In a recent post James suggested I post the new 3 Clock question for review. I thought I had done so but it appears I haven't so here it is. To eliminate numerous argumentative points about acceleration, synchronization, simultaniety, and to correct improper differential velocities used in the first 3 Clock Paradox, etc, the test is designed to use only periods of linear velocity between clocks and is based on VAF calculated differential velocities. http://www.fourmilab.ch/cship/timedial.html Time dilation: t' = t/(1-(v/c)^2)^.5 http://www.scar.utoronto.ca/~tawfiq/Relativity/L13.pdf Velocity Addition Formula gamma = (u+v)/(1+vu/c^2). In the original 3 Clocks I had used space clocks traveling at 0.2c and 0.3c and had taken the delta v between the space clocks as 0.1c. This introduced an error as follows: (.3 - .2)/(1+(.2*.3)) = .1/1.06 = 0.094339 or 6% difference. However, that difference does not resolve the issue. The following tables use the VAF to keep the number correct. TEST: Dan K. McCoin - 3 Clock Paradox Place three clocks in space traveling in a common vector. They will have different velocites and each will monitor the rate of clock ticks of the other clocks in comparison to their own clock for a period of 10 hours per their clock and the respective clock rates are recorded by a counter. The clocks output ticks one per second their local time. At which time they stop their counters for the other clocks and raise a flag to indicate they have stopped their test but leave their clock running until all observers have raised their flag declaring the test is over. Dilation Rate is set by Relativity as t' = t/ (1 - (v/c)^2)^.5 "A's Perspective:........................Ticks Counted for Clocks "A" at rest......., v =0.00c......................36,000 "B" relative to "A" v =0.50c...............31,176 "C" relative to "A" v =0.75c...............13,176 Note: Relavistic differential velocity between "B" and "C" is (.75-.5)/(1+(.5*.75)) = .25/1.375 = 0.1818 c "B's" Perspective: "B"..........(Ref =Rest)..........................36,000 "A".........(0.5000 c Delta)...................31,176 "C".........(0.1818 c delta)...................35,388 "C's" Perspective: "C"...............(Ref = Rest)....................36,000 "A"......................................................13,176 "B"......................................................35,388 Summary of recorded time dilations of clocks by each observer: A total of (4) different times for (3) Clocks 36,000, 35,388, 31,176 and 13,176 counts are recorded. And; "A" must simultaneously display 36,000, 31,176 and 13,176 counts. That corresponds to a local time flow as being 10 h, 8.66 h and 3.66 h. "B" must simultaneously display 36,000, 35,388 and 31,176 counts. Which corresponds to 10 h, 9.83 h and 8.66 h. "C" must simultaneously display 36,000, 35,388 and 13,176 counts.Corresponding to 10 h, 9.83 h and 3.66 h. Now since physical clocks cannot dilate at multiple rates nor simultaneously display three different times, it should be obvious that Relativity is computing perception and not reality. Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
MacM: Ok. Let's look at this again. First, the relative velocities: From A's perspective: A: v = 0 B: v = 0.50 c (gamma = 1.15) C: v = 0.75 c (gamma = 1.51) From B's perspective: A: v = - 0.50 c (gamma = 1.15) B: v = 0 C: v = 0.40 c (gamma = 1.09) From C's perspective: A: v = -0.75 c (gamma = 1.51) B: v = -0.40 c (gamma = 1.09) C: v = 0 Note that the figure 0.40 c comes from a correct application of the relativistic velocity addition formula. Please tell me if you disagree with this. Here's what each observer sees: <u>Observer A</u> Clock A ticks off 10 hours, or 36000 seconds. At the exact second the 36000th tick happens, A puts its flag up to indicate it has finished the test. At that instant, the times A sees on the other two clocks are (ignoring the time it takes light to travel from the other clocks to A): B: 36000/1.15 = 31304 seconds C: 36000/1.51 = 23841 seconds. So, at this time, A has not yet seen either of the other two clocks put its flag up. When that does happen, B's flag goes up before C's, according to A. <u>Observer B</u> Clock B ticks off 10 hours, or 36000 seconds. At the exact second the 36000th tick happens, B puts its flag up to indicate it has finished the test. At that instant, the times B sees on the other two clocks are: A: 36000/1.15 = 31304 seconds C: 36000/1.09 = 33027 seconds. So, at this time, B has not yet seen either of the other two clocks put its flag up. When that does happen, A's flag goes up before C's, according to B. <u>Observer C</u> Clock C ticks off 10 hours, or 36000 seconds. At the exact second the 36000th tick happens, C puts its flag up to indicate it has finished the test. At that instant, the times C sees on the other two clocks are: A: 36000/1.51 = 23841 seconds B: 36000/1.09 = 33027 seconds. So, at this time, C has not yet seen either of the other two clocks put its flag up. When that does happen, A's flag goes up before B's, according to C. <u>Summary</u> In terms of when the flags go up: A sees his own flag go first, followed by B then C. B sees his own flag go first, then A then C. C sees his own flag go first, then A then B. Once again, this shows that the ordering of events is different for different observers in relative motion. You have made a couple of incorrect statements. The first is your calculation of the relative speed of clock C, as seen by B. The second is in your conclusions, where, for example, you say: A can not and does not simultaneously display three different times. For each observer, at any particular time, A only ever shows one reading. However, events which are regarded as simultaneous by one observer are not simultaneous from another observer's point of view. So, for example, when A's clock reads 36000s, A says that at the same time (ACCORDING TO A), B's clock reads 31304 seconds. But when B's clock reads 31304 seconds ACCORDING to B, A's clock DOES NOT read 36000 seconds, according to B. There is nothing at all paradoxical about this, once you accept the relativity of simultaneity. The entire description is logically consistent.
Questions James R., Ok. Let's look at this again. First, the relative velocities: ?: This is not an arguement but what justification is there to apply a (-) sign to "A's" velocity relative to "B". "B" has assumed a rest position and in his view the velocity is "+". ?:Initial conditions: B = 0.5 c and C = 0.75 c (Newronian delta = 0.25 c) VAF is (.75 - .5)/(1 -vu/c^2) = .25/.625 = .4 I agree. I mixed the +'s and -'s. ?:Again I don't see the justification to apply (-) velocities. Relative to "C" all velocities are viewed as +. ANS: Yes ANS: We agree. ANS: I believe "B" will see "C's" flag before "A's" ANS: Again I believe "C" will see "B's" flag fefore "A's" ANS: Yes ANS: No. His own, C then A. ANS: No. C see his flag then "B" then "A". ANS: We agree but that doesn't resolve the issue. ANS: We agree. I used + and I should have used - in the VAF. ANS: Bad choice of words on my part. The clock does only show one time 10 hours but the counters which predict time aboard other flights do not agree with the times those observers record. Other observers claim that "A" shouldn't be 36,000 counts it should be either 31,304 (B's view) or 23,841 (C's view). The same holds true for the other clocks and counters. The problem stems from the fact that relativity requires that each clock see the other as being dilated (running slower than itself). Since both cannot run slower than each other in real time there is a conflict. Simultaneity does not fix the conflict. The purpose of the test is to show that "Rate Differential's" are perception not reality. Simultaneity runs the clocks until they each raise their flag and record 36,000 counts local time. Simultaneity is only valid when you take information at light speed into account. The computed clock "Rates" do not require simultaneity. The "Rates" are simultaneous already. One can stop the clocks by whatever means such that they stop simultaneously in real time and the recorded times on the clocks will show a simular conflict. i.e. If all clocks are stopped when "A" reaches 36,000 counts: If in fact time dilation is reality then B would have only recorded 8.69 hours test time (31,304 counts local) but would think "A" stopped the test prematurely at 27,221 by his counter monitoring clock "A". That is because he sees "A" running at gamma 1.15 relative to his clock which was stopped by "A" local time 36,000 counts and has "B" running 31,304 counts because it sees "B" at a gamma of 1.15 also. This same complexiety runs between each clock and they data does not jive and cannot be reconciled, except to blindly say that is the way it is. "C" will have recorded 23,841 counts locally but thinks "A" terminated the test prematurely at 15,788 counts on its "A" monitor counter. B thinks C should show 19,545 and C thinks B should show 21,872. Lets not overly complicate the analysis. Lets stick with Clock "A" and "B" only. With "A" being the master control clock and stopping the test after 36,000 count local time. Remember the issue of simultaniety is not at play here. "A's" clock instantly stops the test and all counters when "A" reaches 36,000 counts. 1 - What is the counts on the "B" Clock local time? 2 - What does "B's" counter of "A's: clock read? Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
MacM: Sorry, I stuffed up the ordering of the flags going up, as seen by each observer. I agree with your ordering. My displayed times, are, however, correct. You ask: <i>...what justification is there to apply a (-) sign to "A's" velocity relative to "B". "B" has assumed a rest position and in his view the velocity is "+".</i> Suppose I take the direction "to the right" to be positive, and "to the left" negative. Then an object seen moving to the right has a positive velocity, and an object seen moving to the left has a negative velocity. If A sees B moving towards the right, from A's point of view, A's velocity is zero and B's is positive. From B's point of view, A's velocity is negative and B's is zero. According to B, A is moving away from B to the left. The sign is unimportant for the calculations here, but it is correct and consistent. The crux of your confusion comes at the end of your post: <i>The problem stems from the fact that relativity requires that each clock see the other as being dilated (running slower than itself). </i> That is not a problem. <i>Since both cannot run slower than each other in real time there is a conflict.</i> There is no "real time". There is only what is observed by somebody. That's relativity. <i>Simultaneity is only valid when you take information at light speed into account. The computed clock "Rates" do not require simultaneity. The "Rates" are simultaneous already. One can stop the clocks by whatever means such that they stop simultaneously in real time and the recorded times on the clocks will show a simular conflict.</i> That is not true. Let's look at your final, simple example. <i>Lets stick with Clock "A" and "B" only. With "A" being the master control clock and stopping the test after 36,000 count local time. Remember the issue of simultaniety is not at play here. "A's" clock instantly stops the test and all counters when "A" reaches 36,000 counts. 1 - What is the counts on the "B" Clock local time? 2 - What does "B's" counter of "A's: clock read?</i> A stops B instantly when A reaches 36,000 counts. Let's be very clear about this: A stops B simultaneously <b>according to A</b>. This is the important point. When A stops both clocks they will read: A: 36000 counts. B: 31304 counts. Both observers can bring their stopped clocks back together and look at them. They will agree they show the number of counts listed here. You might well ask: what sequence of events does B see in this example. The answer is: <u>A's point of view</u> When A reaches 36000, both clocks stop simultaneously. <u>B's point of view</u> B sees A's clock ticking slowing than B's clock. Suddenly and mysteriously, when B's clock shows exactly 31304 counts, B gets the signal to stop the clock. But B sees clock A continue to count after B's clock has stopped, because the signals which stop clocks A and B are <b>not simultaneous</b> from B's point of view, even though they <b>are</b> simultaneous from A's point of view. So, B sees A's clock continue to tick after B's clock has stopped, until A's clock reads 36000 counts. Then, A's clock stops. The answer to your apparent paradox is quite simple, though difficult to believe at first. The "stop" signal happens at exactly the same time for both clocks according to A, but at different times for each clock according to B. This is the relativity of simultaneity. It completely explains your 3-clock issues. It is self-consistent, and supported by real experimental evidence. It is not a flaw in relativity. Nor is it a patch to make the equations work. It is an inevitable consequence of the theory. But much more importantly, it is the way we observe our world to work.
Not a Problem James R., ANS: Not a problem. If I can muff up the +/- terms in VAF then we can all have slips. ANS: Yes I agree that the signs are not that important (unless you do half and half as I did). But even your explanation seems arbitrary. If the line of motion (Left/Right) were to be (East/West) and I am standing to the North side facing the event and then walk across the line and turn around facing the same event the +/- designaltion reverses also. ANS: This is where we have our disagreement. ANS: Assuming you are correct is where multiple clock rates derive from. To me that is a problem. ANS: This is where I see the failure of the concept. When "A" stops both clocks (irrespective of information delay and simultaneity) both are stopped and there simply is no continuation of "B" seeing "A" continue to run. That only occurs if you do not stop simultaneously as specified. Simultaniety involves the transfer of information at the sped of light. ANS:When "A" stops and stops "B" "Simultaneously", it also stops "B" counter monitoring of clock "A". It no longer sees "A" running. You are re-introducing simultaneity into a circumstance where simultaneity has been eliminated by definition of the test. ANS: Once again as stated above. When "A" stops "B" it also stops "B's" monitoring of "A" and your "Simultaneity" solution is not available to make the adjustments you want to show in clock data. Recall (though it was not well received)I have shown that the clocks can be stopped simultaneously by "A" using precalculated timing using Relativity to conduct the test. Now assuming Clock "A" stops "B" and its monitoring of "A" simultaneously, what time (counts) does "B"'s think "A" stop the test at? Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
MacM: <i>If the line of motion (Left/Right) were to be (East/West) and I am standing to the North side facing the event and then walk across the line and turn around facing the same event the +/- designaltion reverses also.</i> Yes, the direction you call positive is arbitrary. But as long as you're consistent it doesn't matter which way is positive. If you say B moves in the positive direction according to A, then A must be moving in the negative direction according to B. If you go the other way and say that B moves in the negative direction relative to A, then A must be moving in the positive direction relative to B. <i>When "A" stops both clocks (irrespective of information delay and simultaneity) both are stopped and there simply is no continueation of "B" seeing "A" continue to run. That only occurs if you do not stop simultaneously as specified.</i> You're still missing the point. In relativity there is no absolute notion of what is simultaneous, just as there is no absolute notion of what is stationary. Therefore, <b>events which are simultaneous for one observer are not simultaneous for an another observer in motion relative to the first</b>. In this case, as I already said, A's stop signal goes out to both clocks exactly simultaneously according to A, but at different times according to B. A and B have different ideas of which events are simultaneous. This has nothing at all to do with light or signal travel time, by the way. We're assuming all that kind of stuff has already been factored out of the analysis. <i>ANS:When "A" stops and stops "B" "Simultaneously", it also stops "B" counter monitoring of clock "A".</i> When you say <b>when</b> you must specify whose time you're talking about. If you're talking about A's time, then that has no effect on B's time. B's time is different. The moment when A stops clocks A and B according to A's time corresponds to two different moments in B's time. (This is only possible because clocks A and B are separated by some distance in space, by the way.) <i>You are re-introducing simultaneity into a circumstance where simultaneity has been eliminated by definition of the test.</i> No. I'm not introducing simultaneity into anything. Rather, it is you who is trying to impose an absolute simultaneity onto the situation. Such a thing, like the aether, does not exist. Have another go at answering your own question: <i>Now assuming Clock "A" stops "B" and its monitoring of "A" simultaneously, what time (counts) does "B"'s think "A" stop the test at?</i> The answer is the same as I gave before. And now you should understand why.
I've drawn a diagram to explain this a bit better. See the attached file. The diagram is a spacetime diagram of the situation. the blue axes are A's spacetime coordinates. Time is vertical; space is horizontal. The red line shows the path of clock B as seen by A. The blue horizontal lines (which should be parallel to the space axis) connect points in spacetime which are simultaneous according to A. A sends the signal to stop both clocks at 36000 ticks, at which time B is in the position indicted by the dot in the middle of the picture. The slanted lines connect points in spacetime which are simultaneous according to B. Notice that the line connecting B's position when clock A reads 36000 to the axis corresponds to A's time of 31304. In other words, B receives the signal to stop when A sees B as reading 31304. B agrees that his clock reads 31304 at the time the stop signal is received. But what about B's time when A reads 36000? The time 36000 on the axis corresponds to a time after the stop signal is received, according to B. By the time A reads 36000 according to B, B has moved on from its position as indicated on the diagram, to a new position further up and to the right, on the slanted line which goes through 36000 on the axis. In other words, according to B, A's clock continues to run after B has already received the "stop" signal.
James r., ANS: Not so. At least in this case. As I did once before it can be shown that using precalculated relavistic values clocks "A" and "B" can be shut down at the same exact moment. In this case it is arranged that "A" determines the moment of the end of the test. When that signal is activiated (there are several ways of achieving this) both clocks are stopped, including "B's" monitor counter for clock "A". So when "B" see his clock stop he thinks "A" shut down prematurely but he does not continue to see "A" run due to a shift in simultaneity. When "A" clock stops at the very same instant "A's" view "B" stops andhis counter for "A" stops. End of test, end of simultaneity adjustments. ANS: You do jest. Simultaneity has everything to do with information transfer at v = c. Simultaneity is explained by the two lightening strikes and the train. Remember? ANS: Which is simultaniety, which is no longer at issue. Once again when "A" stops "B" also stops at the same instant in "A's" time. Run the numbers of that case. That is what I did above. That is the end of the questions and has nothing to do with "B" continuing to run or see "A" continuing to run/ "B" is shut down by "A" time. What are the readings? At that moment "A's" time: "A" = 36,000 counts "B" = 31,304 counts "B's" monitor of "A's" clock shows "A" = 27,221 counts Yes or No? ANS: This has nothing to do with aether. This has to do with the precalculated or prearranged circumstance of measuring clocks rates from the vantage point of clock "A" controlling the clocks to start and stop simultaneously by "A's" time and the values indicated according to relavistic mathematics. ANS: Unfortunatly that answer doesn't fit the limits of the test as described. You must give the results of the clocks and monitor counter from "A's" perspective, that is all. No simultaneity of B continueing to run or continuing to see "A" run. this is from "A" perspective. Clock "A" _______________ counts? Clock "B" _______________ dounts? "B's" Monitor of Clock "A"_____________counts? Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
James R., ANS: I do appreciate your effort here but you have not properly described the test or the events. ANS: The problem with your presentation is this. 1 -"A" stops "B's" clock and stops "B's" monitoring counter of "A's" clock. B's view of simultaneity has nothing to do with the test. When "A" stops at 36,000 counts and correctly (and you have agreed here) stops "B" at 31,304 counts. that is simultaneous from "A's" point of view. The question now becomes what is the count on "B's" monitoring counter of "A's" clock before it was stopped at the same instant "B's" clock stopped at 31,304? "B's" monitoring counter of "A's" clock_______________counts when "A'" stop all clocks and monitor. I agree "B" has not seen "A" raise his flag (that is why I added that tid bit) and may think "A's" clock is still running and that something went wrong with the test. But he has no way of knowing that since his monitor has been turned off. You now have data to assess the validity of linear velocity time dilation predictions. Does clock "A" and clock "B's" monitor agree as to the passage of time.? Of course not. The point being that clock "A" will physically show or display 36,000 counts. It is also the point that "A" cannot display the counts predicted by "B's" monitor counter of "A's" clock. That is the argument. Time dilation is perception and not reality. A's clock does not and cannot agree with "B's" view of "A's" time dilation. I did look at your graph. But it introduces simultaneity and events after clock "B" was stopped by "A". Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
This string has been a rehash limited between myself and James R. He has made his statements and I have responded. From my perspective the question remains unanswered but it has been (3) days and no other input. Does anybodyelse have a more conclusive contribution or is this horse DOA.? Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
Hi MacM, I've just read the whole thing, and the difference between James and yourself is simply that you assume there is an absolute frame of reference with a "real time" and a "real length". Furthermore you seem to have a problem with the breaking of simultanity. Unfortunately, you wanted an explanation within the realm of special relativity. You disagree with it, saying that "it cannot be so because that is so incredible" (basically it comes down to that). So there are a few possibilities. Either we go over the entire thing ONCE AGAIN in relativity style. Any scientist will produce the same numbers and explanations as James did (I did not verify the calculations, but I assume they are correct), as this is what the theory of relativity predicts. Or we can do it the good ol' Newtonian style, which makes this a very easy problem. Both not good I think. Or maybe you can tell us what it is you would have liked to hear, then we can point out where conventional science disagrees with you. Bye! Crisp
Crisp, ANS: Thanks for the above response. It is actually good and on point. It does seem that James and I agree on the resulting numbers but disagree as to what the signigfigance of what those numbers mean. With respect to your post I wouldn't use that term. "Incredible" I could live with or accept. "Impossible" is another issue. Since the clocks operate on local time and display time accordingly they obviously operate on and record only one time base. Yet via Relativity the claim is that time dilates and is different for different observers in motion. What I am trying to show is that such a proposition is not reality. It is not reality because the clock physically cannot possess different clock rates simultaneously, hence the concept of time dilation is perception not reality. That is what "B" thinks about clock "A" or "A" thinks about "B" is not time itself but perception of time because the physical clock continues to operate on local time in total disregard to any observers view of its operation. I don't know if that made sense. While simultaneity exists and has proper place in physics it doesn't really resolve this issue if you define the problem in sufficient detail. That is that "B's" clock and monitor counter of "A's" clock stop defacto concurrent (simultaneous) with "A" reaching 36,000 counts. You can't claim that "B" continues to run or sees "A" to continue to run via simultaneity. That is like trying to have your cake and eat it too. Claiming it is both stopped and running at the same time becomes Schroders Cat. I don't think the disagreement I have had with other members here is symantics. I am not saying time dilation doesn't appear to occur via Relativity. I am saying that that observation doesn't alter the clocks physical ticking since it continues to run on local time and that the fact its recorded time and the observed time of a moving observer don't match simply means the observers view of dilated time is only perception. Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
Ehr... this is not entirely accurate. We, as outside observers having the ability to calculate it all, know that there is disagreement amongst the observers whether the clock has stopped or not. I think that because the observers cannot transport information to eachother faster than light, in the end, when "B" gets a message from "A" saying "hey, the clock at C stopped", then by the time "B" gets this message, he will also have established that the clock stopped ticking. I should verify this with some calculations, but I'll leave it up to you as an exercise Please Register or Log in to view the hidden image!. Bye! Crisp
MacM: Hi Mac, you make some good points but I want to ask a question. Sure any given observer perceives the time rate of a moving clock-- but he 'perceives' said time rate via measurements. Can we elevate these perceptions to reality if that observer will always measure what SR predicts and if we can show that any other experiment he can do regarding time (of the moving clock) will indicate to him that same 'percieved' time rate? Or, said another way, if that which has been discussed are all just perceptions, what would constitute a measurement of reality? Please Register or Log in to view the hidden image!
Crisp, ANS: What you say is only true if you do not pre-establish the clock "A" control (as I have done) so that information transfer time is voided and the action taken is (from "A's" view simultaneous at "B's" clock. This can infact actually be done several ways. The one anticipated here is to precalculate the time delay in information transfer and send the signal in time to arrive at "B" the instant "A" reaches the 36,000 count. Doing that you are now stuck with "B" showing one time for "A" and no fuether testing on going since "B" is actually stopped but its projected time for "A" does not match "A". So "B's" view is perception and "A's" recorded time is reality. Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
Natural, ANS: Off the cuff I would have to say local time is reality. That is the time the clock actually records. (Although I have to qualify that statement by saying as far as I am concerned "actual" slowing of a clock does not mean time was altered. That is because all clocks are processes and don't actually measure time perse. And as I have said before if you calibrate a pot of water with graduations on the side of the pot and claim evaportation rate is now a water clock and you then heat the pot you will change the evaporation rate hence the recorded rate of time change but time itself didn't change. Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
MacM: <i>Off the cuff I would have to say local time is reality.</i> Then you've just reinvented relativity. The local times of different observers are different, so each has his own reality. End of story.
I have to say that I agree with MacmPlease Register or Log in to view the hidden image! Time doesn't exist! Time is not physical! Time does, however, look like it's running slower or faster depending on your relative speed to an observer. But that's only due to the finite speed of light, and it give an illusion of time variance. Can I just say that if you DO see an illusion of time slowing, then you MUST be travelling at a different speed relative to that ray of light. Oh, that's funny, this means you ARE changing your speed relative to light! But Einstein doesn't allow that. My head just blew upPlease Register or Log in to view the hidden image!
James R., ANS: So then you agree that time dilation is perception and not reality? That is my dilated view of your clock due to relativie motion does not alter your clock? That has been my statement and argument the entire time.Please Register or Log in to view the hidden image! Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe will make you a genius.
We've been over this already, and from the Lorentz transformation point of view, there is more in the game than just "the finite speed of light". Bye! Crisp
