example.

I'm sorry if this seems like questioning the obvious but so many times in these threads do we find someone coming in with statements like 'but those seconds are in different frames so you can't compare them' – or am I just becoming paranoid?:redface:

What do you mean by "identical" in this context? If in ruler A's rest frame, ruler B's markings are only half the size of ruler A's, doesn't this mean the markings are not identical in this frame?

So here again I use identical to compare the proper units in the two frames; i.e. are the proper units of one ruler, in its frame, equal in magnitude to the proper units of the other ruler, in its frame?
(that is if they were being compared by another frame moving with the same relative velocity to each of those frames) – Sorry for going on so much, but I am trying to be explicit so we both understand what I am saying.

So, you can see that starting with the coordinates in frame C and then applying the Lorentz transformation to re-derive the coordinates in frames A and B yields the same result as starting from the coordinates in frame A and applying the Lorentz transformation to find the coordinates in frames B and C (likewise you'd get the same results if you started with the coordinates in frame B and applied the Lorentz transformation to find the coordinates in frames A and C).
Yes, of course! I had forgotten to use the relative velocity addition formula :redface::redface::redface:

Grimble