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## Light Clock Problem

Quote by harrylin 					 				 				Oops I had overlooked that error in his "givens"! Thanks I'll also correct that for consistency.

Quote by altergnostic 					 				 				Please see my answer above and also my reply to Peter.
Your "givens" are according to SR selfcontradictory; when I corrected it, I showed this by two methods of calculation and I explained it with words. Note that both my approaches differed a little from that of Dalespam and PeterDonis; I only used your input.

Quote by harrylin 					 				 				It seems that we were talking past each other, so I'll first only address the first two or three points:
Yes, it is getting harder by the relative minute to keep up with so many posts.

Once more: there is no spatial rotation here. Nothing rotates at all. And once more, that was already explained in the link I gave, and of which you claimed to understand it (but evidently you don't): http://physicsforums.com/showthread.php?t=574757  Please explain it in your own words before discussing further.
The spatial rotation refers to the observer 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.

From the link, I will comment only upon the same issue I have been discussing here: the beam leaves A and reaches B in the rest frame (the frame that is at rest relative to the clock) and moves at c as seen from a detector at B. From the perspective of an observer that sees the clock in relative motion, the brings no data and he can't say anything about its observed velocity, since it is not observed. By including a signal sent from B to A at the moment the beam reaches B, the observer will have the observed time for that event and may determine time of the event as recorded on the detector by taking into consideration the observed time and the predetermined distance between him and the detector at B, just like the standard relative V implies a given distance/unit of time.

Yes that is what I wrote: you clarified that S' is the rest frame, in which the light clock is moving; and S is the moving frame, in which the light clock is in rest. If you agree, then there is no misunderstanding about this.*
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.

Still it may be that the problem comes from confusion between frames, as you next write that "This setup presents a novel configuration, which is light that doesn't reach the observer" - however light always only reaches light detectors!*
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). So you see, we only need to know the predetermined distance from A to B, the given time of detection as seen from the clock and the observed time of event B, to figure everything else

It may also come from a misunderstanding of the light postulate, or a combination of both, as you next write:  "The postulate of the speed of light strictly states that c is absolute for the source and the observer regardlesd of motion".*  That is certainly not the second postulate. Did you study the Michelson-Morley calculation? And if so, do you understand it? Then please explain it.
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. 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. 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.

PS. This forum is meant to explain how SR works. It is not meant to "prove" a theory. As a matter of fact, such a thing is impossible!
This is no theory other than SR itself, since the same postulates apply to detected light, consistent with every experiment ever done. A correction to an incorrect or incomplete diagram or a new thought problem that does not contradict but expands theory is not a new theory, since it is built upon the same postulates. 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. This setup is meant to analyse this realisation, keeping all aspects of Einstein's original theory intact and remaining consistent with all available empirical data.

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.
Under close inspection, you may realise that neither the postulate nor the evidence are in contradiction with my conclusions, but that it is a necessary outcome of relativity. It is only ligical, 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.

To realise this is consistent with SR may be difficult, but it must be, since the conclusion is strictly dependant upon the acceptance of the constancy of the speed of any received or detected light.

Quote by harrylin 					 				 				Your "givens" are according to SR selfcontradictory; when I corrected it, I showed this by two methods of calculation and I explained it with words. Note that both my approaches differed a little from that of Dalespam and PeterDonis; I only used your input.
Why are they self contradictory? If I walked from A to B and measured the distance, wouldn't this distance be the same you would measure by walking from B to A? This is as symmetrical as any relative velocity, as it must be. If there is no symmetric distance between frames, there can be no symmetric speed. Do you disagree? If so, please explain why and show me how to determine relative velocity without a equally symmetric relative distance.

Are there other givens you still disagree with?

A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time. In the proposed setup, the distance is fixed and has been previously marked. It is the symmetric distance between A and B.

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Quote by altergnostic 					 				 				h' is not being measured, it is given
I realize that, which is why I said that your givens are wrong. Givens can be wrong either by being inconsistent with themselves (e.g. given a right triangle with three equal sides) or by being inconsistent with the laws of physics (e.g. given a moving mass 2m which comes to rest after colliding elastically with a mass m initially at rest).

Quote by altergnostic 					 				 				 it is supposed to be the constituent L of the relative V that is usually given in SR problems, which we are not given in this setup and is the unknown we seek.
I thought that v=0.5c was also a given, is it not? If v is not given then you can simply plug your other givens into $c^2 t^2 = y^2 + v^2 t^2$ where t=h'/c and then solve for v.

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Quote by altergnostic 					 				 				A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time.
This is an incorrect understanding of length contraction. I recommend starting with the Wikipedia article on length contraction. Length contraction has nothing to do with the distance between two bodies in motion. It also does not refer to a change in length over time, it refers to a difference in the same length as measured by two different frames.

Quote by DaleSpam 					 				 				I realize that, which is why I said that your givens are wrong. *Givens can be wrong either by being inconsistent with themselves (e.g. given a right triangle with three equal sides) or by being inconsistent with the laws of physics (e.g. given a moving mass 2m which comes to rest after colliding elastically with a mass m initially at rest).  I thought that v=0.5c was also a given, is it not? *If v is not given then you can simply plug your other givens into $c^2 t^2 = y^2 + v^2 t^2$ where t=h'/c and then solve for v.
I have reestaded many times that you could ditch the speed of the light clock, since it is useless to this problem. What speed would you insert into gamma in the train and embankment problem? We would have to insert the speed of the projectile/beam and align our x axis with AB in my setup, do you see? If you were given the projectile speed and the times you could calculate the distance AB, this is simply the other way around: given the distance and times you can calculate the speed. The speed of the light clock is completely trivial and you can solve even if the observer at A doesn't know it, since he is given distances and knows all times. From my analysis you can actually derive both the speed of the beam and of the clock, since after finding the total speed for the beam, you can subtract c from that and find the speed of the clock.

Quote altergnostic
A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time.

Quote by DaleSpam 					 				 				This is an incorrect understanding of length contraction. *I recommend starting with the Wikipedia article on length contraction. *Length contraction has nothing to do with the distance between two bodies in motion. *It also does not refer to a change in length over time, it refers to a difference in the same length as measured by two different frames.
Yes, but it is usually the difference measured by two frames in relative motion (you could have length contraction between two observers at rest also). But the main point is your final sentence: it is a difference between two distances being measured by different frames. The given distance h' is not being measured at all in my setup. It has been previously marked with lightsecond signs, remember? This distance is visual data, not vt (how could it be measured with no v, after all?).

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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.

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??

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.

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.

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.

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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.

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light clock, special relativity, time dilation

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