What Are the Paradoxes of the Light Clock Problem in Special Relativity?

[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?

 

[harrylin]

[..] 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.

 

[altergnostic]

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.

 

[altergnostic]

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.

 

[nitsuj]

altergnostic, what are you suggesting with post #66.

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

 

[Dale]

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.

 

[Mentz114]

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.

 

[altergnostic]

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?

 

[PeterDonis]

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.

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.

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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).

 

[altergnostic]

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.

 

[altergnostic]

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.

 

[Dale]

B1_b: 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.

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.

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.

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.

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.

 

[Dale]

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.

 

[Dale]

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.

 

[harrylin]

[..] 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.

 

[altergnostic]

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?

 

[altergnostic]

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)

 

[Dale]

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.

 

[harrylin]

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?