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Natural Science Forum / Physics / Relativity / October 2006yet another paradox (?)
Please bear with me again. Consider a spaceship with two people at front and back ends. The person at the front resets his clock and sends a light signal to the person at the back, who resets his clock when he receives the signal. Then, they both move to the center of the ship to compare clocks. Due to the delay of the signal, there will be a certain offset in time that the clocks show. Next, the same scenario is repeated, except it's the person at the back who initiates the reset. If one way speed of light is the same in both directions, the offset between the clocks must the same. The effect of moving the clocks to the center of the ship is the same in both cases, only difference can come from one way speed of light. Paradox is, according to a stationary observer, the light signal in two directions is not the same relative to the spaceship, therefore the offsets in two cases cannot be the same, but the people on spaceship cannot explain this difference. How is this resolved? Thanks in advance. | Please bear with me again. No. | Consider a spaceship with two people at | front and back ends. The person at the front resets his clock and sends [quoted text clipped - 11 lines] | same, but the people on spaceship cannot explain this difference. How | is this resolved? Thanks in advance. It isn't resolved, Einstein was a fuckwit. You'll find any number of pair-o'-dorkses. Move on. http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm Androcles > Please bear with me again. Consider a spaceship with two people at > front and back ends. The person at the front resets his clock and sends [quoted text clipped - 11 lines] > same, but the people on spaceship cannot explain this difference. How > is this resolved? Thanks in advance. This is not too hard to see qualitatively: Moving the clocks to the centre of the ship requires a non-zero velocity, which implies time dilation. The effect of time dilation in the ship frame is not the same as in the stationary observer. In the ship frame, the time dilation is entirely due to the motion of the clocks. In the stationary frame, there is already a time dilation effect, and the motion of the clocks is compounded with it (the rear clock is moving faster than the ship so it is dilated more, while the front clock is moving slower, so it is dilated less). The overall effect can be shown to exactly compensate for the difference perceived by the stationary observer. > > Please bear with me again. Consider a spaceship with two people at > > front and back ends. The person at the front resets his clock and sends [quoted text clipped - 27 lines] > to exactly compensate for the difference perceived by the > stationary observer. Yes, but this can only cause a fixed amount of offset between the two clocks, it cannot compansate in both cases with this fixed amount. > > > Please bear with me again. Consider a spaceship with two people at > > > front and back ends. The person at the front resets his clock and sends [quoted text clipped - 30 lines] > Yes, but this can only cause a fixed amount of offset between the two > clocks, it cannot compansate in both cases with this fixed amount. sorry, I think you're right Dear nobody1357: > Please bear with me again. Consider a spaceship > with two people at front and back ends. ... > If one way speed of light is the same in both > directions, the offset between the clocks must the > same. The act of moving the clocks erases any offset. Experiment dies. Nature will not allow you to measure OWLS. ... > Paradox is, according to a stationary observer, > the light signal in two directions is not the same > relative to the spaceship, It is always c. Spaceship measurements are ancillary. > therefore the offsets in two cases cannot be the > same, It is, because all measurements of c are TWLS. > but the people on spaceship cannot explain this > difference. How is this resolved? The stationary observer will also determine that the spaceship adhered observers will measure c. That is the second postulate, after all. David A. Smith > Please bear with me again. Have you understood your last paradox yet? Why not try learning about relativity so you can answer your own paradoxes? Martin Hogbin Learn tau = (t-vx/c^2)/sqrt(1-v^2/c^2) xi = (x-vt)/sqrt(1-v^2/c^2) Now learn where it comes from. http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm Now that you've learnt about Santa Claus, grow up. Ignore Hogbin, the imbecile doesn't take is own advice. | > Please bear with me again. | [quoted text clipped - 3 lines] | | Martin Hogbin <nobody1357@operamail.com> /: > Please bear with me again. Consider a spaceship with two people at > front and back ends. The person at the front resets his clock and sends [quoted text clipped - 11 lines] > same, but the people on spaceship cannot explain this difference. How > is this resolved? Thanks in advance. How, do you think, come both clocks ticking "in pace"? Right, by synchronising them with each other or with one or more other clocks belonging to their system. And synchronisations are two-way procedures. And once the clocks tick in pace, the reset delays are the same. -- guido http://home.scarlet.be/~pin12499/paratwin.htm "wugi" : > How, do you think, come both clocks ticking "in pace"? Right, by > synchronising them with each other or with one or more other clocks > belonging to their system. And synchronisations are two-way procedures. > And once the clocks tick in pace, the reset delays are the same. ....so that, remembering the relativity of simultaneity, when the signal emitted at the front arrives "in advance" at the back, the clock there will compare it with an "earlier" event, and when the signal emitted at the back arrives "retardedly" at the front, the clock there will compare it with a "later" event, earlier and later to be understood "than emission time", from the viewpoint of a "rest" system, and resulting in equal transmission intervals to and fro, in the moving system. > -- > guido > http://home.scarlet.be/~pin12499/paratwin.htm same:-) |
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