Mike-from-the-Bronx

If you have a PC, I would like to suggest trying my application at http://www.relativitysimulation.com/MainWebPage.html It has a lot of the capability you seem to be looking for. I would be glad to share the code with you.

Oh, I didn’t read the whole thread. Now I see why it is called a triplets paradox. There’s one doing the observing and two being observed. This could be an extra credit homework problem in an introductory course in SR. Let me restate the problem in more precise terms. One of a set of triplets, the observer, is at rest with respect to the source of a distant light pulse which is far off in the x-direction. The second triplet has a measuring stick 1 light-second long with a clock attached at either end and is moving at a speed of .866c toward the source of the light pulse. The third triplet, also with a measuring stick and two clocks is moving at a speed of .866c away from the source of the light pulse. How can the observer triplet believe that the other two, moving in opposite directions, will measure the same value for the speed of the light pulse? Time Dilation, Length Contraction and Relativity of Simultaneity all play a role again. But the key to understanding what is going on is this: The leading clock has the trailing time. The trailing clock has the leading time. (That’s just another version of Relativity of Simultaneity.) Start with the triplet moving toward the light pulse. The right side clock is its leading clock and the left is its trailing clock. The right side clock records the passage of the light pulse first. Say it records the passage at time t=10.000sec. The left side clock reads .866sec ahead of the right side clock. Familiar number? So, according to the observer triplet, at the moment the light pulse passes the leading clock (10.000sec) the left side clock already reads 10.866sec. (I’ll leave it for another day to show how that’s calculated) How long will it take the light pulse to reach the left side clock? Because of length contraction the distance between clocks is only half a light second and the left side clock and light pulse are closing on each other at 1.866c. So the observer triplet calculates that it will take .5c/1.866c = .2679sec for the light pulse and left side clock to meet. Since the clocks moving with the second triplet are ticking at 50%, only .2679 / 2 = .134sec will elapse on the left side clock. So, starting time of 10.866 +.134 gives 11.000sec elapsed. The second triplet will calculate 11.000 -10.000 or 1sec for the time for the pulse to traverse the 1-light second long measuring stick and the second triplet will conclude the light pulse is traveling at speed “c”. Now, for the third triplet, the one moving away from the light pulse. (I’m getting tired) Again the passage of the light pulse will be recorded by the right side clock first. But now the right side clock is the trailing clock and has the leading time. So this time according to the observer triplet, when the right side clock records the passage of the light pulse at, say t=100.000sec, the left side clock will only read 99.134sec. How long for the light pulse and left side clock to meet? Well, according to the observer triplet, the light pulse is closing on the left side clock at a speed of only (1c -.866c) = .134c. (.5c/.134c) = 3.731sec for the light pulse and this left side clock to meet. Dividing by 2 again to account for 50% time dilation gives 1.866sec elapsed on the left side clock. (99.134 + 1.866) = 101.000sec at the meeting. The third triplet will calculate 101.000 -100.000 or 1sec for the time for the pulse to traverse the 1-light second long measuring stick and the third triplet will also conclude the light pulse is traveling at speed “c”.

I don’t usually post to these forums because I don’t have time to do a back and forth dialog. But I have to reply to this thread. The “paradox” presented by the author that esbo referenced is so basic it could be given as a homework problem to a student taking “Introduction to Special Relativity”. The key to the resolution, as has already been mentioned, is Relativity of Simultaneity. Let me walk thru the homework problem. Given: You are sitting on the surface of the Earth monitoring a box of decaying muons with a clock. I am monitoring an identical box of decaying muons with my own clock. But I am at an altitude of 10 light-seconds above hurtling toward you at .866c. (10 light-seconds is the distance light travels in 10 seconds). We both start monitoring at the same time according to you. As far as you are concerned, when I reach you, 11.55 seconds will have elapsed on your clock (10c/.866c). But only 5.77 seconds will have elapsed on my clock since my clock is running at 50% of yours. Since less time has elapsed for me, you expect that I will have more muons in my box than you have in your box. Assignment: Analyze the problem from my reference frame. What do I expect? Solution. From my point of view, the distance to you is only 5 light seconds (50% length contraction). So I expect to meet you when 5.77 seconds have elapsed on my clock (5c/.866c) and I expect to have an amount of muons in my box consistent with that time. That’s the same time (and the same amount of muons) as you expected for my clock and my box. So far, no contradiction. Now for the hard part. I believe your clock is running at 50% the rate of my clock. So if 5.77 seconds elapsed on my clock, only 2.89 seconds will have elapsed on your clock between the time I started monitoring and the time we met. But it was given that 11.55 seconds elapsed on your clock when we met. I’m standing right next to you. I have to see the same number on your clock or there is a paradox. How do I resolve that? Well, you must have started monitoring 17.33 seconds before I started monitoring. Since your clock was running at 50%, 17.33 /2 means 8.67 seconds already elapsed on your clock before I even started my clock. Add 2.89 more seconds while both of us were monitoring and I will expect 11.55 seconds elapsed on your clock when we meet. That eliminates the contradiction. If you’re thinking I invented my answer just for this problem, I didn’t. It’s part of the theory. To Summarize: From your point of view we both started monitoring our boxes of muons at the same time. From my point of view you started monitoring your muons 17.33 seconds before I started monitoring mine. What is simultaneous to you is not simultaneous to me. That’s how Length Contraction, Time Dilation and Relativity of Simultaneity work together to eliminate any contradiction. Can I show that using actual calculations? Sure. I could post some equations with the right numbers on the other side of the “equals” sign. Would you then be convinced? I doubt it.

Here's mine. There are no graphs. The effects are presented in 3d graphics; http://www.relativit...ainWebPage.html

May I suggest the following short analogy: http://www.relativit...relyAnalogy.htm

Excellent. I am going to try to add that to my 3d simulation program.

Well, since my post was addressed to Dr Rocket and you are not Dr Rocket I would not expect you to know.

Dr Rocket; I've read many of your posts and I'm confused about your opinion with regards to SR and Acceleration. Which side of the fence are you on? Some references: From the book "Special Relativity" by A. P. French Copyright 1966 Massachusetts Institute of Technology Chapter 5, Relativistic Kinematics, page 153 "Because Einstein developed a whole new theory (his general theory of relativity, published in 1916) based upon dynamical equivalence of an accelerated laboratory and a laboratory in a gravitational field, it is sometimes stated or implied that special relativity is not competent to deal with accelerated motions. This is a misconception. We can meaningfully discuss a displacement and all its time derivatives within the context of the Lorentz transformations." From the book "Basic Relativity" by Richard A Mould Copyright 1994 Springer-Verlag Chapter 8, Uniform Acceleration, page 221 "Until the relationship between mater and metric is explicitly stated, we cannot be said to have left the domain of special relativity, even when working with non-inertial frames of reference."

Most textbooks use the same examples. In mine the example for derivation of relativistic momentum goes like this. Mary and Frank are moving in opposite directions horizontally with respect to each other. Mary throws a ball straight down and Frank throws a ball straight up. The balls are perfectly rigid and perfectly elastic. The distances, speed and mass of the balls are symmetrical so that the balls bounce off each other and back into their hands. To prove that this must happen in Newtonian physics, you need to use the laws of Conservation of Energy and Conservation of Momentum. If you still want to have a Conservation of Momentum in SR, you can't use the Newtonian definition of p=mv. So what definition will work? If that's the example you have in your book, I can help.

in Special Relativity: To an inertial observer time for any inertially moving body or any accelerating body always runs SLOWER. To an accelerating observer time for any inertially moving body always runs FASTER. To an accelerating observer time for any other accelerating body may run FASTER or SLOWER. It's more complicated.

I'm sorry if my post offended you. I was really trying to be helpful. I know the answers to your issues but I don't care to participate any more in this thread.

The Original Poster of this thread was asking for examples of the measurement of the speed of light. If anyone posted the opinion that MMX measured the speed of light they are wrong. Unless I have misunderstood the intent of the OP, any further discussion of MMX is really off-topic. You seem to have an important issue that you would like to discuss with regards to MMX. I have seen such issues resolved in the course of discussing another subject. But, frustratingly, in this particular case that is not happening. So I would suggest starting a thread of your own and in that thread asking posters to stay on your topic.

I have to disagree with the opinion being expressed here. I have half a dozen physics text books that deal with SR in my bookcase. Not a single one says that MMX ratifies SR. What some say is that the results of MMX are consistent with what SR predicts. Not the same thing. From the book "Special Relativity" by A.P. French Copyright 1966, by Massachusetts Institute of Technology Page 73 "This null result is consistent with the proposition that the speed of light is the same in all directions with respect to a reference frame having an arbitrary (but unknown) motion through space." From the book "Modern Physics for Scientists and Engineers" Copyright 2006 by Brooks/Cole Page 77 "Efforts by Michelson and Morley proved in 1887 that the either the elusive ether does not exist or there must be significant problems with our understanding of nature." P.S. I can't speak for books and articles written by laymen. I don't pay any attention to them.

The following link describes several different methods for measuring the speed of light. http://en.wikipedia..../Speed_of_light Scroll down to section 5 Mesurement.

I stand corrected. I'm the one who got lost and confused. As has been pointed out, two events that are not simultaneous in one reference frame can be simultaneous in another. However the point I was trying to make still holds. Two event cannot switch chronological order. And again I jump the gun. That's not right either. I'm just trying to say that all observers must agree on who is younger once "A" and "B" are in the same reference frame again

Ouch. Just like with the Twins Paradox, calculate who is youger from the point of view of one inertial observer and all other inertial observers must agree. If that's not true there will paradoxes with causality. Let's not get lost here folks. Yes it is true. Think about it. Relativity of simultaneity means two observers in two different inertial reference frames will determine ONE event to occur at two different times. But you must never get the two observers thinking TWO events occured in a different sequence. Also see my last post.

I think granpa has misunderstood the problem. Since "A" and "B" do not take off at the same time, the first traveler will always be younger. The problem can easily be solved from the point of view of a third observer who is always stays on earth. The logic is as follows: "A" and "B" start out at the same place on earth (an inertial reference frame). Phase 1: "A" accelerates away from earth until he reaches some another inertial reference frame. "B" stays on earth. Therefore "A" has lost some time on his clock relative to earth. "B" has not. Phase 2: "B" takes off with the same acceleration until he reaches the same inertial reference frame as "A". "A" stays inertial. Therefore "B" has lost the same amount of time on his clock (compared to earth) as "A" did in phase 1. "A" has lost some additional time. They are now in the same inertial reference frame and their clocks now run at the same rate. Conclusion from the point of view of Earth: "A" is younger than "B". If you can't draw the same conclusion by analyzing the problem from the point of view "A" or "B" you are doing something wrong.

Actually vuquta, granpa has it right. The solution you set up in your last post is wrong. The hyperbolic formula you posted assumes constant acceleration and the starting positions of the inertial and accelerating observer coincide. That works for the initial acceleration away from earth but not the turnaround because the acceleration is now in the opposite direction. So for the turnaround, we need to reset the parameters and now the starting positions don't coincide. For the turnaround we must use the more complete version of the formula you posted, the one that includes the difference in starting position. Regardless, your conclusion is still correct. To accelerating observer, an inertial clock runs faster. I'm not sure what you mean by the phrase "logically undecidable". We are always able to determine who will be younger.