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## Light Clock Problem

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.

Mentor

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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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  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

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

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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.

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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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.

Quote by DaleSpam 					 				 				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.
A beam describing ABA would indeed take 2.24s for the round trip, as is confirmed by experiment, but this is not what is happening here. The observer at A is not the emitter, and from the point of view of the emitter, the path is CBC, not ABA. The observer at A only has the distance AB and a signal from B to A, but he has no direct information on the path from A to B, he must calculate it from the time he receives the signal from B, and we are disagreeing on that reception time, so we have to clear this up. Would you agree that, if a stationary clock placed at B is synchronized with the clock at A, we can find the reception time by taking the time of the event as shown on the clock at B and adding the time it takes the signal from that event to get to A? Or, likewise, we can find the time of the event on the stationary clock at B by taking the reception time and subtracting the time it takes for light to cross the distance AB? Isn't this how we find the times of stationary synchronized clocks? For instance, to synchronize the clocks at A and B we would take a roundtrip measurement (2.24s, like you said) and half that value to get the time separation of the clock at B, so we know that a tick of the clock at B is observed 1.12s later at A. After that synchronization, we can calculate the time of any event at B from A simply by subtracting 1.12s from the reception time, don't you agree? Since the observer at A receives the signal from B 1.12s later than the time of emission, and the time of emission (according to the clock at B - which is stationary and synchronized with the clock at A) is 1s, the observer at A must receive that signal at 2.12s. I can't see it any other way. If the reception time was 2.24s than the emission time would have to be 1.12s and that is not in agreement with what is shown on the clock at B. If you are right, either the clocks at A and B are out of sync, or the clock at B is time dilated (and I don't see how that can be true) or the event at B occurs at 1.12s from the point of view of B, which would then be in disagreement with the light-clock time for that event, which is insane since the event at B happens at the same spacetime coordinates for both the light-clock and the clock-signaler at B.

Please post the Lorentz transform from A to B then. You will see that your claims are inconsistent with the Lorentz transform.
You misunderstand me (or I wasn't clear). What I meant is that this frame alignment is what would allow us to apply standard lorentz transforms. I am simply requiring that both frames x-axis are aligned with AB, no calculations needed for this requirement, it is standard procedure in most SR problems and it helps to simplify things a bit.

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Quote by altergnostic 					 				 				The observer at A is not the emitter
Doesn't matter. The observer at A is co-located with the emitter at the instant of emission. So the path of the light, in the unprimed frame, *is* ABA.

Quote by altergnostic 					 				 				and from the point of view of the emitter, the path is CBC
No, it's C'B'A', because "from the point of view of the emitter" means in the primed frame, *not* the unprimed frame. The geometry in the primed frame is *different*; it does *not* look like the triangle diagram you drew.

Quote by altergnostic 					 				 				Would you agree that, if a stationary clock placed at B is synchronized with the clock at A, we can find the reception time by taking the time of the event as shown on the clock at B and adding the time it takes the signal from that event to get to A?
Yes, *if* the clock at B is at rest relative to the clock at A and synchronized with it.

Quote by altergnostic 					 				 				Or, likewise, we can find the time of the event on the stationary clock at B by taking the reception time and subtracting the time it takes for light to cross the distance AB?
For a *stationary* clock at B, yes.

Quote by altergnostic 					 				 				the time of emission (according to the clock at B - which is stationary and synchronized with the clock at A)
No, it isn't. Your "clock at B" that reads 1s when light pulse 1 arrives is moving with the light clock; it's the "clock" on the light clock's mirror that records when light pulse 1 arrives. It's not stationary and synchronized with the clock at A. A *stationary* clock at B, that was synchronized with observer A's clock, would record a time t = 1.12s when light pulse 1 arrives at the mirror.

Quote by altergnostic 					 				 				the event at B happens at the same spacetime coordinates for both the light-clock and the clock-signaler at B.
I don't know what you mean by "the clock-signaler at B". Do you mean an observer *stationary* at B, who emits light pulse 2 back to A? If so, then your statement is incorrect; the clock-signaler's clock will read t = 1.12s when light pulse 1 arrives at the mirror, as I said above. The clock-signaler's clock is not synchronized with the moving mirror's clock, nor are his spatial coordinates the same as those of the light clock. The *event* is the same, but it is described by different coordinates (i.e., a different set of 4 numbers t', x', y', z') in the primed frame than in the unprimed frame.

You apparently do not understand how frames in relative motion work. The only event that has the same spacetime coordinates (i.e., the same set of 4 numbers t, x, y, z) in both frames is the origin, the event at which light pulse 1 is emitted by the light clock source/detector at the instant it is spatially co-located with the observer at A. Every other event in the spacetime has *different* coordinates in the two frames. I suggest a review of a basic relativity textbook, such as Taylor & Wheeler's Spacetime Physics.

Quote by harrylin 					 				 				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.
Harry, that analysis does not apply to this setup at all, that is my point. The only possible way to do that is like it is always done: giving c to AB, so by doing it my arguments and givens are useless. You see, I am putting that very assumption into question with my setup, so I can only do what you ask me to do by giving up what I am concerned about in the first place. Also, it is done all over the place and we can simply google "light clock time dilation interferometer, etc" and we will have tons of results. You can't use that analysis to answer "what is the speed from A to B" since that is a given in that analysis in the first place. Also, MM analysis don't have an observer at A like I have here, and also no possible communication from reflection events to that observer (or any other moving observer for that matter). So I can't possibly relate the interferometer analysis to my setup. If we assume that light travels at c as seen from an absolute ether or in any direction for any frame, my analysis is meaningless since it is intended to check those assumptions.

If we rerun my analysis with a projectile going 0.5c from A to B instead of the beam, would you agree that it stands? Do you agree on A's reception time of 2.12s for the signal from B? If not, are you saying that the time of emission of the signal from B shown on the clock at B is 1.12s? If so, how can that be since at t'=1s the light-clock is reflecting the beam at B? Is that event happening at t'=1s or t'=1.12s, as seen from B? Are you saying we can have two different times for the same event, as seen from B? If so, please explain how, because I can't see how that is possible in this problem.

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