Interference Pattern versus SR

[Adel Makram]

I posted this thought experiment in a previous thread before 4 months or so, but I would like to reiterate it now:
A frame of reference (FOR) has double slits moves relative to a ground FOR. Let`s make an arrangement so that when the 2 ends of FORs coincide, 2 small slits of moving FOR are opened at the same time relative to the ground observer for a brief moment to allow just 2 photons to enter from an electromagnetic source put on the opposite side of him. Let `s make the distance between the 2 slit small enough comparable to the wave-length of the photons to cause an interference pattern.
For the ground observer, he sees 2 slits open at the same time and therefore the 2 photons entering the 2 slits and create an interference pattern on a screen on his frame.
But according to SR interpretation, the moving FOR`s observer sees the front slit opens for a brief moment and then shuts before the rear one opens,,, so at one time, only one slit opens and therefore no interference pattern could ever occur. But when he looks at the screen from his window, he will see an interference pattern on the ground screen.
Can the train observer now explain why this interference pattern occurs when just the slits open one at a time?

 

[Mentz114]

Nothing strange will be observed because interference depends on phase, which is a scalar and therefore Lorentz invariant. Both observers will see the same pattern ( or non-pattern ).

 

[Adel Makram]

Nothing strange will be observed because interference depends on phase, which is a scalar and therefore Lorentz invariant. Both observers will see the same pattern ( or non-pattern ).

How will the moving FOR see the same pattern as the ground one! the idea is the interference pattern will not happen unless the 2 slits open at the same time. If this the case for the ground one, it will not be the case for the moving one, so he will not see the pattern. The moving FOR will document the near slit open and shut before the rear one, so at any time only one slit open and no interference pattern occurs !

 

[Mentz114]

Relative simultaneity. If two events are at the same time in one frame they are not necessarily simultaneous in another.

Interference patterns are cause by the coming together of worldlines, and so are made up of events. The existence of an event cannot be altered by a Lorentz transformation.

 

[Adel Makram]

Relative simultaneity. If two events are at the same time in one frame they are not necessarily simultaneous in another.

Interference patterns are cause by the coming together of worldlines, and so are made up of events. The existence of an event cannot be altered by a Lorentz transformation.

but the sequence of events can! at least according to SR. if 2 slits open simultaneously according to ground observer, they will not according to the slits observer

 

[Mentz114]

but the sequence of events can! at least according to SR. if 2 slits open simultaneously according to ground observer, they will not according to the slits observer

Your point being ? You've just restated my first sentence. It is not possible to construct a causal paradox based on inteferference for the reasons given. If you think you have, you've made a mistake.

 

[Adel Makram]

if the 2 slits open one at a time relative to slit`s FOR, the slit observer will not see the interference pattern from his window. There will not be even a wave-phase to talk about! MY question, how the physical reality, the interference pattern, can be matter of conflict between the 2 observers?

 

[Adel Makram]

Y It is not possible to construct a causal paradox based on inteferference for the reasons given. If you think you have, you've made a mistake.

I think it is possible to create a paradox based on my experiment,,,

 

[Mentz114]

if the 2 slits open one at a time relative to slit`s FOR, the slit observer will not see the interference pattern from his window. There will not be even a wave-phase to talk about! MY question, how the physical reality, the interference pattern, can be matter of conflict between the 2 observers?

There won't be any conflict between the observers.

Certainly if the slits and the screen are oriented at right angles to the direction of relative motion, there will be no relativistic effects.

 

[Adel Makram]

the period of opening of slits can be chosen arbitrarily so as to make the front slit opens and closed before the back slit open relative to the slit observer. But for the ground observer, they are opened at the same time

 

[Doc Al]

the period of opening of slits can be chosen arbitrarily so as to make the front slit opens and closed before the back slit open relative to the slit observer. But for the ground observer, they are opened at the same time

What matters is not whether the slits are open at the same time or not, but whether the light from each slit reaches the same spot on the screen at the same time. All observers must agree on that.

 

[Mentz114]

the period of opening of slits can be chosen arbitrarily so as to make the front slit opens and closed before the back slit open relative to the slit observer. But for the ground observer, they are opened at the same time

Please draw a diagram of the setup and someone will show you why there's no paradox or conflict.

 

[Adel Makram]

What matters is not whether the slits are open at the same time or not, but whether the light from each slit reaches the same spot on the screen at the same time. All observers must agree on that.

They will reach at the same time, that is why there is an interference pattern, which I am not in doubt for. But, according to the quantum M interpretation, it is just a single slit experiment for the slit observer

 

[Adel Makram]

for a slit observer, if there is some way to label which slit the photon enters, then no interference pattern will form

 

[bahamagreen]

I like this thought problem. Let's see if we can find the problem (the incorrect assumptions) by breaking down the question some more...

The way it is presented is that the stationary FOR observes photons going through both slits together and sees an interference pattern; then the question is if the moving FOR would observe that pattern after seeing the photons pass through single slits in sequence.

Since no objection has been made to the problem, may we assume that the assumptions there are correct - that the stationary observer will see photons go through two slits and make a pattern, and the moving FOR will see photons go through single slits?

One assumption is that a FOR should expect to see a pattern if the observation is that the photons go through two slits at the same time. Is this correct? Is this not the result of experiments to examine this very thing?

Another assumption is that the relative motion of a FOR does not impact the result... all FOR will see a pattern if they see the two photons go through two slits at the same time. Is this correct? Is this not what SR expects?

Another assumption is that a FOR should not expect to see the pattern is it is observed that the photons go through single slits in sequence. Is this correct? Is this not the observation from experiment?

Another assumption is that the relative motion of the FOR has no effect on observing the lack of pattern if the photons go through single slits in sequence. Is this also not correct?

So the assumptions so far are:
For any FOR one should expect to see a pattern if one observes the photons going through two slits together, and one does not expect to see a pattern if one observes the photons going through single slits in sequence.

The paradox is that the two FOR have opposing observations about whether the two photons go through a pair of slits together or whether they go through single slits in series. The paradox is that these two observations and their resulting expectation for a pattern or not are different... that based on the assumptions above, one will see a pattern and the other will not.

Menzt114 is stating that both will see the same thing (pattern or not pattern), but has not stated which of these is the observed result. Either way maintains the paradox if the above assumptions hold.

The presentation could be made the other way around... it could be stated first that the moving FOR sees the photons pass through single slits in sequence thereby seeing no pattern, but then the paradox is that the stationary FOR will see them go through two slits at the same time and will see a pattern.

Either way, the question is all about how both must see the same thing... and exactly which observation will be that same thing, the pattern or no pattern? And then, why the moving FOR should see sequential photons through single slits make a pattern, or why the stationary FOR should see photons go through two slits together and not see a pattern.

Menzt144, if both must see the same thing (pattern or no pattern), which is it, based on the presentation of the problem? As just above, certainly the order with which the observations are presented cannot have any bearing on the answer?

What assumptions are incorrect? Or are there some incorrect assumptions that have not been revealed yet?

Reversing the problem does not seem to make a difference (the moving FOR sees photons go through two slits together and result in a pattern, the stationary FOR sees photons go through single slits in sequence and sees no pattern).

 

[Dale]

They will reach at the same time, that is why there is an interference pattern, which I am not in doubt for. But, according to the quantum M interpretation, it is just a single slit experiment for the slit observer

Closing a slit doesn't instantaneously change the wave function over all space. Similarly, opening a slit doesn't instantaneously change the wave function over all space. Saying "QM" doesn't magically change the problem significantly. Doc Al's post 11 is applicable.

 

[Mentz114]

Mentz114, if both must see the same thing (pattern or no pattern), which is it, based on the presentation of the problem? As just above, certainly the order with which the observations are presented cannot have any bearing on the answer?

I don't know what the observers will see because I can't work out the arrangement of slits, screen and emitters. But all observers will see the same pattern, whether it is dots or bars.

 

[Adel Makram]

 

[Mentz114]

Only one pair of slits is necessary. Have a look at this

http://en.wikipedia.org/wiki/Double-slit_experiment

If the ground observer sees an interference pattern on his screen, then so will the train observer.

 

[Adel Makram]

Only one pair of slits is necessary. Have a look at this

http://en.wikipedia.org/wiki/Double-slit_experiment

If the ground observer sees an interference pattern on his screen, then so will the train observer.

again my question is not whether the slit-observer sees a pattern or not! but how he interprets what he sees if he consider the experiment a single-slit experiment not a double one?

 

[Mentz114]

again my question is not whether the slit-observer sees a pattern or not! but how he interprets what he sees if he consider the experiment a single-slit experiment not a double one?

I don't understand the setup. Why are there 4 slits ? How are the slits opening and closing ?

However, from the principles of relativity and simple ray-tracing, whatever is seen by one observer ( say the one at rest wrt the slits) will be seen by all.

The reason is that if a beam goes through a slit in one frame it will do so in all frames, because the coincidence in space and time of the beam and the slit is absolutely invariant. This applies to any arrangement of light paths, slits screens and whatever.

How many times do you have to be told ?

 

[Adel Makram]

you are responding to a question: will the slit-observer see a pattern or no? You are still did not get the point. sorry

 

[Adel Makram]

There is a difference between the physical phenomena and the observer interpretation. As in Einstein thought experiment. The observer in the middle of the moving train receives the light signal from the front end before the back end. This is a physical phenomena which is also seen by the ground observer. The interpretation is different, for the train observer, the event of light striking the front end happens before the back end. while for the ground observer, the mid-train observer moves toward the light from the front end faster than from the back end. that is the interpretation which causes the principle of relativity of simultaneity ,,, just try to apply the same logic in my experiment.

 

[Adel Makram]

and forget about the 2 slits near the source, let's make them just one big window :)

 

[Adel Makram]

the difference between the 2 thought experiments, is that in the Einstein`s one, the time of events in both ends of the train does not depend on the physical phenomena of receiving the light from both ends by the mid-train observer. But in y experiment, the double slit experiment tells us that unless you let the 2 slits open at the same time, you will never get a pattern. The pattern is a fixed phenomena but the opening of 2 slits at the same time is different. It is a single slit experiment for the slit-observer, I hope you get it this time

 

[Adel Makram]

https://www.physicsforums.com/attachment.php?attachmentid=42989&stc=1&d=1327254325

 

[Mentz114]

the difference between the 2 thought experiments, is that in the Einstein`s one, the time of events in both ends of the train does not depend on the physical phenomena of receiving the light from both ends by the mid-train observer. But in y experiment, the double slit experiment tells us that unless you let the 2 slits open at the same time, you will never get a pattern. The pattern is a fixed phenomena but the opening of 2 slits at the same time is different. It is a single slit experiment for the slit-observer, I hope you get it this time

The interference pattern needs only that the slits are both open when the light reaches them, not on the perception of their simultaneous opening by another observer.

Thus, the pattern will be there whether or not an observer sees the slits open at the same time.

 

[Adel Makram]

The interference pattern needs only that the slits are both open when the light reaches them, not on the perception of their simultaneous opening by another observer.

Thus, the pattern will be there whether or not an observer sees the slits open at the same time.

I guess the slit-observer will disagree with you about that statement. For him, the pattern will form only when the 2 slits open at the same time

 

[Adel Makram]

Put yourself instead of the slit-observer. You see A open and closed before B open. You will also see a pattern on an outer screen from your window. For you, the slit is an inertial FOR and the law of physics will apply there perfectly. Would you conceive the pattern is formed on the outer screen because may be some other observer sees the 2 slits open at the same time and therefore the pattern forms?

 

[Mentz114]

I guess the slit-observer will disagree with you about that statement. For him, the pattern will form only when the 2 slits open at the same time

The diagram shows a 2 slit setup. P and Q are bright spots, where the interfering beams are in phase. A and B are observers. The slits open for a small time simultaneously from A's point of view. Observer B does not see the slits open at the same time¹. Will B see the same pattern as A, or something different ?

Nothing is moving in this setup.

[1] if the slits are open for time T, then if T< (SB-S'B)/c then B will see one slit open and close before the other opens.

 

[Adel Makram]

yes B will see what A saw. But as long as A and B are not moving, they must see the 2 slits open at the same time

 

[Mentz114]

yes B will see what A saw. But as long as A and B are not moving, they must see the 2 slits open at the same time

No, B will not see the slits open at the same time because of his asymetrical position.

 

[Adel Makram]

but the slit observer is in the middle of the train inbetween the 2 slits and must be in symmetric location

 

[Mentz114]

Forget relativity and relativistic effects. Do you agree that both the observers will see the same pattern in my setup, despite the fact that B might not see the slits open at the same time ?

 

[Adel Makram]

B knows that he is not in epicenter of the slits, in fact this the idea behind the interference pattern, that the different time of arrival of top and bottom of the wave depends on their location on the screen

 

[Mentz114]

I assume you agree that B will see the same pattern as A - but your reason is irrelevant.

Let's go back to first principles of relativity. If two objects collide, this means their world lines have coincided.

Do you agree that no Lorentz transformation can change this fact ? This means that any observer will agree that the objects collided.

If light collides with a screen and makes a bright spot, then this is also a coincidence of world lines, and will be seen as a bright spot in all frames.

Similarly with two coherent beams colliding with a screen, given that the phase difference is also invariant, the result of the interference will be the same in all frames.

 

[my_wan]

Interference patterns as a result of moving frames is well established with the Sagnac effect, though technically there is acceleration involved in that case. However, in the particular arrangement you described (if I visualized it properly) the "ground screen" IS the observer in question which determines what all observers see. The motion of the screen relative to the slits determines the outcome, and not any motion of the observer looking at the screen.

When you are think about reference frames and observers you cannot think of yourself as an observer that determines what the result look like. Any recording device, that records results, is the observer that determines the outcome for all observers of an outcome determined by that recording.

 

[jartsa]

In this experiment a wide photon is needed. How is a wide photon created? One side first, then the other side, as seen from some suitable FOR.

 

[Dale]

But in y experiment, the double slit experiment tells us that unless you let the 2 slits open at the same time, you will never get a pattern.

I don't believe there is any experiment that shows this. There are experiments that show that two slits which have been open for a long time show interference, and there are experiments which show that if one slit has been open a long time there is no interference, but I don't think there are any which show that if you have two slits and you quickly close one that you instantaneously lose the interference. And that statement is certainly not justified by theory.

 

[JDoolin]

I posted this thought experiment in a previous thread before 4 months or so, but I would like to reiterate it now:
A frame of reference (FOR) has double slits moves relative to a ground FOR. Let`s make an arrangement so that when the 2 ends of FORs coincide, 2 small slits of moving FOR are opened at the same time relative to the ground observer for a brief moment to allow just 2 photons to enter from an electromagnetic source put on the opposite side of him. Let `s make the distance between the 2 slit small enough comparable to the wave-length of the photons to cause an interference pattern.
For the ground observer, he sees 2 slits open at the same time and therefore the 2 photons entering the 2 slits and create an interference pattern on a screen on his frame.
But according to SR interpretation, the moving FOR`s observer sees the front slit opens for a brief moment and then shuts before the rear one opens,,, so at one time, only one slit opens and therefore no interference pattern could ever occur. But when he looks at the screen from his window, he will see an interference pattern on the ground screen.
Can the train observer now explain why this interference pattern occurs when just the slits open one at a time?

Interesting question, I think. Are you looking for the right answer or the right question? I am really curious about this question as well; It's puzzling me, as questions should. I don't exactly know what the question is; I'm looking for the correct question. But I think it has something to do with how does a diffraction pattern Lorentz Transform.

The invariant parameter of a photon, as near as I can tell, is proper-time between source event and destination event divided by frequency. τ/ν; though I'm basing that off a calculation I haven't finished.

I don't think just because something is a scalar, it is Lorentz Invariant. There are some particular calculated quantities that happen to be invariant. Maybe I don't understand what is meant by scalar... I was taught anything that just had a number and unit was a scalar. But meters aren't invariant, seconds aren't invariant. Maybe the \small \sqrt{c^2\Delta t^2-\Delta x^2} is invariant, but that's not because it's a scalar, but because it gives a way of numbering consecutive hyperbolas and Lorentz Transformation keeps things on the same hyperbola.

The mysterious thing is, the number of wavelengths of the photon between source and destination is NOT invariant. It goes right down to zero, if you use a reference frame that follows the photon. Because what happens with a photon? It is a ZERO space-time interval. A null interval. There are no events between the source and destination.

So HOW does it interfere? How does it interact with those two slits, or diffraction grating? It must be interacting laterally. Perpendicular to its direction of motion! Anyway, like I said, I don't know the answer; I'm still searching for the right question.

 

[Adel Makram]

The idea behind my experiment lies not just on the expected classical explanation based on the light transmission but instead on the expected puzzle enforced by Quantum Mechanics Interpretation. If the interference pattern forms because of the invariance of wave-phase or the coincidence of 2 worlds line representing the photons and the screen as in Mentz work, the issue might be easier, yet not solved still

The challenge is what QM predicts that no pattern could ever be formed if one knows which slit the photon passes through. This can be inferred from the famous Wheeler`s Delayed Choice Experiment or Delayed Choice Quantum Eraser Experiment which both of them have been confirmed. Seeing the slit-A opens and closed before slit-B opens relative to the slit-observer is the exact way to know which slit allows the light photon to go through and consequently will destroy the pattern. There is no counterpart in the classical physics, therefore attempt to explain the pattern based on light transmission is either incomplete or wrong

Furthermore, the QM allows the experiment to be done on electrons for example and still has the same result regardless the need to use photons because of its possible exploitation in synchronization of the time the slit open

 

[Dale]

Seeing the slit-A opens and closed before slit-B opens relative to the slit-observer is the exact way to know which slit allows the light photon to go through and consequently will destroy the pattern.

I don't think that this is correct. Can you derive the wavefunction and show this?

 

[jartsa]

In this experiment a wide photon is needed. How is a wide photon created? One side first, then the other side, as seen from some suitable FOR.

Can you follow my reasoning? The first created side goes through the first opened slit ... and so on.

Now a narrow photon: This photon meets the slits at a small angle, one side of the photon goes through a hole, other side would go through the other hole that is further away, if the hole wasn't moving even further away.

Now a photon that is born narrow and then widens: This photon's different sides are directed towards different slits. They have different distances to travel too.

 

[JDoolin]

The idea behind my experiment lies not just on the expected classical explanation based on the light transmission but instead on the expected puzzle enforced by Quantum Mechanics Interpretation. If the interference pattern forms because of the invariance of wave-phase or the coincidence of 2 worlds line representing the photons and the screen as in Mentz work, the issue might be easier, yet not solved still

The challenge is what QM predicts that no pattern could ever be formed if one knows which slit the photon passes through. This can be inferred from the famous Wheeler`s Delayed Choice Experiment or Delayed Choice Quantum Eraser Experiment which both of them have been confirmed. Seeing the slit-A opens and closed before slit-B opens relative to the slit-observer is the exact way to know which slit allows the light photon to go through and consequently will destroy the pattern. There is no counterpart in the classical physics, therefore attempt to explain the pattern based on light transmission is either incomplete or wrong

Furthermore, the QM allows the experiment to be done on electrons for example and still has the same result regardless the need to use photons because of its possible exploitation in synchronization of the time the slit open

What is meant by wave-phase?

You can't really view a photon from the side, because such viewing would imply there are multiple events associated with a photon. I think that's part of the problem of trying to deal with the situation "classically"

A photon is only "observed" twice; once at the emitter and once at the destination. Treating it as an extended object existing at multiple points in space at the same time would be a non-null space-like spacetime interval. But the photon only follows a null spacetime interval. It does not exist as an extended object along its own path. It can only exist as an extended object perpendicular to its path.

However the classical treatment of interference patterns, conceptually, always treats a wave as an extended object along its path, having peaks and troughs that superimpose on each other, creating an interference pattern.

We need to look at two other formulas; a formula for the time dilation effect, and the formula for relativistic doppler effect. (And also determine whether we trust the equations, i.e. how to interpret the meaning's of the variables.) The relativistic doppler effect http://en.wikipedia.org/wiki/Relativistic_Doppler_effect is
f_0=\gamma (1-\beta)f_s=\sqrt{\frac{1-\beta}{1+\beta}}f_s
where f_s = frequency of the photon in the source's reference frame and f_0 =frequency of the photon in the receiver's reference frame.

An observer traveling toward the source will see a greater frequency. An observer traveling away from the source will see a lesser frequency.

While the time dilation factor http://en.wikipedia.org/wiki/Time_dilation is
\frac{\Delta t&#039;}{\Delta t}=\frac{1}{\sqrt{1-v^2/c^2}}\equiv \gamma
where Δt'=the time passed by the receiver during the photon's transmission and Δt = the time passed by the emitter during the photon's transmission.

An observer traveling along a vector from the receiver toward the source will see a greater time interval. While an observer traveling in the opposite direction will see a shorter time interval.

The point is, though, that the frequency goes up as the distance goes up, and the frequency goes down as the distance goes down. So the number of wavelegths is not invariant.

If there is a Lorentz-invariance, it's in a quantity like \tau/\nu where the frequency and time both increase or decrease together, and cancel out. The emitter and the receiver, moving at different speeds, will agree on the numerical value of this quotient, but they will not agree on the time interval, the frequency, or how many wavelengths there are.

I'm sure that this doesn't answer your question. But I think it is important to be specific about what, precisely, is the "invariance of wave-phase" and think fairly deeply about the topic. Because it is a different thing for different types of waves.

 

[Mentz114]

The point is, though, that the frequency goes up as the distance goes up, and the frequency goes down as the distance goes down. So the number of wavelegths is not invariant.

For a light beam traveling from one observer to another, the number of wavelengths on the path is invariant. The diagrams show this happening from the perspective of both frames. It would be the same under an arbitrary LT.

The diagrams are coventional ST plots with time on the vertical axis ( you've used my program so you'll know how I made them )

 

[JDoolin]

For a light beam traveling from one observer to another, the number of wavelengths on the path is invariant. The diagrams show this happening from the perspective of both frames. It would be the same under an arbitrary LT.

The diagrams are coventional ST plots with time on the vertical axis ( you've used my program so you'll know how I made them )

Yes. I think I see what you are doing.

Are you showing 7 explicit events on the null-path interval, and how those events do not disappear when you do a Lorentz Transformation?

Are those events are tied to nodes and peaks of the wave-form of a single photon?

Or are those nodes and peaks marking the locations of 7 individual photons along the same null-interval?

 

[Mentz114]

Yes. I think I see what you are doing.

Are you showing 7 explicit events on the null-path interval, and how those events do not disappear when you do a Lorentz Transformation?

Are those events are tied to nodes and peaks of the wave-form of a single photon?

Or are those nodes and peaks marking the locations of 7 individual photons along the same null-interval?

No photons, please. We're doing classical wave optics. Yes, those events are equally spaced along the null direction representing complete wavelengths.

[edit] removed a non-sequitur. Phase must be invariant because it is a scalar.

 

[JDoolin]

No photons, please. We're doing classical wave optics.

With classical wave optics we treat light as waves, right? What do we call it when we treat light as particles? What do we call it when we treat light as neither wave, nor particle or both wave and particle?

Yes, those events are equally spaced along the null direction representing complete wavelengths.

When you place events equally spaced along the null direction, are those wavelengths? I don't think they are wavelengths. They seem like wavelengths at first, but...

A wavelength involves a span over a distance. A span over a distance implies a space-like interval. We have no space-like interval here. We have a null interval.

[edit] removed a non-sequitur. Phase must be invariant because it is a scalar.

What do you mean by scalar? Anything with a single number and a unit is a scalar, right? Aren't measurements of meters and seconds scalars? Aren't they ...variant? I'm using the Math 101 meaning of scalar here. Did the definition of scalar change for General Relativity?

-------------

Oh, I should be asking what you mean by phase. Because the interference pattern is going to be the same, whatever the reference frame, so wherever peaks are meeting, you get a bright line, and wherever peak meets trough, you get a dark line. This must be invariant, because you get the same bright and dark lines regardless of reference frame. Is this sort of what you mean by phase?

 

[Mentz114]

When you place events equally spaced along the null direction, are those wavelengths? I don't think they are wavelengths. They seem like wavelengths at first, but...
A wavelength involves a span over a distance. A span over a distance implies a space-like interval. We have no space-like interval here. We have a null interval.

The spatial separation ( look at the x-axis) between the events is equal ( and so is the temporal because its a null direction). There could be a set of observers at those events.

Oh, I should be asking what you mean by phase. Because the interference pattern is going to be the same, whatever the reference frame, so wherever peaks are meeting, you get a bright line, and wherever peak meets trough, you get a dark line. This must be invariant, because you get the same bright and dark lines regardless of reference frame. Is this sort of what you mean by phase?

That is exactly what I meant. Interference between waves is determined by phase differences. In the 2-slit experiment, the phase difference is because the path lengths of the 2 beams is different.

I think this thread is done. The OPs question has been answered by several people.

 

[JDoolin]

I think I've found a clearer way to make my point

The number of wavelengths between the source event and destination event are not associated with the events marked in red in the diagram.

The number of events marked in red would be invariant, but the number of wavelengths between emission event and absorption event are observer dependent; they're not really events.