# Light Clock Problem

by altergnostic
Tags: light clock, special relativity, time dilation
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P: 4,774
 Quote by altergnostic I prefer this analysis over the previous one, but the problem remains: you are not given any v
You are missing the point of this version of my analysis; I let v be unknown. It is the speed of the light clock relative to the observer, but I assumed no value for it. My whole point was to show that, even if you do not know the speed of the light clock relative to the observer, you can still show that the velocity of the "projectile" inside the clock is different relative to the observer than it is relative to the clock, *if* the "projectile" moves slower than light. But if the "projectile" is a light beam, then its speed is 1 (i.e., c) in both the clock frame and the observer frame. My understanding was that that was your point of confusion: you couldn't understand how a projectile's velocity could change from frame to frame if it moves slower than light, but yet the velocity of a light beam does *not* change from frame to frame. I've now given you an explicit formula that shows why that's true for the clock scenario.

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 in--why 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 $v$ relative to the observer who remains motionless at A (in the unprimed frame).

The "projectile" within the clock moves at some speed $v_p$ 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 $v_p$, 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 co-located 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:

$$v_p = \frac{AB}{T_{AB}} = \frac{BD}{T_{BD}} = \frac{AB}{T_{BA}}$$

$$v = \frac{AC}{T_{AC}} = \frac{CD}{T_{CD}} = \frac{AC}{AB} v_p$$

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.
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 Quote by altergnostic I have reestaded many times that you could ditch the speed of the light clock, since it is useless to this problem.
Yes, you have restated many times many incorrect things in this thread. Obviously, whenever you are interested in the effect of speed on the operation of a clock then the speed of the clock is important.

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.

 Quote by altergnostic given the distance and times you can calculate the speed.
Certainly. And as long as your givens are both self-consistent and consistent with physics then you will always get that light travels at c.

 Quote by altergnostic (you could have length contraction between two observers at rest also).
No, you cannot.

 Quote by altergnostic The given distance h' is not being measured at all in my setup. It has been previously marked with lightsecond signs
This is yet another self-contradictory statement. The lightsecond signs are a measuring device, and comparing anything to them is performing a measurement.
P: 3,179
 Quote by altergnostic Why are they self contradictory? [..]
I trust that that has been solved now; so we simply ditch v and find v=0.45 from your 1.12.
 Quote by altergnostic [..] server line of sight. I propose that he places his x axis in line with AB so he can give all the motion to beam and solve without knowledge of the clock's velocity.
- spatial rotation is useless. Standard is to orient the system speed along X (don't you know that the Lorentz transformation uses that convention?)
- 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.
 From the link, [http://physicsforums.com/showthread.php?t=574757] I will comment only upon the same issue I have been discussing here: the beam leaves A and reaches B in the rest frame [..]
Sorry, the first thing you discussed here and which was not solved were the speed and direction - and those also appear in your last example, so you must be sure to get it right...
 Maybe my terminology is incorrect? The rest frame is meant to be the frame at rest relative to the clock. The moving frame is supposed to be the observer who sees the moving clock.
That is exactly how I understand it; thus you disagree about the calculation method, which is identical to the method to explain the Michelson-Morley experiment.
 The beam never reaches the observer at A. Only the signal from B does. He then marks the observed time of event B and plots it against the given L. He can then find the time of event B as marked on the detector's clock by subtracting the time it takes light to cross the distance between the coordinates of the event at B from the observed time. From that, he can calculate the distance from source to receiver as seen from the clock itself by transforming the calculated speed of the beam into c (the speed of light as seen locally, which is the speed supported by a huge amount of evidence). [..]
That is wrong: it appears that you heard some stuff about SR and some stuff about GR and mixed them up. There is nothing to "transform"; as you earlier claimed there is no transformation to make, as all distance and time measurements are made with the single reference system S', the "rest" system.
 [MMX:] The interferometer was supposed to test the existance of the aether and find a variance of the speed of light relative to the ether. The setup was built so we had perdicular light paths.
Note: the essential point of the experiment is that the light paths in the ether are not perpendicular; I hope that that is clear.
 As the apparatus (and the Earth) revolved, no fringe displacement was detected (actally, no significant displacement). So it was concluded that the speed of light was constant regardless of the orientation of the beams or the detectors relative to an absolute frame.
Not exactly, no... There was only one detector. And no effect was detected from changing orientation regardless of the velocity (although apparently Michelson only measured it at one velocity; others repeated it at other velocities).
 Notice that in the equations applied to this experiment, V was the relative velocity wrt the aether, so there was only one possible V, as the aether was absolute. Since then, the speed if light was measured with ever increasing accuracy, always in the same manner: by noting the return time if light as it went forth and back a specified distance. The time of detection after emission is always proportional to c. This is exactly the premise of my setup: light can't be detected to move at any V other than c. Just notice that the beam is not detected by A, only the signal from B is. There's no return, no detection, nothing that relates the path of the beam with the experiments that tested the speed of light. [..]
I still can't make any sense of that last part. Sorry. But there are sufficient glitches in your descripton, and sufficent lack of equations, to make me confident that before anything else it will be better to go through a special relativity calculation example of MMX. And this thread is already too long. Please start a new one on MMX. I do think it likely that this whole topic here will disappear after that exercise. Especially because:
 This is merely the outcome of a realisation that this thought problem has never been done before and that the postulate of the speed of light applies to source and receiver, but not necessarily to a non-receiver. It is the preconception that the postulate of the speed of light also applies to undetected light that keeps you from aknowledging the possibility I intend to discuss here. [..] the conclusion is strictly dependant upon the acceptance of the constancy of the speed of any received or detected light.
No. Once more: there are detectors where you like, even non-detected light is assumed to go at c, and such experiments as the one you describe have been done, both in theory and in practice. As a matter of fact, radar uses that same set-up. And the basic calculations are similar as with MMX.
 if light reaches us at a constant speed, distance events will be seen at a later time than the time of the event determined locally (for an observer in close proximity and at rest relative to the event). Hence, if the observed time is delayed and distances are given, the observed speed must be smaller.
Do you think that the traffic police will observe speeding cars further away as going slower - or that GPS will tell that you are going slower as the satellite is further away??
 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 projectile-clock) 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.
P: 3,179
 Quote by altergnostic [..] 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. [...]
Just one more remark: you are mixing up reference systems, just as we already discovered - in fact you try to use t instead of t' for S'. That doesn't work. This will become clearer when you discuss MMX, which is the "mother" of all such calculations.
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 Quote by altergnostic 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.
You are correct that you could set up the scenario so that the distances AB, AC, and BC were pre-determined; then you would have to control the speed of the light clock so that the beam actually hit the light clock's mirror at point B, instead of at some other point along the mirror's trajectory. That's fine, but it doesn't change anything about my analysis; my analysis is still correct, because even if you don't need the light clock's speed to do the analysis (which you actually do--see below--but for the moment I'll assume for the sake of argument that you don't), that is not the same as saying that an analysis which does use the light clock speed (even if it is left unknown) is incorrect.

 Quote by altergnostic 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 projectile-clock) you could find the distance AB.
Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified $v_p$, but I should have said that you specified $v'_p$. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.)

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.

 Quote by altergnostic 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.
Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment.

 Quote by altergnostic Later you state that T_AB = T_BA, but that is an assumption.
Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for *both* endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that *receiving* a beam counts as directly observing it, then *emitting* a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam.

 Quote by altergnostic 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.
But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c? If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam?

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 round-trip distance by round-trip 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.

 Quote by altergnostic We know T_BA as seen from A from experiment, the speed of any directly detected light must be c.
But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light?

 Quote by altergnostic We don't know T_AB as seen from A because T_AB is only observed after T_BA
No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed.

 Quote by altergnostic T_AB must be calculated in this setup.
So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction.

 Quote by altergnostic 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.
Unbelievable; you now *admit* this, yet you were arguing that we could *not* assume this before.

 Quote by altergnostic 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').
Which it is; the time of event B, *in the unprimed frame*, *is* 1.12s (if we allow v, the velocity of the light clock relative to the observer, to be set appropriately to 0.45 instead of 0.5, per the comments of DaleSpam, harrylin, and myself). The time of event B, in the *primed* frame, is 1s; but that's not what the observer at A is interested in. He's interested in the time of event B in his frame, the unprimed frame, and that time is different from the time of event B in the primed frame because of time dilation. Which, of course, requires you to know the velocity of the light clock relative to the observer, contrary to your repeated erroneous claim that you don't.
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 Quote by altergnostic 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
No matter how you try to get rid of it, it is there anyway since both the distance traveled and the observed time depend on the speed of the clock.
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 Quote by altergnostic 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.
You keep on mixing up quantities in different frames. When we say that the "time" for the beam to travel from A to B is 1.12s, we mean in the unprimed frame; i.e., t = 1.12s. That is perfectly consistent with that time being 1s in the primed frame; i.e., t' = 1s.

 Quote by PAllen 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
No, it can't. Where are you getting that from? The time of event B (event B1 in my nomenclature) is 1.12s in the unprimed frame (observer's frame), and 1s in the primed frame (clock frame). No one has said that the time of event B1 is 1.12s in the primed frame.

 Quote by PAllen 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 does "recorded from B itself" mean? Who is doing the recording? If the observer at B doing the recording is at rest relative to the observer at A, then he will record t = 1.12s. To record t' = 1s, he would need to be moving with the light clock, i.e., at an angle to the x-axis with the orientation of axes you are using. That means such an observer is not "at B" except at the instant when he records the arrival of the light beam there.

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.

 Quote by PAllen 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.
But the observer at A is co-located with the bottom mirror when it emits the light beam, so he can directly observe its time of emission by any reasonable definition of "directly observe". By your extremely strict definition of "directly observe", practically no events are ever directly observed.

 Quote by PAllen 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.
And that, all by itself, is enough to show that *all* the other light beams in your scenario also move at c. That was one of the points of my various posts analyzing the scenario. But I think you are right that, if we haven't reached agreement on that point by now, we're not likely to.
P: 119
 Quote by PeterDonis Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified $v_p$, but I should have said that you specified $v'_p$. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.)
No problem, that was an honest mistake. Don't bother correcting it.
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.

 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.
You don't need to do this at all, since we are calculating speeds from given distances and times, not from different speed vectors.

 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
Once again, the distance does change between frames. The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'. For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data. Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A, not the speed of the beam - and that is the unknown we are seeking.

 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.
But you don't have to transform upward components to find the observed speed of the beam! All you have to do is plot the given distance over the observed time, and since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A. You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that.

 Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment.
But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove. If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B.
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?

 Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for *both* endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that *receiving* a beam counts as directly observing it, then *emitting* a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam.
You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission. And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates.

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!

 But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c?
Yes!

 If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam?
Finally the fundamental question!
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.

 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 round-trip distance by round-trip 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.
Assuming the speed is the same as seen from A is the problem. You can't make that assumption. You can assume the speed from A to B as seen fron B is the same as the speed from B to A as seen from A, though, but this is not what the setup demands.

 But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light?
See above.

 No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed.
That is what I meant. I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB.

 So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction.
Correct, but he knows the distance AB and he knows that directly observed light must travel at c. T_BA is simply AB/c. This is where the postulate of SR makes the problem possible to solve.

 Quote by PeterDonis Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified $v_p$, but I should have said that you specified $v'_p$. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.)
No problem, that was an honest mistake. Don't bother correcting it.
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.

 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.
You don't need to do this at all, since we are calculating speeds from given distances and times, not from different speed vectors.

 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
Once again, the distance does change between frames. The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'. For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data. Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A, not the speed of the beam - and that is the unknown we are seeking.

 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.
But you don't have to transform upward components to find the observed speed of the beam! All you have to do is plot the given distance over the observed time, and since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A. You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that.

 Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment.
But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove. If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B.
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?

 Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for *both* endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that *receiving* a beam counts as directly observing it, then *emitting* a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam.
You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission. And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates.

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!

 But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c?
Yes!

 If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam?
Finally the fundamental question!
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.

 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 round-trip distance by round-trip 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.
Assuming the speed is the same as seen from A is the problem. You can't make that assumption. You can assume the speed from A to B as seen fron B is the same as the speed from B to A as seen from A, though, but this is not what the setup demands.

 But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light?
See above.

 No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed.
That is what I meant. I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB.

 So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction.
But he knows the distance AB and he knows that directly observed light must travel at c. T_BA is simply*

 Unbelievable; you now *admit* this, yet you were arguing that we could *not* assume this before.
This applies if the observer and the point B are at rest wrt each other.
See above.

 Which it is; the time of event B, *in the unprimed frame*, *is* 1.12s (if we allow v, the velocity of the light clock relative to the observer, to be set appropriately to 0.45 instead of 0.5, per the comments of DaleSpam, harrylin, and myself). The time of event B, in the *primed* frame, is 1s; but that's not what the observer at A is interested in. He's interested in the time of event B in his frame, the unprimed frame, and that time is different from the time of event B in the primed frame because of time dilation. Which, of course, requires you to know the velocity of the light clock relative to the observer, contrary to your repeated erroneous claim that you don't.
I think I have answered this above, but just for good measure, notice I said it must be 1s as seen from B, which would be in the primed frame. But we know that the mirror at B must detect the beam at T=1s measuring from its internal clock, and si the event B must be seen by the observer at A at T=T'AB+TBA.
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 follow-up.
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P: 15,617
 Quote by altergnostic 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.
You are mixing up quantities in different frames. B receives it at t=1.12 (wrt the frame where the clock is moving). B receives it at t'=1 (wrt the frame where the clock is stationary). That is the whole point of the exercise.

 Quote by altergnostic 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.
Yes, that is true. However, since the universe disagrees with your position, I will stick with mine for now. I would rather disagree with you than disagree with the universe.
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P: 4,774
 Quote by altergnostic Once again, the distance does change between frames.
Only distances parallel to the relative motion are affected by length contraction. The distance A'B' (in the primed frame) and CB (in the unprimed frame) are perpendicular to the relative motion, so they don't change, and A'B' = CB by the geometry of the problem.

 Quote by altergnostic The distance marks are clear and visible for both frames. From the clock's measurements, AB = CB = y'.
You don't measure distance directly from clock measurements. You have to have a speed. You are basically allowing the speed of light beams to be c when you want it to be, but insisting that the speed of other light beams is not c when you don't want it to be.

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.

 Quote by altergnostic For the distant observer, event B1 occurs h' lightseconds away from A. This is direct visual data.
Huh? Who is the "distant observer"? Is he located at B and at rest relative to the observer at A (i.e,. at rest in the unprimed frame)? If so, why are you assigning a distance in the primed frame to what he observes?

 Quote by altergnostic Also, time dilation effects in this setup occur because of the constant speed of the signal from B to A
Time dilation is not a "travel time delay" effect. It is what is left over *after* you have subtracted out all effects from signal travel time delay.

 Quote by altergnostic since we know the time of the event B as seen from the light clock's frame, we can add the time the signal takes to reach A from B to find the observed time of event B as seen from A.
You can't add times in different frames and get a meaningful answer. How many times does this need to be repeated?

 Quote by altergnostic You don't need to separate the motion into vector components at all, you have the time and distance of the event at B, you can calculate the observed speed straight from that.
In the primed frame, yes. In the unprimed frame, no.

 Quote by altergnostic But you can't assume both beams travel at the same speed! That's the point of the setup. You have to SHOW that. Otherwise you are assuming what you are trying to prove.
This really gets to the issue of what counts as a "directly observed" light beam, since you agree that any directly observed beam does travel at c. I claim that both beams in this case are directly observed; but you want to say that only one is. See below for further comment.

 Quote by altergnostic If you are at rest in A relative to B and send a beam towards B, it will take 1.12s to get there, but if you are at A moving along with a mirror 1 ls away and send a beam towards the mirror, it takes 1s to reach it, even if it is coincidentally at B.
Unbelievable. Now you *agree* that T_AB (in the unprimed frame) is 1.12s, and T'_AB (in the primed frame) is 1s. But you were arguing before that this is *not* true. Do you read your own posts?

 Quote by altergnostic 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?
*I* would find the observed time for event B as seen from A using the standard SR formulas, which I have already done several times in this thread. You don't appear to like that method, so let's try this one: the beam from B to A is directly detected by A; therefore its speed should be c, as seen by A, according to your own claim that any "directly detected" light beam travels at c. A also knows the distance from B to A; it is 1.12 light seconds (for the numbers we are currently using). Therefore the time of event B, according to A, is 1.12s before the time he receives the light beam.

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?

 Quote by altergnostic You can't calculate the speed of light from the time of emission, you need a distance, a time of emission and a time of reception. The reception is the observation, not the emission.
In other words, you are claiming that it is impossible to directly observe the emission of light?

 Quote by altergnostic And although emission gives you knowledge of the coordinates of one endpoint of a beam (since you can determine the place and time of emission as you please), it is far from enough to determine any speed, any distance travelled and any timr of travel, so we absolutely need the reception coordinates.
The same would be true if we only had the reception coordinates, wouldn't it? We would need *both* sets of coordinates, in the *same* frame, to calculate the speed, right? Oh, wait:

 Quote by altergnostic 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.
And yet you are claiming that the observer at A somehow "directly knows" that the beam he receives from B took 1.12s to get to him, even though he doesn't know the coordinates of emission. In other words, again you are picking and choosing when you can assume light travels at c and when you can't; but yet you also claim that you can't calculate a speed at all unless you know both the emission *and* the reception coordinates. You seem to me to be contradicting yourself.

 Quote by altergnostic 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!
But you did not state what the coordinates of event A2 (the event where the observer at A receives the light signal from B) were; you have claimed to calculate them by adding 1s to 1.12s. You need to justify this calculation. Alternately, if you change your ground and claim that T_A2 = 2.12s is part of your statement of the problem, you need to show how that is consistent with the other givens. You haven't done any of that. Again, you're picking and choosing numbers to suit your claims, without backing them up.

 Quote by altergnostic 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.
What does "measured locally relative to the point of detection" mean?

 Quote by altergnostic We are always at rest relative to the point of detection (this is also true for emission).
Then why is emission treated differently from detection?

 Quote by altergnostic 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?
For a "projectile" moving slower than light, yes; it amounts to saying that the observed speed of the projectile depends on your motion relative to its source.

For light, no; the speed of light is independent of the motion of the source. That is what the null result for the Michelson-Morley experiment means; and that experiment has been repeated with greater and greater accuracy, and the null result continues to hold.

If you ran a "Michelson-Morley experiment" using slower-than-light 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.

 Quote by altergnostic I just point out that the roundtrip can't be split in half to determine any speed here. Theory and experiment clearly shows that the speed from B to A must be c, the rest must be given to AB.
But you have to determine what "the rest" is, and you are trying to do it by mixing numbers from different frames.

 Quote by altergnostic I think I have answered this above, but just for good measure, notice I said it must be 1s as seen from B, which would be in the primed frame.
Yes.

 Quote by altergnostic But we know that the mirror at B must detect the beam at T=1s measuring from its internal clock, and si the event B must be seen by the observer at A at T=T'AB+TBA.
No; you're adding numbers from different frames, which is meaningless. What you should be doing is T=T_AB + T_BA = gamma (T'_AB) + T_BA, where gamma is the gamma factor corresponding to the light clock velocity relative to the observer.

 Quote by altergnostic 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.
This applies to T'_AB, but *not* to T_AB; the mirror can't "directly measure" a time in the unprimed frame, only in the primed frame.

 Quote by altergnostic The times are different indeed, but I don't know if I should call the reason "time dilation".
Time dilation *is* the reason that T'_AB does not equal T_AB; it *is* the reason why I wrote T_AB = gamma (T'_AB) above. Working things out from the geometry of the problem that you gave shows that gamma (T'_AB) = T_AB = T_BA, which means that both light beams (A to B and B to A) travel at c as seen from the unprimed frame. We already know that the beam from A to B travels at c as seen from the primed frame; we then have to calculate the coordinates of event A2 (where the beam from B to A is received by A) in the primed frame to show that the beam from B to A travels at c as seen from the primed frame.

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?
 Mentor P: 15,617 altergnostic, several people have repeated the claim that the experimental evidence supports SR. In fact, if you are uncomfortable with the light-speed 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
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 Quote by altergnostic 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.
Oh, really? You mean I can freely use that assumption when analyzing any of these scenarios? Great! That makes things a *lot* simpler.

 Quote by altergnostic 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.
The proper time of the event only matches coordinate time in one particular frame, so you still can't add it to a coordinate time in some other frame and expect to get a meaningful answer.

 Quote by altergnostic 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.
Only if the clock at B and the clock at A are mutually at rest and synchronized. But the "clock at B" that you have been referring to is moving relative to the observer at A.

 Quote by altergnostic Likewise, if the time of the event is 1s at B, the observed time will be 2.12s. That is all I am saying.
And it's still wrong no matter how many times you repeat it. See above and further comments below.

 Quote by altergnostic 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.
Yes, the last is a given. No, the former is *not* an assumption. It's a calculation. Since you have said you agree that the speed of light in an inertial frame is c regardless of the motion of the source, the calculation is simpler than what I have posted before. The givens are:

(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.
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On re-reading, there is one other error that seems to me to be worth commenting on:

 Quote by altergnostic 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 light-second away. The path of detected light in the primed frame is CB.
No, it isn't. The triangle diagram you drew is a diagram of the spatial geometry in the unprimed frame, *not* the primed frame. The point you have labeled C is the location that the light clock source/detector is at, in the unprimed frame, at the time when light pulse 1 hits the mirror. It is *not* the location of the source/detector in the primed frame, because in the primed frame the source/detector and the mirror (i.e., the entire light clock assembly) are at rest.

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.
P: 3,179
 Quote by altergnostic [..] It has been said that if I simply do the MM interferometer analysis I will have solved this setup without any contradiction. I disagree. [..]
Please do so - even if I was wrong and one or two issues remain, it will enormously clear up this discussion. It still appears that you disagree about almost everything, which makes it difficult to untangle the mess.
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P: 15,617
 Quote by altergnostic 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.
If the distance is 1.12 ls and the speed is c regardless of direction then the round trip time is clearly 2.24 s.

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.

 Quote by altergnostic I am simply requiring B's frame of reference to be obtained from A's frame by a standard Lorentz transformation.
Please post the Lorentz transform from A to B then. You will see that your claims are inconsistent with the Lorentz transform.

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