How can we explain the behavior of interference patterns in different frames?

[Adel Makram]

A train is moving from the left to the right direction. There is a light source emitting 2 beams of lights toward 2 slits A and B at both ends of the train. The source is put near the slit A Than B.

A ground observer is watching the scene and sees that the 2 light rays reaching A and B simultaneously and in-phase to form a pattern.
However, the train observer sees that the light toward A takes a shorter path than light toward B which makes the ratio between those 2distance and the wave length of the light not necessarily the same in order that the 2 beams arrive in phase!
So how the phase become a physical invariant value?
https://www.physicsforums.com/attachment.php?attachmentid=46226&stc=1&d=1334428798

 

[Dale]

A ground observer is watching the scene and sees that the 2 light rays reaching A and B simultaneously and in-phase to form a pattern.

You don't get any interference pattern unless you recombine the beams somewhere.

 

[Adel Makram]

You don't get any interference pattern unless you recombine the beams somewhere.

The ground obs, does not need to recombine any thing as long as he sees the 2beams reaching at 2 slits simultaneously and in phase.

 

[Dale]

He does if he wants an interference pattern.

http://en.wikipedia.org/wiki/Interference_(wave_propagation)

"interference is a phenomenon in which two waves superimpose to form a resultant wave of greater or lower amplitude"

 

[Mentz114]

For the interference pattern to form, the rays must also reach in phase.

( from the PDF)
Not true. Changing the phase of one of the interfering beams just shifts the pattern. And you need to combine two sources to get interference, as DaleSpam has pointed out.

 

[Adel Makram]

He does if he wants an interference pattern.

http://en.wikipedia.org/wiki/Interference_(wave_propagation)

"interference is a phenomenon in which two waves superimpose to form a resultant wave of greater or lower amplitude"

The interference pattern needs 2 waves of same phases exits 2 slits and projecting onto a screen, all those cretieria are met for the ground observer!

 

[Dale]

The interference pattern needs 2 waves of same phases exits 2 slits and projecting onto a screen, all those cretieria are met for the ground observer!

Where is the screen that both beams are projecting onto?

 

[Mentz114]

The interference pattern needs 2 waves of same phases exits 2 slits and projecting onto a screen, all those cretieria are met for the ground observer!

Not true ! Will you stop talking nonsense, please. The waves going through interference slits do not have to be in phase ! And you don't have any slits in any case.

 

[Adel Makram]

Not true ! Will you stop talking nonsense, please. The waves going through interference slits do not have to be in phase ! And you don't have any slits in any case.

Before u accusing my words of being nonsense, you should ask more details about the experiment set up in order to understand it if it woes not clear from the first instance,,,,,, what do u mean by not having slits in any cases! The wave leaving slits must be in the same phase (wave anrounds or the ground FOR , no matter what will be the superposition on the ground screen,,, What appears on ground screen is a superposition of two waves of different phases, only on the screen, because of the relative difference between the two light paths,,, this is not my topic, my topic is clear from the first post! Will be the phase when leaving slits an invariant physical quantity ?

 

[Adel Makram]

Where is the screen that both beams are projecting onto?

In the ground

 

[Adel Makram]

( from the PDF)
Not true. Changing the phase of one of the interfering beams just shifts the pattern. And you need to combine two sources to get interference, as DaleSpam has pointed out.

What I meant by "reach" is reaching slits not screen of course

 

[Dale]

In the ground

Can you draw it, because it is not at all clear from your drawing. In fact, the ground isn't shown in your drawing. I don't know if this picture is looking from the side, the top, or the front. Also, the only way I can think to combine the beams given your arrangement is more like a mirror than like a slit, so I am not sure how that is going to work. You do understand that slits diffract, not reflect, right?

Also, Mentz114 is correct. It is not required that the two waves be in phase, merely that they be coherent. See the first paragraph of the Wikipedia link I posted earlier.

my topic is clear from the first post!

No, it wasn't. If it were we wouldn't be at post 12 still with no idea how there can even be interference here.

 

[yuiop]

For the sake of discussion here is a suggested set up which I think is in the spirit that the O.P. intended. At each end of the rocket we have a mirror at 45 degrees that deflect the light rays towards the ground. After reflecting off the mirrors, the light rays are passed through their respective slits in the lower side of the rocket. The ground is the screen. In the rest frame of the rocket the source is exactly one half wavelength forward of the centre of the rocket. In the rest frame of the rocket it does not matter that the ground is moving relative to the rocket as far as the interference pattern is concerned as it will appear exactly the same as if it was projected onto a co-moving screen.

Some complications to consider:

1)Doppler shift.
In the ground rest frame, light going to the rear of the rocket is red shifted and light going forward is blue shifted. If these were reflected straight back to the source the Doppler shifts would cancel out as the source and mirrors are co-moving, but when the rays are deflected at right angles there will probably be Doppler effects measured in the ground rest frame. The frequencies at the light rays pass through the slits should however be equal when measured in the rocket rest frame.

2)Aberration.
When light reflects of relativistic mirrors, the angle of the incident ray is not equal to the reflected ray relative to the norm of the mirror measured in the ground frame. Neither is the angle of the mirrors as seen in the ground frame the same as seen in the rocket frame. These effects are not too difficult to calculate. To transform the light paths as seen in the rocket frame to the ground frame, we only have to take relativistic aberration into account.

3)Interpretation.
In the rest frame of the rocket, differences in phase at the slits results in the interference pattern shifting relative to the rocket. This is fairly easy to measure. In the ground frame the interference pattern is moving relative to the ground. Measuring the shift of pattern is is difficult in this frame as the rocket will have moved during the time interval it takes the light to travel from the slits to the ground. Finding a meaningful reference might be difficult.

There are additional difficulties regarding the feasibility of whether any interference pattern is observed or not. Some texts state that photons only interfere with themselves and individual photons pass through both slits. This will be difficult to arrange in the rocket with the slits so far apart, but might be possible with entangled photons. The visibility of an interference pattern also requires the slits to be very close together so it is debatable whether or not the pattern would be visible even if it did form in the first place.

As you can see, this seemingly simple experiment is actually quite complex and may not be possible even in principle.

 

[ghwellsjr]

A train is moving from the left to the right direction. There is a light source emitting 2 beams of lights toward 2 slits A and B at both ends of the train. The source is put near the slit A Than B.

Are there a total of 4 slits, 2 at A and 2 at B, or are there a total of 2 slits, 1 at A and 1 at B?

And where does the light go after it passes through the slit(s) at each end of the train?

 

[Adel Makram]

Are there a total of 4 slits, 2 at A and 2 at B, or are there a total of 2 slits, 1 at A and 1 at B?

And where does the light go after it passes through the slit(s) at each end of the train?

There r only 2 slits, one at each end

Light goes to a ground screen

 

[Adel Makram]

Can you draw it, because it is not at all clear from your drawing. In fact, the ground isn't shown in your drawing. I don't know if this picture is looking from the side, the top, or the front. Also, the only way I can think to combine the beams given your arrangement is more like a mirror than like a slit, so I am not sure how that is going to work. You do understand that slits diffract, not reflect, right?

Also, Mentz114 is correct. It is not required that the two waves be in phase, merely that they be coherent. See the first paragraph of the Wikipedia link I posted earlier.

No, it wasn't. If it were we wouldn't be at post 12 still with no idea how there can even be interference here.

The blue line is the ground screen, the train moves in the right direction,,, it does not matter wether I put mirrors or the slit emission is done by diffraction https://www.physicsforums.com/attachment.php?attachmentid=46253&stc=1&d=1334500883

 

[Adel Makram]

3)Interpretation.
In the rest frame of the rocket, differences in phase at the slits results in the interference pattern shifting relative to the rocket. This is fairly easy to measure. In the ground frame the interference pattern is moving relative to the ground. Measuring the shift of pattern is is difficult in this frame as the rocket will have moved during the time interval it takes the light to travel from the slits to the ground. Finding a meaningful reference might be difficult.

There are additional difficulties regarding the feasibility of whether any interference pattern is observed or not. Some texts state that photons only interfere with themselves and individual photons pass through both slits. This will be difficult to arrange in the rocket with the slits so far apart, but might be possible with entangled photons. The visibility of an interference pattern also requires the slits to be very close together so it is debatable whether or not the pattern would be visible even if it did form in the first place.

1) we can arrange the set up by making the rocket /train moves in a quite large circle ( ignoring the effect of GR) and repeat the emission of light at exactly the same point over and over again to complete the pattern, or alternatively, by making a very long train and multiple slits replicating A and B so as to complete the pattern over many cycles
2) the wavelength can be chosen To match the distance between the 2 slits

 

[Dale]

The blue line is the ground screen, the train moves in the right direction,,, it does not matter wether I put mirrors or the slit emission is done by diffraction

Actually, you need both mirrors and slits for this configuration to work. So you could have the light go out from the source above the slits, then the mirrors reflect the light in and down to the slits, then the slits diffract it to the ground making an interference pattern.

OK, so now that I understand your scenario, what is your concern?

 

[Adel Makram]

OK, so now that I understand your scenario, what is your concern?

will the wave- phase when light exiting slits be an invariant physical value?

 

[Dale]

Yes. But simultaneity is not.

 

[Adel Makram]

Yes. But simultaneity is not.

I Know that simultaneity part, but Would u please explain the " yes"?

 

[Dale]

Yes, the phase of a signal at any given event is a relativistic invariant. I.e. all frames agree on the phase at any event.

 

[Adel Makram]

Yes, the phase of a signal at any given event is a relativistic invariant. I.e. all frames agree on the phase at any event.

I know the statement, I want the proof

 

[Mentz114]

I know the statement, I want the proof

I found this fairly convincing. If you have any questions, please ask.

 

[Dale]

I know the statement, I want the proof

Mentz114 has provided one proof. I prefer to simply note that (in units where c=1)

k^{\mu}=( \omega, k_x, k_y, k_z ) is a four-vector and
x^{\mu}= (t,x,y,z) is also a four-vector. So then
\phi=\omega t - k_x x - k_y y - k_z z = k_{\mu} x^{\mu} is manifestly a scalar.

But that does require previous knowledge of the wave four-vector.

 

[Adel Makram]

I found this fairly convincing. If you have any questions, please ask.

The ratio ΔL/λ is a direction-sensitive and uses a relativistic Doppler effect which depends on whether u apply it to slit A or B and naturally, the Lorentz factor cancels each other out. But for the train observer, the phase is a location sensitive which depends merely on the location of the source relative for the slits. For example, if I displace the location of the source by an amount = 1/2λ, of course i need to change the velocity of train for the 2 wave fronts to appear at the same time, still the phase is the same for him, but for the train one it differed by 1/2λ

Your calculation does not describe the whole picture

 

[Adel Makram]

One more thing, I think the pattern will not form any way in this particular set up. Although the phase is always equal at Slits A and B relative to the ground observer because the arrival of the wave front at the same time, yet the Doppler shift at slits makes the interaction of those waves of different wave length after leaving the slits not possible.
Still my concern not about the pattern but about the relative phase at A and B for the train observer!

 

[Mentz114]

The ratio ΔL/λ is a direction-sensitive and uses a relativistic Doppler effect which depends on whether u apply it to slit A or B and naturally, the Lorentz factor cancels each other out. But for the train observer, the phase is a location sensitive which depends merely on the location of the source relative for the slits. For example, if I displace the location of the source by an amount = 1/2λ, of course i need to change the velocity of train for the 2 wave fronts to appear at the same time, still ΔL/λ will not change, and the phase is the same for him, but for the train one it differed by 1/2λ

Your calculation does not describe the whole picture

I have no idea what you're talking about or what your problem is. I don't think you understand the concept of Lorentz invariance and I suggest you go and do some learning before you make ridiculous claims.

One thing we can be certain about is that if any observer sees a certain phase relationship ( like an interference pattern ) then all observers will see it.

[DaleSpam's proof is showing that the amplitude of a plane-wave at any point in spacetime ( i.e. phase) is an invariant.]

 

[Dale]

the Doppler shift at slits makes the interaction of those waves of different wave length after leaving the slits not possible.

The Doppler shift at the slits is not the only Doppler shift. There is also a Doppler shift at the source, at the mirror (you are probably including this one with the slit), and at the screen.

Still my concern not about the pattern but about the relative phase at A and B for the train observer!

Suppose that we set up a pair of clocks at each slit. Suppose further that one clock reads 12:00:00 when the phase is 0 at one slit and the other clock reads 12:00:01 when the phase is π at the other slit. Then, in all reference frames when the first clock reads 12:00:00 the phase is 0 at the first slit and when the other clock reads 12:00:01 the phase is π at the other slit. This is what it means for the phase to be a relativistic invariant.

 

[Adel Makram]

I have no idea what you're talking about or what your problem is. I don't think you understand the concept of Lorentz invariance and I suggest you go and do some learning before you make ridiculous claims.

One thing we can be certain about is that if any observer sees a certain phase relationship ( like an interference pattern ) then all observers will see it.

[DaleSpam's proof is showing that the amplitude of a plane-wave at any point in spacetime ( i.e. phase) is an invariant.]

You should think of the experiment set up in a more close up instead of doing general math not particularily applicable for every case. I Was not talking about the pattern at all, so stop reiterating it over and over

 

[Adel Makram]

The Doppler shift at the slits is not the only Doppler shift. There is also a Doppler shift at the source, at the mirror (you are probably including this one with the slit), and at the screen.

Suppose that we set up a pair of clocks at each slit. Suppose further that one clock reads 12:00:00 when the phase is 0 at one slit and the other clock reads 12:00:01 when the phase is π at the other slit. Then, in all reference frames when the first clock reads 12:00:00 the phase is 0 at the first slit and when the other clock reads 12:00:01 the phase is π at the other slit. This is what it means for the phase to be a relativistic invariant.

Ok, so the n phase shift at the other clock may depend on the relative path taken by the light to travel from the source toward the clock. But for that particular location of the source relative to the train, there is a velocity that an external observer can see the train with the 2 wave fronts arrive simultaneously with no n- shift!

 

[Dale]

the n phase shift

I am talking about phase, not phase shifts.

 

[Adel Makram]

I am talking about phase, not phase shifts.

The world lines of meeting between the light and the slits are invariant, so do the phase values and the phase shift ( or difference)!. So how the n read by one clock still be so on another FOR which should detect it as 0

 

[Dale]

Worldlines are usually not considered invariant. Worldlines transform as vectors, not scalars.

Note above, I am talking about phase, not phase shift. A phase shift is a difference between two phases, for example the difference between the phase at the emitter and the phase at the slit at the same time. Phase differences may not be invariant, particularly if the difference is defined using notions of simultaneity.

 

[Adel Makram]

Also, what if I change the color of the laser I use but keeping the location of the source,,, so for the ground observer, he still sees both wavefront arising simultaneously with a phase value of n for both waves,,, but the train one should see different phase value depending on the ratio between the lengths of the two paths/ λ

The detection of the wave front that defining the phase relative to the ground observer is always the same so long as he sees them simultaneously, though with different Doppler shift, but not the case for the train which the phase depends on the location of the source and the λ used

 

[Dale]

Also, what if I change the color of the laser I use but keeping the location of the source,,, so for the ground observer, he still sees both wavefront arising simultaneously with a phase value of n for both waves,,, but the train one should see different phase value depending on the ratio between the lengths of the two paths/ λ

Again, you are talking about phase shifts at the slits. The phase shift between the two slits at the same time is a value which has no physical signifcance whatsoever. The phase at slit A at cannot affect anything at slit B at the same time.

The only phase shift which has any physical significance is the phase shift between the two waves where they overlap (i.e. on the screen). That phase shift determines if there is a bright spot or a dark spot on the screen.

 

[Mentz114]

You should think of the experiment set up in a more close up

The experimental setup is irrelevant if you're only talking about phase.

... instead of doing general math not particularily applicable for every case.

Phase invariance is applicable to every case ! You should find out what invariance means.

I Was not talking about the pattern at all, so stop reiterating it over and over

You should stop trying to construct situations where you think this breaks down, because it never does. If you subject phase to a Lorentz transformation it remains the same.

 

[Adel Makram]

Again, you are talking about phase shifts at the slits. The phase shift between the two slits at the same time is a value which has no physical signifcance whatsoever. The phase at slit A at cannot affect anything at slit B at the same time.

The only phase shift which has any physical significance is the phase shift between the two waves where they overlap (i.e. on the screen). That phase shift determines if there is a bright spot or a dark spot on the screen.

I am talking about the absolute value of phase at each slit. As the wave front determines what phase will be and as the wave front reach both slits at the same time for the FORg, then they are same for him too. The case is different for FORt,,, the thing is that the light propagation determines what the phase will be for the ground observer, but the content of the wave which is the λ does so for the train one,,, Du u agree that the scenario can be seen as marbles ejected at both slits, let's say when the marble exit the slit, the phase is 1 and when there is no marble the phase is 0, so as long as the marbles reach the slots simultaneously relative the ground observer, he always see them in phase,,, 1 . For the train observer, he the relative location of the marble gun does not necessarily affirm the phases are always 1 at slits!

 

[Adel Makram]

The experimental setup is irrelevant if you're only talking about phase.

Phase invariance is applicable to every case ! You should find out what invariance means.

You should stop trying to construct situations where you think this breaks down, because it never does. If you subject phase to a Lorentz transformation it remains the same.

U are not think about the problem, u left it and kept talking about the math,,, not sure that discussion with u is ganna work

 

[Dale]

I am talking about the absolute value of phase at each slit.

This is phase, and it is frame invariant and has a physical significance.

As the wave front determines what phase will be and as the wave front reach both slits at the same time for the FORg, then they are same for him too. The case is different for FORt,,,

This is a phase difference at two spatially separated locations. You are correct that it is different. It is also frame variant and has no physical significance.

Du u agree that the scenario can be seen as marbles ejected at both slits, let's say when the marble exit the slit, the phase is 1 and when there is no marble the phase is 0,

Agreed, this is a phase (in weird units) and is frame invariant.

so as long as the marbles reach the slots simultaneously relative the ground observer, he always see them in phase,,, 1 . For the train observer, he the relative location of the marble gun does not necessarily affirm the phases are always 1 at slits!

Again, this is a frame variant phase difference. It has no physical significance.

 

[Dale]

U are not think about the problem, u left it and kept talking about the math,,, not sure that discussion with u is ganna work

Please use correct English. "U" is a letter, "you" is a person, "ganna" is not a word in English, and "are not think" should be "are not thinking". I realize that some of this may be a language barrier, and I try to make allowances, but if there is a language barrier then it is even more essential that you do the best you can to communicate clearly and correctly.

Mentz114 is correct, however, the whole point of doing a general derivation is that it applies generally. The details are not relevant.

 

[Adel Makram]

This is phase, and it is frame invariant and has a physical significance.

This is a phase difference at two spatially separated locations. You are correct that it is different. It is also frame variant and has no physical significance.

Agreed, this is a phase (in weird units) and is frame invariant.

Again, this is a frame variant phase difference. It has no physical significance.

I think I got ur point, for the phase of marbles to be determined, only twins marbles ejected from the gun at one time and exiting slits , though different times for the train, should determine what their phases will be. So it does not matter whether ratio L/λ is same or not at both direction in the train, as long as only the marbles ejected from the gun are taken into account

 

[Dale]

That isn't how I would say it, but as long as you get the point that is good.

I would say something more like it doesn't matter if the twin marbles ejected from the gun at one time arrive at the slits at the same time or not, but only if they arrive back at a given point on the screen at the same time (bright spot, constructive interference) or not (dark spot, destructive interference). There is no physical significance to their arriving at the slits at the same time or not.

 

[Adel Makram]

That isn't how I would say it, but as long as you get the point that is good.

I would say something more like it doesn't matter if the twin marbles ejected from the gun at one time arrive at the slits at the same time or not, but only if they arrive back at a given point on the screen at the same time (bright spot, constructive interference) or not (dark spot, destructive interference). There is no physical significance to their arriving at the slits at the same time or not.

This supports that Mentz calculation does not fit for the marbles-case. On contrarily, it complicated it by introducing L/λ,,, So, It is all about relativity of simultaneity

 

[Adel Makram]

That isn't how I would say it, but as long as you get the point that is good.

I would say something more like it doesn't matter if the twin marbles ejected from the gun at one time arrive at the slits at the same time or not, but only if they arrive back at a given point on the screen at the same time (bright spot, constructive interference) or not (dark spot, destructive interference). There is no physical significance to their arriving at the slits at the same time or not.

My concern was about the phase not about the pattern. The pattern just put into the experiment to provide a physical evidence for the interaction based on phase difference. But you can stop at the stage when light is just exiting slits

 

[Mentz114]

As the wave front determines what phase will be and as the wave front reach both slits at the same time for the FORg, then they are same for him too. The case is different for FORt

This supports that Mentz calculation does not fit for the marbles-case. On contrarily, it complicated it by introducing L/λ,,, So, It is all about relativity of simultaneity

Why should it fit marbles ? I was talking about light. It is not about simultaneity.

The phase is determined by the number of wavelengths between the source and receiver and does not depend on simultaneity. The number of wavelengths between source and receiver is invariant ( so is the number of marbles ).

 

[Adel Makram]

The phase is determined by the number of wavelengths between the source and receiver and does not depend on simultaneity. The number of wavelengths between source and receiver is invariant.

The relativity of simultaneity is what makes the 2 photons of the same phase emitted by the same source at one time leaving slits at the same time for FORg and at different time for FORt with the same phase.

L/λ applies for a continuous wave and does not fit for my example

 

[Mentz114]

The relativity of simultaneity is what makes the 2 photons of the same phase emitted by the same source at one time leaving slits at the same time for FORg and at different time for FORt with the same phase.

L/λ applies for a continuous wave and does not fit for my example

If λ is the (distance between marbles)+(diameter of marble) then L/λ is the number of marbles between the source and receiver. And all observers will see the same number of marbles.

 

[NotAName]

Wow, this conversation could really do with another drawing or some sort of simplification.

Adel, would it be fair to reduce your experiment to the following?

If a light emitter is placed in the center of a moving train with mirrors at each end, an observer on the train will detect the first wavefront of each arriving back at the center at the same time. Whereas an observer on the ground will not agree?

If you are asserting this, the answer is no. The observer on the ground will agree. Even in classical terms and substituting sound, you'll find that the rearward and forward beams both have to undergo the effect of both an upstream and downstream path.

If they reflect at a 45 degree angle, the effects of the angles cancel each other and you are back at the situation above.

Does that help?


If you would like to create an experiment which would be different between classical and relativistic and only be a first order effect unlike the michelson-morely I can suggest one for you.

 

[Adel Makram]

If λ is the (distance between marbles)+(diameter of marble) then L/λ is the number of marbles between the source and receiver. And all observers will see the same number of marbles.

My initial question was not how many marbles or λ should be seen by each observer, as if I was asking how many units of length should be recorded by different observer. Of course that will be constant no matter what will be the length measured by different observers for the unite length. You still did not pick up my question which was, will the wave fronts of a spherical wave for example, or encoded marbles, detected on slits by an observer match with the observation made by a different observer!
The answer was just simply, yes. The wave fronts or encoded marbles when exiting slits should be reordered the same for all FORs. No need to divide L over λ or you will get a phase of a different wave front at one slit when compared with the other slit!