An immortal fumble by Androcles (Hexenmeister) (25-Apr-2006)
Negative time means that a clock runs backwards
"Paul B. Andersen" <paul.b.andersen@hiadeletethis.no> wrote in message news:e2lapo$5qo$1@dolly.uninett.no... | eleaticus wrote: | > | > The real question: | > | > is your post because you didn't take some (prescribed) drug or because | > you did take some (illegal) drug? | > | > Try again when you are (a) back on your meds (b) off the illegal (c) | > have removed your head from up your ass. | > | > eleaticus | > ee-lee-aT-i-cus | | Since you respond in such a friendly manner, I will | take the time to go through it again. | | Eleaticus wrote: || Let there be at your stationary system origin a hundred mile-long highway || extending along your x-axis to, of course, x=100. || || As the moving system moves by at v~.866c (so g=gamma= 2), when its origin || is at your x=0 you set your clock to zero and calculate what the moving || observer will see: x' = g(x-vt). t=0 and g=2 so: x'=200. | | And this is quite correct. | I, being in the stationary system, can calculate | that when the moving observer "see" the event | "the end of the road x = 100, at the time t = 0", | then the coordinates of this event in the moving frame are: | x' = g(x - vt) = 200 | t' = g(t - xv/cc) = -173.2/c | | That is, as you so correctly stated, the moving observer | will "see" the end of the road at his x' = 200. | He will also see that his own clock is t' = -173.2/c. ahahaha..HAHAHAHA...hahaha... | You didn't mention this, but I am sure that a clever guy | like you were aware of this. ahaha...HAHAHA... hahaha... Can Eleaticus have a diploma? | | Since the moving system moves in the positive x direction | with the speed v, it is pretty obvious that seen from | the moving system, the stationary system moves in the negative | x' direction with the same speed v. So the clock runs backwards... ahaha...HAHAHA... hahaha... | | So since the end of the road (x = 100) moves in the negative | x' direction with the speed v, it is easy to calculate where | the end of the road will be at t' = 0: | x' = 200 - (173.2/c)(0.866c) = 50 | | That is, when the moving observer's clock shows t' = 0, | he will "see" the end of the road at x' = 50. | | Isn't this rather obvious, Eleaticus? Sure... clocks run backwards all the time.. ahaha...HAHAHA... hahaha... | Or is it something you don't agree to so far? No, of course not. Everyone knows clocks run backwards. ahaha...HAHAHA... hahaha... | || But consider the 'actual' moving system measurement. He too has set his || clock to zero and he too sees gamma=g as being 2. Our x=100 is his x' = || 100 = g(x'-vt'), so with t' =0 he gets, x'=50. | | So the question is, when the moving observer sets his clock to zero, | t' = 0, where does he then "see" the end of the road (x = 100)? | We need only one of the transform equations to answer that question: | x = g(x'+ v t') where x = 100, t' = 0 and g = 2. | So the answer is, as you so correctly stated, x = 50 | | When the moving observer's clock show t' = 0, he "sees" the end | of the road (x = 100) at x' = 50. | | So we agree on everything. | Or don't we? | | Is there a problem somewhere? Maybe... "He will also see that his own clock is t' = -173.2/c." -- might be slightly questionable. ROFLMAO! Androcles. |
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| Fumble Index | Original post & context: Dsr3g.12689$tc.5404@fe2.news.blueyonder.co.uk |
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