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Time difference: light emitted vs observed

  1. Mar 11, 2006 #1
    Hi. I have a question that I suppose is related to the Lorentz transformation, but I think that this situation is more simple: instead of relating the position of a light source between two observers, how would one relate the time difference between when a light signal is actually emitted versus when it is observed between an observer and an emitter that have a relative velocity?

    So, if the two systems originally coincided and they are moving away from each other at a constant velocity "v", and at time "t", light is emitted from the emitter, at what time (t') relative to t, will the observer receive the signal?

    I'm assuming that the Lorentz transformation, as it is, doesn't apply here since it's a significantly different situation (unless we can just set x=0?).

    Thanks in advance, for any insight
     
  2. jcsd
  3. Mar 11, 2006 #2

    jtbell

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    Yes, you have to take into account the fact that the distance between the source and the observer changes while the signal is in flight.

    To find the time difference in any particular reference frame, you don't need to take time dilation into account at all. Simply set up equations for the trajectories of the signal and the observer, taking into account their initial positions and their velocities, and solve them together to find the time and position at which the trajectories intersect. So long as you refer all times and positions to the same inertial reference frame, this is no different from a non-relativistic problem involving two cars traveling along a highway at different speeds, except of course that the signal always moves at speed c and the observer can never move at a speed greater than c.
     
  4. Mar 12, 2006 #3

    robphy

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    Bondi k-calculus

    Certainly, jtbell's method will work.

    However, it might be worth mentioning that one can analyze this as a Doppler Effect followed by a Time Dilation effect... efficiently solved in a few lines using the Bondi k-calculus (although the motivation in words may take a few more lines).

    When two inertial observers meet at an event have them reset their stopwatches. At some later event, one observer (the emitter E) sends a light signal to the other observer (the receiver R). The receiver-stopwatch-time of reception (T) is proportional to the emitter-stopwatch-time of emission (t), with proportionality constant k. (This is Doppler, as you shall see.)

    [What you seek is the emitter-stopwatch-time (you called t') when the receiver receives the signal.] (This is time-dilation.)

    So, T=kt. (k depends on v, which will be shown shortly.) If R immediately responds with a signal (i.e. reflects the light signal back), then (by symmetry of the observers, "relativity") E receives the radar-echo at E's-stopwatch-time of kT, which is equal to k2t.

    E just completed a radar-measurement of the reception event on R's worldline:

    Half of the sum of E-stopwatch-times (the average) is the time that E says is simultaneous with the reception event. That is, dt=(k2t+t)/2. This is your t' (but we still need to relate k to v).

    Half of the difference of E-stopwatch-times is the average round trip time for the light signal. Multiply by c to the get the spatial displacement of the reception event according to E. That is, dx=c (k2t -t)/2.

    The constant relative velocity v is (dx)/(dt)=(k2-1)/(k2+1). If you solve this for k, you'll find that k is the Doppler factor.

    Use this k to find your t'.

    The result can be written elegantly in terms of t, Bondi-Doppler factor k, and the time-dilation factor [tex]\gamma[/tex].

    (You can also do this problem with rapidities, using the connecting relation: [tex]v=c\tanh\theta[/tex]. The result can be written elegantly in its own right as a function of the rapidity.)
     
  5. Mar 12, 2006 #4
    no time dilation

    OK thanks, this is interesting.

    Alright, I'm not sure, but if I understand correctly, then the equation relating the two times would be:


    [tex]
    t' = \frac{t}{\sqrt{1-\beta}}
    [/tex]


    where t' is the time on the observer's watch when the signal was received and t is the time on the same observer's watch that the signal could be deduced (with the above equation) to have been sent.

    So, this is essentially the same as the equation for time dilation, except, in this case, it's not really dilation, but only represents the extra time of flight because of the extra distance that the light has to travel to reach the observer due to the relative velocity (duration)?

    Does that sound correct?

    Thank you again
    Rich
     
  6. Mar 12, 2006 #5

    pervect

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    I think this is a bit muddled :-(.

    The reception of transmitted signals is governed by the doppler shift factor. This is raw, unprocessed "time-of-reception" data.

    When one puts a further framework on this raw data, one gets time dilation. This framework "subtracts out' the extra signal delay. After one takes the raw data, and subtracting out the effects of signal delay mathematically (using the fact that the speed of light is always constant), one winds up with the result that the moving clock must be running slower (i.e. time dilation).
     
  7. Mar 12, 2006 #6

    robphy

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    This may help

    [tex]
    \]

    \unitlength 1mm
    \begin{picture}(55,90)(0,0)
    \linethickness{0.3mm}
    \put(20,10){\line(0,1){80}}
    \linethickness{0.3mm}
    \multiput(20,90)(0.12,-0.12){250}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,30)(0.12,0.12){250}{\line(1,0){0.12}}
    \linethickness{0.3mm}
    \multiput(20,10)(0.12,0.2){250}{\line(0,1){0.2}}
    \put(15,30){\makebox(0,0)[cc]{t}}
    \put(14,60){\makebox(0,0)[cc]{\gamma kt}}
    \put(20,60){\circle*{2}}
    \put(15,90){\makebox(0,0)[cc]{k^2t}}

    \put(55,60){\makebox(0,0)[cc]{kt}}

    \end{picture}
    \[
    [/tex]
     
    Last edited: Mar 12, 2006
  8. Mar 14, 2006 #7

    pervect

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    A classic space-time diagram. Did you type in the generating tex code manually, or is there some software that let you use a graphical interface?

    I hope people with questions have taken the time to look at your post #3 and see how this diagram is a graphical depiction of what you said in text - either everyone has gotten the point, or they've given up.
     
  9. Mar 14, 2006 #8

    robphy

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    I used http://jpicedt.sourceforge.net/ .
    Later, I did a few corrections manually.
     
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