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#127
Nov1612, 02:31 PM

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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. 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. So already I have spotted two wrong premises in your argument; no wonder you're getting erroneous answers. We are given the following event coordinates: [tex]A0 = D0 = A0' = D0' = (0, 0, 0)[/tex] [tex]B1a' = B1b' = (2, 1, 0)[/tex] [tex]D2' = (4, 0, 0)[/tex] This assumes a "projectile" traveling at 0.5c in the primed frame, in the xdirection. 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: [tex]t' = \gamma ( t  v y )[/tex] [tex]x' = x[/tex] [tex]y' = \gamma ( y  v t )[/tex] Primed to Unprimed: [tex]t = \gamma ( t' + v y' )[/tex] [tex]x = x'[/tex] [tex]y = \gamma ( y' + v t' )[/tex] This yields the following coordinates for events in the unprimed frame: [tex]B1a = B1b = ( 2 \gamma, 1, 2 \gamma v )[/tex] [tex]D2 = ( 4 \gamma, 0, 4 \gamma v )[/tex] The velocity of the projectile in the unprimed frame on each leg can then be computed like this: [tex]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}[/tex] 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): [tex]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[/tex] 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. 


#128
Nov1612, 02:57 PM

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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 [itex]v_p'[/itex] in the rest frame of the clock. We are given the following event coordinates (the projectile travels in the x direction): [tex]A0 = D0 = A0' = D0' = (0, 0, 0)[/tex] [tex]B1a' = B1b' = (1 / v_p', 1, 0)[/tex] [tex]D2' = (2 / v_p', 0, 0)[/tex] The transformation equations are as before, and they yield the following coordinates for events in the unprimed frame: [tex]B1a = B1b = ( \gamma / v_p', 1, \gamma v / v_p')[/tex] [tex]D2 = ( 2 \gamma / v_p', 0, 2 \gamma v / v_p')[/tex] The velocity of the projectile in the unprimed frame on each leg is then: [tex]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)}[/tex] For [itex]0 < v_p' < 1[/itex], this gives [itex]v_p' < v_p < 1[/itex] for any [itex]0 < v < 1[/itex]. For [itex]v_p' = 1[/itex], this gives [itex]v_p = 1[/itex] for any [itex]0 < v < 1[/itex]. QED. 


#129
Nov1612, 03:54 PM

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#130
Nov1612, 05:12 PM

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#131
Nov1712, 06:06 AM

P: 119

^RED mine. 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. 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.* 


#132
Nov1712, 07:13 AM

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It seems that we were talking past each other, so I'll first only address the first two or three points:
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. 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 MichelsonMorley 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! 


#133
Nov1712, 09:39 AM

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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. 


#134
Nov1712, 11:52 AM

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#135
Nov1812, 04:24 AM

P: 119

[/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 nonobserved light, in the sense that this is light that doesn't reach a detector directly. 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. 


#136
Nov1812, 04:27 AM

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#137
Nov1812, 04:28 AM

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#138
Nov1812, 05:38 AM

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#139
Nov1812, 06:13 AM

P: 119

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. 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. 


#140
Nov1812, 06:26 AM

P: 119

Are there other givens you still disagree with? 


#141
Nov1812, 08:11 AM

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.



#142
Nov1812, 10:24 AM

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#143
Nov1812, 10:28 AM

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#144
Nov1812, 11:05 AM

P: 119

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. 


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