Interference Pattern versus SR

In summary, the conversation discusses a thought experiment involving a moving frame of reference (FOR) with double slits opening at the same time for a brief moment, causing an interference pattern on a screen for a ground observer. However, according to special relativity, the moving FOR's observer would only see one slit open at a time and no interference pattern. The conversation then delves into the concept of relative simultaneity and the idea that the sequence of events can be altered by a Lorentz transformation. Ultimately, it is concluded that there is no conflict or paradox between the two observers, as they must both agree on the timing of the light reaching the screen for an interference pattern to occur.
  • #71
Mentz114 said:
If one observer sees the slits opening and the light goes through, then it will be so for all observers. It's the worldlines coinciding and that is absolute. If you add invariance of phase differences, it's obvious that all observers see the same pattern.

Thanks for being back again!

So can i understand from that the simultaneous opening of A & B when the light goes through relative to the ground observer will be the same for the slit one? then where is the relativity?

And where is the answer to the question, which slit will open first for the slit-observer?
 
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  • #72
Adel Makram said:
https://www.physicsforums.com/attachment.php?attachmentid=43147&stc=1&d=1327565161

It is a bit different from the original experiment, but still need to work it out.
The source emits 2 beams of light at the same time. For the ground observer, her sees the slit A and B open simultaneously, so the pattern will form on his screen at the defined point to the left side. For the slit observer, the slit B opens before slit A initially to receive the beam from its right hand labeled in red. While, ultimately, the slit A should open first to release it to go to screen.
So which slit should open first?

Your image is wrong. See attachment for corrected version. You are forgetting that source and screen are in motion.
 

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  • #73
Adel Makram said:
Thanks for being back again!

So can i understand from that the simultaneous opening of A & B when the light goes through relative to the ground observer will be the same for the slit one? then where is the relativity?

And where is the answer to the question, which slit will open first for the slit-observer?
Within certain constraints, it is possible to get interference if the slits are never simultaneously open in the ground frame nor the slit frame. It is also possible to see interference if either slit opens first.

So it doesn't matter which slit opens first, there would still be interference if the other slit opened before all the light from the early opener reached the screen.

If these constraints are satisfied all observers will see interference, maybe only for a short time.

If the openings are too far apart in time, every observer sees 2 spots of light.
 
  • #74
zonde said:
Your image is wrong. See attachment for corrected version. You are forgetting that source and screen are in motion.

you are right, i was about to change it any way,,, but there is still another challenge:
The accumulated phase between the source and the screen is the same for both ground and slit-observer, yet the phase at the slits is different. For the ground observer, the phase at A and B must be the same, while it is not the case as recorded by the slit-observer!

So if the phase is a scalar quantity, its value should be invariant at the time the light leaves the slit?
 
  • #75
https://www.physicsforums.com/attachment.php?attachmentid=43153&stc=1&d=1327589224

The phase will be the same at 2 slits ( A & B) only and only if:

λ/c = d/v

where λ is the wavelength of the light, d is the distance between 2 slits relative to the ground observer and v is the velocity of the slits

But this is a special case only which ensures the phase of 2 light beams will be equal at the time they hit the 2 slits, while for a ground observer, the phase will always be equal
 

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  • #76
Adel Makram said:
It is a bit different from the original experiment, but still need to work it out.
The source emits 2 beams of light at the same time. For the ground observer, her sees the slit A and B open simultaneously, so the pattern will form on his screen at the defined point to the left side. For the slit observer, the slit B opens before slit A initially to receive the beam from its right hand labeled in red. While, ultimately, the slit A should open first to release it to go to screen.
So which slit should open first?
And are you sure that the phase-difference is zero for the slit observer considering the whole paths?
The slits should open simultaneously in the reference frame of othe slits in order for the interference pattern to appear on the screen. The observers will then see what's displayed on the screen and all observers will see the interference pattern. It is irrevelant whether they see if the slits open at the same time or not. The observers do not observe the light directly but after the light is reflected from the screen in the bright spots.
 
  • #77
Adel Makram said:
The accumulated phase between the source and the screen is the same for both ground and slit-observer, yet the phase at the slits is different.
So what? The interference pattern depends on the phase at the screen, not at the slit.

Adel Makram said:
So if the phase is a scalar quantity, its value should be invariant at the time the light leaves the slit?
The phase is invariant, but the time is not.
 
  • #78
DaleSpam said:
So what? The interference pattern depends on the phase at the screen, not at the slit.

The phase is invariant, but the time is not.

If, the phase of the wave is represented by a single wave front, then the difference between the time of slit opening relative to the slit-observer depends on the the distance between the slits, the velocity of the source and the perpendicular distance between the slit and the source. While the difference in time according to Lorentz transformation depends on the distance between the slits and the velocity of the source only!

So how that could be conceived?

look at a similar nice demonstration from Mike

http://www.relativitysimulation.com/Documents/DoubleSlit.htm
 
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  • #79
Tantalos said:
The slits should open simultaneously in the reference frame of othe slits in order for the interference pattern to appear on the screen. The observers will then see what's displayed on the screen and all observers will see the interference pattern. It is irrevelant whether they see if the slits open at the same time or not. The observers do not observe the light directly but after the light is reflected from the screen in the bright spots.

As i understood from this discussion that the slits need not to be open simultaneously for the FOR of slits in order to see the pattern. The original experiment emphasize that the slits should open simultaneously relative to the ground observer only
 
  • #80
Adel Makram said:
If, the phase of the wave is represented by a single wave front, then the difference between the time of slit opening relative to the slit-observer depends on the the distance between the slits, the velocity of the source and the perpendicular distance between the slit and the source. While the difference in time according to Lorentz transformation depends on the distance between the slits and the velocity of the source only!

So how that could be conceived?

look at a similar nice demonstration from Mike

http://www.relativitysimulation.com/Documents/DoubleSlit.htm

That doesn’t read like what I was showing. I my reference, the slits would have to be opened at the same time (according to the FOR of the slit observer). That’s because I chose a wave front that was created when the source was directly opposite the slits and equidistant from both (according to the FOR of the slit observer). Of course the geometry is different in the FOR of the source.

Also the Lorentz transformation does not apply to light. In my reference the geometry/speed/time of source and slit are transformed using the Lorentz transformation, but not the geometry/speed of the light wave.

And, in general, I avoided writing about interference patterns because I don’t believe an observer would see anything on the other side of the slits. The material from which the slit is made is not reflective. The instrument would not work if it were. The material is supposed to be completely absorbent, no? So only photons with a clear line-of-sight path through the slit will make it out the other side. Trying to get an interference pattern from your setup where the source is moving with respect to the slits would seem to be impossible.
 
  • #81
MikeLizzi said:
Also the Lorentz transformation does not apply to light.
Yes, it does. The whole point of the Lorentz transform is that Maxwells equations re invariant under it.
 
  • #82
Adel Makram said:
If, the phase of the wave is represented by a single wave front, then the difference between the time of slit opening relative to the slit-observer depends on the the distance between the slits, the velocity of the source and the perpendicular distance between the slit and the source. While the difference in time according to Lorentz transformation depends on the distance between the slits and the velocity of the source only!
I haven't checked Mike's math, but again, so what? Differences in time are not frame invariant, so the fact that they are different in two frames is not surprising.
 
  • #83
MikeLizzi said:
That doesn’t read like what I was showing. I my reference, the slits would have to be opened at the same time (according to the FOR of the slit observer). That’s because I chose a wave front that was created when the source was directly opposite the slits and equidistant from both (according to the FOR of the slit observer). Of course the geometry is different in the FOR of the source.

Also the Lorentz transformation does not apply to light. In my reference the geometry/speed/time of source and slit are transformed using the Lorentz transformation, but not the geometry/speed of the light wave.

And, in general, I avoided writing about interference patterns because I don’t believe an observer would see anything on the other side of the slits. The material from which the slit is made is not reflective. The instrument would not work if it were. The material is supposed to be completely absorbent, no? So only photons with a clear line-of-sight path through the slit will make it out the other side. Trying to get an interference pattern from your setup where the source is moving with respect to the slits would seem to be impossible.

https://www.physicsforums.com/attachment.php?attachmentid=43180&stc=1&d=1327671188

Here is the setup of my experiment
 

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  • #84
DaleSpam said:
I haven't checked Mike's math, but again, so what? Differences in time are not frame invariant, so the fact that they are different in two frames is not surprising.

If the accumulative phase given the whole paths and the absolute value of the phase at A & B are invariant, then the different time of opening of slits would depend also on the geometry of the setup including the perpendicular distance to the source which is not a variable in a LT
 
  • #85
Perhaps you can show mathematically what your concern is, because I don't see any problem from your English descriptions.
 
  • #87
DaleSpam said:
Yes, it does. The whole point of the Lorentz transform is that Maxwells equations re invariant under it.

Ooops. So what is the formal way of saying "you can't transform the position or shape of a light sphere" using the Lorentz transformation because it is traveling at the speed of light and gamma is undefined".
 
  • #88
Adel Makram said:

Oh, your statement on that diagram is wrong. Your diagram shows the sequence of events if the slit is moving with respect to the observer. A does not open before B relative to the slit observer. A opens before B relative to the light source observer.

Disregard above. I misunderstood your diagram. Something is wrong with it. I'll get back to you.
 
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  • #89
MikeLizzi said:
Ooops. So what is the formal way of saying "you can't transform the position or shape of a light sphere" using the Lorentz transformation because it is traveling at the speed of light and gamma is undefined".

The Lorentz transform applies between frames moving at relative speed less than c. There is no frame for the light itself. The Lorentz transform is exactly how you you transform the the position or shape of a light sphere between observer A and observer B, who might be moving with relative speed .9c. The gamma relates to the .9c. The speed of light is invariant (between the two frames). Its direction and shape of a wave front will be different between A and B.
 
  • #90
PAllen said:
The Lorentz transform applies between frames moving at relative speed less than c. There is no frame for the light itself. The Lorentz transform is exactly how you you transform the the position or shape of a light sphere between observer A and observer B, who might be moving with relative speed .9c. The gamma relates to the .9c. The speed of light is invariant (between the two frames). Its direction and shape of a wave front will be different between A and B.

Are you sure about your wording? If I want to transform a scene having many spheres moving at different velocities to an observer traveling at .9c with respect to the current observer, I transform all the spheres using the Lorentz Transformation. The position, proper time and geometry (contraction) of all the spheres change. But, if there is a light sphere in the scene, I don't bother transforming it because it's stays a sphere. No? Or am I cheating?
 
  • #91
MikeLizzi said:
Oh, your statement on that diagram is wrong. Your diagram shows the sequence of events if the slit is moving with respect to the observer. A does not open before B relative to the slit observer. A opens before B relative to the light source observer.

Disregard above. I misunderstood your diagram. Something is wrong with it. I'll get back to you.

Update to my previous comment about your recent diagram. You have a “v” with an arrow that seems to indicate that the Double Slit is moving. So the two diagrams appear to me to be a sequence showing a single light wave at different times from the FOR of the light source. And yes, the single light wave will intersect the two slits at two different times in the FOR of the light source. I don’t have to know physics to recognize that.

But, as DaleSpam said, the differences in times are not frame invariant. You are not entitled to declare that the intersections take place at two different times in the FOR of the Double Slit. You have to calculate that. And what if it turns out that they do? Again, as DaleSpam said, “so what”? You conclude that this fact contradicts an equation. I don’t know what equation that is, but I doubt it was derived for the condition where the light source was moving perpendicular to the orientation of the double slit.
 
  • #92
MikeLizzi said:
Are you sure about your wording? If I want to transform a scene having many spheres moving at different velocities to an observer traveling at .9c with respect to the current observer, I transform all the spheres using the Lorentz Transformation. The position, proper time and geometry (contraction) of all the spheres change. But, if there is a light sphere in the scene, I don't bother transforming it because it's stays a sphere. No? Or am I cheating?

For the moving and the ground observers, the light sphere remains a complete sphere, but the ground observer watches the moving one sees the sphere in his FOR contracted in the direction of motion
 
  • #93
Adel Makram said:
for sake of simplicity, consider the attached diagram
the time to A = 1/c (sA)
the time to B = 1/c (√(sA^2 + (AB – vt)^2))

where AB is the distance between the 2 slits

In LT, Δt= AB * v/c^2 (√(1-v^2/c^2))
That would be correct if slit A were at rest and slit B were moving, but I thought that you intended for the slits to be moving in the same direction at the same speed. If the location of the source is the origin, and the location of slit A is (vt, s) and the location of slit B is (ab + vt, s) then the flash from the source at t=0 reaches A at
[itex]0=c^2 t_A^2 - (vt_A)^2-s^2[/itex]
and it reaches B at
[itex]0=c^2 t_B^2 - (ab+vt_B)^2-s^2[/itex]
 
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  • #94
MikeLizzi said:
Update to my previous comment about your recent diagram. You have a “v” with an arrow that seems to indicate that the Double Slit is moving. So the two diagrams appear to me to be a sequence showing a single light wave at different times from the FOR of the light source. And yes, the single light wave will intersect the two slits at two different times in the FOR of the light source. I don’t have to know physics to recognize that.

But, as DaleSpam said, the differences in times are not frame invariant. You are not entitled to declare that the intersections take place at two different times in the FOR of the Double Slit. You have to calculate that. And what if it turns out that they do? Again, as DaleSpam said, “so what”? You conclude that this fact contradicts an equation. I don’t know what equation that is, but I doubt it was derived for the condition where the light source was moving perpendicular to the orientation of the double slit.

the derivation of the difference between the time when the light front hits the 2 slits depends on the geometry of the setup shown. But as the ground observer watches the slits open simultaneously, the LT tells that there should be difference in the time of slits opening relative to the moving one by an amount related only to the distance between the 2 slits and the velocity of the slit not the geometry of the setup.
 
  • #95
DaleSpam said:
That would be correct if slit A were at rest and slit B were moving, but I thought that you intended for the slits to be moving in the same direction at the same speed.

no, both of them moving,,, actually you should imagine the diagram as a projection of 2 diagrams in 2 different time where the source is fixed in location
 
  • #96
Adel Makram said:
no, both of them moving,,, actually you should imagine the diagram as a projection of 2 diagrams in 2 different time where the source is fixed in location
Understood, so your math is incorrect. See the corrected math I just posted above in post 93.
 
  • #97
DaleSpam said:
Understood, so your math is incorrect. See the corrected math I just posted above in post 93.

it is the same like mine but with different orientations,,, any way (s) appears in your equation as well while it should be s-free by LT
 
  • #98
Adel Makram said:
it is the same like mine but with different orientations,,, any way (s) appears in your equation as well while it should be s-free by LT
Why? The LT certainly does not imply that light travels instantaneously in the y direction.
 
  • #99
of course no, but LT does not consider (s) when calculating the time difference between 2 slits. It considers only ab and v
 
  • #101
so the time difference is different in the 2 calculation
 
  • #102
MikeLizzi said:
Are you sure about your wording? If I want to transform a scene having many spheres moving at different velocities to an observer traveling at .9c with respect to the current observer, I transform all the spheres using the Lorentz Transformation. The position, proper time and geometry (contraction) of all the spheres change. But, if there is a light sphere in the scene, I don't bother transforming it because it's stays a sphere. No? Or am I cheating?

It depends. If the only thing you are interested in is a complete spherical wave front, you can just LT its emission event, and then track the sphere expanding at c from the emission event. However, if you ask about a section of spherical wave front, the size of solid angle as well as the orientation change. Applying LT to the light front itself is the most direct way to account for this.
 
  • #103
Adel Makram said:
so the time difference is different in the 2 calculation
What 2 calculations? So far there is only one.
 
  • #104
DaleSpam said:
What 2 calculations? So far there is only one.

:)),,, ok let's see it from side of the ground observer this time,,, he watches 2 slits open simultaneously, but the slit-rider records 2 different times,,, this is the first calculation

From the source side now: when the light front reaches the 2 slits in different time ( given the same phase),,, the slit calculate that difference considering (s),,, this is the second calculation
 
  • #105
Show your math, please. Just because the Lorentz transform doesn't involve s doesn't mean that it doesn't show up in the equations. That is why it is important to actually work things out and not simply make assumptions.
 

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