# Light Clock Problem

by altergnostic
Tags: light clock, special relativity, time dilation
P: 50
 Quote by bahamagreen 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.

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?
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P: 4,773
 Quote by altergnostic 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.
 Mentor P: 15,610 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: $r(t)^2=c^2 (T-t)^2$ 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.
P: 119
 Quote by DaleSpam 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.
P: 119
 Quote by harrylin ...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.
P: 119
 Quote by PeterDonis 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.
P: 119
 Quote by DaleSpam 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: $r(t)^2=c^2 (T-t)^2$ 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?
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P: 4,773
 Quote by altergnostic 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.

 Quote by altergnostic 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.

 Quote by altergnostic 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

$$t' = 1.16 ( t - 0.5 y )$$
$$x' = x$$
$$y' = 1.16 ( y - 0.5 t )$$

Primed to Unrimed Frame

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

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):

$$A0 = A0' = D0 = D0' = (0, 0, 0)$$

$$B1a' = B1b' = (1, 1, 0)$$

$$D2' = (2, 0, 0)$$

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

$$B1a = B1b = (1.16, 1, 0.58)$$

$$D2 = (2.32, 0, 1.16)$$

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:

$$\Delta t^2 - \Delta x^2 - \Delta y^2 = 0$$

or

$$\Delta t = \sqrt{ \Delta x^2 + \Delta y^2 } = \sqrt{ 1 + (0.58)^2 } = 1.16$$

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:

$$A2 = (2.32, 0, 0)$$

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

$$A2' = (2.68, 0, -1.34)$$

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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P: 4,773
 Quote by altergnostic 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:

$$D1 = (1.16, 0, 0.58)$$

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:

$$D1' = (1, 0, 0)$$

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

 Quote by altergnostic 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.
P: 3,179
 Quote by altergnostic 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!
 P: 119 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.
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P: 15,610
 Quote by altergnostic 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.
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P: 15,610
 Quote by altergnostic 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.
P: 119
 Quote by harrylin 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.
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P: 15,610
 Quote by altergnostic 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.

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.
P: 3,179
OK, I think that we are getting somewhere:
 Quote by altergnostic [little mutual misunderstanding ...] 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.
A bullet doesn't have the same speed in both frames; I'll stick to the topic instead, which allows much better clarity.
 [..] 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.
I already illustrated that it is not necessarily to use that light for the time data. So, if you mean that it is light or radio waves or other means that brings us the clock synchronisation, then I completely agree.
 [..] 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.
In post #8 I gave you the answer (in the link). However, you did not comment on it.
[EDIT: In fact, my answer took care of the spatial rotation misconception]
 [..]You showed it can be done with a rigorous setup, but I still contend the numbers are different from current diagrams.
Of course, we narrowed down the problem as stemming from a calculation error - either made by "all" textbooks and students, or made by you. My purpose was (and still is) to get rid of the bug, after which you can be an asset to this forum - and you will feel better.
 [..]I actually learned a lot from your scattering setup
Thanks - such feedback is helpful to remain motivated with this voluntary work!
 so it is not like I changed the subject, I only went where the discussion led me. [..]
OK. I will next look into your calculation example with the light beams, and give my analysis.
P: 119
 Quote by DaleSpam 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.
My premise has nothing to do with the operation of the clock, it has to do with apparent effects. Like doppler and such things.

 Why are we looking for T? You already agreed that it cannot affect the operation of the clock in any way. It is irrelevant.
Because T is the time shown in the unprimed frame when the observer receives the signal from B, which is different from the primed time, and which is the time that determines the perceived speed of the beam from the observer at the origin. It is the time the beam seems to take to cross the distance AB. It is far from irrelevant.

 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.
There's no contradiction whatsoever. Observed/detected light is going c in all frames. Einstein never once thought about how light would look like at a distance since it seemed ridiculous to do so, since we can't have any knowledge of undetected light. He and everyone else always addressed this issue by placing another observer in the path of light, but that says nothing about the observed behavior of light receding from the observer (I really believe that taking harrylin setup of scattering light from the beam's path plus my suggestion that we take this scattered light and bring it to one single point - the observer - is a novelty).
How do you measure the speed of the train? You take the distance it covers over the time it takes to cover it. This velocity is not the same as marked on the trains own speedometer, for instance. You are on the train and it marks .5c on the speedometer. You cross 1 light second, from your perspective, in 2 seconds.
An observer at the embankment would see the train cross that distance in 2 seconds + the time it takes light to reach back to him from that distance (which would be the time marked on the observers watch at the moment he sees the train at that distance). The speed will seem to be slower relative to the speed as seen on the onboard speedometer!

I have a diagram on this from a discussion I had a few months ago:

But I'm afraid this will divert us from the topic, maybe this needs a thread itself.
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