Exploring the Paradoxical Light Clock Problem in Special Relativity Analysis

[Nugatory]

OK... so I think what you are saying is that the light clock diagram must be viewed as an inferred and possibly flawed interpretation of what is remotely happening, not an existential representation.

If so, and the paradox stems from the current diagram, can the diagram be saved by altering it or is there no proper diagram possible for representing the remote behavior of the light path?

It seems clear that the diagram is "an inferred and possibly flawed interpretation" - it's a picture for crissakes! - How could it be anything else?

However, it can be saved easily. Just erase the lines that purport to show the path of the light, so that only three events appear in the picture.
A) Flash of light is emitted from bottom mirror in conformance with Maxwell's equations applied locally at the moment of emission.
B) Flash of light hits upper mirror and is reflected, again in conformance with Maxwell's equations applied locally at the moment of emission.
C) Flash of light is is detected at the lower mirror.

We eliminate the concerns (especially Altergnostic's concern in #24 that we cannot assign coordinates of events we are not aware of) about how these events are observed by using the procedure described in Taylor and Wheeler's "Spacetime Physics" (this procedure is, BTW, one of the better ways of avoiding the "what is a frame?" quagmire). We scatter observers, at rest relative to each other and carrying synchronized clocks, throughout space. Whenever anything happens right under an observer's nose, so only local considerations apply, the observer writes down what happened and the time it happened on a slip of paper. At some later time, we collect all the slips of paper and piece together a complete global picture.

We do all of this, and where's the paradox? The local angle each observer reports is exactly what you get by Lorentz-transforming from what observers at rest relative to the mirrors see.

Do those lines in the diagram that purport to show the path of light between these events represent anything "real"? I think so, for the same reason that if I see a mouse enter one end of a length of pipe, and later see a mouse emerge from the other end, I'm inclined to think that I've observed a mouse scamper from one end of the pipe to the other. But I have no quarrel with anyone who tells me that without observations of the mouse's passage through the pipe I'm just offering an interpretation.

 

[Austin0]

Please explain how photons are never given momentum from the source but a change in frequency due to the momentum of the source yet can be given the lateral momentum of the sideways motion of the source to cause them to continue across the moving lab and still hit the mirror and come back to the detector.

I certainly can't explain it , any more than it can be explained in Newtonian mechanics.
But it appears as if photons act just like Newtonian ballistic particles only with a certain difference due to the limit of c.
In low velocity conservation (of massive particles) the conservation is complete and directly additive.
This cannot be the case with light (or relativistic massive particles), as is clear if you look at a photon emitted along the vector of motion. The forward motion of the source cannot add to the forward velocity of the photon which is already maxed out.
I guess it might be looked at as the forward momentum being expressed as Doppler blueshift (or redshift for anti-parallel emission) but I don't know if that is an accepted view.
On the other hand, photons emitted orthogonally to motion conserve the forward momentum,( as motion ),completely, while the change of momentum through frequency shift of the intrinsic emission frequency is at a minimum (zero at some point close to 90deg.) .
SO the percentage of forward momentum of the source conserved as forward motion is a function of angle.
I hope this might be helpful.

 

[marty1]

But it not just the maximum they cannot exceed in the direction of travel. They also cannot be slowed down when emitted from a source moving away. They are instead red shifted. Momentum in every other direction seems to be completely Newtonian. Do they "know" which direction they are heading and follow different laws? Do they have a head and a tail?

 

[Nugatory]

Please explain how photons are never given momentum from the source but a change in frequency due to the momentum of the source yet can be given the lateral momentum of the sideways motion of the source to cause them to continue across the moving lab and still hit the mirror and come back to the detector.

No such thing happens - the lateral component of the momentum doesn't change during the reflection. The vertical component changes sign, switching directions.

 

[Austin0]

But it not just the maximum they cannot exceed in the direction of travel. They also cannot be slowed down when emitted from a source moving away. They are instead red shifted. Momentum in every other direction seems to be completely Newtonian. Do they "know" which direction they are heading and follow different laws? Do they have a head and a tail?

They are not slowed down in any direction. The forward momentum is conserved wrt photons emitted to the rear with the same restrictions and angle dependence as applies to forward emissions. That forward momentum does not change the speed in either frame but it changes the angle of propagation. Put another way it changes the x component of the velocity vector , reducing the motion in the -x direction but increasing it in the y and z directions relative to the lab frame but with a constant speed.
The result is the vector is tilted forward in the direction of motion in the lab frame.

So the amount of motion conservation is zero at 0 and 180 deg. emissions in the source frame, smoothly increasing to maxima (1) at 90 and 270 deg.
While the Doppler shift is roughly the inverse. Maximal at 0 and 180deg. to minimal around 90 and 270deg.

 

[Dale]

It is real in all frames, but an "observer" can't assign coordinats of events he is not aware of. Those light paths are not seen by a distant observer, so he can't diagram them. If you want him to know the times of the reflections, you have to send him a signal at each reflection event.

Sure, but I have a hard time taking this as a serious objection. That should be an obvious and trivially understood part of the operation of any clock. Any clock has an oscillator of some sort (Caesium atom, quartz crystal, pendulum, etc.) and a mechanism for signalling the results (dial with hands, digital display, broadcast signal, paper printout, etc.). When you analyze the physics of a specific clock, you analyze the oscillator part, as is traditonally done with a light clock. It is understood that the oscillation can be reported in any number of ways.

Once you know that the detection occurred then you may use any reference frame to analyze the clock. There is no requirement to use a frame where the clock or the observer is at rest. Everything else follows.

As it is, the light clock diagram assumes he can actually see those light paths, which easily leads to paradoxes.

No true paradoxes have been found in over a century of careful scrutiny by the most brilliant minds in the world. All that have been found are unintuitive things which confuse students, many of which are hyperbolically advertised as paradoxes.

 

[marty1]

The result is the vector is tilted forward in the direction of motion in the lab frame.

Ok I understand how that all works out nice mathematically, but a photon changing direction of emission depending on velocity of source?

 

[altergnostic]

This is not correct. The observed diagonal paths in HarryLin's proposed set up would always produce c as a direct calculation of the distance and detection times (D/dt5) in the LAB frame.
No transformation whatsoever required in that frame.

Exactly as would be the case if the mirror was removed and the photon continued on to a detector at rest in the lab frame.

I agree, see post #25.

 

[altergnostic]

We scatter observers, at rest relative to each other and carrying synchronized clocks, throughout space. Whenever anything happens right under an observer's nose, so only local considerations apply, the observer writes down what happened and the time it happened on a slip of paper. At some later time, we collect all the slips of paper and piece together a complete global picture.

We do all of this, and where's the paradox? The local angle each observer reports is exactly what you get by Lorentz-transforming from what observers at rest relative to the mirrors see.

I don't know if I have followed your reasoning, are you saying that the final diagram will agree with the stationary frame diagram?

 

[marty1]

This really all made sense to me before, but something is cause me a problem with the photon's momentum in the direction of travel of the lab. If I think classically I can see that setup behaving like that if the observers are moving in the opposite direction. I just can't seem to fit in my brain the moving lab except a small one not moving fast. A photon shot sideways could get to the other side after the mirror is gone without the sideways momentum of the lab given to the photon.

 

[Austin0]

Ok I understand how that all works out nice mathematically, but a photon changing direction of emission depending on velocity of source?

It does not change direction relative to the source frame.

It is only a change of direction relative to outside inertial frames. Exactly like a bullet fired laterally in a moving frame has an oblique angled trajectory relative to an observing frame.

perhaps a better way of saying it is it does change the direction relative to the emission angle but this is not perceived in the source frame because that frame is moving right along with the forward displacement due to conserved momentum.

Again comparable to the bullet situation.

 

[marty1]

Again you are imparting the sideways momentum of the lab onto the photon (sideways meaning the labs toward direction which is sideways to the sideways fired photon). That is where I get lost.

 

[Austin0]

Again you are imparting the sideways momentum of the lab onto the photon (sideways meaning the labs toward direction which is sideways to the sideways fired photon). That is where I get lost.

I am sorry if there has been confusion. I interpreted your question as related to a moving clock (source) as observed from a stationary lab.
All my remarks were addressed to a moving source.

If the clock is considered stationary then of course the forward momentum of a moving observer can have no effect on the propagation of the photon.
In this case it is simply the purely kinematic effect of relative motion.

Once again completely analogous to moving by a ball bouncing straight up and down . If the ball is covered with red ink and you could move a strip of paper along it as you pass the resulting motion chart would be a zig zag right? Well actually a sine wave but with lights instantaneous acceleration in it's case there are sharp turns.

 

[Nugatory]

I don't know if I have followed your reasoning, are you saying that the final diagram will agree with the stationary frame diagram?

The final diagram will agree with the stationary frame diagram if the speed of all these observers happens to be the speed of the light clock. Otherwise it will agree with one of the moving frame diagrams.

But nothing prevents me from doing this "scatter observers, at rest relative to each other and carrying synchronized clocks" thing multiple times using a different speed for each flock of observers... And I can still gather up all those slips of paper and correlate them. So I'll get to see the exact same light-hits-mirror event as reported from both the stationary frame and as many different moving frames as I please.

I cannot say this strongly enough: get hold of and understand the relevant chapter of that Taylor/Wheeler book. It will extract you from the "what's a frame?" quagmire that you've fallen into like nothing else can.

 

[marty1]

So the emitted photon does carry the momentum of the source in its movement 90 degrees from its direction of travel.

 

[altergnostic]

Sure, but I have a hard time taking this as a serious objection. That should be an obvious and trivially understood part of the operation of any clock. Any clock has an oscillator of some sort (Caesium atom, quartz crystal, pendulum, etc.) and a mechanism for signalling the results (dial with hands, digital display, broadcast signal, paper printout, etc.). When you analyze the physics of a specific clock, you analyze the oscillator part, as is traditonally done with a light clock. It is understood that the oscillation can be reported in any number of ways.

Well, it should be obvious, but (as you point out at the end of your post) many "less brilliant" minds don't find it so obvious, so why not diagram the setup more realistically? Einstein usually tried to do so, and usually using many different thought problems and examples to explain the same thing. He is the one who said that if you can't explain something to a six year old, you have not understood it yourself. The point is that the propagation of the signals, or, the times of observations of each tick of the clock from the moving frame point of view, must enter the equations. As it is, they are left out. There's no consideration of the distance between the light clock and the observer in the moving frame, and neither any consideration of the time it takes for each event to be observed by the moving frame. As it is, the observer is assumed to instantaneously see each event, which is very not-SR.

Once you know that the detection occurred then you may use any reference frame to analyze the clock. There is no requirement to use a frame where the clock or the observer is at rest. Everything else follows.

Exactly! So exactly at what x,y,z,t coordinates does the observer know the detection (reflection in the light clock) has occurred? In other words, at what time and relative position does the observer actually observes the events? Is this irrelevant to this problem? Why is it irrelevant here and relevant in problems like the train and embankment?

No true paradoxes have been found in over a century of careful scrutiny by the most brilliant minds in the world. All that have been found are unintuitive things which confuse students, many of which are hyperbolically advertised as paradoxes.

I actually almost agree with this. There is no true paradox if you remember to send signals to the observer, since the final result is correct regardless. But as it is the assumptions are unreal and impossible. You can't diagram unseen light, it is simple as that. Of course you are bound to find a lot of unintuitive things and confused students since the diagram is unintuitive and confused to start with. It is at least incomplete, and it is never diagrammed realistically and complete. We never see any analysis of how the observer knows about those paths, how he sees them, from what distance, and at what times, etc. Even though we have a lot of confused students, we keep the diagram incomplete. Why? Why not remember them that they need to signal the events to the observer somehow? Why do we allow students to wonder how the motion of the emitter can affect the direction of the beam without changing the beam's velocity (very counter-intuitively), in seemingly contradiction with the postulate of SR that the motion of emitter doesn't affect the speed of light? How are they not to get confused? We have many animations on the web showing how the motion of an emitter compresses the frontal waves and stretches the trailing waves, creating Doppler, and those diagrams show no change in the direction of the emitted light due to the motion of the source, so how are they to believe both diagrams without confusion?

You admitted that we need to send signals at each tick of the light clock if we want the observer to diagram anything. So if just remembering that reflection events are not seen directly helps to make some sense of the diagram, why not do so? If you actually need to signal the reflection events to the moving observer, those signals have to travel some distance until they reach the observer, after some time from the moment of emission. The students would immediately relate the problem to the famous train and embankment, for example, and it would be easier to comprehend. And also, the moving observer wouldn't diagram those diagonal paths in such a setup at all. You would apply the transforms on the incoming signals and the light clock's beam would be diagrammed straight up and down. No angles. No confusion.

If you still think the diagram is perfect and it is the students that fail, you are not trying to make a comprehensive diagram, or to make SR easier to understand, you are just trying to flunk them.

 

[Austin0]

So the emitted photon does carry the momentum of the source in its movement 90 degrees from its direction of travel.

Yes. If it is emitted at 90 deg. relative to motion in the source frame the forward momentum will result in the path pointing at some angle forward in an observing frame.

 

[Austin0]

Well, it should be obvious, but (as you point out at the end of your post) many "less brilliant" minds don't find it so obvious, so why not diagram the setup more realistically? Einstein usually tried to do so, and usually using many different thought problems and examples to explain the same thing. He is the one who said that if you can't explain something to a six year old, you have not understood it yourself. The point is that the propagation of the signals, or, the times of observations of each tick of the clock from the moving frame point of view, must enter the equations. As it is, they are left out. There's no consideration of the distance between the light clock and the observer in the moving frame, and neither any consideration of the time it takes for each event to be observed by the moving frame. As it is, the observer is assumed to instantaneously see each event, which is very not-SR.



Exactly! So exactly at what x,y,z,t coordinates does the observer know the detection (reflection in the light clock) has occurred? In other words, at what time and relative position does the observer actually observes the events? Is this irrelevant to this problem? Why is it irrelevant here and relevant in problems like the train and embankment?



I actually almost agree with this. There is no true paradox if you remember to send signals to the observer, since the final result is correct regardless. But as it is the assumptions are unreal and impossible. You can't diagram unseen light, it is simple as that. Of course you are bound to find a lot of unintuitive things and confused students since the diagram is unintuitive and confused to start with. It is at least incomplete, and it is never diagrammed realistically and complete. We never see any analysis of how the observer knows about those paths, how he sees them, from what distance, and at what times, etc. Even though we have a lot of confused students, we keep the diagram incomplete. Why? Why not remember them that they need to signal the events to the observer somehow? Why do we allow students to wonder how the motion of the emitter can affect the direction of the beam without changing the beam's velocity (very counter-intuitively), in seemingly contradiction with the postulate of SR that the motion of emitter doesn't affect the speed of light? How are they not to get confused? We have many animations on the web showing how the motion of an emitter compresses the frontal waves and stretches the trailing waves, creating Doppler, and those diagrams show no change in the direction of the emitted light due to the motion of the source, so how are they to believe both diagrams without confusion?

You admitted that we need to send signals at each tick of the light clock if we want the observer to diagram anything. So if just remembering that reflection events are not seen directly helps to make some sense of the diagram, why not do so? If you actually need to signal the reflection events to the moving observer, those signals have to travel some distance until they reach the observer, after some time from the moment of emission. The students would immediately relate the problem to the famous train and embankment, for example, and it would be easier to comprehend. And also, the moving observer wouldn't diagram those diagonal paths in such a setup at all. You would apply the transforms on the incoming signals and the light clock's beam would be diagrammed straight up and down. No angles. No confusion.

If you still think the diagram is perfect and it is the students that fail, you are not trying to make a comprehensive diagram, or to make SR easier to understand, you are just trying to flunk them.

I think you are right in that the fact that the speed of light is independent of the motion of the source should be expanded to include the fact that the direction of propagation is not also independent. With the elaboration of its effects, the relativistic aberration, that is certainly relevant to the light clock and it's understanding.
As for the rest; you may have a certain point but it should be clear that certain unrealistic aspects of most thought experiments are not really relevant to their aid as constructs for understanding. IMO

 

[marty1]

Yes. If it is emitted at 90 deg. relative to motion in the source frame the forward momentum will result in the path pointing at some angle forward in an observing frame.

So I post my question again. What is special about the "sides" of a photon that permit the source to give it momentum but any attempt to do that in the direction of propagation results only in red or blue shift?

It seems that from the side a photon is no different from any other particle.

 

[altergnostic]

The final diagram will agree with the stationary frame diagram if the speed of all these observers happens to be the speed of the light clock. Otherwise it will agree with one of the moving frame diagrams.

But nothing prevents me from doing this "scatter observers, at rest relative to each other and carrying synchronized clocks" thing multiple times using a different speed for each flock of observers... And I can still gather up all those slips of paper and correlate them. So I'll get to see the exact same light-hits-mirror event as reported from both the stationary frame and as many different moving frames as I please.

I cannot say this strongly enough: get hold of and understand the relevant chapter of that Taylor/Wheeler book. It will extract you from the "what's a frame?" quagmire that you've fallen into like nothing else can.

This setup will yield the same results as harrylin's setup.

Only the observers in direct line with the path of the beam will detect anything, the others can only copy what they wrote down. This is true either if they are moving or at rest relative to the beam.

If they are at rest relative to the light clock, the diagram looks like the stationary one. Also, any observer in line with the path of the beam will measure the same period of time between two detections, and they will know the light is bouncing up and down since they will have as many successive detections as they want, as long as they keep their positions wrt the setup. Notice that each observer in line only sees the beam as it passes through him. They detect the beam by contact. They don't know the times of reflections directly, they only know their local detections. This means that if you stick with one observer, you will diagram the beam up and down like in the stationary diagram, and you don't even need a flock of observers, since you can use the time between detections to calculate the distance between mirrors and your relative position between them.

On the other hand, if the observers are moving, each observer may get only one detection, so no single observer knows the light is bouncing up and down directly. Obviously, no single observer directly knows the time between ticks also, and each observer makes his detection at a later time than the previous observer (who will be some x ahead and some y below). There's a true x distance between the first observer (who detected the beam at y=bottom) and the last (who detected the beam at y=top). This x distance implies a time separation. If you want to diagram what a single observer would see, you have to consider this time separation.

If you want to compress all the observers' detections into one diagram, you will get a zig-zag path only because you are compressing observations from many different observers into one diagram, which is very different from diagramming what a single observer sees and very little SR-like. If you choose one observer (A) from the flock from where to do measurements, you will only detect the beam directly once. Any detection after that is done by another observer (B) and occurs some time later. If you want to know about the detection of B, B has to send a signal to you, and you must consider the time it takes for this signal to reach A, so you can calculate the time of detection on B's clock. Only after you do this you will be able to diagram the beam's path properly. The coordinates of the beam are accessible to you only through information detected by the other observers, and it takes time to receive any information from other observers, so you have to apply the transforms. Since the only direct information you have from the beam is a single detection at a single (x,y,z,t) point, you have to receive data from other observers to calculate the beam's path, and you must apply the transforms to find the coordinates measured by the other observers. In the end, you will graph the local x,y,z,t coordinates from every observer and you will find the path to be just like the stationary diagram.

----------

Either if you want your observers to communicate between them with light signals or with written notes, you can't transmit information faster than c, and the time it takes for that information to reach you must be accounted for.

And you must have some sort of transmission of information between the event and the observer. If the event is light being reflected from one mirror to another, this light doesn't reach the distant observer and he can only know about it by secondary means, either light signals, a flock of observers or scattered light. All of which are more realistic than the original diagram, which I take that many of you have already shown to be at the very least incomplete, since it has been admitted and demonstrated over and over that we need some sort of transmission of information (usually light) between the event and the observer to make sense out of it...

 

[altergnostic]

...
The only thing some of you still seem to disagree (?) with me is that once you use information transmitted directly to you, you must apply the transforms using this directly received light, and not on the bouncing undetected light. And once you apply the transforms, you will find the light clock system's coordinates and use them to diagram the path of a beam between the mirrors, which can only yield on stationary-like diagram.
What some seem to be claiming is that you can actually see those diagonal paths, or that they have some intrinsic reality.
Most seem to be missing my argument that if light is reflected at a right angle wrt the mirrors in one system, it must do so in all frames. And I still can't save this diagram without light signals being emitted at each moment of reflection, and if we do that, we have to apply transforms on those light sources, and after that, the coordinates will be the stationary coordinates, and the beam will bounce like in the stationary frame.

 

[altergnostic]

I think you are right in that the fact that the speed of light is independent of the motion of the source should be expanded to include the fact that the direction of propagation is not also independent. With the elaboration of its effects, the relativistic aberration, that is certainly relevant to the light clock and it's understanding.
As for the rest; you may have a certain point but it should be clear that certain unrealistic aspects of most thought experiments are not really relevant to their aid as constructs for understanding. IMO

OK, I can live with that, since we have agreed on the physics, we can have differing opinions on how thought experiments should be done. And yes, most unrealistic aspects of most thought problems are completely acceptable and actually very important as they simplify the problem. I don't think this is the case with the light clock diagrams, though. I really think that leaving a mechanism of transmission between the reflection events and the observer out of the picture is bad, because it allows us to draw the diagonal paths as if they are real data observed by a distant observer in relative motion, and this is not a simplification, it is a wrong assumption. From my experience, wrong assumptions are virulent, and can spread like diseases. We should never allow people to even consider that distant light can be seen. I don't care if the resulting equations are correct. I rather achieve correct equations with correct diagrams.

 

[Austin0]

So I post my question again. What is special about the "sides" of a photon that permit the source to give it momentum but any attempt to do that in the direction of propagation results only in red or blue shift?

It seems that from the side a photon is no different from any other particle.

I thought that was clear. Directly aligned with the motion vector the forward momentum of the source cannot affect the motion of the photon in that direction because that speed is limited by the intrinsic properties of spacetime.
In this regard a photon is no different than a massive particle. For instance an electron accelerated laterally would conserve the forward momentum completely but if accelerated along the direction of motion it would not. For example say it was accelerated to 0.99c in a frame that was going 0.8 c relative to the lab. it could not conserve that 0.8c momentum in the forward direction completely and so have a velocity of 1.79 c in the lab frame as would be the case with a low velocity bullet. It instead would have a velocity of .9988 in the lab.
Exceedingly little contribution from the motion of the source. 0.0088c

 

[bahamagreen]

You know... part of the problem (the part about the motion of the emitter and wondering why the remote observer wouldn't expect the light go 90 degrees from the point of emission and miss the moving mirror) is that the translation of Einstein's writing confounds two terms.

The translations say that "the velocity of light is not influenced by motion of the emitter", but in German their word for velocity only means speed, not speed and direction like we use velocity as a vector in English.

If you read "velocity", and assume it is a vector sourced at the emission, the remote observer would expect the light to ignore the emitter's motion and travel 90 degrees from the emission point with respect to the rest frame of the remote observer, and miss the mirror...

 

[marty1]

Angle is the end result, but the truth is that the sideways component of the momentum of a photon is complete under the influence of the source. I guess this is obvious but something I have never seen. Photons are only special along their axis of propagation.

 

[harrylin]

[..] Ok. Where are these plates? If they are on the stationary system, they will measure the path straight up. If they are moving relative to the beam, they will draw a straight vertical line on each (considering a long exposure), but the scattered light will reach each plate at a right angle at each moment of detection.[..]
Do you see my point now?

No, I only see what appears to be a colossal error.
It's useless to develop this discussion further with times or other things as long as this is not taken care of. Evidently my example is still not clear to you (although it seems clear to the others), for you ask me where the plates are while I thought to have depicted it very precisely. I will make it more colourful then, with a technically extreme scenario.

Perhaps you know the kind of metro trains and airport shuttle trains with glass doors on the train and facing glass doors on the platform. Put the platform doors dangerously close to the tracks, so that they risk touching the train doors when the train passes by. Stick a high tech, high power solid state laser against the bottom of the glass door in the train, aiming straight upwards.
Turn the train into a riding cloud chamber. And stick a huge photographic plate, which we will call plate S, against the glass doors of the platform.
You may also place another photographic plate - let's call it S' - in the train, pushing it against the laser, with the laser in the middle; however I think that there are no issues about that one.

Now speed the train at an impossibly high speed past the platform, giving off a nanosecond light pulse at the exact moment that the laser is next to the middle of the plate S. That point we give the plate coordinates x=0, y=0 for S, and x'=0, y'=0 for S'.

Note: at this point we don't care about any length units; merely a qualitative description will do, in order not to get distracted by non-issues at this point of discussion.

You imply that the horizontal position of the light scattering water molecules at different heights wrt the photographic plate S on the platform is x=0 at any height y. That is exactly what I disproved in my post #14, and which you did not really answer, probably because you did not understand it. I'm confident that now, with the more concrete example of the pulse laser in the train, it cannot be misunderstood. And as it's a few pages back, I'll copy it back here:

Take a light ray going straight up from bottom to top, as depicted on the left, but in a cloud chamber with glass walls, and to which we attach the label S'; necessarily scattered light from halfway up (at Y=0.5L) is also at the same horizontal position in S'.*

However, what if this cloud chamber S' is moving at very high speed to the right as observed by a stationary system S, as depicted in the sketch on the right?

The scattering water molecules at the bottom in S' will be detected at for example x=0 in system S.
However, while the light moves up in S', S' moves to the right. Necessarily the scattering water molecules at 0.5L in S' are not at x=0 in S, but are slightly more to the right. And the scattered light at the top is even more to the right.

IOW, by geometric necessity this is what must be measured in S.

However, according to your answer, the water molecules that scatter must be at rest relative to both plates! But that is not possible, for that corresponds to zero train speed.
If we make the train high enough then the laser light can already be past the platform but still be leaving a trace on the photographic plate on the platform, following your analysis.

Thus, once more: please explain in detail why you disagree with my above analysis.

Next we can discuss your light angle question, which I thought to have answered in post #8.

 

[altergnostic]

:
You imply that the horizontal position of the light scattering water molecules at different heights wrt the photographic plate S on the platform is x=0 at any height y.

I don't know why we are not in sync... I do not imply that. In post 17 I wrote:
"It is at the same horizontal y position at the moment of emission. But when it is detected some x distance away, the beam has moved in the y direction. So you will have the same y coordinate, but not the same t."
Cleay the two x's here are not the same, as I was talking about a x,y graph here just to visualize the detection of the scattered light.
In your setup, I realize now, you are trying to make that x=0.
But it doesn't matter, I fully accept that the setup will describe a diagonal. If the beam is up in the z direction, the distance between the plate and the beam is y and the chamber moves in the x direction, there must a change in x as time elapses due to the motion of the chamber and a change in y due to the motion of the beam. That is clear to me, and I thought I made it clear later on in many posts that I don't disagree with this. Just notice that by taking notes of the times of each x,y position, and the length of the line imprinted on the plate, you will directly calculate light to travel faster than c, but let's go on.

Take a light ray going straight up from bottom to top, as depicted on the left, but in a cloud chamber with glass walls, and to which we attach the label S'; necessarily scattered light from halfway up (at Y=0.5L) is also at the same horizontal position in S'.*

This is what I was talking about previously, if you consider the distance between the beam and the plate, the detection will mark the same x position but not the same y at each instant of time, since at each detection the beam has moved up a bit.

However, what if this cloud chamber S' is moving at very high speed to the right as observed by a stationary system S, as depicted in the sketch on the right?
The scattering water molecules at the bottom in S' will be detected at for example x=0 in system S.
However, while the light moves up in S', S' moves to the right. Necessarily the scattering water molecules at 0.5L in S' are not at x=0 in S, but are slightly more to the right. And the scattered light at the top is even more to the right.

IOW, by geometric necessity this is what must be measured in S.[/I]

Agreed.

However, according to your answer, the water molecules that scatter must be at rest relative to both plates! But that is not possible, for that corresponds to zero train speed.

what? It is not the scattering mecules that are at rest relative to the plates! It is the scattered light, the light that impacts the plate, that is "at rest". It has no x component. It moves straight from the point of emission towards the plate, regardless of the motion of the train. At each instant, light is scattered in all directions from the position of the beam. The beam can keep moving in any direction after that, but that will not influence the direction of the scattered light. The scattered light that reaches you directly has no x or y component, even assuming the beam does.

If we make the train high enough then the laser light can already be past the platform but still be leaving a trace on the photographic plate on the platform, following your analysis.

Of course it can. Remember, it is not the beam that is actually leaving the trace, it is light scattered from the beam. It is very possible that during the time between a certain point of emission and a point of detection the beam moved away from the platform. SR: from the beam's point of view, it leaves the platform at t. From the moving plate frame, it leaves at t + the time between emission of scattered light and detection on the plate.

Thus, once more: please explain in detail why you disagree with my above analysis.

Next we can discuss your light angle question, which I thought to have answered in post #8.

I don't disagree with it, i think that you are not paying close attention to the fact that it is just the cattered light that is detected, not the beam. You must collect your data with this light and than calculate the coordinates of the beam from that. From this point I must introduce SR to make my arguments. Just think about how would you normally apply the transforms in SR. You have to apply the transforms because your data is data from a moving frame, you don't know the primed times and positions of the points where light was scattered from. Once you do, you will get the left diagram. You can do this realistically using data brought by scattered light or you can do it fictitiously using the presumed path of the beam. Both get the same results, but the first is free from incorrect assumptions, the second allows you to believe that it is possible to directly observe light move faster than c, and that you can bring this back to normality with time dilation. But you can really never observe light to move faster than the speed of light.

At least we seem to agree that there must be a mechanism of data transmission between the plate / observer and the beam.

I realized it is crucial to state the the observer is NOT the scientist that processes the plate, it is the plate itself. You have already added scattered light to the setup, so the observer could see anything. The observer, placed in the same situation as the plate, will receive light scattered from a different point at each instant, and he can't use the observed times and positions to calculate the time it takes for the beam to go from y'=0 to y'=n (the primed space-time coordinates), he has to calculate the coordinates in the primed system first, using the observations brought by the scattered light, but not the "observations" brought by the beam.

 

[harrylin]

[..] Cleay the two x's here are not the same, as I was talking about a x,y graph here just to visualize the detection of the scattered light.
In your setup, I realize now, you are trying to make that x=0.
But it doesn't matter, I fully accept that the setup will describe a diagonal. If the beam is up in the z direction, the distance between the plate and the beam is y and the chamber moves in the x direction, there must a change in x as time elapses due to the motion of the chamber and a change in y due to the motion of the beam. That is clear to me, and I thought I made it clear later on in many posts that I don't disagree with this. [..]

Good!
So you agree that the observed trajectory in the train is straight up, as depicted on the left sketch in the diagram, and the observed trajectory on the platform is diagonal, as depicted on the right sketch in the diagram.

However, if indeed you agree that these are the traces that can be observed, then it is very puzzling to me why you have a problem with that so that you stated that this not what will be observed:

"Where are these plates? If they are on the stationary system, they will measure the path straight up. If they are moving relative to the beam, they will draw a straight vertical line on each" (post #25, bold mine)

Here, just as apparently before, you seemed to claim that there would be a straight vertical line on both a moving and a stationary plate, instead of, as you now agree, a diagonal like in the sketch on the right.

[..] I don't disagree with it, i think that you are not paying close attention to the fact that it is just the cattered light that is detected, not the beam. You must collect your data with this light and than calculate the coordinates of the beam from that.

I was so far only discussing the picture of a straight upward trajectory as observed by one observer and a diagonal trajectory as observed by another observer. It appeared from your earlier remarks that you thought that that picture was wrong, and so I first tried to make clear that it is necessary that straight up for the one must be diagonally for the other.

From this point I must introduce SR to make my arguments. Just think about how would you normally apply the transforms in SR. You have to apply the transforms because your data is data from a moving frame, you don't know the primed times and positions of the points where light was scattered from. [..]

That is wrong of course: these can be communicated by the train observer to you and vice-versa. In my original version I did not have photographic plates but detector arrays; these can signal detection events directly by radio to both observers in the train and on the platform. Thus both data sets can be directly available to both, just as this is commonly depicted in SR textbooks. And why did you think that it matters?

Note: there appears to be a misunderstanding about relativistic effects which you try to "fix" by means of mistaken arguments. There is definitely no problem/contradiction/paradox involved; to the contrary, understanding how the two data sets can be observed by both observers is the key to correct understanding. Or, in other words: denial of this reveals a misunderstanding of SR. It was because I noticed that sign of misunderstanding that I did not let you get away with it.

[..] I realized it is crucial to state the the observer is NOT the scientist that processes the plate, it is the plate itself. You have already added scattered light to the setup, so the observer could see anything.

In SR, "observations" are merely the observed phenomena, such as the diagonal on the photographic plate and the sequence of light propagation from left-bottom to right-up wrt that plate such as can be provided by detectors. And of course, also clock readings and ruler measurements can be observations. Often "observers" are introduced which represent operators who read instruments or look at such a photographic plate. Sometimes even an observer's eye can be used, however that is not a suitable instrument for precise (x, y, z, t) recordings. In physical experiments one uses detectors, oscilloscopes and photo plates. I will thus replace your "observer" by "photo-detector" and comment with that reinterpretation:

The [STRIKE]observer[/STRIKE]photo-detector, placed in the same situation as the plate, will receive light scattered from a different point at each instant, and the observer can't use the [STRIKE]observed[/STRIKE] recorded times and positions to calculate the time it takes for the beam to go from y'=0 to y'=n (the primed space-time coordinates), he has to calculate the coordinates in the primed system first, using the [STRIKE]observations[/STRIKE] recordings brought by the scattered light, but not the "observations" brought by the beam.

The recorded times and positions of S differ from those of S'. That is correct. I don't understand your last remark, or why you think that this simple fact somehow invalidates the recorded trajectories on the photographic plates, as depicted in the figure which you claim to be wrong for a reason that now completely escapes me. Also (from earlier):

Just notice that by taking notes of the times of each x,y position, and the length of the line imprinted on the plate, you will directly calculate light to travel faster than c.

Perhaps it's time for you to present a calculation, based on either that figure or the photographic plates, and which you think would produce a speed of light greater than c. Probably an interpretation error will show up (but perhaps a calculation error).

 

[Dale]

The point is that the propagation of the signals, or, the times of observations of each tick of the clock from the moving frame point of view, must enter the equations.

Why? How does the manner of transmitting the information to a remote observer change the operation of the clock in any way? How does broadcasting a tick using light or sound or paper printouts after the tick change the time it takes for the clock to tick? That seems to violate causality.

As it is, the observer is assumed to instantaneously see each event, which is very not-SR.

That is certainly not an assumption. It is an irrelevant detail, so it is omitted along with other irrelevant details like the price of pork bellies.

EDIT: Actually, on further reflection I realized that the whole concept of the external observer is irrelevant, all that is relevant is the clock and the reference frame chosen to analyze the clock.

 

[harrylin]

PS I had overlooked a point (bold mine):

[..] I don't disagree with it, i think that you are not paying close attention to the fact that it is just the cattered light that is detected, not the beam. You must collect your data with this light and than calculate the coordinates of the beam from that.

What you still seem to miss is the main purpose of my illustration: the (x, y) and (x', y') coordinates of the trajectory are here not calculated but directly measured on the plates (or, if done electronically, by the detector cell position in the detector array). Perhaps you meant that the time values are a matter of calculation? Yes, of course, on the plates they are a matter of calculation (with time proportional to speed and distance).

However, that is not a matter of principle, as you seem to assert. Each scatter event is a detection of the light beam at that point. With electronic detection the time labels can be directly provided by internal clocks next to each detector cell, so that to each scatter event an (x,y,t) label can be given by these enhanced detectors. As z is already known, this gives a full (x,y,z,t) measurement of the light beam wrt both S and S'. The "trick" here is that the time labels are in part a matter of convention, different for each system. The trick is not that these data points cannot be obtained by measurements in both systems; I think to have this sufficiently illustrated by now.

So, I think that now the time has come for you to present your calculation to us (see my former post).

 

[altergnostic]

However, if indeed you agree that these are the traces that can be observed, then it is very puzzling to me why you have a problem with that so that you stated that this not what will be observed

In your setup, yes, that is what is recorded on a plate, but notice that if you apply the same setup for the train and embankment problem, your observer (plate) will cover the entire length of the tracks, from the origin to the end.
Multiple detector cells imply a timespace separation between them, which is not consistent with the notion of a single observer, since an observer can't be at any distance from himself. Every single cell must therefore be a unique observer.

That is wrong of course: these can be communicated by the train observer to you and vice-versa. In my original version I did not have photographic plates but detector arrays; these can signal detection events directly by radio to both observers in the train and on the platform. Thus both data sets can be directly available to both, just as this is commonly depicted in SR textbooks. And why did you think that it matters?

Note: there appears to be a misunderstanding about relativistic effects which you try to "fix" by means of mistaken arguments. There is definitely no problem/contradiction/paradox involved; to the contrary, understanding how the two data sets can be observed by both observers is the key to correct understanding. Or, in other words: denial of this reveals a misunderstanding of SR. It was because I noticed that sign of misunderstanding that I did not let you get away with it.

Of course they can be communicated, that was my first argument, actually, that everything from the primed frame has to be signaled to the observer, orherwise he can't observe anything and we can't even begin to make the setup logical. And it matters, as you will see later on, because the times and distances of these signals must enter the calculations. In an array of detectors or a big plate this is not clear, but that is only because we actually have a large number of "observers" in such a setup - each cell or detector is at a different position and the distance scattered light has to travel from the beam to the detector right next to it is the same for all detectors. If consider a single point where scattered light converges, maybe you can start to see my point before I even begin the math.

[/QUOTE] Also (from earlier):
"Just notice that by taking notes of the times of each x,y position, and the length of the line imprinted on the plate, you will directly calculate light to travel faster than c."

Perhaps it's time for you to present a calculation, based on either that figure or the photographic plates, and which you think would produce a speed of light greater than c. Probably an interpretation error will show up (but perhaps a calculation error).[/QUOTE]

Excuse me for this long delay. The above has been written over a month ago. I've been extremely busy with other things over the past month, but let's get back to it.

The first thing i want to point out is that a photosensitive plate must actually be thought of as a set of spatially separated observers. The very fact that you need multiple detector cells, or that each detection occurs at a different place in the plate, is enough to demonstrate this.

The setup should have the observer fixed at some origin. Consider this:

All lengths and distances are primed because they are all locally measured numbers. If you like, imagine that before the experiment we have walked all the distances and measured them with a measuring rod. You can even imagine that we have marked each km with a big sign with the km distance painted on it, so even in the moving system S we know the primed spatial coordinates from direct observation.

Fix an obsever at A, his frame denoted by S.


V=c/2= the velocity of the light clock in the x direction, as shown on the onboard speedometer. Also, this velocity has been previously agreed upon, so there's no doubt that's the speed of the light clock.

BC=y'=1 lightsecond = the distance between mirrors (measured locally, remember) and this is a given.

AC=x't=0.5 lightseconds = distance traveled by the light clock in the x direction between reflection events T_A and T_B

T_A=T'_A= 0 = time of first recorded reflection, at both S and S' origins, which are both at A.

T'_B=1s = the time of the second reflection event as calculated in the rest frame S' (I say calculated because it is half the time between two consecutive observed reflections in the bottom mirror, where we can assume the observer inside the light-clock would be fixed at)

T_B=T'_B + h' = the moment a reflection signal (i.e.: scatter event) from B is seen in S, which is the the time light takes to travel from the reflection event T'_B to the observer at A.

This triangle represents the first two reflection events as seen in the moving frame S, where one mirror is seen at A when T'=T=0 and the other at B when T'=1, which are separated by y'=1 lightsecond.
The light clock is going at .5c, and this is given by an onboard speedometer, and has been decided from the begining, so the observer in S can assume it as given. But this velocity is only useful in this setup to determine AC. Since we are trying to get numbers from the beam going AB, that's where S should fix the x axis, just like in the train problem.

AB = h' = 1.12
T_B = T'_B + h = 2.12s
V_AB = h'/T_B = .53c

That's what the observer at A would conclude. You will say that we can't do it this way because we are using primed values for distances, and while S would really see them compressed, he can use primed numbers because they are already marked, so we can skip all the transforms.
It is also curious to notice that V_AB is different from V'_AB, and that this would seem to imply that the speed of light is not the same for all observers. This is the most important place to reestate that the beam is NOT observed in the same manner that we observe light. In this setup, the observations are of events (scatter events, radio signals, etc), and we receive the data from these events through EM waves, and it is those waves that always travel at c: detected light. Seeing light by secondary means can very easily change the relative speed of light, that's why I believe it is of great importance to remember always that we must consider light detection as a local event in all relativistic problems.

As a footnote, I want to point out that doing the inverse processo, using the unprimed values and transforming them into primed values, or correcting the speed of the beam to c, will result in the beam covering a straight line with the same length as y', and since we know from experiment that light travels at c, it must do so inside the lightclock as well, and that's how we would correctly diagram the behaviour of the beam, if we want to keep it's velocity as c. And that's why I have been saying that the only way we would correctly diagram the path of the beam is in the stationary frame. Light detection always occurs locally, and any .

My last remark is that it seems impossible to use the lorentz transformations consistently in this setup. First, it is hard to decide which V to use, since they are not the same for each observer, and secondly, the values don't add up either way. And if we plug in c for the speed of the beam, gamma vanishes, as we all know.
As it is, we have no clear position for the observer. As has been proposed in this discussion, we may include an observer that is large enough to cover the entire distances with, but that can only be consistent with SR if it is thought of as an array of observers. A single observer in SR must be thought of as a point at the origin of its own frame of reference, where observations are made from - or we should take the spatial separation between detections into account.

If you don't agree with the above, even if you're able to correct my analysis, I would like to see how you would deal with the setup such as I presented, with the same givens.

 

[PeterDonis]

I would like to see how you would deal with the setup such as I presented, with the same givens.

I haven't been following this thread closely, but I'll take a crack at re-stating the problem and then giving a quick analysis.

You have basically defined four events, which I'll label A0, B1, C1, and D2:

A0: The light clock pulse is emitted.

B1: The light clock pulse bounces off the mirror.

C1: The spatial location of the light clock source/detector "when" the light clock pulse bounces off the mirror. We'll defer discussion of this event for a bit.

D2: The light clock pulse is received.

In the primed frame (i.e., the rest frame of the light clock), three of these events have coordinates (if I've understood you correctly), given as (t', x', y'):

A0': (0, 0, 0) -- the origin of the frame

B1': (1, 0, 1) -- the mirror is 1 light-second away from the origin in the y direction, so light takes 1 second to get to it. No distance is traveled in the x-direction.

D2': (2, 0, 0) -- light takes another second to get back to the spatial origin, where the source/detector sits.

Now we want to transform to the unprimed frame, in which the light clock is moving at v = 0.5 in the positive x direction. That means our transform from the primed to the unprimed frame uses a v of *minus* 0.5 in the x direction. The transformation formulas are thus:

t = 1.16 (t' + 0.5 x')

x = 1.16 (x' + 0.5 t')

y = y'

where 1.16 is the gamma factor associated with v = 0.5. This gives for the three event coordinates in the unprimed frame (t, x, y), where I have assumed that event A0 is the common origin of both frames:

A0: (0, 0, 0)

B1: (1.16, 0.58, 1) -- notice that the y-distance is still 1, because the relative motion is in the x-direction, so distances in the y direction are not affected. However, the mirror has traveled 0.58 light seconds in the x direction, in a time of 1.16 seconds according to this frame's clock (the light clock only registers 1 second, but this is the "time dilated" time from the viewpoint of the unprimed frame, i.e., it is the 1.16 seconds that elapse in the unprimed frame, times the time dilation factor).

D2: (2.32, 1.16, 0) -- similar remarks here, 2.32 seconds have elapsed in the unprimed frame, and the light clock has moved 1.16 light seconds in the x direction.

Now let's talk about event C1, the one I deferred discussion of. The way you have defined it, its coordinates in the unprimed frame should be:

C1: (1.16, 0.58, 0) -- the source/detector in the light clock has moved along with the rest of the light clock in the x direction, but it remains at y = 0.

But notice that this means the light clock has traveled 0.58 light seconds, according to the unprimed frame, between the events you are calling T_A and T_B (and I am calling A0 and B1). If we now transform back to the primed frame, we find that the primed coordinates of event C1 are:

C1': (1, 0, 0) -- the source/detector stays at the spatial origin at all times in the primed frame.

The above analysis looks correct to me, and I don't see any paradox anywhere; it just requires being careful about defining events and frames.

 

[altergnostic]

A0: The light clock pulse is emitted.

B1: The light clock pulse bounces off the mirror.

C1: The spatial location of the light clock source/detector "when" the light clock pulse bounces off the mirror. We'll defer discussion of this event for a bit.

D2: The light clock pulse is received.

Using your terms, the events would actually be:

A0: The light clock beam is emitted/reflected from bottom mirror.

B1_a: The light clock pulse bounces off the top mirror - in the direction of the bottom mirror (towards D in the unprimed frame).

B1_b: The mirror emits a signal towards A.

A2: The signal reaches the observer at A.

In the primed frame (i.e., the rest frame of the light clock), three of these events have coordinates (if I've understood you correctly), given as (t', x', y'):

A0': (0, 0, 0) -- the origin of the frame

B1': (1, 0, 1) -- the mirror is 1 light-second away from the origin in the y direction, so light takes 1 second to get to it. No distance is traveled in the x-direction.

D2': (2, 0, 0) -- light takes another second to get back to the spatial origin, where the source/detector sits.

D2' doesn't enter the problem, since we are seeking just the time of the reflection on the top mirror and the behavior of the light beam from A to B from the point of view of the observer at A. We could include a third event in place of D2' which would be the time the observer receives the signal as seen from the light clock, but that is not what we are concerned with.

Now we want to transform to the unprimed frame, in which the light clock is moving at v = 0.5 in the positive x direction...

t = 1.16 (t' + 0.5 x')
x = 1.16 (x' + 0.5 t')
y = y'

...1.16 is the gamma factor associated with v = 0.5...

That can't be right, although I see your reasoning. Your are calculating gamma using the speed of the light clock, but we must place the x-axis in line with the beam (which would be h'), because we are seeking values for the beam itself. All your subsequent coordinates are then jeopardized.

Now let's talk about event C1, the one I deferred discussion of. The way you have defined it, its coordinates in the unprimed frame should be:

C1: (1.16, 0.58, 0) -- the source/detector in the light clock has moved along with the rest of the light clock in the x direction, but it remains at y = 0.

But notice that this means the light clock has traveled 0.58 light seconds, according to the unprimed frame, between the events you are calling T_A and T_B (and I am calling A0 and B1). If we now transform back to the primed frame, we find that the primed coordinates of event C1 are:

C1': (1, 0, 0) -- the source/detector stays at the spatial origin at all times in the primed frame.

Of course the conclusion that the detector stays at the origin is correct. But gamma is not correct here. Also, you are using 0.58 for the distance traveled from the point of view of the observer at A, but remember the only way the observer has to know this distance would be by knowing the velocity, but he knows only the speed of the light clock, not of the beam (he has to calculate it). You may say that he does know the speed of the beam, since it must be c, but that's an assumption I want you to hold for a second and imagine how that would actually be measured. That's why I placed km signs at every distance, so the observer can use the primed spatial coordinates to get the velocity and do the transforms.

The above analysis looks correct to me, and I don't see any paradox anywhere; it just requires being careful about defining events and frames.

The problem with your analysis and what I have been saying from the start is that the reflection event at T'=1s is not directly observed, it can only be calculated from some sort of signal emitted from the top mirror (B). And that has to be achieved by taking the primed space and time coordinates of that event and adding the time it takes for the signal to reach the observer.

Also, you are using the velocity of the light clock in the x direction, but that is only useful to determine the length of the opposite side of the triangle, so we find the hypotenuse (h') and calculate the time it takes the signal to reach A. I know this is confusing, but bare with me. Remember the train and embankment problem? We fix the x-axis in line with the tracks, because we are seeking numbers for the train. Here, we are seeking numbers for the beam, so our x-axis must be fixed in line with the path of the beam (h'), and plug in the speed of the beam. You can't plug in the speed of the light clock in the x direction, since that would only work if we were seeking numbers for the light clock's motion, which is not the case. Just as we plug the speed of the train to find gamma, we must use the speed of the beam here in the same way.

Another remark, it seem you are imagining the beam returning to the unprimed origin or something like that, but we don't need to worry about the path of the beam after this reflection. But if we were to take this into account, the beam moves back to the bottom mirror some further distance away (in the unprimed frame). The observer never sees the beam directly, the beam is always moving away from A, so it never reaches him. We need another light source reaching that observer bringing information from the beam if we want him to see anything. If we send a signal from the top reflection event (at T'=1s), and we are given the primed coordinates of that event, we just have to calculate how much time it takes for a light signal from B to reach the observer at A and calculate the speed of the beam from that. Remember that distances are given and times are observed in my setup, and we are given the speed of the light clock, but the speed of the beam can only be c over an assumption, otherwise it becomes clear that c only applies to light that reaches the observer, which is what I believe the postulate of the speed of light really means. We never see Einstein talking about the speed of light at a distance, since that light is plainly undetectable. It is a constant relative to the source and the observer, but that says nothing about the speed of light relative to a non-observer.

At the very least, even if you don't agree with the bulk of my analysis, I think it should be clear that the present light clock diagrams are illogical, or incomplete, since the position of the observer is never defined, nor the process by which the beam's path is determined in the unprimed frame.

 

[Mentz114]

We never see Einstein talking about the speed of light at a distance, since that light is plainly undetectable. It is a constant relative to the source and the observer, but that says nothing about the speed of light relative to a non-observer.

Sneaky stuff, that light. When we're not watching, it speeds up or slows down ?

Do what Einstein would have done - imagine a set of observers with clocks and rulers arranged along the path of the light.

 

[Dale]

At the very least, even if you don't agree with the bulk of my analysis, I think it should be clear that the present light clock diagrams are illogical, or incomplete, since the position of the observer is never defined, nor the process by which the beam's path is determined in the unprimed frame.

The price of beans in Botswana is also not shown on the diagram. Does that make them incomplete?