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Concept of the constant speed of light

  1. Mar 8, 2007 #1
    I am having trouble wrapping my head around the concept of the constant speed of light.

    Imagine I am moving in a ship at a certain speed (let's say 1/4 the speed of light). I send two probes out at the same speed, one directly ahead of me, one directly behind me. They are set to activate a timer when they separate from me, move an identical distance relative to me, and await a beam of light from me. I send a beam of light in each direction which stops each probe's timer.

    The probes then return to my ship and I compare timers. Will the timers have stopped at the same values, or will one probe's time value be higher than the other's?

    Thanks for any feedback.
  2. jcsd
  3. Mar 8, 2007 #2
    If they are moving at the same (though opposite) speed relative to you, then this is no different from the case where you're stationary (since, relative to you, you're stationary). The timers would stop at the same time.
  4. Mar 8, 2007 #3
    Edit: Never mind...misunderstood the question. :redface:
    Last edited: Mar 8, 2007
  5. Mar 8, 2007 #4
    NeoDevin is correct in the sense you probably intended by the question. But the wording leaves the possibility of an alternative interpretation. When you say you are in a moving spaceship, you are implying that you are not in the rest frame. Then when you say the two probes are moving at the same speed in opposite directions, you could mean as measured in the rest frame. The answer is then, of course, different.
  6. Mar 9, 2007 #5
    Thank you for your answers, I understand now.

    I have another question: Why is it said that one cannot attain the speed of light given that our frames of reference are relative to each other? For example, if I was in a ship deep in the middle of space, and I made the ship accelerate up to 99% the speed of light (a number that large for argument's sake), then I turned off the engines and had the ship coast. Say then I hit my head hard upon a beam within the ship, had amnesia, and thought, since there was no frame of reference, that the ship was not even moving. Would I not be perplexed when I found I could not accelerate more in the direction my ship was pointing?

    I described this example in simplistic terms for illustration's sake, but this question directly relates to me asking myself "How fast is the galaxy moving through space? And since there is a speed limit, couldn't we know the speed of the galaxy by how quickly we attain the barriers presented us by the speed of light when we accelerate in a particular direction from our galaxy?"

    Thanks for any feedback.
    Last edited: Mar 9, 2007
  7. Mar 9, 2007 #6


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    Staff: Mentor

    No, you wouldn't, because to you it would still seem like you were accelerating. You'd feel the forces and never notice any difference (you have to: that's what the principle of relativity means). But if you look out your window, you'd never measure your speed relative to anything else to be above C.

    Both this and your next questions are based on the belief that there is absolute time, space, and therefore speed. There isn't.
  8. Mar 9, 2007 #7
    I understand that there is no such thing as absolute time, space, and speed, and have trouble fully grasping that concept, which is why I ask the questions.

    So after my ship's engines are turned off at 99% the speed of light from my original point, and since there is no acceleration on me or the ship, I think that the ship and I are at rest, you're saying that I could accelerate a little less than 299,792,458 more meters per second?

    I guess I have trouble understand why light speed is measured in measures of time and space (meters per second) if it's supposed to be so relative.

    Last edited: Mar 9, 2007
  9. Mar 10, 2007 #8


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    Gold Member

    DocZ, I can see how you're thinking but I should point out the your state of rest or motion does not depend on whether your engines are running.

    Relativity is all about choosing who's moving or not. Suppose we have 2 trains coasting in opposite directions on parallel tracks. We now have 3 choices to say who's moving or not. We could say the ground isn't moving, but both trains are. Or we could choose one of trains to be not moving and let the the other train and the ground move. The point is that nothing really changes whichever frame you choose.
  10. Mar 10, 2007 #9


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    Staff: Mentor

    Yes, from the point of view of your new reference frame (after the first acceleration), you can do that. Nevertheless, from the point of view of your original reference frame (before the first acceleration), your velocity remains less than c. Velocities don't "add" in the same way in relativity as they do in classical physics.

    For example, suppose in the first acceleration you increase your velocity to 0.9c. Then in the second acceleration, you increase it by 0.8c relative to your new reference frame. Your net velocity relative to your original reference frame is not

    [tex]0.9c + 0.8c = 1.7c[/tex]

    but rather

    [tex]\frac{0.9c + 0.8c}{1 + \frac{(0.9c)(0.8c)}{c^2}} = 0.9884c[/tex]
  11. Mar 10, 2007 #10
    Correct, and, except for trivial cases, they are not even commutative or associative.
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