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Fitzgerald -Lorentz contraction |
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| Sep2-08, 01:52 PM | #18 |
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Fitzgerald -Lorentz contractionIf you call the rest length of the train (in its own frame) to be L', the time for the light to travel from one end to the other--according to train observers--will be t' = L'/c. In the track frame, the light travels a distance L + vt, where L is the length of the train as measured by the track observers. Thus the time for the light to travel from one end to the other--according to track observers--must satisfy: ct = L + vt, thus t = L/(c-v). Of course, relativity tells us that the train is contracted, thus [itex]L = L'\sqrt{1 - v^2/c^2}[/itex]. If you are interested in experimental evidence for relativity, read the sticky at the top of this forum: FAQ: Experimental Basis of Special Relativity |
| Sep2-08, 01:56 PM | #19 |
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yes that would work but you didnt look up light clock like I suggested. time dilation is the only way to explain the different travel times of the light in the light clock (perpendicular to the motion of the object). only length contraction is left to explain the difference in travel time parallel to the objects motion.
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| Sep2-08, 06:00 PM | #20 |
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I had some misunderstandings on some of Michelson's points of view. For the image of the river,he speaks of currents (being a resistance to flow) which is not found in the M&M experiment. The first swimmer remains beside the bank and swims to and fro with and against the current...ok this I can take. But instead of having the second swimmer aiming 90 degrees from the bank, he writes: "It won't do simply to aim directly for the opposite bank-the flow will carry the swimmer downstream. To succeed in going directly across, the swimmer must actually aim upstream at the correct angle (of course, a real swimmer would do this automatically)." But within the experiment, the sources of light are indeed separated by 90 degrees. He goes on to write : "Thus, the swimmer is going at 5 feet per second, at an angle, relative to the river, and being carried downstream at a rate of 3 feet per second". However, if the swimmer points 90 degrees to the other bank (which is what the experiment denotes) then, he crosses the bank in 20 seconds or the distance (100 feet) / velocity (5 feet/sec).However, due to the influence ofthe second velocity, that of the river, he will arrive further away along the bank. Andre |
| Sep2-08, 07:15 PM | #21 |
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[/QUOTE] If you are interested in experimental evidence for relativity, read the sticky at the top of this forum: FAQ: Experimental Basis of Special Relativity [/QUOTE] Andre |
| Sep2-08, 07:41 PM | #22 |
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I'm sorry granpa; was there a thread in particular you wanted me to look at for light clocks?Or just the internet? Andre |
| Sep2-08, 07:48 PM | #23 |
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a light clock is simply 2 mirrors with a pulse of light bouncing between them. each time it bounces off the mirror the clock ticks. the light path is perpendicular to the motion of the object (in the frame of the object). an observer at rest sees the light take a much longer path. time dilation is the ONLY way to explain it.
time dilation alone cant explain both the light path perpendicular and parallel to the motion of the object. |
| Sep2-08, 08:04 PM | #24 |
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As far as I know, there is currently no direct experimental confirmation of length contraction (it's just too small to detect), but the experimental evidence for relativity as a whole (which requires length contraction) is overwhelming. |
| Sep2-08, 08:13 PM | #25 |
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Andre |
| Sep2-08, 08:28 PM | #26 |
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the M&M experiment showed that the speed of light is independent of the observer. simple thought experiments suffice to show that time dilation, length contraction, and loss of simultaneity must follow |
| Sep2-08, 08:56 PM | #27 |
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Andre |
| Sep2-08, 09:13 PM | #28 |
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Was this experiment performed? If yes, where could I find such an experiment? Andre |
| Sep2-08, 09:21 PM | #29 |
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the important thing about the M&M experiment is that the earth is moving around the sun and therefore changes speed constantly. if by some miracle the apparatus had been stationary at one point it certainly wasnt later on. also as I understand it the apparatus was rotated 90 degrees without producing any result. thats impossible unless the speed of light is constant for all observers. what do you mean 'light shifts' and 'arrive at the eye'? there as only one pulse and there as no eye. |
| Sep2-08, 09:44 PM | #30 |
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Newton's particle theory of light would have predicted the same nul results as Relativity did...except in the case where "changes in speed"or accelerations are implied, something that even Relativity would not have predicted as well. You wrote: The idea was that the two light signals would arrive at the same time according to the observer within the frame of the experiment, because both light paths are equal, according to this observer. However, the observer on the moving frame would observe the light signals as arriving at different time since, in his frame of reference, one path is longer than the other. Andre |
| Sep2-08, 09:53 PM | #31 |
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| Sep2-08, 10:09 PM | #32 |
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Actually, not really, granpa. Here, again, I apologize for my lack of skills in explaining things. On the frame of reference of the experiment, the observers will measure both lengths of the light's paths (horizontal and vertical) as being the same. If two light signals are sent at the same time from a source, one taking the horizontal path, the other taking the vertical path, the two light signals would return at the same time due to the invariant speed of light. Now, for an observer on a moving frame, the horizontal path has been contracted, according to the theory of Relativity, while the vertical path is not contracted. Therefore, Relativity would predict the two light signals would not return at the same time; this could be detected by observing a shift in the light spectrum. However, If the horizontal path in not affected by any contractions, then even the observer on the moving frame will agree that both light pulses arrived at the same time. Andre |
| Sep2-08, 10:15 PM | #33 |
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you've forgotten to take the objects motion into account.it is contracted but it is moving. you have to do a little algebra to determine the amount of time required for a round trap. it works out. trust me.
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| Sep2-08, 10:27 PM | #34 |
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From my point of view, if the light speed remains constant , the time will be t = distance (contracted) / c for the horizontal path, and t= distance (non contracted) / c. So I cannot see it other than the observer as measuring a time difference between the two light signals arriving. Andre |
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