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Determined c in S v. Result of the Michelson_Morley Exp.

  1. Jun 10, 2008 #1
    1. Let S' be an x'y'-coordinate system. Let the x'-axis of S' coincide with the x-axis of an xy-coordinate system S. Let the y'-axis of S' be parallel to the y-axis of S. Let S' move in system S along the x-axis of S with constant velocity v in the direction of increasing x, and let the origin of S' coincide with the origin of S at the time t = t' = 0s.

    2. 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'1 at the time t' = t'1, and let it be reflected at x'1 back to x' = 0m, reaching 0m at the time t' = 2*t'1.

    3. Let the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 be, in the moving system S', the length L of a rigid rod.

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

    5. Let the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 be also (independently), in the moving system S', the length L of the rigid rod.

    6. 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 x1 at the time t = t1, and let it be reflected at x1 back to x = 0m, reaching 0m at the time t = 2*t1.

    7. Let the length of the path of the ray of light emitted by the stationary system S from x = 0m to x = x1 be also (independently), in the stationary system S, the length L of the rigid rod.


    8. Let a ray of light emitted by the stationary system S depart from y = 0m at the time t = 0s towards y = y1 and reach y1 at the time t = t1, and let it be reflected at y1 back to y = 0m, reaching 0m at the time t = 2*t1.

    9. Let the length of the path of the ray of light emitted by the stationary system S from y = 0m to y = y1 be also (independently), in the stationary system S, the length L of the rigid rod.

    10. Let the time (t'1 - 0s) = (t1 - 0s) = L/c.

    11. Let the Length of the path of the moving system S' from x = 0m to x = v*L/c be, in the stationary system S, the length D of a rigid rod.

    12. In the moving system S', the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

    2*L.

    13. In the moving system S', the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

    2*L.

    14. In the stationary system S, the length of the path of the ray of light emitted by the stationary system S from x = 0m to x = x1 and back to x = 0m is

    2*L.

    15. In the stationary system S, the length of the path of the ray of light emitted by the stationary system S from y = 0m to y = y1 and back to y = 0m is

    2*L.

    16. In the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

    D + L + D + (L - 2*D) = 2*L.

    17. In the stationary system S, the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

    sqrt(L^2 + D^2) + sqrt (L^2 + D^2) = 2*sqrt(L^2 + D^2).

    18. The ray of light emitted by the moving system S' moves in the stationary system S with the determined velocity c.

    19. The time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1 and back to x' = 0m is

    2*L/c.

    20. The time in the stationary system S the ray of light emitted by the moving system S' takes to move from y' = 0m to y' = y'1 and back to y' = 0m is

    2*sqrt(L^2 + D^2)/c.

    21. By the result of the Michelson-Morley Experiment,

    2*L/c = 2*sqrt(L^2 + D^2)/c, or

    L = L*sqrt(1 + (D/L)^2).

    22. The problem with the result of the Michelson-Morley experiment is that (1) L in both sides of the equation "L = L*sqrt(1 + (D/L)^2)" is the length of the rigid rod in the stationary system S, and (2) the ray of light emitted by the moving system S' moves in the stationary system S with the determined velocity c.

    23. The problem with the ray of light emitted by the moving system S' moving in the stationary system S with the determined velocity c is that (1) L in both sides of the equation "L = L*sqrt(1 + (D/L)^2)" is the length of the rigid rod in the stationary system S, and (2) the equation "L = L*sqrt(1 + (D/L)^2)" is the result of the Michelson-Morley experiment.
     
    Last edited: Jun 10, 2008
  2. jcsd
  3. Jun 10, 2008 #2

    Dale

    Staff: Mentor

    Hi arbol,

    I am not going to go through each step in a 23-point derivation. Let me just say that you are correct that the emitter is moving at the same velocity as the detector in any frame.

    However, remember that light (especially in this experiment) is a wave. Most waves that had been studied up until the M&M experiment propagated through a medium. If you were to build a similar apparatus for acoustical waves you could use it to detect the motion of the medium wrt your device. This is the velocity the the M&M experiment surprisingly failed to detect: the velocity of the medium through which light propagates, aka the aether.
     
  4. Jun 10, 2008 #3

    I was going to suggest the following:

    I propose (1) that in the stationary system S, the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 moves with the determined velocity c and (2) that in the stationary system S, the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 moves with velocity sqrt(c^2 + v^2).

    The purpose for my proposition (if it has not been proposed already) is to have

    (1) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1 and back to x' = 0m be

    2*L/c, and

    (2) The time in the stationary system S the ray of light emitted by the moving system S' takes to move from y' = 0m to y' = y'1 and back to y' = 0m be

    2*sqrt(L^2 + D^2)/sqrt(c^2 + c^2) = 2L/c = 2*t'1 (the result of the Michelson-Morley experiment).
     
  5. Jun 11, 2008 #4

    I think the following is a more appropriate and a more comprehensive proposition:

    I propose (1) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 is

    L + D, then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
    to be

    (L + D)/(c + v), and

    the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 to move in the stationary system S with velocity (c + v),

    (2) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = x'1 back to x' = 0m is

    L - D, then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = x'1 back to x' = 0m
    to be

    (L - D)/(c - v), and

    the ray of light emitted by the moving system S' from x' = x'1 back to x' = 0m to move in the stationary system S with velocity (c - v),


    (3) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

    2*L, then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
    and back to x' = 0m to be

    2*L/c, and

    the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity c,

    (4) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

    2*t'1*(c^2 - v^2), then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
    and back to x' = 0m to be

    2*t'1*(c^2 - v^2)/(c^2 - v^2), and

    the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity (c^2 - v^2), and

    (5) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' (1) from y' = 0m to y' = y'1 or (2) from y' = y'1 back to y' = 0m is

    sqrt(L^2 + D^2), and

    the length of the path of the ray of light emitted by the moving system S' from y' = 0m to y' = y'1 and back to y' = 0m is

    2*sqrt(L^2 + D^2), then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move (1) from y' = 0m to y' = y'1, (2) from y' = y'1 back to y' = 0m, or (3) from y' = 0m to y' = y'1 and back to y' = 0m to be

    sqrt(L^2 + D^2)/sqrt(c^2 + v^2), sqrt(L^2 + D^2)/sqrt(c^2 + v^2), or 2*sqrt(L^2 + D^2)/sqrt(c^2 + v^2) respectively, and

    the ray of light emitted by the moving system S' (1) from y' = 0m to y' = y'1, (2) from y' = y'1 back to y' = 0m, or (3) from y' = 0m to y' = y'1 and back to y' = 0m to move in the stationary system S with velocity sqrt(c^2 + v^2),
     
    Last edited: Jun 11, 2008
  6. Jun 11, 2008 #5
    This part ought to be written as

    (4) that if in the stationary system S, the length of the path of the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m is

    t'1*(c^2 - v^2), then


    the time in the stationary system S the ray of light emitted by the moving system S' takes to move from x' = 0m to x' = x'1
    and back to x' = 0m to be

    t'1*(c^2 - v^2)/(c^2 - v^2), and

    the ray of light emitted by the moving system S' from x' = 0m to x' = x'1 and back to x' = 0m to move in the stationary system S with velocity (c^2 - v^2),
     
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