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X' relative to S' and relative to S

  1. Jan 25, 2008 #1
    S = a stationary one-dimensional coordinate system. The x'-axis of another one-dimensional coordinate system S' coincides with the x-axis of S. S' moves along the x-axis of S in the direction of increasing x with velocity v. A ray of light is emitted from the origin of the moving system S' at the time t' = 0s to x' = 299,792,458m, arriving there at the time t' = 1s, and is reflected back to the origin, arriving there at the time t' = 2s.

    x' = c*sqrt(1 - sq(t' - 1)) (1), where t' is in the closed interval [0s. 2s], describes the motion of the ray of light along the x'-axis of the moving system S' by the observer stationed on the x'-axis of the moving system S'.

    x' = x - v*t (2), where t is in the closed interval [0s, 2s], describes the motion of the ray of light along the x'-axis of the moving system S' by the observer stationed on the x-axis of the stationary system S.

    Adding each side of equation (1) to each side of equation (2) respectively, we get

    2*x’ = c*sqrt(1 – sq(t’ – 1)) + x – v*t.

    If x' = x = 299,792,458m, and

    t' = t = 1s, then

    v = 0m/s, provided we keep c constant.

    If x' = 299,792,458m, and x = 299,822,258m, and

    t' = t = 1s, then

    v = 29,800m/s, provided we keep c constant.
     
  2. jcsd
  3. Jan 25, 2008 #2

    JesseM

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    Where did you get equation 1? From t'=0 to t'=1, the position of the light ray would be described by the equation x' = c*t', and from t'=1 to t'=2 (after the light ray is reflected), the position of the light ray would be described by the equation x' = 299,792,458 - c*t'.
     
  4. Jan 29, 2008 #3
    if x' = c*t', then c*(2s) = 2*(299,792,458m), and not 0m.

    In a general and algebraic way

    x' = c*sqrt(sq(t') - sq(t' - t)), where t' is in the closed interval [0s, 2*t].

    with this algebraic and general equation, if t = 1s, then at t'= 2s,

    x' = 0m,

    at t' = 0s,

    x' = 0m,

    at t = 1s,

    x' = 299,792,458m.

    if c*t is the semi-major axis of an ellipse, and t is the semi minor axis, then

    (sq(x')/sq(c*t)) + (sq(t' - t)/sq(t) = 1,

    where the center of the ellipse is at (0m, t).
     
  5. Jan 29, 2008 #4

    JesseM

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    But I said that I was using two different equations for the light's position, one for the time interval t'=0 to t'=1 s (that was the x' = c*t' equation), and another for the time interval t'=1 to t'=2 s. So, to find the position at t'=2 s you have to use the second equation, which I actually wrote incorrectly, it should be x' = 299,792,458 - c*(t' - 1). If you plug t' = 2 s in there you correctly find that the light is at x' = 0.
    This is even more confusing. You originally had the equation for the light's path as x' = c*sqrt(1 - sq(t' - 1)), why did you replace the first 1 with sq(t') and the second 1 with t? Why would you need to bring in the S frame's t coordinate at all if you just want to know position x' as a function of time t' in the S' frame?
    Why? Where are you getting these strange equations? Can you give any sort of physical argument for them? (By the way, have you noticed that in your original equation x' = c*sqrt(1 - sq(t' - 1)) the light's position x' will be a complex number when t' is smaller than 1? For example, at t'=0 this equation becomes x' = c*sqrt(1 - sqrt(-1)), and sqrt(-1) is the imaginary number i.)
     
    Last edited: Jan 29, 2008
  6. Feb 7, 2008 #5
    In this reply to you, I am quoting from Einstein's work, On the Electrodynamics of Moving Bodies.

    If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighborhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an A-time and a B-time. We have not defined a common time for A and B, for the latter cannot be defined at all unless we establish by definition that the time required by light to travel from A to B equals the time it requires to travel from B to A.
    Let a ray of light start at the A-time tA = 0s from A towards B; let it at the B-time tB be reflected at B in the direction of A, and arrive again at A at the A-time t"A.
    In accordance with definition the two clocks synchronize if
    tB - tA = tB = t"A - tB.
    The following reflexions are based on the principle of relativity and on the principle of the constancy of the velocity of light. These two principles we define as follows:
    1. The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion.
    2. Any ray of light moves in the stationary system of coordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.
    Let S = a stationary x-coordinate system, and let system S be provided with a rigid measuring rod and a number of clocks, and let all the clocks of the system be in all respects alike.
    Let the point A be located at the origin of S and let x be the location of the point B on the x-axis of S.
    c = B/tB, and
    c = B/(t”A - tB), which implies that
    c*tB = B, and
    c*(t”A – tB) = B respectively, which implies in turn that
    t”A = 2*tB.
    Let S' = a stationary x'-coordinate system, and let the x'-axis of S' coincide with the x-axis of system S, and let a clock be located at the origin of system S'. Let this clock of system S' be in all respect like all the clocks of system S. Let system S' move along the x-axis of system S with a constant velocity v in the direction of increasing x. Let the point B on the x-axis of the stationary sytem S coincide with the origin of the moving system S' at the time t = t' = 0s.

    According to the Galilean transformation equations,
    x = x' + v*t', where x' = 0m, and x' and x (the location of the point B on the x-axis of the stationary system S) coincide initially at t = t' = 0s.

    Let the ray of light now start at the A-time tA = 0s from A towards (B + v*t'B); let it at the (B + v*t'B)-time t'B be reflected at (B + v*t'B) in the direction of A, and arrive again at A at the A-time t"A.

    c = (B + v*t'B)/t'B, and
    c = (B + v*t'B)/(t"A - t'B), which implies that
    c*t’B = B + v*t’B, and
    c*(t”A – t’B) = B + v*t’B respectively, which implies in turn that
    t"A = 2*t'B.
     
  7. Feb 7, 2008 #6

    JesseM

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    All right, all this is from section 1 of On the electrodynamics of moving bodies.
    I don't see this part in the above paper--if it's in there, what section is it in? In any case, if the symbols are interpreted properly I don't disagree with any of the equations here, but I don't see what these equations have to do with your own equations x' = c*sqrt(1 - sq(t' - 1)) and x' = c*sqrt(sq(t') - sq(t' - t)).
     
  8. Feb 8, 2008 #7
    I meant that I quoted him (zzzzz...). Nevertheless,


    S' = a stationary x'-coordinate system. System S' is provided with a meter stick and a number of clocks, and all the clocks of system S' are synchronous to each other. A ray of light is emitted from the origin of system S' with velocity c to a point at x'1 along the x'-axis of system S' in the direction of increasing x' at the time t'0 = 0s, arriving there at the time t'1, and is reflected back from the point at x'1 to the origin, arriving there at the time t'2.

    That the clock located at the origin of system S' is synchronous to the clock located at the point x'1 is confirmed by the time t'1 - t'0 (that the ray of light takes to travel from the origin to the point at x'1) being equal to the time t'2 - t'1 that the ray of light takes to travel back from the point at x'1 to the origin. In other words, the clock located at the origin of system S' is synchronous to the clock located at the point x'1 on the x'-axis of system S' if and only if t'2 - t'1 = t'1 - t'0.

    c*t'1 = x'1, and

    c*(t'2 - t'1) = x'1; therefore,

    t'2 - t'1 = t'1, which is equivalent to

    t'2 = 2*t'1.

    S = a stationary x-coordinate system. System S is also provided with a meter stick and a number of clocks. The meter stick of system S is in all respects like the meter stick of system S', and all the clocks of system S are synchronous to the clocks of system S'. The x'-axis of system S' coincides with the x-axis of system S, and system S' moves along the x-axis of system S with velocity v in the direction of increasing x. The origin of the moving system S' coincides with the origin of the stationary system S at the time t'0 = t0 = 0s.

    According to the Galilean transformation equations,

    x'1 = x1 - v*t1, where the clock located at the point x1 - v*t1 on the x-axis of the stationary system S marks the time t1 the ray of light arrives at the point x'1 on the x'-axis of the moving system S'.

    That the clock located at the origin of the stationary system S is synchronous to the clock located at the point x1 - v*t1 is confirmed by the time t1 - t0 (that the ray of light takes to travel from the origin of system S' to the point x'1) being equal to the time t2 - t1 that the ray of light takes to travel back from the point x1 - v*t1 on the x-axis of the stationary system S to the origin of the stationary system S . In other words, the clock located at the origin of the stationary system S is synchronous to the clock located at the point x1 - v*t1 on the x-axis of the stationary system S if and only if t2 - t1 = t1 - t0.

    c*t1 = x1 - v*t1, and

    c*(t2 - t1) = x1 - v*t1; therefore,

    t2 - t1 = t1, which is equivalent to

    t2 = 2*t1.

    Since x'1 = x1 - v*t1, we can substitute the left side of this equation in the right side of

    c*(t2 - t1) = x1 - v*t1 and say that

    c*(t2 - t1) = x'1.

    Since c*t'1 = x'1, we can infer that

    t2 - t1 = t'1, which is equivalent to

    t2 = t'1 + t1.

    We can also substitute the left side of x'1 = x1 - v*t1 in the right side of c*t1 = x1 - v*t1 and say that

    c*t1 = x'1.

    Since c*t'1 = x'1, we can also infer that

    t1 = t'1. If we substitute t'1 in the second term of the right side of t2 = t'1 + t1,

    we can say that

    t2 = 2*t'1, which confirms our initial condition (that all the clocks of the stationary system S are synchronous to all the clocks of the moving system S').
     
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