0 Members and 1 Guest are viewing this topic.
Mods,Why didn't my reply post. It says "click here to view og's post," but nothing happens. More censorship or just a tech glitch?
What exactly does "The frame of the Michelson Morley experiment is in relative motion compared to an observer at relative rest." mean here? If you're at rest with 'something' it will have no motion relative you, if it has, you can't be at rest with it Butch?
As soon as the light is received by both planes then the light is perpetuated ( copied) back into the perspective mirrors.
The observer is at relative rest.The MMX is in relative motion.
The apparatus is in the moving frame.The observer is in the rest frame.
There are no such things as absolute "moving" or "rest frames" in relativity. What matters is what is moving with respect to what else.
If you mean that the observer is moving with respect to the experiment, and along one of it's two arms, then what I said about Doppler shifts is important.
There are 2 inertial frames, K and K’.Relative to frame K, frame K’ is in motion.Relative to frame K’, frame K is at rest.
In the experiment the amount of time is equal for the light to traverse both arms, verified by the lack of significant fringe.The speed of light is constant and the same for all observers.
If K' is in motion relative to K, then K is in motion relative to K'.
I think what you're saying is that K contains the experiment.
An observer in K sees the experiment at rest. An observer in K' sees it in motion.
You're comparing observations by an observer in K and an observer in K'.
I take this to mean that as observed in frame K, the light takes the same time to traverse both arms, with no significant fringes: true.
My point still stands. It is not possible for the light to take the same time.
It is possible, however, to see no significant fringe even if the time taken on each arm is unequal. Fringes occur when the product of frequency times time taken is not equal (or technically is not a multiple of 2*pi). Relative motion with respect to the experiment changes the time taken, but it also changes the frequency of the observed light due to the relativistic Doppler shift, both of which will have to be accounted for to predict fringes. I suspect this is how you resolve the paradox,…
"No significant fringe" does not mean equal time. It means the product of time x frequency is equal.
The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the time remained equal for light to traverse both arms of unequal length judged from relative rest. True?
From relative rest, the arms are of equal length, so the time is obviously the same.
In motion, they appear as different lengths, so the time is different. The frequency will also change, which has to be accounted for.
I'm not sure how much clearer I can be. If you don't understand what I'm getting at here, perhaps someone else can explain better.
an observer in frame K' observes the experiment … sees no fringes.
an observer in frame K observes the experiment… see a bright fringe.
I can’t find any resources that discuss light projected in the direction of motion within a moving frame judged from relative rest. Do you or anyone else know of any? (I’m certain there aren't any)
If an observer in frame K observes the experiment, one arms will appear to be length contracted, so the time taken for the light to travel down back along the two arms will no longer be equal. However, the frequency of the light down the two arms will no longer be equal due to the Doppler shift. Since frequency multiplied by time determines the condition for a bright fringe, this observer will probably also see a bright fringe. Again, I haven't worked out the math.
No observer sees a bright fringe
The easiest way to resolve this thing is to put lots of semi-silvered mirrors into the experiment to reflect some of the light out sideways to the observer observing from another frame. When you do this, that observer will see the progress of the light through the apparatus travel in accordance with the speed limit for light of his own frame, and the frequency of the light will remain constant throughout from his point of view too, though it will be at a lower frequency than would be observed by an observer moving with the apparatus.
If instead of the semi-silvered mirrors we allow our stationary observer to stick his head into the beam as the apparatus goes past him, he will then see the frequency of the light change depending on which direction it's going in relative to the apparatus, but that has no bearing on what happens when the fringe is formed (or not formed) because at the location where that happens (or doesn't happen) we have two light beams merged together again and travelling along the same path with the same frequency.
I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.
Can there be any doubt that the time for light to traverse the perpendicular arms of the apparatus is the same for all observers?
Quote from: David Cooper on 30/10/2012 20:23:55I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.David, the moderators have asked you many times to stop stating your opinions about SR as facts. You have made many good contributions to the forum, so please don't make us moderate you further on this topic.
“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”
It doesn't. They take different times. The reason you still get a null result of the experiment is explained in that link.