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Lorentz contraction of a light pulse

  1. Dec 7, 2009 #1
    Suppose the earth, moon, and an extraterrestial observer(ET) are at the corner of an equilateral triangle. The earth observer points a laser at the moon and emits a powerful laser pulse for exactly .01 seconds. The 1,860 mile long pulse travels to the moon in about 1.3 seconds. The ET knows that the pulse is .01 seconds long because he has been monitoring the earth scientists' conversations, but the ET can't see the pulse in the vacuum of space. Suppose, however, there is a low density of dust particles in space that scatters some of the light in the direction of the ET(similar to the visualization of a searchlight beam on a foggy night). The ET will see a 1,862 mile long "blip" travel to the moon in 1.3 seconds.
    Or will he?
    The pulse is moving at the speed of light and the ET knows the pulse lasted for.01 seconds(1,862 miles long). Knowing that Lorentz contraction always shrinks an object in the direction of motion, the ET was expecting the Lorentz contrcted pulse to be shorter than 1,862 miles.What is the reason the light pulse is not Lorentz contracted? Is it because the pulse is not a material object and is therefore exempt from relativistic effects?
  2. jcsd
  3. Dec 7, 2009 #2
    What is ET's relative motion to all this?
  4. Dec 7, 2009 #3


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    Staff: Mentor

    Length contraction is always relative to the proper length of an object: the length of the object in its rest frame (the inertial reference frame in which it is at rest). But a pulse of light has no rest frame! It moves at speed c in any inertial reference frame.

    The length of the pulse does depend on the reference frame, nevertheless. First consider the rest frame of the source. Let the time interval that the source is "on" be [itex]\Delta t_0[/itex]. Then the length of the pulse in this frame is [itex]L_0 = c \Delta t_0[/itex]. We can't call this the "proper length" of the pulse, but I'm going to call it [itex]L_0[/itex] anyway for convenience.

    Now consider what this looks like in a frame in which the source is moving. Suppose the source is moving in the +x direction, and it emits the light in that direction, too. In this frame, the time duration of the pulse is greater because of time dilation: [itex]\Delta t = \gamma \Delta t_0[/itex] where as usual

    [tex]\gamma = \frac {1} {\sqrt {1 - v^2/c^2}}[/tex]

    Also suppose that in this frame, the source starts emitting when it's at x = 0. It stops emitting after time [itex]\Delta t[/itex]. At this time, the front edge of the pulse is at [itex]x_{front} = c \Delta t[/itex], and the rear edge of the pulse is at [itex]x_{rear} = v \Delta t[/itex] because that's where the (moving) source is at that time. So the length of the pulse in this frame is

    [tex]L = c \Delta t - v \Delta t = (c - v) \Delta t = (c - v) \gamma \Delta t_0[/tex]

    After some algebra we get

    [tex]L = \sqrt { \frac {c - v} {c + v}} (c \Delta t_0) = \sqrt { \frac {c - v} {c + v}} L_0[/tex]

    So the length of the pulse does depend on the velocity of the source, but it's not the length-contraction relationship.
    Last edited: Dec 7, 2009
  5. Dec 8, 2009 #4


    Staff: Mentor

    If the ET is at rest wrt the other observers involved then it will measure the same length as they do.

    FYI, in general I discourage the use of the time dilation and length contraction formulas. I think it is always best to use the full Lorentz transform equations. The length contraction and time dilation formulas then automatically pop out whenever appropriate simply by setting certain terms equal to 0.
  6. Dec 13, 2009 #5
    The ET the earth and the moon are all at rest. The only thing moving is the light pulse.
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