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Is the unsynchronised clock actually at the place where seen to be

  1. Sep 17, 2013 #1
    It all started when I read that different inertial observers from the same place at the same time should see same things.

    Say there are two clocks C1 and C2 in a stationary frame of reference S. C1 is at X=0 and C2 at X=X (some positive X) and both are syncronized in this frame. Say there is an observer named Mary located at X=0

    Let's say Tom is an observer in a moving frame of reference S' located at X' = 0 with a clock C'. Let its relative velocity w.r.t frame S be V and moving towards positive X direction. When Mary and Tom are at the same place i.e., X=0 & X'=0 coincides in space, Clocks C1 and C' are synchronised and the reading is C1=C'=0. The second clock is at X' = X√(1-(V/C)^2) in Tom's frame. Now what the second clock C2 reads to Mary and Tom?

    C2 reads zero to Mary as it is synchronised with C1 in Mary's frame. If Mary sees the second clock reading zero, then Tom should also see the same. But then how Tom reconciles the reading with the concept of unsynchronised clocks C1 and C2 in his own reference frame?

    The point is that the clock C2 is moving towards Tom in his frame and the light from the clock should have left some time before. This "sometime: turns out to be (VX/C^2)/√(1-(V/C)^2) time units in his frame. But Tom also knows only (VX/C^2) time units should have passed in the moving clocks. So Tom concludes that the current time of the clock C2 is (VX/C^2).

    The clock C2 is at X' in Tom's frame but it is showing the time what it was sometimes before when it was not at X'. In that case, should not Tom see the clock itself located at the point where it was located sometimes before? But he is not seeing so. If he sees it located at X' only, then it was the location of clock soemtimes before. So he should also conclude the clock is located more closer to him i.e., it should have had moved some distance towards him in this "sometime". So he should concludes that
    X' = { X√[(1-(V/C)^2)] - V [(VX/C^2)/√(1-(V/C)^2)].

    Please explain me why the above conclusion about C2 location is not right.
     
    Last edited: Sep 17, 2013
  2. jcsd
  3. Sep 17, 2013 #2
    I did some sloppy work by saying C2's reading to Mary and Tom as zero. Since light will take sometime to reach even Mary, C2 should be reading something in negative. But anyway we will end up in the same scenario of C2 at X' showing to Tom, the time what it was sometimes before when it was not at X'. So my question about C2 location is still alive.
     
    Last edited: Sep 17, 2013
  4. Sep 17, 2013 #3
    " If Mary sees the second clock reading zero, then Tom should also see the same."
    They will see the same thing but , as you pointed out, they have to correct what they see for light speed lag, and that correction is different for different observers
     
  5. Sep 17, 2013 #4
    Is the problem with "zero" reading? or with the "same reading" for both of them?

    As I learnt, Mary and Tom when located at the "same place" in space should see same reading in C2. Do you say this is not correct?
     
  6. Sep 17, 2013 #5
    yes the correction is different for different observers. Mary has to correct only for time, as the clock is not moving in her frame. But Tom, in addition to time correction, should he not also correct the space coordinates?
     
  7. Sep 17, 2013 #6
    He corrects it for time only, but the distanced traveled by tight in his coordinates are different than the one in her coordinates, so the corrections are different.
     
  8. Sep 17, 2013 #7
    i agree with you on different time corrections for Mary and Tom. Whatever time C2 that Mary concludes at in her frame is different from what Tom concludes at in his frame. The method I went by for Tom's conclusion of C2, if right, poses a question to me that should he not also correct the space coordinates of C2?
     
  9. Sep 17, 2013 #8
    I see what you mean. Well, the only correction you might need, depending on how you're solving the problem, is that you must use the C2's coordinate at the time it emitted the light signal which will be different than the coordinate at current time because it is moving in Tom's Ref. frame. Of course, that correction is on top of the Lorentz contraction correction.
     
    Last edited: Sep 17, 2013
  10. Sep 17, 2013 #9

    Erland

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    Science Advisor

    Let us consider the following three events:

    1. The event when the light is emitted from C2 which Mary and Tom both see when they pass each other at their common origin.

    2. The event at the position where C2 is located and simultaneous in Mary's frame to the event when Mary and Tom pass each other.

    3. The event at the position where C2 is located and simultaneous in Tom's frame to the event when Mary and Tom pass each other.


    For Event 1: The coordinates in Mary's frame are ##(x,-\frac xc)##. Since C2 is stationary in Mary's frame and synchronized with C1, the reading both Mary and Tom see on C2 is ##-\frac xc##.

    If we apply the Lorentz Transformation to this, we obtain the coordinates in Tom's frame as ##(x\sqrt\frac{1+\frac vc}{1-\frac vc},-\frac xc \sqrt\frac{1+\frac vc}{1-\frac vc})##. This is, according to Tom, the position and the time when the light from C2 was emitted.


    For Event 2: The coordinates in Mary's frame are simply ##(x,0)##. The coordinates ##(x',t')## in Tom's frame can be calculated with the Lorentz Transformation, but this is not of interest for us, although we will get ##t'\neq 0##.


    For Event 3: We need to find ##t## such that ##(x,t)## in Mary's frame transforms to ##(x',0)## in Tom's frame, for some ##x'## which we then calculate, using the Lorentz Transformation both times.

    Then, we easily obtain ##t=\frac{vx}{c^2}##, and then ##x'=x\sqrt{1-\frac {v^2}{c^2}}##, which is consistent with the length contraction formula. The coordinates in Tom's frame are then ##(x\sqrt{1-\frac {v^2}{c^2}},0)##. This ##x'## is the position where C2 is located "right now" according to Tom.
     
    Last edited: Sep 17, 2013
  11. Sep 17, 2013 #10

    Dale

    Staff: Mentor

    Ok, let's use units where c=1 and X=1. So when Tom passes Mary they both visually see that C2 reads t=-1. According to Mary, that is because the clock was located 1 unit of distance away and correctly synchronized. According to Tom, the clock was located a distance of (1+v)γ away and it was not correctly synchronized, it read -1 when it should have read -(1+v)γ.
     
  12. Sep 18, 2013 #11
    Enlightened, Erland :)

    I used these coordinates in Tom's frame to verify the current location of C2. That too gave me the location that is consistent with Length contraction ##x' = x\sqrt\frac{1+\frac vc}{1-\frac vc} + v [ \frac xc \sqrt\frac{1+\frac vc}{1-\frac vc} ] = x\sqrt{1-\frac {v^2}{c^2}}##
     
  13. Sep 19, 2013 #12
    Oops , there is a typo. It should be a minus sign between the two terms, instead I had put a plus sign there by mistake

    ##x' = x\sqrt\frac{1+\frac vc}{1-\frac vc} - v [ \frac xc \sqrt\frac{1+\frac vc}{1-\frac vc} ] = x\sqrt{1-\frac {v^2}{c^2}}##
     
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