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RelativityLC=length contracted or length contractionTD=time dilated

  1. May 25, 2008 #1

    LC=length contracted or length contraction
    TD=time dilated or time dilation
    LOS=loss of simultaneity

    When we measure the speed of light, we find that it is the same in all directions even though we know for certain that the earth is moving. We conclude that moving objects must become LC, TD, and experience LOS.

    Now suppose that rocket A, which is stationary, is at the origin and a line of one million and one stationary, atomic clocks (clocks zero through one million) lie evenly spaced at one light sec apart along the x axis. These atomic clocks send out radio pulses at regular intervals with which they synchronize themselves. when A looks through a telescope at these clocks he will see them out of synch due to the time it takes for the light from the clocks to reach him. but he knows the distance to the clocks and the speed of light so he can calculate the current time on each clock and therefore concludes that they are all perfectly synchronized.

    Clock zero is official rocket A time. clock 1 synchronizeds with clock 0 by sending clock 0 a radio pulse (takes 1 sec). when clock 0 recieves this pulse it sends clock 1 a pulse in return (takes another sec). when clock 1 receives this pulse it simply determines the total time between sending and receiving (two sec)and divides by 2. this is the time for a signal to move between clock 1 and clock 0(one sec). clock 0 then sends its current time to clock 1. clock 1 adds the transit time (one sec) to that time and sets itself to the result.

    at t=0 the middle of rocket B (which is two light seconds long) passes the origin at gamma=10 (v=0.99^0.5). on this rocket are 3 atomic clocks(clock BB in the back and clock BM in the middle and clock BF in the front) identical in every way to the other clocks. clocks BB and BF synchronize themselves with clock BM (clock BM is the official rocket B time) in exactly the same way clocks zero through one million synchronize themselves with rocket A. But since the rocket is moving the radio pulses from these clocks will take longer to travel one direction than the other so these 3 clocks will appear to A to be out of synch. At t=0 clock BM reads zero.

    rocket A will see rocket B as LC, TD, and as experiencing an LOS as we have already shown. Now the question is what does B see? the easy answer is that B considers himself to be stationary and sees exactly what A saw with their positions reversed. but that is getting ahead of ourselves. Lets ask ourselves what would A expect B to see?

    A would expect that when rocket B looks through his telescope at the clocks he will see exactly the same image that rocket A saw. But since A would expect that B would perceive the speed of light to be c relative to himself (even though he is moving) therefore A will expect B to calculate that the line of clocks are out of synch even after B compensates for light travel time. For the same reason A would expect B to calculate that the clocks on board hisown ship are perfectly synchronized.

    Since B is LC and TD it seems reasonable to assume that A would expect B to perceive that A and the line of clocks are longer and ticking at a faster rate than B. We know that isnt the case. So whats wrong. Whats wrong is that we left out the third effect which is LOS.

    B would consider the distance between the position of the front and back of his rocket at one simultaneous moment to be 2 light seconds. If clock BM sends out a pulse then A would expect that B would consider clock BBs reception of that signal and clock BFs reception of that signal to be simultaneous. A would perceive the distance between those events to be 2*gamma (in light seconds). Therefore, A would expect that B would perceive that A is contracted by a factor of gamma.

    A knows that B considers the moment that clock BF meets clock 100 and the moment that clock BB meets clock 80 to be simultaneous. the time that clock BB shows when it meets clock 100 is equal to whatever showed on clock BB when it was at clock 80(just pretend it read zero, its easier) + the time that elapses on clock BB while it moves to clock 100. That equals gamma*2 light seconds/(v*gamma). We know that the time elapsed on clock 100 between the time it meets clock BF and meets clock BB is given by (2 light seconds/gamma)/v. Placing one over the other, A will conclude that B will perceive that A is time dilated by a factor of gamma.

    A knows that B perceives A to move a distance of 2 light seconds in gamma*2 light seconds/(v*gamma). So A concludes that B will percieve A to be moving at velocity v.

    And of course, since B perceives the line of clocks to be LC and TD and perceives the speed of light to be c relative to himself and the line of clocks are synchronizing themselves by radio pulses (he sees exactly what A sees when he looks at B) it follows that B will perceive the line of clocks to be out of synch is exactly the same way and to exactly the same extent that A considers B to be out of synch.

    This shows that B sees exactly the same thing that he would see if he was stationary and A was moving.

    Regarding the concept of aether, since the aether, if it exists, appears to be undetectable it is irrelevant to any discussion of relativity. Furthermore, even if aether does exist that still wouldnt mean that absolute velocity exits. the aether would just be another object that one would move relative to. Philosophically speaking, absolute velocity is a meaningless concept.
    Last edited: May 25, 2008
  2. jcsd
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