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Relativity and light

  1. Jul 27, 2011 #1
    If you are traveling at the speed of light and you were to shine a light the observer wouldn't observe twice the speed of light? Please explain this phenomena to me
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  3. Jul 27, 2011 #2


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    The phrase 'travelling at the speed of light' on it's own doesn't mean much, because ( according to Newton) one cannot distinguish 'rest' from a state of uniform motion. It is only meaningful to talk about the relative velocities of objects.

    So it's customary to talk about two observers, who have a relative velocity. So each one sees the other approaching or departing.

    Given this situation, neither will see the other have a relative velocity equal to the velocity of light (c). This is because the rule for addition of velocities precludes this.

    Another postulate of relativity states that all observers see light travelling at the same speed. If either observer emits light, it will be seen by both to be travelling at c m/s.

    This is counter-intuitive but that's what SR tells us.
  4. Jul 28, 2011 #3


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    If person A is traveling at speed [itex]v_A[/itex] relative to you and throws object B directly away from you at speed [itex]v_B[/itex] relative to himself, that object will have speed, relative to you,
    [tex]\frac{v_A+ v_B}{1+ \frac{v_Av_B}{c^2}}[/tex]

    Note that, whatever [itex]v_A[/itex] is, if [itex]v_B= c[/itex], that will give
    [tex]\frac{v_A+ c}{1+ \frac{v_A}{c}}[/tex]
    multiplying both numerator and denominator by c, we have
    [tex]\frac{c(v_A+ c)}{c+ v_A}= c[/tex]

    So no matter how fast A is moving relative to you, if he shines a light ahead of himself, it will have speed c relative to both of you.
  5. Jul 28, 2011 #4
    Very nice that's the explanation i was looking for.
  6. Jul 28, 2011 #5


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