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Is c exceeded by this logic?

  1. Sep 7, 2012 #1
    Suppose we have two points A and B, separated by ten light years. Now I shoot off a proton at .999c from A to B. From the perspective of the proton the distance between A and B is now about .3 light years. Would it get from A to B in less than four months in the time frame of the proton? If not, why not? If so, it would seem that the proton has traveled from A to B at a speed greater than c. Before being accelerated the proton would agree (this is an intelligent proton) that A and B are 10 light years apart, and it would also agree after it stopped at B. Is it the acceleration that is the problem here? If so, how is it taken into account mathematically?
  2. jcsd
  3. Sep 7, 2012 #2
    The proton's acceleration is not a problem. However, in SR only inertial frames are valid frames for the laws of physics. In no inertial frame is it seen to exceed c; it can only "exceed c" in an improper way, by mixing two reference frames. If you don't believe it, just give one inertial frame according to which you go faster than c.

    However, for practical purposes it is as if things and people can go faster than c because Δx/Δt' can exceed c. Perhaps that is what Einstein meant when he remarked: "we shall, however, find in what follows, that the velocity of light in our theory plays the part, physically, of an infinitely great velocity."
    - http://www.fourmilab.ch/etexts/einstein/specrel/www/
  4. Sep 7, 2012 #3
    The proton believes the distance to be about .3 light years. It sees A and B race past at a speed slightly less than the speed of light, and so the time between A passing it and B passing it is slightly more than .3 years.

    The proton may be intelligent, but it has no reason to believe the initial 10 light-years measurement should be any more absolute than the .3 light-years measurement from when it was done accelerating in the first place.

    Moreover, anyone at rest at A or B would agree the proton took a little more than 10 years to reach B.
  5. Sep 7, 2012 #4
    Intelligent protons are also skeptical. When at rest at A, it says to itself, hey, I'm 10 LY from B. Then after I accelerate it, it says, wow, now I'm only about .3 LY away from B so I'll get there in .3 years. After it gets to B and stops, it confirms that it traveled 10 LY, but it believes it did so in about .3 years. This is a very confused proton!
  6. Sep 7, 2012 #5


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    There is no such thing as a proton "at rest" Protons are traveling at c the instant they are created.
  7. Sep 7, 2012 #6


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    Uh, I think there is some confusion here. I assume the OP is talking about a proton (which is also what you typed), the positively-charged, massive particles found in atomic nuclei. These can certainly be at rest.

    Photons, which have no charge and are massless, cannot be at rest, because they must travel at c in all inertial reference frames. So there is no such thing as a rest frame for a photon.
  8. Sep 7, 2012 #7
    :bugeye: A proton is not a photon! Protons can't travel at c.
  9. Sep 7, 2012 #8
    Yes, I am describing a PROTON, which is at rest at A, travels to B at .999c and is again at rest at B.
  10. Sep 7, 2012 #9


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    NUTS ... my dyslexia is acting up again. That is the second time on this forum where I have read AND typed the word proton when my brain was processing both as photon.

    Sorry and thanks for the correction.
  11. Sep 7, 2012 #10


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    Confused, yes, because somehow point A is a lot farther away than it expected - but there was never a moment when point A was moving faster than light relative to the proton, so what's confusing the poor thing is that point A is so far away when there WASN'T any superluminal travel.
  12. Sep 7, 2012 #11
    PerpStudent, the quantity you are trying to define is called the "celerity," or "proper velocity" of the proton. The proton finds this quantity by dividing the distance in the AB rest frame by the time in the moving frame. The celerity can be as large as you please, and is not bounded by c. But nor is it the actual, measured velocity of anything.
  13. Sep 8, 2012 #12
    Apart of that little misunderstanding, everyone explained you the same in different words. Did you get it?
  14. Sep 8, 2012 #13
    Thanks for all the helpful comments.
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