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Length contraction & light

  1. Nov 17, 2005 #1
    Hey, sorry if this is a stupid question but I'm getting really confused...
    If I am standing on the Earth and watch a spaceship moving past at say 0.8*c from left to right, I will observe the spaceship to contract in its direction of motion. If a person on the spaceship shines a torch also from left to right, what happens to the wavelength of the light as seen from my reference frame? Does it also contract?
  2. jcsd
  3. Nov 17, 2005 #2


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    In the ship's frame, the wave peaks move at c, so the distance between peaks is just c/f, where f is the frequency that the peaks are being emitted by the torch as seen in the ship's frame. In your frame, the peaks still move at c, but the frequency becomes [tex]f \sqrt{1 - v^2/c^2}[/tex] due to time dilation, and meanwhile the torch is moving at velocity v so it will have moved a distance of [tex]v / (f \sqrt{1 - v^2/c^2})[/tex] between emitting successive peaks, so the the distance between peaks should either be [tex]c / (f \sqrt{1 - v^2/c^2}) - v / (f \sqrt{1 - v^2/c^2})[/tex] or [tex]c / (f \sqrt{1 - v^2/c^2}) + v / (f \sqrt{1 - v^2/c^2})[/tex] depending on whether the torch is shining in the direction of the ship's motion or in the opposite direction. This simplifies to [tex](c \pm v)/(f \sqrt{1 - v^2/c^2})[/tex]. You could also get this from the relativistic doppler shift equation, [tex]f_{observed} = f_{emitted} \sqrt{1 - v^2/c^2} / (1 - v/c)[/tex], keeping in mind that the wavelength you observe is [tex]c / f_{observed}[/tex].

    To compare the wavelength you see with the wavelength seen by the ship-observer, just divide [tex](c \pm v)/(f \sqrt{1 - v^2/c^2})[/tex] (the wavelength seen by you) by c/f (the wavelength seen by the ship-observer), which gives [tex](1 \pm v/c) / \sqrt{1 - v^2/c^2}[/tex] for the factor that the wavelength changes in your frame. This is not the same as the Lorentz contraction factor [tex]\sqrt{1 - v^2/c^2}[/tex].
    Last edited: Nov 17, 2005
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