chroot Prosoothus, Actually I think I agree with ch on this one. But the following is my response on the same issue to everneo under topic "3 Clocks.
MacM: You need to be very careful in specifying which rulers you're using in your merry-go-round problem. Are you using rulers rotating with the merry-go-round, or rulers which are stationary in space above the rotating merry-go-round? The answers will be different depending on which rulers you use. Also, even if you use only rulers on the merry-go-round, you need to be aware that rulers near the centre of the merry-go-round are not the same length as rulers near the outer edge. This is due to the curvature of spacetime caused by the rotation or, equivalently, the relative motion of the outer edge with respect to the centre.
Agree James R., We agree. But you didn't address the issue. That is the fact that the ruler at the edge in motion contracts. But so does the circumference, hence no measureable change in Pi. My complaint was not against Lorentz Contraction it was against the claim that the measurement changes. It can't in my view because the rotating disk and the ruler are subjected to the very same influences. ryans, I hope you note that James R., has just stated that the affects are at the rotating edge and not the radius as you claimed.
Re: I Agree Inpart No one is saying that you don't know anything, only that your knowledge of Relativity is incomplete, and that this incomplete understanding is leading you to erroneous conclusions. You need to heed the words of Alexander Pope: "A little learning is a dangerous thing; drink deep, or taste not the Pierian spring: there shallow draughts intoxicate the brain, and drinking largely sobers us again."
James R will not bring up the issue of the radius because you will never understand it. Along the radial vector there is an acceleration. It is as if there is mass along the perimeter of the m-g-r producing a gravitional field(equivalence principle). From the observer on the m-g-r, he thinks space time is warped(i.e. he feels a force) and thus for him the radius will not be the same as for an observer not on the m-g-r. This question here is being answered within the framework of SR, but if an accurate quantitative answer were to be sought, then this effect will have to be accounted for. With out this effect, the observer may as well be in rectilinear motion, for which your case is that both observers measure the same length, which of course is not true. Thus if you fail to take into account radial acceleration, this problem reduces to the simple case of rectilinear motion. Can you see your paradox. Do you belive that 2 observers in rectilinear motion will measure the same, or different lengths of a rod?
Fair Janus58, ANS: I agree with the first half of your post. My education falls short of many on this MSB including yours. But neither am I the many things that ch and ryans like to call me. That doesn't bother me what bothers me is the distraction from answering the questions directly. Sometimes it is cute and funny. At the same time I have to disagree with the later half of your statement. In this case at least my view has not been lacking nor in error. They haven't told me one thing I didn't know or understand. The only thing wrong has been ryans and ch's assumption about my knowledge and understanding and mis-stating what I have said. This topic got started because I said to ryan that I didn;t like Brian Greene's "the elegant universe" because he claimed the ruler being used by a man crawling along the edge of the rotating m-g-r would measure the circumference differently and that therefore changed the Pi ratio. I further stated at the outset that I accept relavistic contraction. I am aware and understand acceleration and gravity affects. But according the ryans, I don't know any of these things and am a complete idiot that can't possibly understand any of it. And that frankly is a bunch of crap from a guy that can't keep his own story straight. But rather than address the point I raised, they chose to change subjects and talk about other issues saying that because I don't know this stuff (wrongfully and selfservingly) that is why I am wrong. The fact is I was right and all this BS is just BS. I don't mind being told I'm wrong when I'm wrong. But these guys are pathetic. Based on some of the assertions they make sometimes one must wonder if they even really understand Relativity. You however, have always been fair and I enjoy your input . Thanks.
HHHAAA.... You don't know how do deal with acceleration and you don't know how to deal with non-flat geometries. Is that a fair statement? THAT is why you cannot come to the conclusion that I do in regard to the m-g-r problem. Every post I submit that deals with acceleration, you ignore. I repeat You don't know how do deal with acceleration and you don't know how to deal with non-flat geometries. No you aren't.
Lets See ryans, ANS: Not in the least. ANS: It is you that don't know how to answer my question. "Regardless of any affect by any theory or geometry at the circumference and/or radius, how do you justify that you claim Pi changes since any such affect by any theory affects the ruler and rotating disk equally ". quote: -------------------------------------------------------------------------------- I am aware and understand acceleration and gravity affects -------------------------------------------------------------------------------- No you aren't. ANS: So now you are a phsychic as well. Try answering the question and knock off you uneffective attacks. Woulds you like me to post a few formulas and run a few calculation for you? How about just try answering the question guy.
Show me the proof that the shortest distance between 2 points in a Euclidian space is a straight line?
Right After ryans, ANS: No more diversions and dodges. I'll address your issue right after you answer the question. This is only about opportunity 18 for you to do so and you continue to bring in new questions. Just answer the question. Thank you.
I've answered your question on several fronts, within several frameworks. My final statement is that the radius changes, the ruler doesn't due to an effective curvature of space-time. Now answer my question.
Understanding Grammer ryans, ANS: Apparently you need somebody to explain to you the meaning of the question. ANS: You have simply made a statement (unfortuantely an impossible and incorrect one). You have not asnwered the question and that is to justify your claim that the ruler is not also affected by the same functions of relativity and in the same amounts as the rotating disk. Once again the m-g-r is fabricated of assembled measuring rods. This is to eliminate yur dodge of the question, the consequences are not hereby altered. Now count the number of rods while the m-g-r is at rest and again while in motion. Clearly since the radius is laid out by mearuring rods you can see if you claim the the radius changes then so does the rulers and no measurement change can be percieved. Try again
You are using the incorrect instrument do define length. An observer on the m-g-r who measured the rods with light (interferometer, time intervals), would find that their length changes as a function of radius. Thus in this frame, the rods no longer become an accurate tool for measuring geometry.
Not on Point ryans, ANS: You are not on point. I think deliberately so. "I " am not doing anything. "I" merely complained about the way others, the relavists were doing things. If you would simply acknowledge that your resistance to stating the truth here (that I was and am right) we could move on but you are so determined that I can never be right that you keep looking for ways to claim that I am wrong. On this one at least you sir blew it. Now that others have acknowledged that I what I have said is fact and that what you have said is wrong, I hereby give you one last opportunity to acknowledge your error and move on. I will not however, keep chasing your rainbows trying to find ways to alter the discussion and distract the MSB from the fact that you simply refuse to admit that "The Ignorant Old Man" has put you down one knotch. Question: "How do you justify claiming the calculation of Pi changes on a rotating disk vs a stationary disk using a ruler by an observer moving along the radius and then the circumference of the disk?" Hint: Your answer should be something like "I don't because there is no measurable difference in the circumference or the radius since the affects of Relativity alter the disk in like manner and magnitude as the ruler being used to measure"
Holy Shit ryans, :bugeye: You damn near made me choke on my coffee. You should have stuck with 3 Clocks. You might have had better luck imposing "Simultaniety" on me. James and Janus58 are currently. I will now go back and find the question you refer to and see what I think. OBTW: Since you stopped posting the question about the 100 Kg of U239, can we assume that my answer was correct?
Mac, I said that you were right, but these people don't think so. http://www.lns.cornell.edu/spr/2000-06/msg0026245.html This one from http://scienceworld.wolfram.com/physics/GeneralRelativity.html giving a description of GR. And here is the best one yet, where radial acceleration has been ignored. http://www.gravitywaves.com/bmyth/ Here p means pi. of course they are wrong Mac, Aren't they.
What? ryans, You still don't get it. Why would I disagree with any of the above? Their affect still change the m-g-r and ruler in motion by the same amounts and do not alter computed Pi. As to the shortest distance between points, there are several methods one being Dijkstra's algorithm but they compute the shortest distance and your questions is "Prove its the shortest". That is I believe a substantially different question. Such proofs tend to be somewhat complex and since it doesn't relate to anything other than mathematical skills and your effort to suggest you are smarter, I don't intend to play that game. I acknowledge that you are most likely more knowledgeable in math and adept to finding processes for such solutions. Having said that I hasten to add should such proofs not involve calculus in all likely hood I could muddle through it. But this I feel is a selfserving exercise by you which has no bearing on any subject being discussed.
The proof is really quite simple Mac, once you bothered to even try to learn some math. I don't know what the f*** you are talking about with this Dijkstra's algorithm. You claim that you have superior reasoning, but you cannot even prove that the shortest distance between 2 points in euclidian space is a straight line. P.S. This is a very much relevant question. I was hoping to extend it to non-euclidian geometries, but since you wish not to answer anybody elses questions, and only converse issues which are at your pitifully poor level, you can get ******. Relativity has been proven right over and over again. Move on.
