Light Clock Problem

by altergnostic
Tags: light clock, special relativity, time dilation
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P: 4,769
 Quote by altergnostic That's right! But you are confusing the coordinates. He uses the primed distances and unprimed times. And that is allowed because he is directly seeing the km marks, you see. I have determined that in the setup, all distances have been previously walked and each km marked with a big sign. The observer can see those directly.
No, he can't. He has to get light signals from them, just like from everything else that's not at his spatial location. Putting km markers everywhere doesn't magically transmit information instantaneously.

In any case, no matter how he gets the numbers, you can't mix numbers from different frames and expect to get meaningful answers. Any valid calculation of the relative velocity has to use either an unprimed distance and corresponding unprimed time, or a primed distance and a corresponding primed time. Using a primed distance and unprimed time is just as meaningless as using an unprimed distance and a primed time.

 Quote by altergnostic But we have to! Just like in a projectile analysis, the velocities are different for each frame.
For an object that moves at less than the speed of light, the velocity of that same object is different in different frames, yes. But that doesn't apply to light; the velocity of a light beam is c in every frame (assuming inertial frames and flat spacetime). I've explicitly proven that to be the case in your scenarios.

Also, you have claimed that the *relative* velocity of two frames is different in one frame than in the other; this is the basis of your claim that you don't know which gamma factor to use. The fact that the velocity of the same object (if it moves slower than light) is different in different frames is irrelevant to that claim, because here you are not comparing the velocity of the same object in different frames; you are comparing the velocities of *different* objects in different frames. You are comparing the velocity of the train in the observer frame, with the velocity of the observer in the train frame. Those two velocities have the same magnitude, but opposite directions; I've proven that.

 Quote by altergnostic I am very aware of this, and this is where the problem lies. It is an assumption, I repeat.
And I repeat, this is *not* an assumption; it is a proven fact. I've proven it in my previous posts. You can prove that all light beams travel at c using only facts about the relative motion of the light clock and the observer (or the train and the ground), without making any assumptions about the speed of the light beam.

 Quote by altergnostic This beam cannot go at c on both frames just as a projectile on that same path can't have the same velocity in both frames. If you disagree, please explain HOW one setup differs from the other, and why should the beam's apparent velocity not change just like the projectile's.
Because the beam is moving at c, and the projectile is moving slower than c. The two cases are different. That's just a fact about how relativistic spacetime works. The only assumption I can see that is worth mentioning here is the isotropy of space (since that's really what underlies the claim that the velocity of the observer relative to the clock is equal in magnitude and opposite in direction to the velocity of the clock relative to the observer). That assumption is supported by a massive amount of evidence, so I don't see the point of questioning it here.

 Quote by altergnostic Remember, this beam is not seen directly, you have to find its velocity from the signals. This is a premise from the setup - how does the observer at A determines the speed of the beam given primed distances, primed times and observed (unprimed) times?
By using known facts about the speed of the light clock relative to him to establish the coordinates of the relevant events in the unprimed frame, independently of the speed of the light signals that travel between those events.

 Quote by altergnostic But you only get c for the beam because you are inserting the speed of the light clock into gamma
Well, of course. Why not?

 Quote by altergnostic and you are not calculating the speed of the beam from the signals at all.
I don't know what "calculating the speed of the beam from the signals" means. I'm calculating the speed of the beam from knowledge of the distance and time intervals between two known events on the beam's worldline.

 Quote by altergnostic Forget the light clock's velocity. Take it that the observer has no velocity information whatsoever. He has only primed and unprimed times and unprimed distances.
Then I can re-write all the formulas in terms of the unknown gamma factor for the relative velocity, and I will *still* get the same answer. All the factors of gamma end up canceling out. You should be able to see that from my analysis, but if not, I can re-write it this way (although you really should be able to do that yourself). In fact, I actually do re-write the formula for the velocity of the light beam in the light clock this way (for comparison with the velocity of a projectile in a "projectile clock") below.

 Quote by altergnostic The only data available to the observer is: t'A = tA = 0s
Yes.

 Quote by altergnostic t'B = 1s
Yes. But you are aware, aren't you, that this implies tB = gamma, right?

 Quote by altergnostic AB = h' = 1.12 lightseconds
Is this supposed to be a distance in the unprimed frame? If so, how do you know it is consistent with the values for tA, t'A, and t'B that you just gave? If you start out with an inconsistent set of premises, of course you're going to get a meaningless answer.

 Quote by altergnostic AC = y'= 1 lightsecond
I don't see how this follows at all. AC is supposed to be the distance, in the unprimed frame, that the light clock source/detector travels during the time of flight of the beam, correct? In that case, it should be gamma, not gamma primed, because the time of flight of the beam, in the unprimed frame, is tB = gamma (see above).

So already I have spotted two wrong premises in your argument; no wonder you're getting erroneous answers.

 Quote by altergnostic Conversely, replace the beam with a projectile, with this set of givens: t'A = tA = 0s
Yes.

 Quote by altergnostic t'B = 2s (this is the moment the projectile impacts B as seen from the primed frame)
Yes. But again, you are aware that this implies tB = 2 gamma, right?

 Quote by altergnostic AB = h' = 1.12 lightseconds AC = y'= 1 lightsecond
Same comment as above. This value of AB is not consistent with the above, and AC is 2 gamma, not gamma primed, or even 2 gamma primed, which is what you should have written by the same reasoning as the above--t'B is twice as large, so AB should be twice as large as well since the light clock spends twice as much time traveling as before. You are aware, aren't you, that using a projectile traveling at 0.5c, instead of a light beam traveling at c, changes the geometry of your triangle diagram, right? It's *not* the same triangle.

 Quote by altergnostic The setup is the same and the operations should be equivalent. If both operations are not equivalent, tell my why. If the reason is something like "because the beam can't vary its speed" then you are assuming what you are trying to prove.
I've already done the light beam case in my previous analysis, but perhaps it would help to do a similar analysis on the "projectile clock" case; we'll assume a projectile traveling at 0.5c in the primed frame, in the direction perpendicular to the relative motion of the light clock and the observer in that frame. I'll orient the x-axis in the direction the projectile travels, and the y-axis in the direction of the relative motion of the "projectile clock" and the observer. (Yes, I know you think that the x-axis has to point directly along the path to where the mirror will be in the unprimed frame when the projectile hits it. I've already shown that that doesn't matter; I don't think it needs to be shown again. But rotating the spatial axes will be easy enough after the analysis I'm about to show, if you really insist on seeing it done that way.)

We are given the following event coordinates:

$$A0 = D0 = A0' = D0' = (0, 0, 0)$$
$$B1a' = B1b' = (2, 1, 0)$$
$$D2' = (4, 0, 0)$$

This assumes a "projectile" traveling at 0.5c in the primed frame, in the x-direction.

The transformation equations are (I'm writing them now in terms of an unknown gamma factor, since you are now saying we don't know the relative velocity of the light clock and the observer):

Unprimed to Primed:

$$t' = \gamma ( t - v y )$$
$$x' = x$$
$$y' = \gamma ( y - v t )$$

Primed to Unprimed:

$$t = \gamma ( t' + v y' )$$
$$x = x'$$
$$y = \gamma ( y' + v t' )$$

This yields the following coordinates for events in the unprimed frame:

$$B1a = B1b = ( 2 \gamma, 1, 2 \gamma v )$$
$$D2 = ( 4 \gamma, 0, 4 \gamma v )$$

The velocity of the projectile in the unprimed frame on each leg can then be computed like this:

$$v_{projectile} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + 4 \gamma^2 v^2}}{2 \gamma} = \frac{1}{2} \sqrt{(1 - v^2) \left( 1 + \frac{4 v^2}{1 - v^2} \right)} = \frac{1}{2} \sqrt{\frac{(1 - v^2)(1 - v^2 + 4 v^2)}{1 - v^2}} = \frac{1}{2} \sqrt{1 + 3 v^2}$$

This will be somewhere between 1/2, which is the velocity of the projectile in the primed frame, and 1 (but always less than 1 for v < 1); but it is only *equal* to 1/2 for v = 0, so you are correct that the velocity of the projectile in a "projectile clock" *does* change if you change frames.

I could leave it to you to spot the difference between this and the case of the light clock, but I suppose I'll go ahead and give it; the corresponding formula from my previous analysis would be (written with an unknown gamma factor to make it clear how it drops out of the analysis):

$$v_{beam} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2}}{\gamma} = \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{1 - v^2} \right)} = \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2)}{1 - v^2}} = 1$$

As you can see, this formula always gives 1, regardless of v (as long as v < 1). So unlike the projectile case, the speed of a light beam in a light clock does *not* change when you change frames.
 PF Patron Sci Advisor P: 4,769 Just to put some icing on the cake, here's an even more general formula which works for arbitrary "projectile" velocities (thus including the case of the light beam as the "projectile" as well). We have a clock using a "projectile" that travels between a source/detector and a reflector. The projectile's velocity is $v_p'$ in the rest frame of the clock. We are given the following event coordinates (the projectile travels in the x direction): $$A0 = D0 = A0' = D0' = (0, 0, 0)$$ $$B1a' = B1b' = (1 / v_p', 1, 0)$$ $$D2' = (2 / v_p', 0, 0)$$ The transformation equations are as before, and they yield the following coordinates for events in the unprimed frame: $$B1a = B1b = ( \gamma / v_p', 1, \gamma v / v_p')$$ $$D2 = ( 2 \gamma / v_p', 0, 2 \gamma v / v_p')$$ The velocity of the projectile in the unprimed frame on each leg is then: $$v_p = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2 / v_p'^2}}{\gamma / v_p' } = v_p' \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{v_p'^2 (1 - v^2 )} \right)} = v_p' \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2 / v_p'^2)}{1 - v^2}} = v_p' \sqrt{1 + v^2 \left( \frac{1}{v_p'^2} - 1 \right)}$$ For $0 < v_p' < 1$, this gives $v_p' < v_p < 1$ for any $0 < v < 1$. For $v_p' = 1$, this gives $v_p = 1$ for any $0 < v < 1$. QED.
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P: 15,576
 Quote by altergnostic The givens are clear: t'A = tA = 0s t'B = 1s h'= 1.12 lightseconds
Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives $\sqrt{1^2+0.577^2}=1.15$ ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.
P: 3,180
 Quote by DaleSpam Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives $\sqrt{1^2+0.577^2}=1.15$ ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.
Oops I had overlooked that error in his "givens"! Thanks I'll also correct that for consistency.
P: 119
 Quote by harrylin BC=y=1 lightsecond = the distance between mirrors (measured with S) and this is a given.* In addition, according to the second postulate (see above) the same must be true as measured with S'. There is no disagreement about the height of the train: y=y'= 1 lightsecond according to both S and S'. I hope that you see that. A light ray that is sent from C to B will take 1 s according to S'.* Agreed, but noticed that this y is measured at the origin (it may not seem important, but it is, because this y will not be the y after the spatial rotation - where we must line up the x axis of the observer to AB) AC=x't'=0.5 lightseconds = distance traveled by the light clock in the x direction in 1 s. * TA=T'A= 0 = time of first recorded reflection, at both S and S' origins, which are both at A. * T'B≠1s. It is the time of the second reflection event as calculated in the rest frame S', based on the fact that CB= 1 light second. As a matter of fact, if according to S', two light flashes are sent simultaneously from A and C, the flash from C will arrive after 1s but the path length AB is longer, so it must take longer. Of course, it is assumed that the speed of light is the same in all directions; else no prediction is possible at all.* The S' frame is the rest frame (not the moving frame), it is the measurements done from inside the light clock itself. So T'B = 1s

^RED mine.

 The second postulate effectively says that you may set up an inertial reference system, pretend that it is at rest in the ether, and everything will work as if your assumption was correct...
No problem here, actually, if that isn't true my analysis fails. I don't know where is the misunderstanding coming from, though... I may need to clarify the problem further, since you have already shown that you misunderstood what I have been calling the primed frame (it is the rest frame).

 Anyway, let's continue the -now trivial- analysis:
Yes!

 T'B2=T'B + T'[SUB]BA[/SUB = the moment a reflection signal (i.e.: scatter event) from B is seen at A in S, which is the time light takes to travel from the reflection event T'B to the observer at A. Of course, the trajectories AB and BA are the same, along h', and are assumed to take the same time.
Ah, the point of the setup in the first place - to check if this remains consistent.

If the observer at rest in A sent a flash of light from A to B, it would take the same time to reach B as the time the signal takes to reach A from B. BUT, we aren't dealing with that here. What we are seeking is not how fast light travels from A to B as seen from B or B to A from A. From A to B, it would seem to travel at c for an observer at B (and it would take 1.12 seconds to do so if h'=1.12 lightseconds). From B to A, it will seem to travel at c for the observer at A. But we are seeking how fast would the beam would seem to travel from A to B as seen from A. That's why we need the signal from B in the first place. As a matter of fact, our givens show that light takes 1s to go from A to B as seen from inside the lightclock (the primed frame), where the distance is 1 lightsecond.

 This triangle represents the first two reflection events as seen in the moving frame S. Pythagoras: Light path in S' AB = √(12 + 0.52)=1.12.
Thus one mirror is seen at A when T'=T=0 and the other at B when T=1, and T'=1.12[/QUOTE]

Actually, its the other way around: T' is the time in the rest frame - the time seen from the lightclock itself - and that is 1s.*

 The light clock is going at 0.5c, as given by an onboard speedometer, useful in this setup to determine AC.* Note that you did not draw the figure for S, which is simply vertical with y=y'.
*I didn't draw it because it is not necessary. Again, you may take that he only knows the primed times, sees the marked distance signs and receives the signal from B at a time TB.

 AB = h' = 1.12 T'B2 = 2T'B = 2.24 s VAB = 2h'/T'B2 = 1 lightsecond/s
You see right there! You just assumed the beam takes 1.12 seconds to go fram A to B as seen from A! You just assumed what had to be proven here. What you need to take into consideration is the primed time of event B and calculate the observed time for that event as seen from A, and then calculate the observed time from A to B.

 In fact, it is quite useless to proceed if you don't master for example a classical analysis of the Michelson-Morley experiment.
But I do, and as I keep saying, this a different case. In MM we were always measuring the speed of received light, or light detected directly, which applies to the received signal from B in this setup. The postulate of the speed of light strictly states that c is absolute for the source and the observer regardlesd of motion, but that is clearly light that is in direct contact, in direct line, with either source or observer. This setup presents a novel configuration, which is light that doesn't reach the observer, and we have never performed any such experiment seeking the speed of distant light, as far as I know, so we have to be very careful with our assumptions and with our interpretation of the postulate of the speed of light. The goal of this setup is to calculate the speed of the beam, and clearly assumed this value so your analysis can't be proof that the speed of the beam as seen from A would actually be c.
P: 3,180
It seems that we were talking past each other, so I'll first only address the first two or three points:
 Quote by altergnostic ^RED mine.
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):

 [..] I may need to clarify the problem further, since you have already shown that you misunderstood what I have been calling the primed frame (it is the rest frame).
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.

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!

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.

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!
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P: 15,576
 Quote by altergnostic The goal of this setup is to calculate the speed of the beam
This stated goal is mathematically impossible. By the Pythagorean theorem for any signal speed C we have the geometrical relationship $C^2 t^2 = y^2 + v^2 t^2$. If you fix y and v then assuming a value for t is the same as assuming a value for C. Here you are incorrectly assuming that t=0.5, which is equivalent to assuming that C≠c.

 Quote by altergnostic and clearly assumed this value so your analysis can't be proof that the speed of the beam as seen from A would actually be c.
Yes. We ASSUME that the speed of the beam is c. That is what a postulate is, a fundamental assumption of the theory. If you are assuming anything contrary (which you are) then you are not doing SR. From that assumption we can then calculate the value of t using the relationship above.

There is no mathematical proof of the validity of the assumption (you can't mathematically prove postulates or axioms, by definition), however an enormous body of experimental evidence supports the assumption and contradicts your alternative assumption.
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P: 4,769
 Quote by altergnostic The goal of this setup is to calculate the speed of the beam
In my analysis, I *calculated* (*not* assumed) the speed of the light pulse traveling between events B1a and D2, neither of which are co-located with the observer, so this is a "distant light" beam by your definition. Its speed is *calculated* to be c. So my analysis meets your goal. What's the problem?
P: 119
 Quote by PeterDonis No, he can't. He has to get light signals from them, just like from everything else that's not at his spatial location. Putting km markers everywhere doesn't magically transmit information instantaneously.
I don't mean that he sees anything instantaneously, this is just how he will measure anything. Length contraction applies to vt, but the markers are just distances and they are at rest relative to the observer. These markers belong to both frames, and they are not being measured by any of them. You could call them x'' or something. They are given and are not being measured in the process. Usually, we are given V and find the distance from that, but since we are seeking V, I have to give L somehow. A given V contains a given L which is the same for both frames, you see. This is the given distance. The time it would take for light to go from A to B, as seen from B, is the same as from B to A seen from A, so this distance must be the same and it is not being measured by any of the frames. It is a given just like the distance implied in the given V from most SR thought problems. If you can have the same velocity between two frames, you already have an equivalent distance between two frames, over the standard unit of time.

 In any case, no matter how he gets the numbers, you can't mix numbers from different frames and expect to get meaningful answers. Any valid calculation of the relative velocity has to use either an unprimed distance and corresponding unprimed time, or a primed distance and a corresponding primed time. Using a primed distance and unprimed time is just as meaningless as using an unprimed distance and a primed time.
See above.

 For an object that moves at less than the speed of light, the velocity of that same object is different in different frames, yes. But that doesn't apply to light; the velocity of a light beam is c in every frame (assuming inertial frames and flat spacetime). I've explicitly proven that to be the case in your scenarios.
You have not proven it, you have assumed it, since the operation you developed is not the same as you would if the beam was a projectile. And you simply stated that it must be different because we are supposed to know that the beam is moving at c from A to B as seen from A, and that's not a proof, that is an assumption. You claim it is an assumption backed up by numerous experiments, but not a single experiment in history has measured the speed of light that does not reach a detector directly, you have to aknowledge this. I state again that c applies to light directly detected, light that reaches and "touches" the observer.

 Also, you have claimed that the *relative* velocity of two frames is different in one frame than in the other; this is the basis of your claim that you don't know which gamma factor to use. The fact that the velocity of the same object (if it moves slower than light) is different in different frames is irrelevant to that claim, because here you are not comparing the velocity of the same object in different frames; you are comparing the velocities of *different* objects in different frames. You are comparing the velocity of the train in the observer frame, with the velocity of the observer in the train frame. Those two velocities have the same magnitude, but opposite directions; I've proven that.
That is correct, and it is obvious anyway. But in this case we ARE measuring the speed of the same "object" in different frames: the setup is meant to give us the oportunity to measure the speed of the beam from both frames.

 And I repeat, this is *not* an assumption; it is a proven fact. I've proven it in my previous posts. You can prove that all light beams travel at c using only facts about the relative motion of the light clock and the observer (or the train and the ground), without making any assumptions about the speed of the light beam.
You haven't shown this at all, Peter. If you carefully check your analysis you will see that you assumed the observed speed of the beam seen from A is c, not shown it.

[/QUOTE]Because the beam is moving at c, and the projectile is moving slower than c. The two cases are different. That's just a fact about how relativistic spacetime works. The only assumption I can see that is worth mentioning here is the isotropy of space (since that's really what underlies the claim that the velocity of the observer relative to the clock is equal in magnitude and opposite in direction to the velocity of the clock relative to the observer). That assumption is supported by a massive amount of evidence, so I don't see the point of questioning it here.[/QUOTE]

Now you do assume it is an assumption. The evidence shows the speed of light is c for directly detected light, as I said above. I am not aware of any experiments that allow us to measure the speed of non-observed light, in the sense that this is light that doesn't reach a detector directly.

 By using known facts about the speed of the light clock relative to him to establish the coordinates of the relevant events in the unprimed frame, independently of the speed of the light signals that travel between those events.
Again, assumptions. And not backed up by evidence at all - evidence shows that detected light moves at c. No one can propose that the signal from B to A moves at any other speed than c, since that would contradict every single experiment ever done to measure the speed of light. Since no experiment has ever been performed with distant and indirectly detected light, the assumption that the beam moves at c is not supported by any evidence whatsoever. Not for, nor against.

 Well, of course. Why not?
Because you must calculate the speed of the beam before assuming it is c, since that is the point of the setup in the first place: the variables given allows us to measure the speed of the beam.

 I don't know what "calculating the speed of the beam from the signals" means. I'm calculating the speed of the beam from knowledge of the distance and time intervals between two known events on the beam's worldline.
It means giving c to the signal and plotting the observed times if events A and B against the given L.

 Then I can re-write all the formulas in terms of the unknown gamma factor for the relative velocity, and I will *still* get the same answer. All the factors of gamma end up canceling out. You should be able to see that from my analysis, but if not, I can re-write it this way (although you really should be able to do that yourself). In fact, I actually do re-write the formula for the velocity of the light beam in the light clock this way (for comparison with the velocity of a projectile in a "projectile clock") below.
But you didn't do it relative to A, or "as the observer at A would see it", as I said above.

 Yes. But you are aware, aren't you, that this implies tB = gamma, right?
You can't assume that since you aren't given the speed of the beam. What V do you insert on gamma?

 Is this supposed to be a distance in the unprimed frame? If so, how do you know it is consistent with the values for tA, t'A, and t'B that you just gave? If you start out with an inconsistent set of premises, of course you're going to get a meaningless answer.
It is a distance measured locally prior to the experiment. None of the frames are measuring it, they are given it. Just like the given distance implied if you were given V.

 I don't see how this follows at all. AC is supposed to be the distance, in the unprimed frame, that the light clock source/detector travels during the time of flight of the beam, correct? In that case, it should be gamma, not gamma primed, because the time of flight of the beam, in the unprimed frame, is tB = gamma (see above).
This is a typo, it was meant to be BC, sorry for this. But I guess you did notice this honest mistake, didn't you? The diagram shows clearly where the y' distance stands for.

 So already I have spotted two wrong premises in your argument; no wonder you're getting erroneous answers.
You spotted a typo and a given that is also given if you were given V, so no wrong premises here.

 Yes. But again, you are aware that this implies tB = 2 gamma, right?
Not really. First you have to find the velocity, otherwise what V do you plug into gamma? It may be, it may not be, you have to show that it is, the setup is supposed to allow us to calculate the speed. Than you can show what is the value for gamma.

 Same comment as above. This value of AB is not consistent with the above, and AC is 2 gamma, not gamma primed, or even 2 gamma primed, which is what you should have written by the same reasoning as the above--t'B is twice as large, so AB should be twice as large as well since the light clock spends twice as much time traveling as before. You are aware, aren't you, that using a projectile traveling at 0.5c, instead of a light beam traveling at c, changes the geometry of your triangle diagram, right? It's *not* the same triangle.
The same typo. Also, although you are right and the triangle does change, the hypotenuse is still AB, the distances are still given and the setup is the same. I haven't done the calculations to find the values, though, but it should be clear that the setup is equivalent, and so the operation should be, else you lose consistency and have to prove why this should be so. Since your reason is experimental evidence, I remind you again that no experiment has measured the speed of undetected or indirectly observed light or distant light or light that does not reach the observer/detector.

 I've already done the light beam case in my previous analysis, but perhaps it would help to do a similar analysis on the "projectile clock" case; we'll assume a projectile traveling at 0.5c in the primed frame, in the direction perpendicular to the relative motion of the light clock and the observer in that frame. I'll orient the x-axis in the direction the projectile travels, and the y-axis in the direction of the relative motion of the "projectile clock" and the observer. (Yes, I know you think that the x-axis has to point directly along the path to where the mirror will be in the unprimed frame when the projectile hits it. I've already shown that that doesn't matter; I don't think it needs to be shown again. But rotating the spatial axes will be easy enough after the analysis I'm about to show, if you really insist on seeing it done that way.) We are given the following event coordinates: $$A0 = D0 = A0' = D0' = (0, 0, 0)$$
I assume that D0 here stands for A0 right? You are specifying that they are the same point in the projectile-clock, right? I'm asking this because D is really unecessary after all, we can focus on A, B and C and find what we seek from that.

 $$B1a' = B1b' = (2, 1, 0)$$ $$D2' = (4, 0, 0)$$ This assumes a "projectile" traveling at 0.5c in the primed frame, in the x-direction. The transformation equations are (I'm writing them now in terms of an unknown gamma factor, since you are now saying we don't know the relative velocity of the light clock and the observer):
I'm saying we don't know the PROJECTILE velocity. We don't need the clock velocity at all, since we are given the basic coordinates that must compose the relative velocity - which is the unknown we seek.

 Unprimed to Primed: $$t' = \gamma ( t - v y )$$ $$x' = x$$
The vy term should be vAB, since the speed from A to B is not the same from A to C - that would be the speed of the clock, but we are transforming the projectile, not the clock.

 $$y' = \gamma ( y - v t )$$ Primed to Unprimed: $$t = \gamma ( t' + v y' )$$ $$x = x'$$
See above

 $$y = \gamma ( y' + v t' )$$ This yields the following coordinates for events in the unprimed frame: $$B1a = B1b = ( 2 \gamma, 1, 2 \gamma v )$$ $$D2 = ( 4 \gamma, 0, 4 \gamma v )$$ The velocity of the projectile in the unprimed frame on each leg can then be computed like this: $$v_{projectile} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + 4 \gamma^2 v^2}}{2 \gamma} = \frac{1}{2} \sqrt{(1 - v^2) \left( 1 + \frac{4 v^2}{1 - v^2} \right)} = \frac{1}{2} \sqrt{\frac{(1 - v^2)(1 - v^2 + 4 v^2)}{1 - v^2}} = \frac{1}{2} \sqrt{1 + 3 v^2}$$ This will be somewhere between 1/2, which is the velocity of the projectile in the primed frame, and 1 (but always less than 1 for v < 1); but it is only *equal* to 1/2 for v = 0, so you are correct that the velocity of the projectile in a "projectile clock" *does* change if you change frames.
Yes, but that is the velocity you should plug into gamma (except in the case where the projectile and the clock motion were in the same line). You must do a vector addition and find the speed of the projectile from the unprimed frame at some point, if you want to use a velocity value, but you don't a variable v at all to solve. So gamma is compromised.

 I could leave it to you to spot the difference between this and the case of the light clock, but I suppose I'll go ahead and give it; the corresponding formula from my previous analysis would be (written with an unknown gamma factor to make it clear how it drops out of the analysis): $$v_{beam} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2}}{\gamma} = \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{1 - v^2} \right)} = \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2)}{1 - v^2}} = 1$$
Again, how do you know what V to use, since you aren't given any? You must find a way to solve without it, since I have only given the L component of the relative velocity. You have to find the observed velocity, since that's what we are seeking. As it is, you have two different velicities in your equations, and you are given none, so the only thing I can conclude from your analysis (if it is correct, and I don't think it is for the reasons above) is that the setup doesn't give enough information to solve. But I know this isn't your conclusion.

 As you can see, this formula always gives 1, regardless of v (as long as v < 1). So unlike the projectile case, the speed of a light beam in a light clock does *not* change when you change frames.
This is only so because you assume Vbeam = c in both frames. You assume what has to be proven.

 Quote by PeterDonis Just to put some icing on the cake, here's an even more general formula which works for arbitrary "projectile" velocities (thus including the case of the light beam as the "projectile" as well). We have a clock using a "projectile" that travels between a source/detector and a reflector. The projectile's velocity is $v_p'$ in the rest frame of the clock. We are given the following event coordinates (the projectile travels in the x direction): $$A0 = D0 = A0' = D0' = (0, 0, 0)$$ $$B1a' = B1b' = (1 / v_p', 1, 0)$$ $$D2' = (2 / v_p', 0, 0)$$ The transformation equations are as before, and they yield the following coordinates for events in the unprimed frame: $$B1a = B1b = ( \gamma / v_p', 1, \gamma v / v_p')$$ $$D2 = ( 2 \gamma / v_p', 0, 2 \gamma v / v_p')$$ The velocity of the projectile in the unprimed frame on each leg is then: $$v_p = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2 / v_p'^2}}{\gamma / v_p' } = v_p' \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{v_p'^2 (1 - v^2 )} \right)} = v_p' \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2 / v_p'^2)}{1 - v^2}} = v_p' \sqrt{1 + v^2 \left( \frac{1}{v_p'^2} - 1 \right)}$$ For $0 < v_p' < 1$, this gives $v_p' < v_p < 1$ for any $0 < v < 1$. For $v_p' = 1$, this gives $v_p = 1$ for any $0 < v < 1$. QED.
I prefer this analysis over the previous one, but the problem remains: you are not given any v, you must find another way to solve (which I believe is only possible if you can find vp - the velocity observed from A, in my diagram), and that was the purpose of my setup, and the reason I gave the L component for the relative velocity(s).

If you still have trouble with my given x' and y' - maybe a better form would be X and Y since they are given for both frames - ask yourself what L is implied in the relative V.
P: 119
 Quote by DaleSpam Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives $\sqrt{1^2+0.577^2}=1.15$ ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.
h' is not being measured, it is given, 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.
P: 119
 Quote by harrylin Oops I had overlooked that error in his "givens"! Thanks I'll also correct that for consistency.
P: 3,180
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.
P: 119
 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.
P: 119
 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?
 P: 119 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.
Mentor
P: 15,576
 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.
Mentor
P: 15,576
 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.
P: 119
 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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