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Lengths and times relative to S' and S

  1. Jun 8, 2008 #1
    I propose a new look at lenghts and times relative to S' and S (at least new to me).

    Let S' be an x'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an x-coordinate system S, and let S' move along the x-axis of S with velocity v in the direction of increasing x.

    Let t' = the time, with respect to S', a ray of light emitted by S' takes to move along the x'-axis of S'.

    Let t = the time, with respect to S, a ray of light emitted by S takes to move along the x-axis of S.

    Let T = the time, with respect to S, the ray of light emitted by S' takes to move along the x'-axis of S'.

    At t' = 0s, the ray of light emitted by S' is located at x' = 0m.

    At t' = t'1, the ray of light emitted by S' is located at x' = c*t'1.

    At t' = 2*t'1, the ray of light emitted by S' is located at x' = 0m.

    At t = 0s, (1) the ray of light emitted by S is located at x = 0m, (2) the origin of S' is located at x = 0m, and (3) t' = t.

    At t = t1, (1) the ray of light emitted by S is located at x = c*t1, (2) the origin of S' is located at x = v*t1, and (3) t' = t.

    At = t = 2*t1, (1) the ray of light emitted by S is located at x = 0m, (2) the origin of S' is located at x = 2*v*t1, and (3) t' = t.

    At T = 0s, the ray of light emitted by S' is located at x = 0m.

    At T = T1, the ray of light emitted by S' is located at x = c*T1 = v*t1 + c*t'1 = c*t'1 + v*t'1 = t'1*(c + v).

    Note: t1 = t'1. It is given that the length of the path of the ray of light emitted by S' is equal to the length of the path of the ray of light emitted by S.

    At T = T2, the ray of light emitted by S' is located at x = 2*v*t1.

    T1 = t'1(c + v)/c.

    c*(T2 - T1) = |2*v*t1 - (c*t'1 + v*t'1)| = c*t'1 + v*t'1 - 2*v*t'1 = c*t'1 - v*t'1 = t'1*(c - v).

    T2 - T1 = t'1*(c - v)/c.

    Note: T1 is not equal to T2 - T1, but this inequality does not mean that the clock located at x = 0m and the one located at x = c*T1 are not synchronous. The clock that marks the time T = 0s is located at x = 0m, the one that marks the time T = T1 is located at x = c*T1, and the one that marks the time T = T2 is located at x = 2*v*t'1 (or x = v*T2).

    T = T1 + (T2 - T1) = T2 = 2*t'1.
     
  2. jcsd
  3. Jun 8, 2008 #2

    Fredrik

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    I haven't read the whole thing, but you're already contradicting special relativity here. If S and S' are inertial frames, the x' axis would intersect the x axis with slope v (when the units are such that c=1).

    You also need to explain what you're trying to do more carefully. I'm having a hard time following your argument.
     
  4. Jun 9, 2008 #3
    Yes. I am even having difficulties following what I am writing. I hope this reply is easier to read.

    Let S' be an x'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an x-coordinate system S, and let S' move along the x-axis of S with velocity v in the direction of increasing x, and let the origin of S' coincide with the origin of S at t = t' = 0s.

    Let a ray of light emitted by the moving system S' depart from x' = 0m at the time t' = 0s towards x' = x'1 and reach x' = x'1 at the time t' = t'1, and let it be reflected at x' = x'1 back to x' = 0m, and reach x' = 0m at the time t' = t'2.

    With respect to the moving system S', the length of the path of the ray of light emited by S' is

    2*x'1.

    t'2 = 2*x'1/c.

    Let a ray of light emitted by the stationary system S depart from x = 0m at the time t = 0s towards x = x1 and reach x = x1 at the time t = t1, and let it be reflected at x = x1 back to x = 0m, and reach x = 0m at the time t = t2.

    With respect to the stationary system S, the length of the path of the ray of light emited by S is

    2*x1.

    t2 = 2*x1/c.

    It is given in this thread that x = x'1.

    With respect to the stationary system S, the length of the path of the ray of light emitted by the moving system S' is

    x'1 + v*t'1 + (x'1 - v*t'1) = 2*x'1 = 2*x1.

    Thus, with respect to the stationary system S, the ray of light emitted by the moving system S' takes the time t2 = t'2 to move from x' = 0m to x' = x'1 and back to x' = 0m.
     
  5. Jun 9, 2008 #4

    Ich

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    Yes, as long as x'=x and t'=t, x'=x and t'=t. That's fact.
     
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