Time dilation: speed relative to what?

[Ibix]

Let us imagine a space station far away from any other body in space with an atomic grandfather clock, where the local time is displayed on a large clock face.
This clock, at the origin of the space station's Frame of Reference is keeping Proper Time.
<snip snip snip>
On the observer's clock, that traveling with him, and also keeping Proper Time, only 6 minutes will have passed..

Yet how can that be? For keeping proper time, both clocks will have be reading the same!

The answer of course is that the observer's measurement of the Grandfather clock's passage of time is COORDINATE time, which at 0.866c is twice as fast as Proper Time.

Grimble, you look to me to be confused about what proper time is. It is not some absolute time that clocks track. "Proper" in this context means "its own", from the Latin proprius. It's related to "appropriate". It does not mean "correct" or "true" or anything like that. A clock always shows proper time ("its own time") - that's more or less the definition of a clock. It does not mean that two clocks have to agree.

You may see texts telling you that proper time is an invariant quantity. This simply means that everyone can agree on what a given clock will show at some event. It does not mean that all clocks tell The Proper ("Right") Time - there's no such thing. For example, if the clock breaks down at the point that it reads 12 noon, clocks passing by at that very instant might show 1pm or 11am. However, everyone will agree that they expect the broken clock to read 12 noon at the time it broke down, based on its motion. As you noted - if that weren't the case something would be badly wrong.

 

[Grimble]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.
And if we add two more observers, D & E, permanently positioned between A and C, and B and C, then each of A, B and C will be moving at 0.2165c relative to D and E... and so on.
Extending this line of logic we can prove that all clocks must be running at the same rate. The 'slowing' of fast moving clocks is all relative to the moving observer. It is an effect caused by their relative motion, where the relative speed/velocity is no more than a factor in that measurement.
All clocks run the same but are subject to the conditions (relative velocity) under which they are measured.
Which is why we talk of taking measurements from one Frame of Reference and transforming them by means of the Lorentz transformation equations to determine the measurement relative to a moving observer.

So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

 

[Orodruin]

with standard clocks are traveling at 0.866c relative to one another,

C, is permanently mid-way between them then each will me moving at 0.433c

No. You are trying to use classical addition of velocities here. C will not see each of them moving at 0.433c.

So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.

Yes, what you need to look up is relativity of simultaneity. Different observers will not agree on which events are simultaneous. In C's frame, the events A's clock shows 1 PM and B's clock shows 1 PM are simultaneous. In the frames of A and B they are not.

Relativity of simultaneity would also seem to be irrelevant here

No, relativity of simultaneity is of very high importance here!

 

[nitsuj]

So how is it that we talk of clocks physically slowing; of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

You use the terms "reciprocal" & "symmetrical", and those terms require a unique concept in SR. "at the same time" or simultaneous is also a unique concept in SR for spatially separated events.

TL:DR - The observation that "two clocks physically slowing each with respect to the other at the same time" is physically meaningless. (assumption that they're spatially separated, otherwise one's broken )

Orodruin beat me to it

Reread the OP and there is also a misunderstanding of "reciprocal", as it lead to an incorrect assumption about RoS. Same could be said for "symmetry".

 

[harrylin]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.
And if we add two more observers, D & E, permanently positioned between A and C, and B and C, then each of A, B and C will be moving at 0.2165c relative to D and E... and so on.
Extending this line of logic we can prove that all clocks must be running at the same rate. The 'slowing' of fast moving clocks is all relative to the moving observer. It is an effect caused by their relative motion, where the relative speed/velocity is no more than a factor in that measurement.
All clocks run the same but are subject to the conditions (relative velocity) under which they are measured.
Which is why we talk of taking measurements from one Frame of Reference and transforming them by means of the Lorentz transformation equations to determine the measurement relative to a moving observer.

So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

I can't follow your reasoning, but perhaps that is due to a lack of definition of what you mean with "running at the same rate". According to all Galilean reference frames except the one in which the clock is not moving, the clock is "running at a slower rate".
Relativity of simultaneity is essential. In fact, the rate at which you "observe" a distant clock to be ticking (as well as the length contraction) is a function of your assumption of the one-way speed of light; simultaneity is defined such that light speed appears to be isotropic relative to your frame of choice.

The only sure and unambiguous way to compare clock rates is to compare clocks side by side. Other people who reasoned that all clocks must be running at the same rate, concluded that Einstein's calculation about an accelerating clock must be wrong (his calculation assumes that acceleration has no direct effect on clock rate):
"by the clock which has remained at rest the traveled clock on its arrival at A will be ½tv²/c² second slow".
- §4 of http://fourmilab.ch/etexts/einstein/specrel/www/

What does your line of reasoning give?

 

[Grimble]

OK, let us say that the relative speed of C to A and to B is v.

I can't follow your reasoning, but perhaps that is due to a lack of definition of what you mean with "running at the same rate". According to all Galilean reference frames except the one in which the clock is not moving, the clock is "running at a slower rate".

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/

So clocks, A & B, are running at the same rate as measured by C,

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/

What is your problem here? If C is permanently at the mid point of the line AB, then AC = BC and the speed of each relative to C will be the same - or the distances AC and BC would not continue to be equal.
If the speed of each relative to c is identical then their Lorentz factors will be the same and the slowing of those clocks, relative to C's clock will be the same, therefore clock A must read the same as clock B as measured by observer C. How is that not symmetry?

So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...

 

[nitsuj]

It seems like a reciprocal arrangement to me...

That's seems like a "privileged" perspective right? Analysis of all the results at the same time is different from when it is playing out in a continuum. This is the same perspective where you mentioned something happening "at the same time".

 

[Dale]

All ideal clocks run at a rate of one second per second of proper time (trivially $d\tau/d\tau = 1$ or barely less than trivially $c\,d\tau/ds=1$). Ideal clocks run at a rate of less than one second of proper time per second of coordinate time in any inertial frame in which they are moving ($d\tau/dt \le 1$).

 

[Nugatory]

From the initial post in this thread:

Extending this line of logic we can prove that all clocks must be running at the same rate.

I must confess that I'm not seeing the proof here. If you could spell it out in a bit more detail how to extend that line of logic to the conclusion that "all clocks must be running at the same rate?

I suspect that when you do, you will find that you are making an assumption:

If an observer finds that clocks A and B are both in motion relative to him and running at the same rate; and a second observer finds that clocks B and C are both in motion relative to him and running at the same rate then there exists some observer for whom all three clocks are running at the same rate.​

This is not correct, except in the uninteresting special case in which at least two of the three clocks are at rest relative to one another. However, every time that you add another clock "in between" (BTW, have you noticed that the physical positions of the clocks is irrelevant? They don't need to go "in between", they can go anywhere as long as they have the right relative velocities) that clock is in motion relative to the clocks on either side of it as well as all the other clocks, so this special case doesn't apply.

 

[Vanadium 50]

Grimble, I expect this is going to go in the exact same direction as all your other threads on this. Making increasingly complicated scenarios, guessing at the solution, and then debating it until the thread gets locked hasn't helped in the past, and is unlikely to help if we try it one more time.

A different approach is needed - taking advantage of physics being a quantitative science. You need to explicitly state what calculation would convince you that clocks run slow, and then we can do it together. For example, X observes Y's clock to read t3 when X's reads t1, and observes Y's clock to read t4 when X's clock reads t2. X will say Y's clock runs slow if (t4-t3)/(t2 - t1) < 1.

Issues about symmetries need to be applied in the solution of the problem, not the posing of the problem. As this (and the other threads) show, starting on the solution while still posing the question only adds to the confusion. Give us a clearly posed question.

 

[pervect]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of
So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

I would recommend the thread "Symmetrical time dilation implies the relativity of simultaneity", https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/

Symmetrical time dilation is what I call the situation where A says B's clock is slow, and B says A's clock is slow. This notion is not consistent with the notion of absolute time.

I rather strongly suspect here that the issue is that Grimble is implicitly assuming the existence of absolute time. This thread is my best attempt to show why absolute time is incompatible with symmetrical time dilation, and how what's known as "the relativity of simultaneity" fixes the issue.

 

[harrylin]

[..] if the 'stationary' observer changed his speed there is no connection to the 'moving' clock [... omit part that I cannot follow...] It can only be that the MEASUREMENT BY the remote observer is affected.
The clocks don't change only the measurements? That can't be right surely?

I agree with that part of your reasoning: the measurements are done with clocks and rulers.

[..]
So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...

Explaining such a complex arrangement with mere words is complicated and will be difficult to do without ambiguity. The best explication is with numbers - and it avoids wasting time on long discussions. Please prepare an example calculation. Then if you do not already find the answer from it yourself, you can present it here.

 

[Ibix]

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.

As has been noted before, the observer C would not be traveling at $\sqrt{3}c/4$ because velocities do not add linearly in a relativistic universe. It turns out that, if A is stationary in some frame in which B is approaching at $\sqrt{3}c/2$, then the observer C must be approaching A at $c/\sqrt{3}$ so that in C's rest frame, both A and B are approaching at the same speed in opposite directions.

So, here is a Minkowski diagram of that setup, as seen by C:

Apologies for the lack of labelling. A Minkowski diagram is basically a displacement-time graph, except that time runs vertically upwards and x position is shown horizontally. The vertical gridlines are one light second apart; the horizontal ones are one second apart.

The world-line of C is the purple line right up the middle. C is at rest in this frame, so the x position never changes. It starts at the origin at time zero (marked by a solid purple square) and stays right there.

The world-line of A is represented by the red line. It starts at x=-8ls at time zero (marked by a solid red square), and moves with velocity $c/\sqrt{3}$ - so as time goes on (up the page) it gets closer to C, and eventually meets at the top of the diagram.

The world-line of B is in blue, and is a mirror image of the world-line of A. It starts at x=8ls and moves with velocity $-c/\sqrt{3}$ and reaches C at the same time as A does.

As you noted, the situation is symmetrical as viewed from his frame. I've also added markers to each line showing when, according to C, clocks moving with A, B and C would tick. The purple ticks are every second; the red and blue ticks occur at the same time as each other, but you can see that they are spaced out more - their clocks tick slower in this frame.

But what does this look like according to A? We can use the Lorentz transforms to calculate the coordinates of each of the points in the Minkowski diagram as seen from a frame in which A is at rest and re-draw the diagram. That turns out to look like:

You can see that this time A (in red) is stationary, so the position is always the same, giving a vertical line. You can see that the red clock ticks are spaced exactly the same as the grid - they are 1s apart. You can see that there are the same number of ticks on each line as there were in the first diagram, and you can see that this time the purple clock is running slow and the blue clock is running really slow. You can also see why this isn't a problem - in this frame the three observers did not start their clocks at the same time.

So, yes, the clocks are running slow - it's not an illusion (although it is only an effect of co-ordinate choice). The different observers disagree about the time the clocks were started so that there is no contradiction when they meet up and can unambiguously compare clocks.

 

[Grimble]

To be sure that I can follow your explanations, that with the number of replies can become a little confusing with so many different replies to consider, let me check a few basic premises to be sure that I am correctly appreciating the concepts.
1. The view of space from any Inertial Frame of Reference is at rest; i.e. there are a set of axes that give a fixed set of coordinates for any point in space.
2. A full set of four coordinates specifying a fixed point in space, at a specific point in time, constitutes an event.
2. Every event can be mapped, that is given a unique set of coordinates, in any Inertial Frame of Reference.
3. Events, being fixed in time cannot move.
4. There must be a fixed point with a set of unique spatial coordinates, midway between any two events, in every Inertial Frame of Reference.
5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous. (Einsteins definition of simultaneity).

 

[harrylin]

Probably you meant it like that, but just to make sure I added some words:

5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous according to that inertial reference system. (Einsteins definition of simultaneity).

PS. this is just the example of Einstein here: http://www.bartleby.com/173/9.html

Personally I find it easier to consider the inverse: a light bulb in the middle of a train gives off a flash of light in both directions. The times that the light rays are thought to reach the ends of the train depend on the used reference system.

 

[Grimble]

Thank you, yes that is correct. I just wanted to be sure that I was understanding this.

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
And there would be a set of spatial coordinates that would define a fixed point in each and every inertial frame of reference that was spatially mid-way between points A and B?
And that the lights emitted at event A and event B would meet at that said mid-point?
If the lights from A and B arrive together, as a single event and be so measured by an observer at rest at the mid points between event A and event B, would that not imply that they could be measured to be simultaneous, in any, indeed in all, inertial frames of reference?
However it must be at a different mid-point in each frame, as the stationary mid-point in one frame would be moving away in any other frame.
So the observer in any frame would declare that he was the only one who could determine simultaneity...

Now, this is my problem: what is it I am getting wrong here? That all seems so very clear and logical and no matter how many times I have gone over it, I cannot see it!
Please help me and explain it?

 

[Mentz114]

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?

(my emphasis)
Events are events. They exist, therefore they exist in all frames. If I meet someone at the station at 8pm that is an unchangeable physical fact. My worldline intersected with the other partys worldline. Everyone will agree that we met, furthermore everyone will agree that our clocks read 8pm when we met.

 

[Nugatory]

those two events A and B would occur in every inertial frame of reference?

Some of your confusion may be that you're thinking that things happen "in" frames. They don't. They happen, and then if it is convenient we assign them time and space coordinates, and a frame is just an arbitrarily chosen convention for assigning these coordinates. When you hear someone saying something like "this event happened in frame F at time t and position x", that's a convenient but sloppy shorthand for the more precise "There's this event. We can choose any frame we want to assign time and space coordinates to that event, and if we choose frame F, we'll end up assigning time coordinate t and space coordinate x to it, but of course if we had chosen a different frame we would have assigned different x and t values".

An event is like a pencil mark on a piece of paper - it's there whether we draw a set of coordinate axes on the paper or not. If it's convenient to describe the location of the point using x and y coordinates, we can draw an x-axis and a y-axis on the sheet of paper as well to create a "frame", and we can choose to draw the axes anywhere we please.

 

[Dale]

And that the lights emitted at event A and event B would meet at that said mid-point?

This can only be true in one frame. In all other frames the light emitted at A and B would not meet at the midpoint.

 

[Ibix]

Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.

 

[Grimble]

Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.

But the flashes of light are events and therefore are fixed. As for timing, if they are triggered by light from a point in the centre of your rod, then they must be simultaneous?
For even measured from a frame that is moving the distances traveled by the light are the same, the movement of the lights are the same, or are we saying that if a light is shone at the centre of the rod and reflected back by two equidistant mirrors, that in one frame alone (where the rod is stationary) will the reflections arrive simultaneously?
And, if the simultaneous arrival of the reflected lights set of a signal, that would only occur in one frame?

Events are events. They exist, therefore they exist in all frames. If I meet someone at the station at 8pm that is an unchangeable physical fact. My worldline intersected with the other partys worldline. Everyone will agree that we met, furthermore everyone will agree that our clocks read 8pm when we met.

Some of your confusion may be that you're thinking that things happen "in" frames. They don't. They happen, and then if it is convenient we assign them time and space coordinates, and a frame is just an arbitrarily chosen convention for assigning these coordinates. When you hear someone saying something like "this event happened in frame F at time t and position x", that's a convenient but sloppy shorthand for the more precise "There's this event. We can choose any frame we want to assign time and space coordinates to that event, and if we choose frame F, we'll end up assigning time coordinate t and space coordinate x to it, but of course if we had chosen a different frame we would have assigned different x and t values".

That is how I see it. A Frame of Reference is merely one particular view of Spacetime. So what is in Spacetime is there in all Frames of Reference, it is only the coordinates that will differ.

This can only be true in one frame.

So that would have to be the 'Privileged Frame' that was so abhorred by Einstein?
You see those two events, the emission of the flashes of lightning, being events cannot be moving. They are moments/points in time whose positions are fixed in every frame - only the coordinates are unique. So in any frame there may be an observer positioned midway between those events, who is STATIONARY in that frame of reference and who therefore may measure simultaneity.
The only question then is whether those two events happen at the same time. But if they were triggered by light traveling equal distances from a single event, and in any frame the distances will be equal and the speed of light c, how can they not be at the same time?

 

[harrylin]

[..] if they are triggered by light from a point in the centre of your rod [..] or are we saying that if a light is shone at the centre of the rod and reflected back by two equidistant mirrors, that in one frame alone (where the rod is stationary) will the reflections arrive simultaneously? [..]

Yes that is exactly what Einstein's example shows; it directly follows from the assumption (in fact it's a convention) that the speed of light is c relative to the chosen reference reference system. With that assumption, if the rod is moving then already the trigger light cannot reach both ends at the same time - it's as simple as that.

PS Note that it is quite different if you choose the opposite and simpler example of a light source in the middle of the rod, with mirrors at each end.
Then the depart of the light pulse in both directions is a single event, that is, we designate to it a single (x,y,z,t) and therefore also a single (x',y',z',t'). It is a useful exercise to verify that according to any inertial reference system the reflected rays will arrive simultaneously back at the centre. That is necessary, for that is also a single event.

 

[Dale]

So that would have to be the 'Privileged Frame' that was so abhorred by Einstein?

No. The laws of physics are the same. Nothing priveliged except that A and B happen to be simultaneous.

In spacetime A and B are events, and the light from A and B are worldlines. The light worldlines meet at an event C. The midpoint between A and B is a coordinate line (that is, a line containing a different set of events in each frame), and there is no guarantee that it intersects with event C. That only happens in the frame where A and B are simultaneous.

But if they were triggered by light traveling equal distances from a single event,

This is never true. A and B are two separate events, even in the frame where they are simultaneous.

 

[Grimble]

PS Note that it is quite different if you choose the opposite and simpler example of a light source in the middle of the rod, with mirrors at each end.
Then the depart of the light pulse in both directions is a single event, that is, we designate to it a single (x,y,z,t) and therefore also a single (x',y',z',t'). It is a useful exercise to verify that according to any inertial reference system the reflected rays will arrive simultaneously back at the centre. That is necessary, for that is also a single event.

But that IS exactly what I described!

 

[harrylin]

But that IS exactly what I described!

Then I misunderstood what you meant with "shone at"; I misunderstood it in the sense of "firing at" while you meant "at the position of". In a second reading I see that it could not have meant what I first thought.
Glad to see that one solved.

 

[Ibix]

In harrylin's example, all frames agree that the pulses meet again at the center of the rod. They do not agree on where the center of the rod is with respect to where it was when the pulses were emitted or reflected.

 

[harrylin]

There remains something funny with "observer who is STATIONARY in that frame of reference and who therefore may measure simultaneity." Observers (sensors) can be moving or not, that doesn't really matter; being stationary in the used reference system merely simplifies things. For example GPS satellites do a great job without ever being stationary in the ECI frame which they relate to.

 

[Grimble]

Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.

Yes, the light sources will be moving, but the events where the light is emitted from those source, being events are fixed, regardless of where those sources go subsequently.
'Halfway between' means the midpoint of a line drawn between those events.

In harrylin's example, all frames agree that the pulses meet again at the center of the rod. They do not agree on where the center of the rod is with respect to where it was when the pulses were emitted or reflected.

But it is not the centre of the rod that it is observers who are at rest at the midpoint between the Events where the lights were emitted, the centre of the rod may no longer be at that point, it will have moved away.

 

[Janus]

Thank you, yes that is correct. I just wanted to be sure that I was understanding this.

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
And there would be a set of spatial coordinates that would define a fixed point in each and every inertial frame of reference that was spatially mid-way between points A and B?
And that the lights emitted at event A and event B would meet at that said mid-point?
If the lights from A and B arrive together, as a single event and be so measured by an observer at rest at the mid points between event A and event B, would that not imply that they could be measured to be simultaneous, in any, indeed in all, inertial frames of reference?
However it must be at a different mid-point in each frame, as the stationary mid-point in one frame would be moving away in any other frame.
So the observer in any frame would declare that he was the only one who could determine simultaneity...

Now, this is my problem: what is it I am getting wrong here? That all seems so very clear and logical and no matter how many times I have gone over it, I cannot see it!
Please help me and explain it?

Maybe this will help.

Consider the following scenario: You have an observer standing along a track and an observer on a moving train car. Two flashes of light originate at points an equal distance from the track observer. The flashes arrive at the track observer at the same instant that the train car passes him. Thus both observers see the flashes simultaneously. these events look like this according to anyone at rest with respect to the tracks.

Now we consider the same events according to the observer on the car. Keeping in mind, the postulate that the speed of light is invariant, meaning that he must measure the speed of each light flash relative to himself as being the same. In other words IF he considered the sources of the flashes as being an equal distance from him at the moment the flashes originated and and he sees the flashes simultaneously, then he can conclude that the flashes originated simultaneously.
However, while he does see the flashes simultaneously, he cannot say that he was an equal distance from the points of origins when the flashes started. He is only at the midpoint between the origins when he sees the flashes. At any moment prior to this he is closer to the left flash's origin point than he is to the right flash's. Since the flashes have to have originated at some time prior to his seeing them, he cannot be at the midpoint when either of them originated, and the flashes, according to him could not have originated simultaneously. Thus event would occur like this according to anyone at rest with respect to the train car.

The flashes still meet at the point where the two observers pass each other, but the flashes do not originate at the same time. If we carry this a bit further we can apply it to the Einstein train example. Here we have observers at the midpoint of a train and on the embankment. Unlike the above example, the flashes originate when the observers are next to each other according to anyone at rest with respect to the embankment. The flashes also originate where the ends of the train and the red dots meet. Thus in the following animation, when the front of the train reaches the left dot a flash is produced and when the rear of the train reaches the right dot a flash is produced.

Thus from the embankment, events occur like this:

Note that the train observer runs into the left flash before the right flash catches up to him.

Now we consider the same events as they occur in according to the train. The first thing to note, is that as measured from the embankment frame, the train is moving and thus has undergone length contraction, in that its measured length is shorter than what it would be as measured by the train itself. It is this length contracted train that fits between the red dots and allows the ends of the train to hit the red dots simultaneously in this frame.

In the train's frame, the train is not moving an not length contracted, but instead, it is the tracks and embankment that is moving and length contracted. So not only does the train measure its own length as being longer than that measured by the embankment frame, but it also measures the distance between the red dots as being shorter than that as measured by the frame. As a result, the train does not, in this frame fit between the two red dots. the front of the train reaches the left dot before the rear reaches the right dot.

For events to be the same in both frames, (such as the flashes originating when the dots and ends of the train align.) the flashes cannot originate at the same time in the train frame and we get the following sequence of the events according to anyone at rest with respect to the train.

The flashes originate at different times, but still when the train ends and red dots meet. The train observer sees the flashes at different times, since he is an equal distance from the train ends and they originated at different times. The embankment observer see the flashes at the same time (just like he did according to his own rest frame).

In fact, every event in both frames matches up perfectly. For example, the same railway car is next to the embankment observer when he sees the flashes in both animations and the train observer is next to the same point of the tracks when he sees each flash in both animations. All the events are the same, it is just that the frames don't agree as to the timing of these events.

 

[Dale]

'Halfway between' means the midpoint of a line drawn between those events.

But that is a different set of events for every reference frame. And there is only one frame where the event C is at the midpoint.