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Break the upper limit: 2x speed of light?

  1. Sep 9, 2008 #1
    I have a very important qs to ask on relativity.

    Picture this:
    2 person "A" and "B" : A is stationary and B is moving in a rocket (lets assume a half transparent and half white painted "rocket") at the speed of light TOWARDS A.
    But B is holding 2 touch light one shining straight at the transparent part towards A and the 2nd touch light at the white painted "wall" inside the rocket.

    So my qs is from A point of view, does the light from B travel 2 times the speed of light towards A ?

    If so, then from B point of view, when he shines the 2nd touch light on the white wall inside the rocket, it will appear at the 1 time speed of light.

    Otherwise if the upper limit of speed is observe, then from A point of view, light from B touch light is only 1 times the speed of light.
    Then from B point of view, shouldn't the light from B to the wall appear 0.5 the speed of light or it will be 1 times the speed of light?

    Can anyone advise

    Thanks a lot
    Last edited: Sep 9, 2008
  2. jcsd
  3. Sep 9, 2008 #2


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    It's impossible for an object with mass, like a rocket, to reach the speed of light--it would take infinite energy. Let's say the rocket is moving at 0.6c in the frame of A instead (also note that all motion is relative, so A cannot be 'stationary' in any absolute sense; in the frame of B, B is stationary and A is moving at 0.6c, and this frame is just as valid as A's).
    Nope, you might think that if B is moving at 0.6c in A's frame, then the light would have to be moving at 1.6c in A's frame, but velocities don't add the same way in relativity as they do in Newtonian mechanics. Each observer defines speed in his frame in terms of distance/time as measured by rulers and clocks at rest in his own frame, and remember that in A's frame, B's rulers are all shrunk due to Lorentz contraction, and B's clocks are slowed down due to time dilation (and there is also something called the 'relativity of simultaneity' that says that if two clocks at different locations are synchronized in B's frame, then they will be out-of-sync in A's frame). So, velocities add according to the relativistic velocity addition formula--if an object is moving at velocity v in B's frame, and B is moving at velocity u in the same direction in A's frame, then the object will not be moving at (v + u) in A's frame as you'd expect in Newtonian mechanics, but will instead be moving at (v + u)/(1 + uv/c^2). You can see that in the special case where v=c, the velocity in A's frame will be:

    (c + u)/(1 + uc/c^2) = (c + u)/(1 + u/c) = (c + u)/[(1/c)*(c + u)] = c*(c + u)/(c + u) = c.

    To understand how this relates to the fact that A sees B's rulers shrunk and his clocks slowed-down and out-of-sync, you have to know that in A's frame, the ruler that B used to measure the distance was shrunk by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex], the time between ticks on the moving clocks is expanded by [tex]1 / \sqrt{1 - v^2/c^2}[/tex], and two of B's clocks which are synchronized in B's frame will be out-of-sync in A's frame by [tex]vx/c^2[/tex] (where x is the distance between the clocks in B's rest frame). On another thread I gave an example of how these factors come together to ensure that both observers measure the light to be moving at c:
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