Simultaneity James R, OK. I have read this many times before but upon your insistance I just read it again. Two strikes of lightening are simultaneous on the bank of a railroad. They strike at point A and B some distance apart. At a location midway between the points A and B, called M, they will converge and appear simultaneous to a stationary observer. If however the observer is in motion toward point B he will see strike B before strike A. They are no longer simultaneous. But that example is a distorted picture. Only if he has moved from point M during the information transit are they no longer simultaneous. That is he can be moving and pass point M at the time an observer on the bank sees the simultaneous strikes and he too will see them simultaneous. It is his location in the example that shifts simultaneity not his velocity. I have added to the given example in the attachment. Now we have two light signaling devices that puts out a stream of changing polarized light. The amount of polarization increases with time from the start of the transmission. The amount of polarization is measured by an observer on the train and converted into numbers 1 - 16. If the observer is at point M and the system is triggered he will see both beam numbers in sync and half way through the test they will read 8/8. If in the above example given for simultaneity he is in motion and bouy #1 triggers the system as he passes by heading from A to B then he will always see the numbers generated by the beams out of sync and when he passes bouy #2 to stop the test and records the last numbers seen they will read 14/2. All this seems to prove my point not yours. The only simultaneity that the clocks and their observers in the 3 Clock Problem care about are their view of clock C when there clock is stopped. If I am looking back at clock C during the test I would see it was only running 66.1% as fast as my clock. Therefore when clock C (by its clock = 10) stops clock B at 6.61 hours, clock B will only see 4.37 hours registered on clock C. I don't think you can have it both ways. Clock B actually sees clock C running slow or it doesn't and when clock B stops the amount of recorded time on clock C in B's view is the 4.37 hours. That is what he expects to see upon return after the test. Not some mathematically adjusted computation to make him thereafter agree upon time dilation mathematically. The arguement all along has been what an observer expects based upon what he sees as a result of the relavistic time dilation formula.
how can polarisation be continually increasing? Do you mean degree of coherence of polarisation? Do you know what polarisation is? What does this mean?
Surely ryans, Surely you jest. I suspect most grade school children understand the post. Further you might try looking at the attachment. If you have trouble understanding how the degree (angle) of polarization is converted into numbers let me suggest a short course in electronics. I'm sorry, I forgot for a moment that it was you and your limited ability to visualize. You place a filter over the projector to polarize the beam and then you simply rotate the filter over time. The angle of the polarized light is then read by a fixed angle filter as a change in intensity (analog) which is converted into numbers corresponding to the intensity of the filtered beam. Hope I didn't lose you this time.
I thought you were really stupid for a sec. You did not say degree of polarisation. You need to use common terminology Mac, not just make up your own terminology. Do the numbers 1-16 divide 360 degrees into 16 intervals?
Arbitrary numbers ryans, The numbers are arbitrary but at 90 degrees orientation to each other the filter is at maximum. Any further rotation would start reversing the analog of intensity vs numerical progression. If you count revolutions you could use the signal intensity as steps in decade for example.
MacM: <i>Two strikes of lightening are simultaneous on the bank of a railroad. They strike at point A and B some distance apart. At a location midway between the points A and B, called M, they will converge and appear simultaneous to a stationary observer. If however the observer is in motion toward point B he will see strike B before strike A. They are no longer simultaneous. But that example is a distorted picture. Only if he has moved from point M during the information transit are they no longer simultaneous.</i> Once again, you haven't understood the example. An observer at M, stationary relative to the railroad, sees light from two lightning strikes at A and B arrive at him simultaneously according to his own clocks. He reasons that, because he is the same distance away from both A and B, and he knows that light travels at the same speed from both points, that the two strikes happened simultaneously. Now consider an observer on a train moving towards B. Just as that observer reaches point M, she observes the light from the two lightning strikes reach her. She knows that the light from both strikes travels at the same speed, and she knows that when the strikes happened she was closer to point A than to point B (because she is continuously moving towards point B as the light travels). Therefore, the only way the light from both strikes could reach her at the same time is if strike A happened after strike B, according to her clocks. Now, MacM, if you think this explanation is incorrect, please point out where you think the mistake is. Edit: corrected which strike came first according to moving observer.
Invarence of Light James R, Your example is different than the ones given to explain Simultaneity, which were as I described and position would be the cause of shift not the velocity since the velocity of light is invarient. But let me see about your rendition. She is just passing point M when the two strikes converge and she concludes that A was first because she knows she was closer to A? Sorry but that doesn't make sense to me. If she arrives at point M and sees them simultaneously, then considering light speed is invarient she should conclude they were simultaneous. Because her velocity relative to the sources didn't alter the speed of the light. They still arrive simultaneously at point M if she is in motion or not.
MacM: There are four different events here: 1. Lightning hits point A. 2. Lightning hits point B. 3. Light from A reaches M. 4. Light from B reaches M. For an observer stationary at M, events 3 and 4 happened at the same time. The reasoning given in my previous post leads the observer to conclude that events 1 and 2 happened simultaneously, though earlier than events 3 and 4. For the observer moving towards point B, events 3 and 4 happened at the same time according to her clocks. But since light had to travel from A to M and from B to M, event 2 must have happened before event 1 according to her. And yes, this relies on the speed of light being invariant for both observers. edit: corrected which event occurs first for moving observer.
Very poor physics This is extremely poor physics. If you want to discuss some problem define it properly, the setup, what is being measured etc. Your definitions are very poor, and your terminology is self defined.
Here's a picture from the point of view of the moving observer: M = middle point. O = observer's location. * = light signals from strikes. Lightning strikes B: --------A------------O-------M----------------*B A little later, lightning strikes A: ------A*------------O-----M-----------*------B The light signals continue to travel: ----A------*--------O---M-------*----------B And a little later, O approaches M: --A------------*----O-M---*--------------B Just as O reaches M, the light signals are received by O. A------------------*O*------------------B
I'm addressing Mac and his explanation of the experimental setup using polarised light. It seems Mac is the only one who is able to understand his own posts.
I Don't Think SO James R, Please bear with me. I am not trying to be difficult here but I don't think I agree with your statements. I'll give it some more thought but what you seem to be saying would be this: 1 - Assume for a moment that we were in a Binary star system and during a trip between the moon and earth, when earh just happed to be midway between the stars with the moon in alignment. 2 - Both stars have simultaneous flare events. 3 - That because we see both flare events at the same time just as we pass earth that we should assume one occurred before the other or that light took more than 8.5 minutes and the other less than 8.5 minutes to reach earth. 4 - That makes no sense what-so-ever. We have gone from a view of "It takes longer due to simultaneity induced by motion" to it takes the same amount of time but we should conclude, that it must have"? 5 - While light will appear invarient in its velocity once it has reached you I have never seen it said that the initial light wave front will reach you sooner or later as a function of your motion. I will try to figure out what it is you are trying to say here but in the meantime let me ask yo a direct question regarding my posts. Question: As B's clock is approaching 6.61 hours and I have been watching C's clock rate running 66.1% of mine. Is not from my view clock C accumulating time which is running behind my clock? And if so when my clock stops at 6.61 hours is not the accumulated time I see 4.37 hours? And finally since I don't see a constructive reply to ryans post let me ask you this. Question: Did you also have a problem understanding the polarized light presentation and where do you think it could have been made more clear? When I post here I assume I am talking with knowledgeable people that can connect the dots without an absolute description using a general description view. To me I would expect others to understand polarization of light, how it works and therefore how it is being used to relay information about time along the light beam without a full explanation about how light is being polarized. Even though there are different polarizations, any can be used to create a signal information transfer system such that details about it are irrelevant.
MacM, Did you try reading that chapter of Feynman's Lectures I posted? The failure of simultaneity is finally shown on page 8. - Warren
Not Yet chroot, Not yet. I have been very busy and will be for at least a couple more days but I do have it in Favorites and will read it. Thanks.
Re: Not Yet So you have time to argue and piss and moan on this website, but don't have enough to read a book that will actually teach you something? You've posted like 20 times since I gave you a copy of that chapter. Where are your priorities, Mac? Really? - Warren
OK chroot, Ok, I took a few minutes before I gotta go but now I want you to answer a simple question. Question: As B's clock is approaching 6.61 hours and I have been watching C's clock rate running 66.1% of mine. Is not from my view clock C accumulating time which is running behind my clock? And if so when my clock stops at 6.61 hours (which we all agree it does) is not the accumulated time I see 4.37 hours? Yes or No? Now qualify your answer please. Thank you.
