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B 2 objects at 0.6c in opposite directions. Can outrun light ?

  1. Jun 12, 2016 #1
    So I'm reading this

    "If you're moving at 0.6 the speed of light in one direction, and your friend is moving at 0.6 the speed of light in the other direction, how could you transmit information to one another? There's no mode that will make up the growing distance between the two of you. So, while many observers could record the information you beam out from your spaceship, your friend can't. It's not that information is or is not, but that information reception is relative."

    Is this true? I thought you cant outrun light .

  2. jcsd
  3. Jun 12, 2016 #2


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    When you friend emits light, it immediately starts moving towards you at c. His speed is utterly irrelevant. Since you are moving at less than c it will catch up with you.
  4. Jun 12, 2016 #3


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    Where is the problem, since the velocity of my friend from the point of view of my reference frame is still smaller than ##c## I can send a message with light signals. In this case the relative velocity is (in units of ##c##)
    $$\beta_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} \sqrt{(\vec{\beta}_1-\vec{\beta}_2)^2-|\vec{\beta}_1 \times \vec{\beta}_2|^2}=\frac{1}{1+0.6^2} \sqrt{1.2^2} \simeq 0.882.$$
    For details, see Sect. 1.7 in

  5. Jun 12, 2016 #4


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    No, the quote is flat out wrong. Two observers, each travelling at .6 c in opposite directions relative to some inertial frame can certainly exchange information via light signals.
  6. Jun 12, 2016 #5


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  7. Jun 12, 2016 #6
    This can be answered easily by pointing at the second postulate of SR. Light always travels at c ~ 300000km/s in a vacuum absent of gravity. That's all you have to know.

    Meaning that no matter how fast you see your friend travelling away of you, whenever he emits a light signal, you will always see that light signal travel at c towards you from the point he emitted it.

    The speed at which you see your friend travelling away is irrelevant. The 2nd postulate alone suffices to answer your question. In reverse, your friend will always see light signals emitted by you, travelling at c towards(and away) of him.
    Last edited: Jun 12, 2016
  8. Jun 12, 2016 #7
    Dear @DirkMan, perhaps I can help you.
    Some other members have post very invaluable information. But, I think I'd like to give you a summary here.
    You move 0.6c to the west wrt (with respect to Bob)
    Alice moves 0.6c to the east wrt Bob.
    So we have three parties here. Dirkman, Alice and Bob.
    With Respect to Dirkman
    Dirkman: 0
    Bob: moves 0.6c to the east
    Alice: moves ##\frac{0.6+0.6}{1+0.6*0.6} = 0.882## as @vanhees71 had pointed out.
    See vanhees post

    wrt Bob:
    Dirkman: 0.6c to the west
    Bob: 0
    Alice: 0.6 to the east

    wrt Alice:
    Dirkman: 0.88c to the west
    Bob: 0.6c to the west
    Alice: 0

    No!, you can't outrun light. Unless you are in a faraway galaxy which moves faster than c because of Hubble Law, but that belongs to cosmology.
    Alice and Bob still can exchange signal, because their speed is below c (0.882)
    Thanks Dale, I have known velocity addition inside out (in 1 dimension only :smile:), but I've never thought the consequences that they can exchange signal.
  9. Jun 13, 2016 #8
    The person who wrote that is confused about the meaning of speed. It makes sense to say say you're moving at a speed of 0.6 c only if you state what you're moving relative to!

    As others have indicated, you can remedy the confusion by stating that they're moving in opposite directions relative to, say, planet Earth. Because speed is the ratio of distance to time, and because measurements of both distance and time are relative, each person will not observe the other moving moving at speed 1.2 c.
  10. Jun 13, 2016 #9
    OK, since you specifically used the word "outrun" rather than "exceed" I'll bite!

    You can in theory outrun a pulse of light by continuously accelerating (you are never going faster than c though). I've tried to find a simple reference but none comes to mind just yet (try searching for "Rindler horizon"). There is an intense mathematical treatment here, but it's not for the faint-hearted.

    The key quotation begins (##a## is the acceleration): "In other words, the event of the emission of the flash of light is always in the present, and always a distance of 1/a behind the ship. As far as the passengers on the ship are concerned, the flash of light isn’t getting any closer."
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