Exploring the Paradoxical Light Clock Problem in Special Relativity Analysis

In summary: No, angles are comparative, it just means that the angle between the emitted light and the observer is the same as the angle between the emitted light and the diagram.
  • #71
altergnostic said:
[..] 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.
Somewhat yes, as I earlier elaborated: every CCD is a separate detector, and the same for the at those points located quartz clocks in my example. An observer collects all the recorded (x,y,z,t) data and constructs either an (x,y,z) trajectory plot or an (x,y,z,t) "space-time" plot. According to SR, if the clocks were synchronised to the rest frame of the position detectors then the laws of nature will work fine for the "space-time" plot.

Which is exactly the contrary of what you claimed:

"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"

And thus your attempts to "solve" the problem as follows:
[..] 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. [..]
However, such an attempt to avoid looking at the collected data is, as I stated, and as probably PeterDonis now clarified (I have not looked at your and his calculations yet), not at all what SR pretends; and what SR pretends is not a problem.

And it matters, as you will see later on, because the times and distances of these signals must enter the calculations.
Yes of course. As you might have seen, I implied that in my post #63 with as many according to you "impossible" information points as you could want. And as for your calculation, I will look into that later - if that still has any use, in view of already someone else having done so.
 
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  • #72
DaleSpam said:
The price of beans in Botswana is also not shown on the diagram. Does that make them incomplete?

Hahaha very funny. Look, this is not trivial, since it directly affects the numbers and the calculated (with signals) or observed (with scattering) speed of the beam! Deciding a position for the observer demonstrates that the speed of the received light is constant and BECAUSE of that the beam's velocity must be relative.
 
  • #73
Mentz114 said:
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.

And? Would that represent what a single observer observes? This is the detection plate all over again. That is useful if we are seeking local data, but if I am given local distances and times, as in the proposed setup, it just confirms the givens.
 
  • #74
altergnostic, what are you suggesting with post #66.

I am unsure how the speed of c could vary. Are you suggesting that it does?
 
  • #75
altergnostic said:
Hahaha very funny. Look, this is not trivial, since it directly affects the numbers and the calculated (with signals) or observed (with scattering) speed of the beam! Deciding a position for the observer demonstrates that the speed of the received light is constant and BECAUSE of that the beam's velocity must be relative.
It may not be trivial, but you are wrong that it is important. It does not affect the numbers. PeterDonis gave a correct analysis above without once mentioning the position of the observer. It is irrelevant to the analysis.
 
  • #76
altergnostic said:
And? Would that represent what a single observer observes? This is the detection plate all over again. That is useful if we are seeking local data, but if I am given local distances and times, as in the proposed setup, it just confirms the givens.
Are you concerned that in the moving frame the light has two components of velocity ? Because that is a mere coordinate effect and can be removed with a spatial rotation.
 
  • #77
DaleSpam said:
It may not be trivial, but you are wrong that it is important. It does not affect the numbers. PeterDonis gave a correct analysis above without once mentioning the position of the observer. It is irrelevant to the analysis.

He clearly fixed the observer at the origin A, and his numbers depend on this position. It can't be irrelevant because it changes all times and distances. Is fixing the observer at the embankment (origin) irrelevant in that famous thought problem?
 
  • #78
altergnostic said:
B1a: The light clock pulse bounces off the top mirror - in the direction of the bottom mirror (towards D in the unprimed frame).

B1b: The mirror emits a signal towards A.

Are you saying there are two light pulses inside the light clock? I thought there was only one (and your picture shows only one), in which case events B1a and B1b are identical.

altergnostic said:
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.

This is not consistent with what you have been saying (at least as I understand it). I understood you to be describing a scenario where the direction the light pulse travels inside the light clock is perpendicular to the direction the light clock is moving (in the frame in which it is moving). Is that correct? If so, my coordinate description is perfectly valid. If instead the light clock is supposed to be moving in the *same* direction as the light pulses within the clock, then please clarify that point so I can re-do my analysis; obviously the analysis I posted above is not applicable to that case, and I wasn't claiming that it was.

I can't respond to the rest of your post until the above questions are answered, since we may be talking about different scenarios.
 
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  • #79
altergnostic said:
He clearly fixed the observer at the origin A, and his numbers depend on this position.

Of course they do; if I put the origin in a different place, the coordinates of events will be different. If that's all you are saying, then of course we all agree, but then what's the point of belaboring it? None of the physics depends on where you put the origin; that's what "translation invariance" means.
 
  • #80
Mentz114 said:
Are you concerned that in the moving frame the light has two components of velocity ? Because that is a mere coordinate effect and can be removed with a spatial rotation.

Not only that, even in the frame where the light's velocity vector has both x and y components, it's easy to verify that the worldlines of the light pulses are null (by calculating the spacetime interval between the event coordinates I gave).
 
  • #81
nitsuj said:
altergnostic, what are you suggesting with post #66.

I am unsure how the speed of c could vary. Are you suggesting that it does?

Not exactly. Anytime you detect light, it is moving at the same constant speed. But light moving away from us (or any direction that doesn't reach us directly) can't be observed, so we can't check it's velocity. Einstein himself has sort of pointed us in this direction actually. He said that if we were to emmit light towards a mirror some distance away and wait for it to come back, we would calculate the speed as c, but we can't know if the receding and approaching speeds were the same, so we just assume it is as a convention, since we will never use receding light to get information. Every SR experiment that has ever been done depends on light reaching some sort of detector directly. But here we are talking about the behavior of light moving away.

I am saying that observed light moves at c. If the approaching signals move at c, and so does the beam from one mirror to another in the primed system, than the apparent (path of the) beam will seem to move slower than light, but that is only because we are not receiving that light directly, we are actually observing signals or scattering events, and the coordinates and magnitudes apply to those events, not to light itself.
 
  • #82
Mentz114 said:
Are you concerned that in the moving frame the light has two components of velocity ? Because that is a mere coordinate effect and can be removed with a spatial rotation.

Not at all, I am proposing the spatial rotation to solve. As observers at A, we have to align our x-axis with the path of the beam (and I fixed the observer at A precisely to make this as simple as possible). I am just cautious to determine how we could ever draw that path of the beam from A to B, since that light is moving away. We must receive some sort of signal, and that is directly detected moving at c. It is because those move at c that the path must seem to move slower than c. Also, because of the spatial rotation, we are no longer allowed to use the speed of the light clock in the former x direction to find gamma.
 
  • #83
altergnostic said:
B1b: The mirror emits a signal towards A.

A2: The signal reaches the observer at A.
Neither of these even happen, they violate the laws of optics. The angle of reflection equals the angle of incidence, so the reflected signal goes towards D, not A.
altergnostic said:
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.
Nonsense, the x-axis is not in line with the beam in either frame. All of his coordinates are correct.

altergnostic said:
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
If you are not using that assumption then you are not doing relativity, but are simply speculating, which is against the rules here.

altergnostic said:
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.
So what? You repeat this trivially true statement that has no bearing on the analysis as though it is some profound insight instead of an obvious distraction. This has no bearing on the operation of the clock itself. You can send signals via EM, sound, or even FedEx. The clock doesn't tick faster or slower.

I reiterate the question I posed back in post 64, which you avoided. 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.

altergnostic said:
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.
All the bulk of your analysis shows is that your analysis skills are pretty poor. You are completely lost in irrelevant details, and don't even know how to account for them.
 
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  • #84
altergnostic said:
He clearly fixed the observer at the origin A, and his numbers depend on this position. It can't be irrelevant because it changes all times and distances. Is fixing the observer at the embankment (origin) irrelevant in that famous thought problem?
Yes. The position of the origin is always irrelevant. That is one of the fundamental symmetries of physics.
 
  • #85
altergnostic said:
Not exactly. Anytime you detect light, it is moving at the same constant speed. But light moving away from us (or any direction that doesn't reach us directly) can't be observed, so we can't check it's velocity.
This is completely illogical. If we can build a device to measure the speed of incoming light then we can build as many such devices as economics permits. We can place them all around and then the light which is moving away from one is still moving towards another. We can then determine it's velocity regardless of its direction. Scientists are not limited to observations made with their own eyes.

Not only is it illogical, it is also contrary to experiment. There is no indication that the laws of physics have a preferred direction, so there is no reason to think that the speed shouldmdepend on the direction. In fact, there is a significant body of evidence showing the opposite.
 
  • #86
altergnostic said:
[..] If the approaching signals move at c, and so does the beam from one mirror to another in the primed system, than the apparent (path of the) beam will seem to move slower than light, but that is only because we are not receiving that light directly, we are actually observing signals or scattering events, and the coordinates and magnitudes apply to those events, not to light itself.
That is wrong of course - and the cause of your attempts to "fix" what needs no fixing.
The whole point of my illustration up to post #63 was that SR predicts that the detected (x,y,z,t) scatter data as well as the (x',y',z',t') scatter data of all the points in S and S' will describe the same lightbeam as propagating at c (more precisely at c/n) wrt each reference system.
I hope that your calculation relates to that according to you impossible case, for else the discussion is not effective.
 
  • #87
DaleSpam said:
Neither of these even happen, they violate the laws of optics. The angle of reflection equals the angle of incidence, so the reflected signal goes towards D, not A.

You are correct, that was a typo from my part, the mirror sends/reflects a BEAM towards B, and B sends a signal towards A at the moment of reflection. Of course the reflected beam moves towards D, that's why the observer at A can't see it and determine it's coordinates directly.

Nonsense, the x-axis is not in line with the beam in either frame. All of his coordinates are correct.

Substitute the beam for the train, and the point A for the embankment. Where would you align your x axis? You can completely remove the light clock itself from the situation, after you find the opposite side of the triangle ABC. We are not seeking numbers for the light clock, we are seeking numbers for the beam, here. Remember, we are NOT seeing the beam directly.

If you are not using that assumption then you are not doing relativity, but are simply speculating, which is against the rules here.

I am not speculating! Just the other way around, I am checking to see if the unprimed frame actually sees that beam go at c. First we agreed that he didn't see it, so we made the setup work so that he did, somehow (through scattering or reflection times signaling). Now, all I am doing is using that received data to check the beam's coordinates from the unprimed frame's point of view. That's what we would do if the beam's path was flown by a rocket, for instance. My assumption is simply to give c to any and all light that we receive, and use that information to check the speed of the beam, since we can't really measure it's velocity directly.

So what? You repeat this trivially true statement that has no bearing on the analysis as though it is some profound insight instead of an obvious distraction. This has no bearing on the operation of the clock itself. You can send signals via EM, sound, or even FedEx. The clock doesn't tick faster or slower.
I reiterate the question I posed back in post 64, which you avoided. 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.

Because it is not trivial, Dale! To answer your question, it doesn't. The manner of transmitting data just delays the time of reception, causing an apparent time dilation. If you turn on a light at some agreed upon time, I will see it lit later, and more the further we are. If I write "it's 9PM now" and give you this note one hour later you will not agree with the note, but will understand it took one hour for the note to reach you, which is all I am doing here. I am taking events in the primed frame and seeing at what times the observer in the unprimed frame will see them, assuming all data reaches him at c. That's it!

Later you said:
This is completely illogical. If we can build a device to measure the speed of incoming light then we can build as many such devices as economics permits. We can place them all around and then the light which is moving away from one is still moving towards another. We can then determine it's velocity regardless of its direction. Scientists are not limited to observations made with their own eyes.

Not only is it illogical, it is also contrary to experiment. There is no indication that the laws of physics have a preferred direction, so there is no reason to think that the speed shouldmdepend on the direction. In fact, there is a significant body of evidence showing the opposite.

I agree. Any detection will yield c. The laws of physics certainly have no preferred direction, since it is hard to understand what "prefered" would be in the first place. The mirror at B detects the beam going at c. It then sends a signal to A. A detects the signal going at c. We are in agreement. Now, the observer at A can only use the data brought by that signal to determine the speed and the coordinates of the beam (replace "beam" with bullet, if it makes you uncomfortable). That's all I am saying. He doesn't detect the beam itself, he may not even know it is a beam moving from A to B, after all, he knows only what he detects, which is a signal sent from a point 1.12 lightseconds away.
All the bulk of your analysis shows is that your analysis skills are pretty poor. You are completely lost in irrelevant details, and don't even know how to account for them.

So please, given primed distances and times, taking the observer to be at A, a beam moving from A to B, and a signal is sent from B to A at the time of reflection (T'=1s), how would you solve? You know the distance between A and B is h' (1.12). What is the observed time for event B in the unprimed frame?
 
  • #88
harrylin said:
That is wrong of course - and the cause of your attempts to "fix" what needs no fixing.
The whole point of my illustration up to post #63 was that SR predicts that the detected (x,y,z,t) scatter data as well as the (x',y',z',t') scatter data of all the points in S and S' will describe the same lightbeam as propagating at c (more precisely at c/n) wrt each reference system.
I hope that your calculation relates to that according to you impossible case, for else the discussion is not effective.

Hi, Harry. That may have been your point, but did you check if it works? If instead of a beam we had a dark (undetectable) bullet going from A to B, with this set of givens:
h'=1.12
T'A=0s
T'B=2s
How would you solve? Is the speed of the bullet as seen from inside the bullet-clock the same as the speed of the bullet as calculated by the observer at A? And how does replacing the bullet back with the beam change our operations?

Just a note, we may take that the primed frame doesn't need any scattering to get data for the beam. Let's just say the primed observer IS the bottom mirror. The beam is seen by direct detection in the primed frame.
(back at the previous setup on post #66, T'B=1s is calculated in the primed frame as half the time it takes the beam to reach back to the bottom mirror at T'=2s, as I said there, precisely to avoid the need for scattering or signaling in the primed frame)
 
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  • #89
altergnostic said:
To answer your question, it doesn't. The manner of transmitting data just delays the time of reception, causing an apparent time dilation.
Since you agree that the manner of transmitting the data to the observer does not affect the operation of the clock, then logically you must conclude that the operation of the clock can be analyzed without considering the manner of transmission. If X doesn't affect Y then Y can be analyzed without considering X.

This refutes your premise.

EDIT: it is, of course, possible to take the results of the analysis of the clock and then use that to analyze the reception of signals by the observer. But it is not necessary in the clock analysis, and the clock analysis must be done before the signal analysis.
 
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  • #90
altergnostic said:
Hi, Harry. That may have been your point, but did you check if it works?
Yes, I checked this several times in the past, enough to make me realize that I was merely verifying that the Lorentz transformation really work. Did your calculation test your claim that it (in fact the Lorentz transformation set) doesn't work?
If instead of a beam we had a dark (undetectable) bullet going from A to B, with this set of givens:
h'=1.12
T'A=0s
T'B=2s
How would you solve? [..]
That is similar to an accelerated electron, which is much more complex than the simple light ray problem that we are discussing - except if we approximate it with Newtonian mechanics. Light is much easier to calculate, as we can simply use c instead of a to-be-solved V. It only distracts from the topic as long as your topic problem is not solved.
Just a note, we may take that the primed frame doesn't need any scattering to get data for the beam. Let's just say the primed observer IS the bottom mirror.
The observer has nothing to with it, as explained by me and others. Instead you can have detectors, rulers and clocks everywhere you like. If I correctly recall, this whole topic came from your claim that it is impossible to detect (x,y,z,t) of light in transit, because if we could (but we can, as I showed), this would according to you cause a self-contradiction of SR. We all agree that there is no problem if there only is a single detector, so why would you discuss that?? Did you try to show that self-contradiction with a calculation based on my detailed scenario, and if so, did PeterDonis' answer suffice to show that it is no problem?
 
  • #91
bahamagreen said:
The OP's language in presenting the paradox is similar to that found here. I'm guessing the OP has read this... others might do so as well to get a sense of the proposed problem.

#28 Has anyone read it?

I read the beginning of this document. My point of view:

"my long paper on Special Relativity" ...didn't read it, will not read it.
"this diagram creates a false visualization" ...maybe.
"A light clock works by emitting a light ray. This ray reflects from a mirror opposite the clock and returns.
One round-trip of the light is a tick of the clock." ...No the light clock is the emitter/reciver + the mirror.
"The diagram is meant to be a visualization of what a distant observer would see." ...No, the clock could fit
in your pocket, it would even be better if it does. It is meant to (visualize)describe the clock of a moving
observer, that is the clock he has pulled out from his pocket, seen by you who see this clock moving with velocity
v.
"The diagram must be from the point of view of a distant observer, since a local observer would not see the clock
moving." ...? Why?
 
  • #92
altergnostic said:
So please, given primed distances and times, taking the observer to be at A, a beam moving from A to B, and a signal is sent from B to A at the time of reflection (T'=1s), how would you solve? You know the distance between A and B is h' (1.12). What is the observed time for event B in the unprimed frame?

You still haven't answered the questions I asked in post #78. Is the motion of the light clock perpendicular to the direction the light pulse goes, or parallel to it? Are there two light pulses, or only one? Nobody can "solve" anything until we know what the scenario is.
 
  • #93
Altergnostic, one further brief explanation why your premise is false. If an object moves along some arbitrary path r(t) in an inertial frame and at some time t emits a light signal towards the observer at the origin who receives the signal at a time T, then the observer can write:
[itex]r(t)^2=c^2 (T-t)^2[/itex]
Which is one equation in one unknown and has a single root where T>t. So we can always obtain t given T, and so the delay from T to t is immaterial.
 
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  • #94
DaleSpam said:
Since you agree that the manner of transmitting the data to the observer does not affect the operation of the clock, then logically you must conclude that the operation of the clock can be analyzed without considering the manner of transmission. If X doesn't affect Y then Y can be analyzed without considering X.

This refutes your premise.

EDIT: it is, of course, possible to take the results of the analysis of the clock and then use that to analyze the reception of signals by the observer. But it is not necessary in the clock analysis, and the clock analysis must be done before the signal analysis.

I'm not sure I follow your reasoning here. What is it that you call clock analysis and signal analysis precisely? If I understand you correctly, isn't the clock analysis simply the primed diagram? And if so, don't you agree that that the observer in the unprimed frame won't agree with that analysis?

Anyway, the maner of transmission diesn't affect the operation of the clock itself, it alters how the clock will seem to operate from a moving frame. It is just an apparent effect.
 
  • #95
harrylin said:
...Did your calculation test your claim that it (in fact the Lorentz transformation set) doesn't work?

That is similar to an accelerated electron, which is much more complex than the simple light ray problem that we are discussing - except if we approximate it with Newtonian mechanics. Light is much easier to calculate, as we can simply use c instead of a to-be-solved V. It only distracts from the topic as long as your topic problem is not solved.

The observer has nothing to with it, as explained by me and others. Instead you can have detectors, rulers and clocks everywhere you like. If I correctly recall, this whole topic came from your claim that it is impossible to detect (x,y,z,t) of light in transit, because if we could (but we can, as I showed), this would according to you cause a self-contradiction of SR. We all agree that there is no problem if there only is a single detector, so why would you discuss that?? Did you try to show that self-contradiction with a calculation based on my detailed scenario, and if so, did PeterDonis' answer suffice to show that it is no problem?

Yes, it showed that the Lorentz transformations are not useful here, which is not a big deal since it is already accepted that you can't use gamma to transform a body moving at the speed of light wrt an observer.

I don't see how replacing the beam with a slower moving bullet is so complex at all. It is the same analysis, with different primed time values, nothing more.

And Peter's analysis is incorrect because he used the speed of the ship while looking for numbers for the beam. By applying a spatial rotation on the observer's frame, the x-axis should be aligned with h'. The main problem here us that everyone is simply assuming that the beam will be observed to move at c from the moving frame's point of view, but Peter is the only
who actually tried to analyse the setup I proposed.

The basic contradiction with your proposed scenario is that you are using several observers to track the beam while I'm discussing how would the beem look like for a single observer. And as you are assuming that it would seem to move at c and refuse to follow my analysis I proposed you replace the beam with a slower projectile to see how that alters the analysis (because it doesn't), but you refused to look into it also.
 
  • #96
PeterDonis said:
You still haven't answered the questions I asked in post #78. Is the motion of the light clock perpendicular to the direction the light pulse goes, or parallel to it? Are there two light pulses, or only one? Nobody can "solve" anything until we know what the scenario is.

I thought this was already clear from the diagram but I'm glad to clarify. The clock moves perpendicular to the direction of the light pulse (the line drawn from one mirror to another in the rest frame). There's a light pulse going from A to B, which we have been calling the beam and is the pulse that makes the clock tick by moving between the mirrors, and there's another pulse from point B to the origin to communicate the time of the reflection event to the observer, since he can't see the beam in this setup. (harrylin has been proposing scattering so that the observer can see the beam also, so you can replace the signal with the scattered light coming from B, but it works just the same).
And since we are looking for numbers for the beam we really need the spatial rotation so that the observer's x-axis is in line with AB.
 
  • #97
DaleSpam said:
Altergnostic, one further brief explanation why your premise is false. If an object moves along some arbitrary path r(t) in an inertial frame and at some time t emits a light signal towards the observer at the origin who receives the signal at a time T, then the observer can write:
[itex]r(t)^2=c^2 (T-t)^2[/itex]
Which is one equation in one unknown and has a single root where T>t. So we can always obtain t given T, and so the delay from T to t is immaterial.

I agree. We are given t'B=1s and we are looking for T. Place AB=h'=1.12 as your arbitrary direction, find T with a given t'B=1s. That's the time the observer will see the object reach B, and the "object" is the beam. What is the apparent velocity?
 
  • #98
altergnostic said:
The clock moves perpendicular to the direction of the light pulse (the line drawn from one mirror to another in the rest frame).

Ok, good.

altergnostic said:
There's a light pulse going from A to B, which we have been calling the beam and is the pulse that makes the clock tick by moving between the mirrors, and there's another pulse from point B to the origin to communicate the time of the reflection event to the observer, since he can't see the beam in this setup.

Ok, this was the part I hadn't understood. Now it's clear. See below for revised analysis.

altergnostic said:
And since we are looking for numbers for the beam we really need the spatial rotation so that the observer's x-axis is in line with AB.

It doesn't matter whether it's the x or the y axis; I had picked the x-axis for the direction of motion of the light clock as a whole since that's a common convention in SR problems. To humor you I'll re-do my analysis with the light pulses moving in the x-direction and the light clock as a whole moving in the y direction relative to the observer.

We have two frames, the "observer" frame (the frame in which the observer is at rest), which will be the unprimed frame, and the "clock" frame (the frame in which the clock is at rest), which will be the primed frame. The relative velocity between the two frames is v = 0.5; the light clock is moving in the positive y-direction in the observer frame (and therefore the observer is moving in the negative y-direction in the clock frame). The gamma factor associated with this v is 1.16 (approximately). So the transformation equations are:

Unprimed to Primed Frame

[tex]t' = 1.16 ( t - 0.5 y )[/tex]
[tex]x' = x[/tex]
[tex]y' = 1.16 ( y - 0.5 t )[/tex]

Primed to Unrimed Frame

[tex]t = 1.16 ( t' + 0.5 y' )[/tex]
[tex]x = x'[/tex]
[tex]y = 1.16 ( y' + 0.5 t' )[/tex]

We have six events of interest (two pairs of events occur at the same point in spacetime and so have identical coordinates, in either frame):

D0 - The light clock source emits a light pulse towards the mirror.

A0 - The observer is co-located with the light clock source at the instant that the pulse is emitted. Thus, events A0 and D0 happen at identical points in spacetime. This point is taken to be the common origin of both frames (moving the origin elsewhere would just add a bunch of constant offsets in all the formulas, making the math more complicated without changing any of the results).

B1a - The light pulse reflects off the mirror.

B1b - A light signal is emitted by the mirror back towards the observer, carrying the information that the light pulse has struck the mirror. Events B1a and B1b happen at identical points in spacetime.

A2 - The observer receives the light signal emitted from event B1b.

D2 - The light clock detector (which is co-located with the source) receives the light pulse that was reflected off the mirror.

We know that the spatial distance between the light clock source/detector and the mirror, in the clock frame, is 1. This, combined with the information that events A0/D0 are at the origin, fixes the following coordinates (primes on the event labels denote coordinates in the primed frame):

[tex]A0 = A0' = D0 = D0' = (0, 0, 0)[/tex]

[tex]B1a' = B1b' = (1, 1, 0)[/tex]

[tex]D2' = (2, 0, 0)[/tex]

Simple application of the transformation equations above gives the unprimed coordinates of B1a/B1b and D2:

[tex]B1a = B1b = (1.16, 1, 0.58)[/tex]

[tex]D2 = (2.32, 0, 1.16)[/tex]

All of this is the same as I posted previously, just with the x and y coordinates switched, since you prefer to have the x-axis oriented in the direction the light pulse travels.

It only remains to calculate the coordinates of event A2. It is easiest to do this in the unprimed frame, since the observer is at rest at the spatial origin in this frame. Therefore, a light pulse emitted towards the observer from event B1b has to travel from spatial point (1, 0.58) to spatial point (0, 0). A light pulse's worldline must have a zero spacetime interval, so the elapsed time in the unprimed frame must satisfy the equation:

[tex]\Delta t^2 - \Delta x^2 - \Delta y^2 = 0[/tex]

or

[tex]\Delta t = \sqrt{ \Delta x^2 + \Delta y^2 } = \sqrt{ 1 + (0.58)^2 } = 1.16[/tex]

Which of course is just the gamma factor. We could have seen this directly by realizing that the time elapsed in the unprimed frame from event A0 to event B1a must be the same as the time elapsed in the unprimed frame from event B1b to event A2; but I wanted to calculate it explicitly to show how everything fits together. [Edit: In other words, I wanted to show that we don't have to *assume* that the light signal travels at c, which is what "zero spacetime interval" means. We can *prove* that it must, by comparing the result we get from the direct method I just gave, with the result we get from the interval calculation I just gave, and seeing that they are the same.]

So we have the unprimed coordinates for event A2:

[tex]A2 = (2.32, 0, 0)[/tex]

Again, a simple calculation using the above transformation formula gives:

[tex]A2' = (2.68, 0, -1.34)[/tex]

What is this telling us? Well, the observer is moving in the negative y-direction in the primed (clock) frame, so the y-coordinate of event A2 is negative in this frame. The time in this frame is *larger* than that in the unprimed frame because it takes extra time for the light signal to catch up to the observer since the observer is moving away from it. We also expect this from the relativity of simultaneity: events A2 and D2 are simultaneous in the unprimed frame (the reason why should be obvious from the discussion I gave above), so they won't be simultaneous in the primed frame; the event that is in the opposite direction from the relative motion (A2 in this case) will occur later in the primed frame.

Once again, this is all straightforward analysis and I don't see a paradox anywhere; it just requires being careful about defining events and frames. I'll put any responses I have to other comments you've made (now that I know we are both talking about the same scenario) in a separate post.

Edit: I suppose I should add that it's easy to confirm that *all* of the light pulses travel at c, in both frames, from the coordinates that I gave above.
 
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  • #99
altergnostic said:
AC=x't=0.5 lightseconds = distance traveled by the light clock in the x direction between reflection events TA and TB

This is not correct; what you are trying to calculate here is the unprimed coordinates of an event that I didn't list in my analysis:

D1 - The event at which the light clock source/detector is located at the instant the light pulse reflects off the mirror, in the unprimed frame.

The unprimed coordinates of this event are obvious from my analysis:

[tex]D1 = (1.16, 0, 0.58)[/tex]

i.e., the same t and y coordinates as B1a/B1b, just x = 0 instead of x = 1. This means, of course, that the distance AC is 0.58, *not* 0.5, in the unprimed frame. And, of course, the velocity of the light clock source/detector, which is just the y-coordinate of D1 divided by the t-coordinate, is 0.5, as it should be.

The transformation equations give for the primed coordinates of event D1:

[tex]D1' = (1, 0, 0)[/tex]

which is obviously what we expect given the primed coordinates of B1a/B1b.

altergnostic said:
AB = h' = 1.12
TB = T'B + h = 2.12s
VAB = h'/TB = .53c

These are incorrect as well, as you can see from my analysis:

- The distance AB, in the unprimed frame, is equal to the gamma factor, 1.16; I don't know where you got 1.12 from.

- The time TB is the t-coordinate of event A2 in the unprimed frame, which is 2.32, i.e., twice the gamma factor; it is *not* obtained by adding T'B and h, which makes no sense since you are adding quantities from different frames.

- The relative velocity VAB, which as I understand it is supposed to be the velocity of the light clock as calculated by the observer, is, as I showed above, 0.5 no matter how you calculate it, as long as it's a valid calculation. Even with a correct value for h' and T'B, dividing them to get a relative velocity is not valid, and I don't understand why you think it makes any sense.
 
  • #100
altergnostic said:
Yes, it showed that the Lorentz transformations are not useful here
Of course the LT or their equivalent are useful here; it's what the light clock example makes clear. You can use Lorentz contraction and clock synchronization to find time dilation with it.
, which is not a big deal since it is already accepted that you can't use gamma to transform a body moving at the speed of light wrt an observer.
There is no body moving at the speed of light in this example...
I don't see how replacing the beam with a slower moving bullet is so complex at all. It is the same analysis, with different primed time values, nothing more.
The velocity transformation equation for bullets is more complex than putting c, especially at an angle. See section 5 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

And Peter's analysis is incorrect because he used the speed of the ship while looking for numbers for the beam. By applying a spatial rotation on the observer's frame By applying a spatial rotation on the observer's frame, the x-axis should be aligned with h'.
Apparently he has corrected it now. However, I missed why anyone would need a spatial rotation - S and S' are moving in parallel and the light ray reflects, it doesn't rotate.

The main problem here us that everyone is simply assuming that the beam will be observed to move at c from the moving frame's point of view, but Peter is the only
who actually tried to analyse the setup I proposed.
See again post #71. Is it needed? Your set-up is supposedly the standard one of textbooks, which everyone including myself analysed (but with pen and paper many years ago).
The basic contradiction with your proposed scenario is that you are using several observers to track the beam while I'm discussing how would the beem look like for a single observer. And as you are assuming that it would seem to move at c and refuse to follow my analysis I proposed you replace the beam with a slower projectile to see how that alters the analysis (because it doesn't), but you refused to look into it also.
Apparently you now mean with "observer", not a non-local observer of instruments but a light detector with a clock next to it (right?). This topic started as follow-up of the other thread with "You can't detect light at a distance. [..] You can't be aware of light moving in any direction other than straight into your eyes (or detectors). So how can a non-local observer see those light rays?" I think (but I did not see you acknowledge it) that that problem has now been solved. Correct? Also, as far as I can see all your questions here have been answered, which were:

- How can we correctly diagram undetected light like that? How does the emitter adjust the angle of emission?
- Conversely, how can light be emitted at an angle if it's speed is not affected by the motion of the emitter?
- Shouldn't we apply SR transforms primarily with light that is actually observed and than use that information to diagram the interior of the light clock?
- Isn't diagraming the light clock in the moving frame like that illegal? Aren't those vectors purely imaginary?

And so you moved on to a different issue, which is, if understand you correctly, that according to you the Lorentz transformation don't work. Once more: if more people need to search for the error in your calculations, just ask!
 
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  • #101
Peter, you made what I believe to be the same mistake I pointed out earlier. You again plugged in the speed of the light clock into gamma. I insist that you can't do it that way. We are not looking for numbers for the light clock, we are looking for numbers for the beam. My analysis is useless if you start assuming the beam going at c frim the beginning. As I said, the speed of the light clock only helps to find the opposite side AC. If you really want to humor me, place the x-axis aling with AB. Imagine the whole light clock as invisible and all you can see is the signal coming from B after a while or something. We actually don't even need the speed of the light clock at all, since we have km marks all iver the place and we know the primed time. Take the primed time for event B (1s) and the given distance AB (1.12).

And the relative velocity Vab is the number we are seeking: the observed speed of the beam.
 
  • #102
altergnostic said:
I'm not sure I follow your reasoning here. What is it that you call clock analysis and signal analysis precisely?
The clock analysis is the analysis of the clock itself, how its mechanism functions in any reference frame. It is, by your own admission, unaffected by how signals about its operation are transmitted to any observer, and so the analysis of those signals is unnecessary for the analysis of the clock mechanism, contrary to your premise.
 
  • #103
altergnostic said:
I agree. We are given t'B=1s and we are looking for T. Place AB=h'=1.12 as your arbitrary direction, find T with a given t'B=1s. That's the time the observer will see the object reach B, and the "object" is the beam. What is the apparent velocity?
Why are we looking for T? You already agreed that it cannot affect the operation of the clock in any way. It is irrelevant.
 
  • #104
harrylin said:
There is no body moving at the speed of light in this example...

But there is! The beam is our object of analysis here, and it is going at c in the primed frame. We are not seeing it directly, so we need the signal from event B to know its speed.

The velocity transformation equation for bullets is more complex than putting c, especially at an angle.

Do it like the train and embankment problem and simply ditch the angle. You are given the primed distance between AB and the primed time for event B, you don't need anything else.



Apparently you now mean with "observer", not a non-local observer of instruments but a light detector with a clock next to it (right?). This topic started as follow-up of the other thread with "You can't detect light at a distance. [..] You can't be aware of light moving in any direction other than straight into your eyes (or detectors). So how can a non-local observer see those light rays?" I think (but I did not see you acknowledge it) that that problem has now been solved. Correct?

I aknowledge that you found a setup that allows us to observe the beam, and you did it only by sending light from the to the detector, so I give you that. But that also proves what I have been saying, that you can't simply take local or primed numbers for the beam and use them as unprimed data. The beam has since acted like an object subject to relative velocities, it is just like a moving train that we see with light reflected from it. The light between the train and embankment acts the same as light between beam and detector, and it is this light that we directly observe that we know moves at c. It is this light that brings us the coordinates for the beam.

- How can we correctly diagram undetected light like that? How does the emitter adjust the angle of emission?
Answered: with a new setup. Current setup is incomplete.

- Conversely, how can light be emitted at an angle if it's speed is not affected by the motion of the emitter?
Actually this hasn't been answered at all, but I already concluded that light that is moving in any direction other than directly at us doesn't have to follow any constancy, since we can't be considered neither source nor observer in that case.
- Shouldn't we apply SR transforms primarily with light that is actually observed and than use that information to diagram the interior of the light clock?
Clearly the answer is yes, otherwise we wouldn't have even mentioned scattering or signaling.
- Isn't diagraming the light clock in the moving frame like that illegal? Aren't those vectors purely imaginary?
You showed it can be done with a rigorous setup, but I still contend the numbers are different from current diagrams.

And so you moved on to a different issue, which is, if understand you correctly, that according to you the Lorentz transformation don't work. Once more: if more people need to search for the error in your calculations, just ask!

That is a conclusion I came to only in recent posts (I actually learned a lot from your scattering setup) so it is not like I changed the subject, I only went where the discussion led me. You refuse to do the analysis with the set of givens I proposed, which are very realistic. And now you claim that substituting the beam with a prijectile is overly complex when I showed that it isn't, you don't even have to take any angle into account since I gave you the distance AB measured locally so you can put the x-axis directly in line with it, and I also gave primed times so you can calculate the speed.
 
  • #105
altergnostic said:
But there is! The beam is our object of analysis here, and it is going at c in the primed frame. We are not seeing it directly, so we need the signal from event B to know its speed.
:rofl:

Besides the funny self-contradiction, it is one of the postulates of SR that it is going at c in all frames. We know its speed according to SR, and alternative theories that disagree with SR are not permitted here.
 
<h2>1. What is the paradoxical light clock problem in special relativity analysis?</h2><p>The paradoxical light clock problem is a thought experiment used to illustrate the effects of time dilation in special relativity. It involves two identical clocks, one stationary and one moving at a high speed, with a beam of light bouncing between them. According to the theory of special relativity, the moving clock will appear to tick slower than the stationary clock, leading to a paradox where the moving clock appears to be both ticking slower and faster at the same time.</p><h2>2. How does special relativity explain the paradoxical light clock problem?</h2><p>Special relativity explains the paradoxical light clock problem by taking into account the concept of time dilation, which states that time moves slower for objects that are moving at high speeds. This means that the moving clock in the thought experiment will actually tick slower than the stationary clock, and the apparent paradox is resolved.</p><h2>3. What are some real-world examples of time dilation in special relativity?</h2><p>Some real-world examples of time dilation in special relativity include the Global Positioning System (GPS) and particle accelerators. GPS satellites orbit the Earth at high speeds, causing their onboard clocks to experience time dilation. This must be taken into account for accurate GPS calculations. In particle accelerators, particles are accelerated to near the speed of light, causing them to experience time dilation.</p><h2>4. How does the paradoxical light clock problem impact our understanding of time and space?</h2><p>The paradoxical light clock problem, and the theory of special relativity in general, has greatly impacted our understanding of time and space. It has shown that time is not absolute, but rather is relative to the observer's frame of reference. It has also led to the concept of space-time, where time and space are interconnected and can be affected by factors such as gravity and velocity.</p><h2>5. Are there any unresolved questions or criticisms of the paradoxical light clock problem in special relativity analysis?</h2><p>While the paradoxical light clock problem has been widely accepted and used to support the theory of special relativity, there are still some unresolved questions and criticisms. Some argue that the thought experiment oversimplifies the concept of time dilation and does not take into account other factors such as length contraction. Others question the validity of the assumptions made in the thought experiment. However, the paradoxical light clock problem remains a valuable tool in understanding the effects of time dilation in special relativity.</p>

1. What is the paradoxical light clock problem in special relativity analysis?

The paradoxical light clock problem is a thought experiment used to illustrate the effects of time dilation in special relativity. It involves two identical clocks, one stationary and one moving at a high speed, with a beam of light bouncing between them. According to the theory of special relativity, the moving clock will appear to tick slower than the stationary clock, leading to a paradox where the moving clock appears to be both ticking slower and faster at the same time.

2. How does special relativity explain the paradoxical light clock problem?

Special relativity explains the paradoxical light clock problem by taking into account the concept of time dilation, which states that time moves slower for objects that are moving at high speeds. This means that the moving clock in the thought experiment will actually tick slower than the stationary clock, and the apparent paradox is resolved.

3. What are some real-world examples of time dilation in special relativity?

Some real-world examples of time dilation in special relativity include the Global Positioning System (GPS) and particle accelerators. GPS satellites orbit the Earth at high speeds, causing their onboard clocks to experience time dilation. This must be taken into account for accurate GPS calculations. In particle accelerators, particles are accelerated to near the speed of light, causing them to experience time dilation.

4. How does the paradoxical light clock problem impact our understanding of time and space?

The paradoxical light clock problem, and the theory of special relativity in general, has greatly impacted our understanding of time and space. It has shown that time is not absolute, but rather is relative to the observer's frame of reference. It has also led to the concept of space-time, where time and space are interconnected and can be affected by factors such as gravity and velocity.

5. Are there any unresolved questions or criticisms of the paradoxical light clock problem in special relativity analysis?

While the paradoxical light clock problem has been widely accepted and used to support the theory of special relativity, there are still some unresolved questions and criticisms. Some argue that the thought experiment oversimplifies the concept of time dilation and does not take into account other factors such as length contraction. Others question the validity of the assumptions made in the thought experiment. However, the paradoxical light clock problem remains a valuable tool in understanding the effects of time dilation in special relativity.

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