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Distance to Light

  1. Sep 22, 2007 #1
    Hey all, I have a quick question about distance and relativity. L=L(proper)/(gamma) Now if you are dealing something that is actually going to speed of light (like light itself), gamma = 1/0. If we bring limits into the mess gamma = infinity. This would make L (the distance of the object to the moving observer) 0. Does this mean that light doesn't "expierience" distance? Such as, to light, is it the same distance from the US to China as it is from one end of the galaxy to the other?
  2. jcsd
  3. Sep 22, 2007 #2


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    This is true as a statement about what happens in the limit as your speed approaches c relative to the Galaxy, but it doesn't really make sense to talk about what light "experiences". Keep in mind that even to say that the distances in your inertial rest frame are what you "experience" is a bit of an oversimplification, what it really means is that if you use rulers and synchronized clocks ('synchronized' using the Einstein synchronization convention) at rest with respect to yourself to give your definition of "distance", then distance contracts according to the Lorentz formula (the synchronized clocks are necessary so that you can locally measure the time that the front and back of a moving object were at particular positions on your ruler, and subtract the position of the front at a given time from the position of the back at the same time). But there's no physical way to generalize this to a photon, since you can't have physical rulers and clocks which are at rest relative to the photon, so considering what the photon "experiences" is more of a pure mathematical game without much physical meaning.
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