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

[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!

 

[Adel Makram]

However, if I leave the OP question for a while, and focus on the pattern again. So far, I am not sure whether a pattern will form or no on my example

I suggested earlier that it will not form because the different wave length of light rays when leaving slits, due to Doppler effect, make the pattern impossible at the ground screen

Now, how if I make the source at rest relative to the ground observer, while I maintain the same location of the source but this time on the ground. This will be the classical version of the interference pattern. In this case, will there be a Doppler effect at slits which exclude the pattern on the ground screen? Or the slits just act as a window of transmission of light waves and allow the pattern to form?
And why there is a difference in the formation of the pattern in two cases, if any? As long as the light is emitted any way no matter what was the state of motion of the source at the moment of emitting the beams?

 

[Dale]

I suggested earlier that it will not form because the different wave length of light rays when leaving slits, due to Doppler effect, make the pattern impossible at the ground screen

And I corrected your suggestion already. You are deliberately neglecting the other Doppler shifts.

 

[Adel Makram]

And I corrected your suggestion already. You are deliberately neglecting the other Doppler shifts.

I can't see but the Doppler effect at slits. Would u please explain, how there will be any effect at the screen relative to the ground observer as long as both the observer and the screen are at rest relative to each other.

 

[Adel Makram]

I would like to know if I am right in this:

There are 2 ways to label the state of motion of the source relative to the ground, either moves or fixed at the time of radiating light beams
There are also 2 ways to see the possible pattern on the ground screen, either seen by the ground observer or by the train observer
So, there are 4 possible ways to have a different state of motion of the source seen by different observers,,, here is the proposed diagram!

https://www.physicsforums.com/attachment.php?attachmentid=46334&stc=1&d=1334675354

 

[Dale]

I can't see but the Doppler effect at slits. Would u please explain, how there will be any effect at the screen relative to the ground observer as long as both the observer and the screen are at rest relative to each other.

I am surprised at this. Following your drawing from post 16. The external observer will see the "forward" light from the source as blueshifted compared to the source, the mirror/slit will receive it as redshifted compared to the external observer, the external observer will then see the light from the mirror/slit as redshifted compared to the mirror/slit, finally the screen will receive the light as blueshifted compared to the external observer.

Clearly there are lots of Doppler shifts. It doesn't make sense to single out one and say "therefore no pattern".

 

[Adel Makram]

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.

The point of having the source located near slit A is to make sure that the light beams arrive at the same time at both slits relative to the ground observer where he wants to see he same classical set up of the pattern experiment. But if you out mirrors at 45degree, there will be no way to combine them at a single point in the screen.

 

[Dale]

I would like to know if I am right in this:

I haven't done the math here, but I suspect that you won't get "no pattern" but rather a "moving pattern".

 

[NotAName]

I suggested earlier that it will not form because the different wave length of light rays when leaving slits, due to Doppler effect, make the pattern impossible at the ground screen

Doppler shift alone will not necessarily eliminate a pattern. If doppler shift were to cause a frequency difference between the two sources then the way the interference pattern would appear would be changed. For instance, since you have light reflecting off mirrors it is polarized light and will create a straight up/down type of interference pattern. If one beam is a slightly different frequency, there will be a constant fringe shift to one side if it is a small change in frequency of only a few percent. Over roughly ten percent difference and the pattern will change so rapidly that there will no longer be one detectable or there may simply be a beat pattern.

Now, how if I make the source at rest relative to the ground observer, while I maintain the same location of the source but this time on the ground. This will be the classical version of the interference pattern. In this case, will there be a Doppler effect at slits which exclude the pattern on the ground screen? Or the slits just act as a window of transmission of light waves and allow the pattern to form?
And why there is a difference in the formation of the pattern in two cases, if any? As long as the light is emitted any way no matter what was the state of motion of the source at the moment of emitting the beams?

Slits will act like a place of "re-transmission" for your beams regardless of their frequency or wavelength. The disturbance of the pattern is because of the difference between wavelengths or because of a difference in the phase.

Interference is all about the way the peaks and troughs line up. With differing wavelengths there is a beat pattern of lining up and not lining up.

The diagrams on this page http://en.wikipedia.org/wiki/Fringe_shift make it easier to understand mechanically.

 

[Adel Makram]

I am surprised at this. Following your drawing from post 16. The external observer will see the "forward" light from the source as blueshifted compared to the source, the mirror/slit will receive it as redshifted compared to the external observer, the external observer will then see the light from the mirror/slit as redshifted compared to the mirror/slit, finally the screen will receive the light as blueshifted compared to the external observer.

Clearly there are lots of Doppler shifts. It doesn't make sense to single out one and say "therefore no pattern".

I don't understand step 2 and 4. What I understood were step 1 and 3. Doppler effect occurs at the receiver who is moving relative to a source. In step 2, you mentioned that the slits receives the light red shifted relative to the external observer, does the external observer emitts any thing? In step 4 the screen is at rest relative to the external observer, so there is no relative velocity plus no emission of lights from the external observer!

 

[Adel Makram]

I haven't done the math here, but I suspect that you won't get "no pattern" but rather a "moving pattern".

I mentioned before that we can make a replicate of the 2 slits by using a long train with multiple slits and the same location of the source so as to complete the pattern at particular location in the ground. So I am concerned about a fixed pattern on a ground

 

[Dale]

I am concerned about a fixed pattern on a ground

Then I would recommend modifying your chart to say "no fixed pattern" (which is correct) rather than just "no pattern" (which may not be correct).

 

[Dale]

I don't understand step 2 and 4. What I understood were step 1 and 3. Doppler effect occurs at the receiver who is moving relative to a source. In step 2, you mentioned that the slits receives the light red shifted relative to the external observer, does the external observer emitts any thing? In step 4 the screen is at rest relative to the external observer, so there is no relative velocity plus no emission of lights from the external observer!

No, the external observer is just the hypothetical observer who measures the frequency of each light pulse in the given reference frame. He never emits anything, and he samples everything without interrupting it. Here are the Doppler related questions:

What is the frequency of:
1) the light from source to A according to the source?
2) the light from source to A according to the frame?
3) the light from source to A according to A?
4) the light from A to ground according to A?
5) the light from A to ground according to the frame?
6) the light from A to ground according to the ground?

 

[Adel Makram]

No, the external observer is just the hypothetical observer who measures the frequency of each light pulse in the given reference frame. He never emits anything, and he samples everything without interrupting it. Here are the Doppler related questions:

What is the frequency of:
1) the light from source to A according to the source?
2) the light from source to A according to the frame?
3) the light from source to A according to A?
4) the light from A to ground according to A?
5) the light from A to ground according to the frame?
6) the light from A to ground according to the ground?

1) no shift because A and the source are at rest relative to each other
3) no shift either
4) red shifted
6) red shifted

I don't understand what do you mean by frame?

 

[Dale]

Reference frame is short hand for the inertial coordinate system associated with some object or observer.

 

[Adel Makram]

Reference frame is short hand for the inertial coordinate system associated with some object or observer.

I know the meaning of the word in physics but I don't know which frame you mean, the train or the ground ?

 

[Dale]

The frame that the drawing in post 16 represents. I don't know how you want to name that frame. I guess it is probably the ground frame.

 

[Dale]

BTW, here is how I would answer those questions I posed above

What is the frequency of:
1) the light from source to A according to the source? =F
2) the light from source to A according to the frame? >F
3) the light from source to A according to A? =F
4) the light from A to ground according to A? =F
5) the light from A to ground according to the frame? <F
6) the light from A to ground according to the ground? <F

 

[Adel Makram]

BTW, here is how I would answer those questions I posed above

What is the frequency of:
1) the light from source to A according to the source? =F
2) the light from source to A according to the frame? >F
3) the light from source to A according to A? =F
4) the light from A to ground according to A? =F
5) the light from A to ground according to the frame? <F
6) the light from A to ground according to the ground? <F

The frequency (f) from the source to a hypothetical external observer >f and from that observer to A < f so they canceled each other out. What remains is only from A to the screen according to the screen which is <f. This is the same as my first suggestion that the effective Doppler shifts is only from slits. So any way, we have now light from A <f and light from B>f, and they are traveling the same paths to the a mid-screen point ( concern about a fixed pattern). This makes the formation of a pattern not possible.

That would be the same conclusion according to the train observer. He sees different paths from A and B, with the same f from A and B, to a point on the screen which makes the pattern not possible ( i don`t think that Doppler effect at the screen relative to the screen will affect the case because the light rays have already reached it and interacted, so the different in D-shifts at the point of screen will not count!)

 

[Adel Makram]

Let`s modify the experiment according to the last suggestion ( which is: no pattern will form)
Think of 2 trains moving in the opposite directions toward each other. That will be our train with velocity v to the right direction and another one with -v to the left direction. The first train has slits; A and B and the second one has; C and D as shown in the diagram. Their mid-points will coincide at a corresponding point in a ground screen when rays leaving slits ( we can make them very thin trains with slits from them at 2 sides are so closed together)
This will make the D-shift from A matchs that from C and B with D,,, ultimately the pattern will form! ( the color of lines in the diagram indicates the Doppler shift)
If so:
1) will the 2 trains move in opposite directions With their sources moving altogether is equivalent to the classical pattern where There is only one fixed source relative to the ground and nothing moving?
2) According to QM, there would be no pattern if more than 2 slits are there and we got 4 slits now! ( forget this point for a while as the number of slits should be in one direction)
https://www.physicsforums.com/attachment.php?attachmentid=46357&stc=1&d=1334750489

 

[Dale]

The frequency (f) from the source to a hypothetical external observer >f and from that observer to A < f so they canceled each other out. What remains is only from A to the screen according to the screen which is <f. This is the same as my first suggestion that the effective Doppler shifts is only from slits. So any way, we have now light from A <f and light from B>f, and they are traveling the same paths to the a mid-screen point ( concern about a fixed pattern). This makes the formation of a fixed pattern not possible.

Yes. Note the clarification I added here and below.

That would be the same conclusion according to the train observer. He sees different paths from A and B, with the same f from A and B, to a point on the screen which makes the fixed pattern not possible ( i don`t think that Doppler effect at the screen relative to the screen will affect the case because the light rays have already reached it and interacted, so the different in D-shifts at the point of screen will not count!)

This is incorrect. For the train observer there is a fixed pattern that the ground is moving under. Obviously a pattern which is fixed relative to the train observer is not fixed relative to the ground observer. You cannot use the non-existence of a fixed pattern in one frame to show the non-existence of a fixed pattern in another frame since the property of being fixed or not is frame variant.