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Light Clock Problem 
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#145
Nov1812, 12:46 PM

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You have said several times now that you can "ditch the speed of the light clock" in the analysis. I don't understand what this means. The light clock itself is part of the scenario, so you have to model its motion to correctly analyze the scenario. If you just mean that you can't assume you *know* what its speed is, that's fine; as I said above, I let that speed be unknown in my latest analysis, and showed how it doesn't matter for the question I thought you were interested inwhy the projectile's velocity changes from frame to frame while the light beam's does not. If you mean that we can somehow model the scenario without including the light clock at all, I don't see how. The motion of the parts of the light clock gives a critical constraint on the motion of the projectile/light beam inside the clock. Also, the light clock does not change direction in the scenario, so you can define a single inertial "clock frame"; but the projectile *does* change direction, so there's no way to define a single inertial "projectile frame". That means we can't just focus on the "velocity of the projectile", because that velocity *changes* during the scenario; it's not a "fixed point" that we can use as a reference. As for "assuming" that the light beam travels at c, I have not assumed that. The only assumptions I have made are translation and rotation invariance, plus the principle of relativity, plus an assumption about how the light clock's "projectile" reflects off the mirror. It's probably futile at this point to walk through the chain of reasoning again, but I'll do it once more anyway. I'll focus on your "triangle diagram" since it's a good illustration of the spatial geometry in the unprimed frame. We have a "projectile clock" consisting of a source/detector, which moves from A to C to D in the triangle diagram, and a mirror/reflector, which moves on a line parallel to the source/detector that passes through B at the same time (in the unprimed frame) that the source/detector passes through C. The clock as a whole moves at speed [itex]v[/itex] relative to the observer who remains motionless at A (in the unprimed frame). The "projectile" within the clock moves at some speed [itex]v_p[/itex] in the unprimed frame (which we take to be unknown at this point), along the line A to B, then B to D. At the instant that this "projectile" (call it P1) reaches B, a second "projectile" (call it P2), moving at the *same* speed [itex]v_p[/itex], is emitted back towards A, to carry the information to the observer at A that the first projectile has reached B. Here are some key facts about the geometry that follow from the above:  Angle ABC equals angle CBD.  Distance AB equals distance BD.  Distance AC equals distance CD.  Line BC is perpendicular to line AD. We can also define the following times of interest (in the unprimed frame): T_AB = the time for P1 to travel from A to B; T_BD = the time for P1 to travel from B to D; T_BA = the time for P2 to travel from B to A. It is then easy to show from the above that all three of these times are equal: T_AB = T_BD = T_BA. We also have T_AC = the time for the light clock source/detector to travel from A to C, and T_CD = the time for the source/detector to travel from C to D. And we have T_AC = T_CD = T_AB = T_BD, because projectile P1 and the source/detector are colocated at A and D and both of their speeds are constant. Thus, we have the following spacetime events: A0 = D0 = the spacetime origin; the light clock source/detector passes the observer at A at the instant projectile P1 is emitted from the source/detector. B1 = P1 reaches the mirror/reflector and bounces off; P2 is emitted back towards A. C1 = the light clock source/detector passes point C. D2 = P1 reaches the light clock source/detector and is detected. A2 = P2 reaches the observer at A. The above facts about the times and the motion of the light clock imply that B1 and C1 are simultaneous (in the unprimed frame), and D2 and A2 are simultaneous (in the primed frame). Finally, we have formulas about the speeds: [tex]v_p = \frac{AB}{T_{AB}} = \frac{BD}{T_{BD}} = \frac{AB}{T_{BA}}[/tex] [tex]v = \frac{AC}{T_{AC}} = \frac{CD}{T_{CD}} = \frac{AC}{AB} v_p[/tex] The last equality follows from the fact that T_AC = T_AB. Do you see what it says? It says that, if you know v_p, AC, and AB, you know v, the "light clock velocity". You stipulated v_p, AC, and AB in your statement of the problem; therefore you implicitly gave a value for v as well. But the value for v that you stated was a *different* v; it did not satisfy the above equation, which is enforced by the geometry that you yourself gave for the problem; that's why your "givens" were inconsistent. I'll stop at this point since this post is getting long. What I'm trying to say is that the scenario you proposed has more constraints in it than you think it does; it already *contains* the information about how the projectile/light beam's velocity has to transform between frames. 


#146
Nov1812, 01:57 PM

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However, if it is not a given in the problem then everything becomes easier. Simply plug the givens into the equation I posted above and we find that v = 0.45 c. At 0.45 c the time dilation factor is 1.12, so in 1.12 s the clock travels .5 ls and the light travels a distance of 1.12 ls. 1.12 ls / 1.12 s = c. 


#147
Nov1812, 03:36 PM

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 he can give [B]no[B] motion to the beam; and how he places his coordinates can have no effect on the beam! Seriously, nothing of that makes any sense to me. 


#148
Nov1812, 04:01 PM

P: 119

Peter, thanks for your patiance and effort. I'll make a couple observations regarding your reply.
You misinterpret my assertion that we can ditch the velocity of the light clock to mean that it is an unknown that must enter the equations nonetheless. No, I mean that we don't need it to solve at all. To determine the speed of the beam all the observer needs to know is the distance teavelled over the observed time  this will be the apparent velocity. Conversely, if you were given the relative velocity of the beam/projectilr from A to B (the vector addition of the upward speed of the projectile and the perpendicular speed of the projectileclock) you could find the distance AB. Given the distance AB you can directly calculate the observed velocity as seen from A, just plot it over the observed time. You have to put yourself in the observer shoes and really imagine how he calculate the speed of the beam. Since we are given the time of event B as measured by the clock itself, we can easily find the time that event is observed at a distance AB. Note this is the case for any scenario, we wouldn't need a beam travelling fron A to B at all. If the blinker at B simply turns on at t=1s (as measured with the blinker's own clock), you just have to consider the spatial separation from the event and the observer to get the observed time. Assuming the event is caused by a beam going from A to B, he only has to plot the given distance over the observed time. The mistake I keep pointing you to is the fact that you assumed the velocity of the clock enters the equations. But it would enter only if we were doing some kind if vector addition, but since the clock is going from A to D and the observer sees the event from B (at a time T), there's no reason to take that velocity into consideration, you are only looking for the time separation between events A and B to figure out at what speed something has to travel this distance to cause the observed times. Later you state that T_AB = T_BA, but that is an assumption. The observer has to calculate T_AB from the observed time of event B, which is the local or proper time of event B plus the time it takes for that event to be seen at A. We know T_BA as seen from A from experiment, the speed of any directly detected light must be c. We don't know T_AB as seen from A because T_AB is only observed after T_BA, orherwise the observer doesn't see event B at all. T_AB must be calculated in this setup. Of course, if the observer at rest were to send a beam of light towards B, that would take the same time as a beam would need to come back from B to A. But notice the very important fact here that, if that was the case, the time of event B as seen from B would NOT be 1s (the time light takes to cross y'), but actually 1.12s (the time it would take for light to cross h'). Your last remark reassures me that you are assuming what you are trying to prove. The setup has no information on the beam, only on the events, which are good to calculate the speed of the beam in each frame. 


#149
Nov1812, 04:29 PM

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#150
Nov1812, 04:29 PM

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So before you do this vector addition, you have to first transform the upward speed of the projectile from the primed to the unprimed frame. The upward distance (AB in the primed frame; BC in the unprimed frame, since the clock is moving in that frame) does not change when you change frames, but the *time* does, because of time dilation, so the upward speed of the projectile (i.e., the upward *component* of its velocity) is different in the unprimed frame than in the primed frame. So you do need to know the relative velocity of the light clock and the observer; without that you can't transform the upward velocity in the primed frame to the upward velocity component in the unprimed frame. By contrast, I am only assuming that the two beams (A to B and B to A) travel at the *same* speed, *without* assuming what that speed is (we *calculate* that by dividing roundtrip distance by roundtrip time, as above). That seems like a much more reasonable approach, since it does not require assuming that there is any difference between an emitted beam and a received beam. 


#151
Nov1812, 06:24 PM

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#152
Nov1812, 09:41 PM

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Giving 1.12 to the time it takes the beam to travel from A to B as seen from A is a mistake. That would only be true if the time of event B were also 1.12s. If you send a light beam from A to B, B would receive it at t'=1.12s. But, in the proposed setup, the time of this event in the clock's frame is t'=1s. Folliwung what has been stated in this thread, the time of event B can be either 1s or 1.12s as seen from the clock's frame and in both cases the speed of the beam as seen from A would be c, but this is only possible if light crosses the distance h' (the hypotenuse) as seen in the clock's frame in one case and the distance y' in the other case, but it is clear that in this setup the beam crosses y' in the clock's frame.
I really don't know how else to state this, but as a last resort, I will ask you to ignore the beam completely and just analyse the events as seen from A, when x is in line with AB, as if no beam was causing the events at all: A0 = A'0 = 0,0,0 B1 = B'1+X = 2.12,1.12,0 While B happens at t'=1s as recorded from B itself. What would be the presumed speed an object would need, as seen from A, to travel from event A0 to B1, if it was to leave A at t=0s and reach B at the time of event B (TB1 = 2.12s)? TB1 = 2.12 X = 1.12 VAB = X/TB1 = 0.528 The observer can subtract the delay caused from BA (the time light takes to travel from B to A) to find the time of the event as measured by a clock placed at B: T'B = TB  TBA = 2.12  1.12 = 1s If you calculate the speed of the beam straight from the times of events and the given distances, that's what you would find, and as the observer standing at A only has access to the events themselves, that's how he would calculate it. We can only disagree on two points, I think: 1: The given distance is not correct or useful or consistent with SR 2: The observed time for B1 as seen from A is not the local or proper time of the event plus the time separation between A and B. Regarding 1, I remind you that any given V implies a given L, which L I'm giving as the distance from A to B (or B to A) measured locally  a proper distance? I don't see why wouldn't we be allowed to be given this length if we are given a relative V in other situations. Regarding 2, I can't see how this isn't the case. The only argument against this is that the primed time is not 1s but 1.12s, which is inconsistent with the light clock's own measurements. This thread is way too long already and I think that if we haven't reached an agreement yet, we won't reach it anytime. Maybe there's an experiment out there that actually determines the speed of an undetected (or indirectly detected) light beam so we could check this, but I believe that this hasn't been done at all. Every experiment built to determine the speed of light that I know about works by measuring the return time or some other setup that relies o direct detection or emission. I repeat that the observer at A is not the source nor the receiver of the beam. The source is the bottom mirror and the receiver is the top mirror. The observer at A is the receiver of the signal travelling at c from B1 to A0. The mirror at B is the receiver of the beam travelling at c from A'0 to B'1 (which equals CB). The postulate of the constancy of the speed if light necessarily applies to those light paths. 


#153
Nov1812, 11:04 PM

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Also, I don't understand your coordinate values. Is the first number supposed to be time? If so, where does 2.12 come from? I can't even make sense of the rest of your analysis, because I don't understand where you're getting the initial numbers it's based on. 


#154
Nov1912, 12:08 AM

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And I didn't give the upward velocity in the unprimed frame because the goal is to find the speed of the beam in the first place. And we can find it since I gave the distances and times you need to do so. The bottom line is that the time of event B in the primed frame is 1 and that same event is observed at A when light from that event reaches A, after crossing BA. If the event at B = 1s was self generated (if it wasn't the outcome of any reflection event, like manually turning on a flash of light), how would you find the observed time for event B as seen from A? You see, the observer moving along with the bottom mirror in the clock only knows T'B because it is half the return time (T'D), so you need both the coordinates of emission and reception to determine anything. Now, the observer at A only knows the emission coordinates for the beam (0,0,0). The next piece of information he receives is the light coming from *event B, from which he must calculate the coordinates of reception! What makes it different is the operation. When we determine the speed if light, we take the distance from the emitter over the time of detection, which is measured locally relative to the point of detection. We are always at rest relative to the point of detection (this is also true for emission). But the detector at B is in relative motion wrt the observer at A, so he is not ar rest relative to the point of detection. This is the fundamental reason, I think, that the speeds are not the same. If you are teavelling along with the clock, you are at rest relative to the point of detection and of emission, so light is always going at c, since all motion is given to the light. If the detector is in relative motion, you are not at rest relative to the point of detection, so you can't give all motion to light. Does this make any sense? Do you ser how this does not violate the light postulate? Light is constant relative to source or detector, but the observer at A is neither source nor detector of the beam going from A to B  the source is the bottom mirror and the detector is the top mirror, and they are both moving relative to A. And I didn't give the upward velocity in the unprimed frame because the goal is to find the speed of the beam in the first place. And we can find it since I gave the distances and times you need to do so. The bottom line is that the time of event B in the primed frame is 1 and that same event is observed at A when light from that event reaches A, after crossing BA. If the event at B = 1s was self generated (if it wasn't the outcome of any reflection event, like manually turning on a flash of light), how would you find the observed time for event B as seen from A? You see, the observer moving along with the bottom mirror in the clock only knows T'B because it is half the return time (T'D), so you need both the coordinates of emission and reception to determine anything. Now, the observer at A only knows the emission coordinates for the beam (0,0,0). The next piece of information he receives is the light coming from *event B, from which he must calculate the coordinates of reception! What makes it different is the operation. When we determine the speed if light, we take the distance from the emitter over the time of detection, which is measured locally relative to the point of detection. We are always at rest relative to the point of detection (this is also true for emission). But the detector at B is in relative motion wrt the observer at A, so he is not ar rest relative to the point of detection. This is the fundamental reason, I think, that the speeds are not the same. If you are teavelling along with the clock, you are at rest relative to the point of detection and of emission, so light is always going at c, since all motion is given to the light. If the detector is in relative motion, you are not at rest relative to the point of detection, so you can't give all motion to light. Does this make any sense? Do you ser how this does not violate the light postulate? Light is constant relative to source or detector, but the observer at A is neither source nor detector of the beam going from A to B  the source is the bottom mirror and the detector is the top mirror, and they are both moving relative to A. See above. Notice that T'AB = CB and TAB is not even measurable from A. TAB is strictly a measurement made in the clock's frame, or more precisely, in the top mirror's frame, since this detection only happens there. The observer at A detects the signal from B to A, and from the given distance and the light speed postulate, he can subtract the time it took light to reach him from event B and find the time of the event in the primed frame. The times are different indeed, but I don't know if I should call the reason "time dilation". Anyway, I'll wait for your followup. 


#155
Nov1912, 06:24 AM

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#156
Nov1912, 09:03 AM

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Also, shouldn't this be A'B' = CB? If both of these distances are in the unprimed frame, they are *not* equal; they can't be, because AB is the hypotenuse of a right triangle and CB is one of its legs. I don't understand what you're doing here. As far as I can tell, you agree with the above; but then you want to go on and claim that, if A starts his clock at zero when the first light beam (from A to B) is *emitted*, he will receive the second light beam (from B to A) at t = 2.12s, because the first light beam will take 1s to travel, *according to A*. But that beam also covers 1.12 light seconds of distance, according to A, so it covers 1.12 light seconds in 1 second, according to A; so you are claiming that light can travel faster than light. Is this what you're claiming? For light, no; the speed of light is independent of the motion of the source. That is what the null result for the MichelsonMorley experiment means; and that experiment has been repeated with greater and greater accuracy, and the null result continues to hold. If you ran a "MichelsonMorley experiment" using slowerthanlight projectiles in the apparatus instead of light, you would *not* get a null result; you would observe different speeds for the two projectiles (moving on perpendicular trajectories) if you were in motion relative to the apparatus. Having said all that, I want to go back to your original claim. Your original claim is different from what you appear to be claiming now. Your original claim was that the standard SR picture of the light clock is inconsistent. I have now produced several analyses that show that the standard SR picture *is* consistent. Now you have shifted your ground, and you are saying that the standard SR picture makes unwarranted assumptions (that the speed of all light beams is c, not just directly detected ones). I have also shown that the assumptions aren't what you are claiming they are (they are things like translation and rotation invariance; things like the speed of all light beams being c are *derived* claims, not fundamental assumptions, if you start with the assumption of translation and rotation invariance). But regardless of the assumptions, the fact is that SR matches experiments, as DaleSpam pointed out. Your claims, such as the light beam from A to B taking only 1s (in the unprimed frame) to travel 1.12 light seconds (in the unprimed frame), do *not* match experiments; if we actually ran a light clock experiment in which the distance from A to B, in the unprimed frame, was 1.12 light seconds, we would *not* get the timing results you have claimed; we would get the results I derived using the standard SR formulas. So I'm not sure where you are going with this thread. Your original claim has been shown to be wrong; the standard SR model of a light clock is consistent. Your claims about the event coordinates are wrong, because they don't match the standard SR model, which agrees with experiment. What now? 


#157
Nov1912, 02:43 PM

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altergnostic, several people have repeated the claim that the experimental evidence supports SR. In fact, if you are uncomfortable with the lightspeed postulate of SR then you could easily NOT assume it, make a general theory of all possible transformations between inertial frames, and use experimental data to set the parameters in such a general theory.
In Robertsons famous paper he did exactly that and demonstrated that SR could be deduced to within about 0.1% from the Michelson Morely, Ives Stillwell, and Kennedy Thorndike experiments even without making Einstein's assumptions. See: http://rmp.aps.org/abstract/RMP/v21/i3/p378_1 I strongly recommend that you read that paper as well as the wealth of experimental evidence listed here: http://www.physicsforums.com/showthread.php?t=229034 


#158
Dec212, 10:41 PM

P: 119

I hold my claim that the standard light clock diagram is inconsistent.
If I understand every claim correctly, the basic disagreement is that, if the distance AB is 1.12ls, then light should take 1.12s to travel that distance regardless of direction, which is basically saying that the speed of light is c regardless of the speed of the source. What seems hard to grasp is that I am not advocating against this assumption. It has been said that if I simply do the MM interferometer analysis I will have solved this setup without any contradiction. I disagree. It has also been said that the speed from A to B has been calculated repeatedly in this thread, but I disagree, and in many places it has been admitted that that speed was assumed based on the distance AB (not calculated), but the purpose of my setup is precisely to have the necessary givens so we can calculate that speed and check that assumption. Another argument is that I am mixing frames by adding primed and unprimed times and that that is not allowed. Although that is generally true, sometimes it is allowed. Of course we can't mix variables that are being measured, but we can add a given with an observation. We are simply taking the proper time of an event and taking into account how long it takes for that event to be seen at another point AB away. If the time (as shown on a clock at B) of an event at B is 0s and light takes 1.12s to get to A, than the observed time for that event as seen from A is 1.12s. Likewise, if the time of the event is 1s at B, the observed time will be 2.12s. That is all I am saying. It has been said that B receives the beam at 1.12s in the observer at A's frame, and 1s in the clock's frame (or the frame of the mirror at B). But that last is a given, and the former is an assumption. If we have an observer at B, at t'=1s the beam appears to come from the stationary mirror at C. At t=1s, B sees the light of the beam on the mirror at C, 1 lightsecond away. The path of detected light in the primed frame is CB. But from the point of view of the observer at A, it is the signal, not the beam, that comes from B, 1.12 lightseconds away. The path of the detected light in the unprimed frame is BA. This is what is observed. The speed of whatever is describing AB, so far, is unknown, as a matter of observations. The question is if the observer at A can assume that the time the beam takes to go from A to B is 1.12s based on his observations. And what are his observations? He knows the distance AB and the time light takes to reach him from any event at the point B. He receives a direct signal from B, but what is the observed time of the event at B? In a real experiment, we would find that number directly, but here we have to use logic to determine it. I say that, from A, the observed time time of event B is simply the time light takes to cross from B to A (tBA) + the proper time of emission: tB = tBA + t'B or t'B = tB – tBA which is t' = t – ct which is simply a way to find the time on a distant synchronized clock: t'B = 0.5(tA1 + tA2) t'B = 0.5(0 + 2.24) = 1.12s Hence every event at B will be observed 1.12s later at A: t' = t – AB and if the reflection event at the point B occurs at t'=1s, then it will be observed in A at: 1 = t – 1.12 t = 2.12s So I hope you see there is no frame mixing. We are not adding quantities that are being measured in different frames. t'B is not (only) being measured in the primed frame, but it is the time of the clock at B given by the standard synchronization method. The observer at A can only conclude that the time of the detection at B is 1s, not 1.12s. Please note that event B is observed at A when t=2.12s and that the beam is reflected from the mirror at B when t=1s. This method applies to anything traveling at any speed between the mirrors (or to nothing traveling at all) because it is related only to the observation of events, regardless of the cause of these events. For an observer at B: V'beam = tCB = c For an observer at A: Vsignal = tBA = c and apparent Vbeam = AB/(tB – tA) = 1.12/2.12 = 0.528c Now, he knows that the time of detection at B is not 2.12s, but 1s, so he could try to calculate the speed of the beam like this: calculated Vbeam = AB/(t'B  t'A) = 1.12/1 = 1.12c But that is a mixing of frames after all. We are no longer using given instants in time of events or observations, but elapsed periods between events, and a period is a measurement, and that measurement does not belong to the observer A  he has transformed an observed period into a directly detected period, or the period as measured by A into the period as measured by B  but hasn't transformed the distance. Therefore, AB no longer applies, since he has to use the distance as observed by B also. That distance can be easily found either by directly using the given primed distance (CB or y' or 1ls) or by turning back to the light postulate and finding the distance from the primed time: x' = ct' x' = 1 = y' So everything's double checked. If at point B there was a clock and a camera, the signal from B to A could carry a photograph of the clock at B and its surroundings and the observer at A would receive direct visual information showing the clock at B marking t'=1s and the mirror reflecting off the beam, and everything would be triple checked. Now, concerning the interferometer, the path AB is not observed nor calculated like in our setup at all. AB was assumed to be the path described by the beam in the ether frame, where light travels at c. It is the path of the beam as seen from an absolute frame. But of course, from the interferometer's point of view, light always reaches the detector directly, like I have been saying. The beam's speed is not being determined by distant events, but by local detections. If our setup was equivalent to the interferometer's, the detected beam would describe CBC and we would have no evidence of path AB at all (just like Michelson and Morley didn't), since we wouldn't be able to determine relative motion. And if we wanted to, we would have to place an observer in a moving frame, just like we are doing here, and we fall back to my analysis. We can't simply assume that path AB is being described by the beam at c as seen by the observer at A because, unlike the ether, he is not present everywhere and must directly receive incoming light to determine any coordinates. Received light is always measured to travel at c, which was thought to imply an absolute ether, but it sounds much more Einsteinian to specify c as the speed of light as measured by any observer (i.e.: by direct detection). Einstein's original claim, that the speed of light is the same regardless of the speed of the source, sounds perfect to me, and I think it is absurd to conclude that the postulate applies to the speed of undetected or indirectly detected light, since Einstein never even tried to describe it that way. He actually required that we place observers everywhere, so that light going in any arbitrary direction was actually directly detected, and it is to that light that we assign c. Objects and events are seen with light and it is with this light that we determine their velocities. If a beam traveling in an arbitrary direction is to be observed with light bouncing off of it (just like an ordinary object) and not directly detected as usual, it is not your standard light anymore, it is the v in v/c, since only the incoming light is that c  it is being detected, not being measured. That c in the denominator is the tool with which we measure other things, and the tool with which we are measuring the speed of the beam from A to B. And we can't have c/c. The denominator is the light with which we see and the numerator is the speed of what we are seeing with that light. The last thing that I think may need clarification is why I ignore the speed of the light clock in my setup. If we are given that number, we will tend to solve like it is always done and not really think the whole problem through. My givens are analogous to a given velocity, since we have fixed times and distances, but I require that the observer's frame is aligned with the frame of the source at B and that both frame's xaxis are in line with AB, since that is the easiest way to find unknowns for AB (we can simply ignore y and z and use the coordinates of each event to find AB). I am simply requiring B's frame of reference to be obtained from A's frame by a standard Lorentz transformation. I hope I made it clear how this setup differs from the assumptions and calculations of the standard lightclock and MM diagrams. 


#159
Dec212, 11:24 PM

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(1) Light pulse 1 is emitted by the light clock source/detector at the instant that it passes the observer at A. That event occurs at time 0 in both the observer frame and the light clock frame. (2) The distance from A to B is 1.12 light seconds in the observer frame. Therefore, we can immediately calculate: (3) Light pulse 1 reaches B at time t = 1.12s in the observer frame, since it is covering a distance of 1.12 light seconds and travels at c. Since you already agree that light pulse 2 takes 1.12s to travel from B back to A in the observer frame, that makes it clear that light pulse 2 arrives at A at time t = 2.24s in the observer frame, *not* t = 2.12s. I won't even bother commenting on the rest of your post; the error you are making should be clear from the above. 


#160
Dec312, 12:32 AM

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On rereading, there is one other error that seems to me to be worth commenting on:
The spatial geometry in the primed frame would be a *different* diagram, in which the positions of the observer at A would lie on a line going out to the left, the spatial path of light pulse 1 would be a line straight up, and the spatial path of light pulse 2 would be the hypotenuse of a right triangle whose legs are the path of light pulse 1 and the path of the observer at A. Any reasoning about spatial paths and lengths in the primed frame would have to be done using this other diagram, *not* the diagram you drew. 


#161
Dec312, 06:57 AM

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#162
Dec312, 07:03 AM

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Furthermore, even without assuming the speed is c, experiments confirm the round trip time (aka the two way speed of light) so any value you choose other than 2.24 s is contrary to experiment. See the Robertson paper I posted earlier. 


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Light Clock  Special & General Relativity  7 