# Interference pattern again!

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
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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.

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.

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

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Dale
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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".

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
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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".

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.

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 dont 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!

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
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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).

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I dont 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?

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 dont understand what do you mean by frame?

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

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 ?

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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.

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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

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 cancelled 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 dont 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!)

Lets 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

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Dale
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The frequency (f) from the source to a hypothetical external observer >f and from that observer to A < f so they cancelled 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 dont 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.

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.

Fixed is fixed to the ground. What i meant by fixed is the isocenter of the pattern corresponds to a mid-point of the train at the time rays exiting slits relative to the ground observer. So I am not interested in a pattern that spread across a long distance in the direction of the train

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Fixed is fixed to the ground. What i meant by fixed is the isocenter of the pattern corresponds to a mid-point of the train at the time rays exiting slits relative to the ground observer. So I am not interested in a pattern that spread across a long distance in the direction of the train
You need to learn to be more clear, talking with you is an exercise in frustration.

OK, there is no fixed pattern in either frame, using your meaning of fixed. There is, however, an interference pattern in both frames. In one frame the pattern is stationary and the ground is moving, in the other frame the ground is stationary and the pattern is moving. The pattern is there whether or not you are interested in it.

Dale
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Lets modify the experiment according to the last suggestion
I am not interested in analyzing new experiments for you. You have been given enough input and instruction that it is now time for you to take responsibility for your own analysis. If you get stuck at some point then please post your work up to that point, and if you want me to check your work I will be glad to. Once you gain some skill in working these problems then you will be able to answer many of your own questions, and hopefully those of other people.

However, you need to know one thing in advance. The behavior of interference patterns are completely governed by Maxwell's equations, and Maxwell's equations are invariant under the Lorentz transform. Therefore, you will never find a situation where interference in one frame is not interference in another frame.