Interference Pattern versus SR

[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'}{\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.

 

[JDoolin]

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.



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.

I think I understand what you meant now. The phases (peaks, troughs, etc.) of the wave each have their own worldlines, and where those phases intersect with each other, are events, which happen in all reference frames.

I see now the OP's question is basically answered, but I was still confused about what quantities are actually invariant, and trying to figure out what you meant by "scalars" being always invariant. I can see that an event that happens in one reference frame must happen in every reference frame, so an event that represents, for instance, the superposition of phases of a wave, and a wall must happen in every reference frame.

But I didn't understand for sure what is meant by "scalar," though I've seen it used like that in other places, too, i.e. "all scalar's are invariant." Now I see though. They did change the definition:

On Wikipedia http://en.wikipedia.org/wiki/Scalar_(physics )

It says: In physics, a scalar is a simple physical quantity that is not changed by coordinate system rotations or translations (in Newtonian mechanics), or by Lorentz transformations or space-time translations (in relativity).

So the number of wavelengths is not a scalar, the period is not a scalar, the distance is not a scalar, the time is not a scalar; I was trying to figure out what was a scalar?

The existence of the phase is Lorentz Invariant, but what numerical quantity about the phase is invariant? And now that I ask the question, I smack myself in the head, and say, AH, of course, the phase, itself, is a numerical quantity, somewhere between zero and 2*Pi.

So the phase of the wave-fronts meeting the surface are invariant. I agree with you, you are totally right.

 

[Mentz114]

Proof of phase invariance ?

Yes, you're right about my diagrams. I made a mistake for which I apologise. I also found my Doppler scripts which have a similar diagram to yours. Memory failure as well.

But I think I've found a mathematical proof.

We can define the phase difference between the sender and reciver like this

phase = ΔL/λ

where ΔL is the path difference and λ is the wavelength. The problem is that the proper length is zero, so we have to reparameterise with an affine parameter so that the path length L is not null. Suppose we define our path length ( with c = 1)

L² = t² + x²

A beam of light goes from P to Q with

P = < t₀, x₀, 0, 0> and Q=<t₀+dx, x₀+dx,0,0>

In this frame ΔL = √2 dx , from (Q-P).(Q-P).

Now we transform P and Q to boosted coords (t', x') and get

P' = <x₀γβ+t₀Y, t₀γβ+x₀Y, 0, 0> and
Q' = <(x₀+dx)γβ+(t₀+dx)Y, (t₀+dx)γβ+(x₀+dx)Y, 0, 0>

and the length L' ²= (Q' - P').(Q' - P') is

2dx² γ²(1+β)².

Now λ transorms according to the Doppler formula

λ' = λ √(γ²(1+β)²)

and so

λ/ΔL = λ'/ΔL'

This is not a general proof, but I'm working on it. Any help appreciated.

[edit] It looks like a good proof, so I made a neater PDF which is attached.

 

[PhilDSP]

Actually number of wavelengths is considered a scalar along with the period, distance and time. A vector will have both a magnitude and a direction. A scalar has a magnitude but no direction.

Velocity is a vector because it carries not only speed (the magnitude of the rate of movement) but the direction of movement as well. Speed is a scalar. What may seem confusing in performing a LT is that the scalar values don't change. But the location of the time-space event in which the scalar value applies may change in performing the LT. The scalar values get re-mapped to a different place and time in the new coordinates.

Edit: You could interpret "number of wavelengths" as wavenumber, which is a vector and which gives an irrational number count of periods of the wave within a unit length (in each orthogonal spatial direction)

http://en.wikipedia.org/wiki/Wave_number

 

[JDoolin]

Thanks, Mentz.

Thank you for providing the mathematical proof, though. It is helpful in understanding your meaning.

I would have expected the length and wavelength to be given as shown in this diagram...

I see you have defined, at least, the ΔL parameter in an interesting way I wouldn't have expected. You actually use ΔL^2 = Δx^2 + Δt^2.
I would have used just ΔL = Δx.

Edit: Whoops. You defined ΔL= k (Δx^2 + Δt^2.), as it is a 45-45-90 triangle, so k is calculable constant. My bad. I have it printed out, now; it's easier to read.

 

[JDoolin]

I am making myself late for work, maybe, but in equation 4 of your pdf, you have +/- signs. I'll verify myself when I have time, but are you sure you aren't turning them around?

Because you can see in the animation:

The wavelegth goes down as the ΔL goes up, even if we use YOUR definition for ΔL. So the quantity ΔL/λ can't be constant.

That is, unless you've also defined λ in a way I'm not familiar with.

 

[Adel Makram]

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

I don`t think so :)
As per your calculation about the invariance of phase difference ΔL/λ = ΔL`/λ`, the total path length that coming from the slit A is different from that path from slit B relative to the slit-observer. But they are always equal in length relative to the ground observer ( especially regarding the central peak ). Therefore, the ground observer will always sees the phase-difference between the 2 paths = 0 at the spots of light forming peaks ( the Central peak is the ideal one), while for the slit-observer, the 2 paths are different by amount depends on the distance between the 2 slits and the velocity of the ground

 

[Adel Makram]

And even if you consider the path of light from A is shortened than the distance between the location of slit A and the location of the peak on the screen because of light photon has to leave slit A earlier than slit B, then the point of calculating the path ΔL between the source ( slit A in this example) and the receiver ( peak on the screen) in your calculation becomes irrelevant

 

[Mentz114]

I don`t think so :)
As per your calculation about the invariance of phase difference ΔL/λ = ΔL`/λ`, the total path length that coming from the slit A is different from that path from slit B relative to the slit-observer. But they are always equal in length relative to the ground observer ( especially regarding the central peak ). Therefore, the ground observer will always sees the phase-difference between the 2 paths = 0 at the spots of light forming peaks ( the Central peak is the ideal one), while for the slit-observer, the 2 paths are different by amount depends on the distance between the 2 slits and the velocity of the ground

Yes, and the phase difference is the same for all observers.

And even if you consider the path of light from A is shortened than the distance between the location of slit A and the location of the peak on the screen because of light photon has to leave slit A earlier than slit B, then the point of calculating the path ΔL between the source ( slit A in this example) and the receiver ( peak on the screen) in your calculation becomes irrelevant

Path length is never irrelevant in interference calculations.

I'm done here.

[JD - PM me if you have a problem with the calculation]

 

[Adel Makram]

Path length is never irrelevant in interference calculations.

I'm done here.

[JD - PM me if you have a problem with the calculation]

true, therefore, the slit observer should considered that parameter when calculating the phase of 2 paths on the ground screen. But the 2 paths are different in length with no grantee that that difference will ensure the phase of the 2 paths are the same when reaching the ground screen

 

[Adel Makram]

The only thing the slit-observer may be sure of, is the phase of the 2 lights are the same at the time they are leaving the slits even if that time is not the same