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Light Clock Problem 
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#91
Nov1512, 07:14 AM

P: 112

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 roundtrip 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
Nov1512, 08:45 AM

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#93
Nov1512, 12:40 PM

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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 (Tt)^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. 


#94
Nov1512, 11:17 PM

P: 119

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
Nov1512, 11:42 PM

P: 119

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
Nov1512, 11:52 PM

P: 119

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
Nov1612, 12:19 AM

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#98
Nov1612, 12:48 AM

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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 ydirection in the observer frame (and therefore the observer is moving in the negative ydirection 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 colocated 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 colocated 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 ydirection in the primed (clock) frame, so the ycoordinate 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. 


#99
Nov1612, 01:26 AM

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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 ycoordinate of D1 divided by the tcoordinate, 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.  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 T_{B} is the tcoordinate 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 V_{AB}, 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
Nov1612, 03:01 AM

P: 3,181

 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! 


#101
Nov1612, 04:27 AM

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. 


#102
Nov1612, 06:13 AM

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#103
Nov1612, 06:17 AM

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#104
Nov1612, 06:20 AM

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#105
Nov1612, 06:31 AM

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Besides the funny selfcontradiction, 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. 


#106
Nov1612, 07:00 AM

P: 3,181

OK, I think that we are getting somewhere:
[EDIT: In fact, my answer took care of the spatial rotation misconception] 


#107
Nov1612, 09:16 AM

P: 119

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. 


#108
Nov1612, 09:31 AM

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