Interference Pattern versus SR

[JDoolin]

The spatial separation ( look at the x-axis) between the events is equal ( and so is the temporal because its a null direction). There could be a set of observers at those events.



That is exactly what I meant. Interference between waves is determined by phase differences. In the 2-slit experiment, the phase difference is because the path lengths of the 2 beams is different.

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

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

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

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

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

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

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

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

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

 

[Mentz114]

Proof of phase invariance ?

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

But I think I've found a mathematical proof.

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

phase = ΔL/λ

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

L² = t² + x²

A beam of light goes from P to Q with

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

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

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

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

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

2dx² γ²(1+β)².

Now λ transorms according to the Doppler formula

λ' = λ √(γ²(1+β)²)

and so

λ/ΔL = λ'/ΔL'

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

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

 

[PhilDSP]

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

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

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

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

 

[JDoolin]

Thanks, Mentz.

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

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

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

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

 

[JDoolin]

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

Because you can see in the animation:

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

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

 

[Adel Makram]

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

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

 

[Adel Makram]

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

 

[Mentz114]

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

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

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

Path length is never irrelevant in interference calculations.

I'm done here.

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

 

[Adel Makram]

Path length is never irrelevant in interference calculations.

I'm done here.

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

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

 

[Adel Makram]

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

 

[MikeLizzi]

Adel Makram, I believe I understand the issue with which you are struggling.
I hesitated to post until other members had their say. See the referenced link.

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

 

[Adel Makram]

Thanks a lot Mike, I may need some time to analyse your post, Thanks once again for everyone

 

[zonde]

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

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

This actually seems quite confusing because what is scalar in classical physics (3D-scalar) is vector in relativity (where scalars are defined in respect to 4D-spacetime) as we can describe 3D-space using (timelike) vector that is normal to that space and module of vector being scalar in that 3D-space.

And seems even more complicated about 4D-scalars in 3D-space plus time.

 

[JDoolin]

This actually seems quite confusing because what is scalar in classical physics (3D-scalar) is vector in relativity (where scalars are defined in respect to 4D-spacetime) as we can describe 3D-space using (timelike) vector that is normal to that space and module of vector being scalar in that 3D-space.

And seems even more complicated about 4D-scalars in 3D-space plus time.

Part of the confusion comes from the fact that we should have three different definitions for length, namely,

(length1) the length of the meter-stick measured as the space-time interval between simultaneous events in the meter-stick's reference frame

(length2) the length of the meter-stick measured as the space-time interval between simultaneous events in the observer's reference frame

(length3) the coordinate distance Δx between two events

AND we seem to have two different definitions for scalar

(scalar1) Any single numeric quantity; i.e. can be represented in one-dimension.

(scalar2) Any quantity that is not changed by rotation or Lorentz Transformation

The first definition of length qualifies as a scalar either way, because it is a single number, AND it is Lorentz Invariant, but the second and third definitions of length are observer dependent, and will change under Lorentz Transformation.

Edit: I left off one definition of length:

(length4) The space-time interval Δs between two events
\Delta s=\sqrt{\Delta x^2+\Delta y^2+\Delta z^2-c^2 \Delta t^2}
This one is also a scalar according to both definitions of scalar.

 

[JDoolin]

Thanks, MikeLizzi.

You've given me a couple of ideas to play around with,

and Thanks Mentz

I will get back to you.

I hope I'll be able to put something together to clarify these ideas. I'm dreaming of a full-blown two-slit interference demo with arbitrary velocities of source, slits, and screen, complete with cross-sectional space-time diagram, showing just exactly why and how that interference pattern is invariant,

 

[Adel Makram]

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?
And are you sure that the phase-difference is zero for the slit observer considering the whole paths?

It was a quick hit, I know, so i have to think about it again,,,

 

[Adel Makram]

By the way, the slit is moving downward in the direction of the short black arrow

 

[spikenigma]

Picking through the sarchastic replies, I can't seem to parse the answer to this question?

Does the ground observer see an interference pattern through the one open slit on the train?, if so, isn't that disturbingly close to a forbidden preferred frame of reference?

 

[Mentz114]

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?
And are you sure that the phase-difference is zero for the slit observer considering the whole paths?

It was a quick hit, I know, so i have to think about it again,,,

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.

 

[Mentz114]

Picking through the sarchastic replies, I can't seem to parse the answer to this question?

Does the ground observer see an interference pattern through the one open slit on the train?, if so, isn't that disturbingly close to a forbidden preferred frame of reference?

It's been stated several times that all observers will see the same thing. If the observer at rest in the frame of the apparatus sees a pattern, so will all observers. It's the worldlines ...

 

[Adel Makram]

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?

 

[zonde]

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.

 

[Mentz114]

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.

 

[Adel Makram]

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?

 

[Adel Makram]

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

 

[Tantalos]

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.

 

[Dale]

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.

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.

 

[Adel Makram]

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

 

[Adel Makram]

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

 

[MikeLizzi]

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.

 

[Dale]

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.

 

[Dale]

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.

 

[Adel Makram]

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

 

[Adel Makram]

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

 

[Dale]

Perhaps you can show mathematically what your concern is, because I don't see any problem from your English descriptions.

 

[Adel Makram]

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))

https://www.physicsforums.com/attachment.php?attachmentid=43181&stc=1&d=1327674446

 

[MikeLizzi]

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".

 

[MikeLizzi]

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

Here is the setup of my experiment

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.

 

[PAllen]

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.

 

[MikeLizzi]

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?

 

[MikeLizzi]

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.

 

[Adel Makram]

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

 

[Dale]

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
0=c^2 t_A^2 - (vt_A)^2-s^2
and it reaches B at
0=c^2 t_B^2 - (ab+vt_B)^2-s^2

 

[Adel Makram]

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.

 

[Adel Makram]

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

 

[Dale]

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.

 

[Adel Makram]

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

 

[Dale]

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.

 

[Adel Makram]

of course no, but LT does not consider (s) when calculating the time difference between 2 slits. It considers only ab and v

 

[Dale]

Agreed. So what?