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V > c for beta-rays?

  1. May 11, 2012 #1
    Hello, I repeat my question
    As is well known that radioactive decay is a random, automatic phenomenon that nothing can stop it. This is unaffected by any variable such as temperature, pressure, acceleration. Each radioactive element has a specific fixed as half-life, wavelength and period, which is fully stable and unaffected.
    Imagine a spaceship made of depleted uranium in the earth is moving at a speed approaching that of light.
    B-rays emitted in the direction that moves beyond the speed of light?
    No, says the special theory. But then is the paradox. A Geiger counter in the spacecraft detects radiation and another mounted on the outside 'in front of the window' was not detected. Both are stationary observers on the craft. To illustrate this, suppose that the craft is made of non-radioactive materials and moving towards earth with a constant speed. Inside there's a piece of uranium, a Geiger counter in front of A is in relation to the direction of craft speed and a second B behind it. If the speed is low, both counters detect radiation. Recall here that the speed of light is constant independent of the observer and inaccessible. So if the speed is equal to that of light, A did not detect radiation. Which of the two would happen if there was no earth or other reference point to know the speed? That is, whether the rings A or not, depends on the observer. The theory introduces a new conception of reality, where a natural phenomenon does not happen by itself, but depends on the existence of observers.
    But you say, nobody can approach the speed of light. But perhaps one core of radioactive isotope in a large accelerator can do. You just have to try it, if there at CERN read our articles. We will see then if the speed prevents the decay of the nucleus, which is, as we said, a random phenomenon. And how the core 'knows' that moving relatively to the observer to 'postpone' its breakdown of?
     
  2. jcsd
  3. May 11, 2012 #2

    Ich

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    Re: Quick question on Special Relativity

    The postulates of relativity:

    1) principle of relativity means that there is no preferred state of motion, which directly means that the outcome of your rocket experiment is independent of its velocity.
    2) speed of ligth the same in every reference frame means that there is no reference frame where the speed of light is zero as in your "So if the speed is equal to that of light, A did not detect radiation.". Which means again that the rocket can't go at the speed of light.

    So, even without any math, your conclusions directly contradict the postulates that the theory is built on. You sure you're talking about relativity? Where did you learn it from?

    As to your questions,
    I don't understand what you're asking.
    Already answered.
    In the core's system, there is no time dilation. Time dilation is a relation between different frames, like core frame and laboratory frame. It has nothing to do with your notion of "slowing processes".
     
  4. May 11, 2012 #3
    Re: Quick question on Special Relativity

    I am not talking about time dilation,what I am asking for is that if the moving core decays,b-rays emmited beyond the speed of light?
     
  5. May 11, 2012 #4

    Ich

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  6. May 11, 2012 #5
    Re: Quick question on Special Relativity

    Thank you,I'll see it,but what about the isotope core emission relative to lab?
     
  7. May 11, 2012 #6

    Ich

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    Re: Quick question on Special Relativity

    Either the nucleus is going v<c, then you're talking about time dilation, which I already answered. Or it is going >=c, then you're talking about something impossible.
     
  8. May 11, 2012 #7

    HallsofIvy

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    Re: new question on Special Relativity

    No one has said you cannot approach the speed- you just can't ever get there. And, no, the speed does not "prevent" decay of the nucleus. You are just using the wrong formula for combination of speeds. If an atom on a space ship, moving at, say, .9c relative to me, emits a beta ray moving at .9c relative to the atom, directly toward me, I would measure the beta ray as moving at
    [tex]\frac{.9c+ .9c}{1+ \frac{(.9c)(.9c)}{c^2}}= \frac{1.8c}{1.81}= .99c[/tex]
    still less than the speed of light.
     
  9. May 11, 2012 #8

    Dale

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    Beta radiation consists of electrons or positrons. It is always slower than c.

    If you believe otherwise then please provide a mainstream scientific reference which states that.
     
  10. May 11, 2012 #9
    I think what (s)he's saying is the very well known newtonian formula for adding velocities, which is invalid in this case.
     
  11. May 11, 2012 #10
    Thank you all
    What I ask in my first question is that if A counter rings,is this for an observator on earth means that b-rays beyond the speed of light?
     
  12. May 11, 2012 #11
    Again, no, the formula isn't [itex]\vec{v}_{1,3}=\vec{v}_{1,2}+\vec{v}_{2,3}[/itex].
     
  13. May 11, 2012 #12

    Dale

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    Also, a Geiger counter does not measure speed of a radioactive particle, it only detects its presence. To measure speed you will need something else, like a Doppler radar.
     
  14. May 11, 2012 #13

    Mentz114

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    In this setup, the detectors are stationary wrt the source. Therefore they will both count beta rays at all times. Furthermore, every observer will agree on the counts.

    No, this is wrong. The counters operation does not depend on a relative velocity with some other frame.
     
  15. May 12, 2012 #14
    Ok,but what about γ-rays also emitting by the nucleus?
     
  16. May 12, 2012 #15
    Since those are light, they travel at c in all reference frames.

    Remember that if we use the relativistic formula for adding speeds, if something's moving at c in one reference frame, it's moving at c in all reference frames (in fact, ignore the relativistic formula for adding speeds, this is what Special Relativity comes from), and if something's moving at <c in one reference frame, you get a speed of <c in all reference frames.
     
  17. May 12, 2012 #16

    Mentz114

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    As Whovian has said, γ-rays are light so they always travel at c in all frames.

    If you placed your apparatus ( the emitter and the detectors ) in any non-accelerating lab it will give the same results. The laws of physics are the same in all inertial frames. You should be aware that there is no absolute motion. The lab may be moving close to c with respect to some other frame but that is irrelevant. In the lab frame there are no relativistic effects.

    If the emitter was in motion with respect to the detectors, there would relativistic effects like time dilation, but the counts of the detectors would be the same in all frames, although the rate of detection might be different.
     
  18. May 12, 2012 #17
    Forgive me but I have doubts until this could be experimentally assured.I asked for it:
    "But perhaps one core of radioactive isotope in a large accelerator can do. You just have to try it, if there at CERN read our articles."
    The core is moving close to c and emitts γ-rays.The frame is the lab.
    Supposed that math produce physics you know Zenon's parodox.
     
  19. May 12, 2012 #18
    More importantly, nothing emitted of any sort will appear to travel faster than light from any inertial reference frame. Either the radiation is EMR and therefore traveling at v=c in all reference frames, or it is traveling v<c in which case as per the relativistic velocity addition formula, no reference frame will observe it moving faster than light.
     
  20. May 12, 2012 #19

    Dale

    Staff: Mentor

    This has been experimentally validated for decades. Particle accelerators produce a plethora of unstable particles which radioactively decay. Their decay rates can be measured and compared to the decay rates of the same particles at rest. My favorite are the muon decay experiments of Bailey, but here is a broad overview:
    http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

    Zeno's paradox is completely unrelated to this topic.
     
  21. May 14, 2012 #20
    HallsoIvy wrote:

    "No one has said you cannot approach the speed- you just can't ever get there. And, no, the speed does not "prevent" decay of the nucleus. You are just using the wrong formula for combination of speeds. If an atom on a space ship, moving at, say, .9c relative to me, emits a beta ray moving at .9c relative to the atom, directly toward me, I would measure the beta ray as moving at
    .9c+.9c1+(.9c)(.9c)c2=1.8c1.81=.99c
    still less than the speed of light."

    If the spaceship moving at c (as its supposed in my statement;impossible?) from the formula above you would measure the beta ray as moving at c.That means that it could not reach counter A so A not rings,or the atom could not emit (speed prevent decay).
    What would be happened in fact?
     
  22. May 14, 2012 #21
    Yes, it is impossible for it to move at exactly c, but it could move at very close to c as has been discussed.

    If the particle is emitted moving in the same direction as the ship, the speed of the particle in the example is increased from 0.9c to 0.99c so it hits the detector more quickly than if the ship were at rest.

    It could only "not reach counter A" if it were emitted in the opposite direction. In that case, if the ship is moving at 0.9c, the particle reaches the detector slowly if it is emitted at 0.89c, it goes off in the opposite direction and never reaches the detector if emitted at 0.91c relative to the ship, and just stays at a fixed distance from the detector if emitted at exactly 0.9c, the two speeds cancelling.

    The above is obvious and you don't even need relativity to work that out so perhaps I have misunderstood what you are asking, can you make your question any clearer?
     
  23. May 14, 2012 #22
    I think you may have got the numbers or the wording a little mixed up here.

    I think you meant to say something along these lines:

    It could only "not reach counter A" if it were emitted in the opposite direction. In that case, if the ship is moving at 0.9c (relative to Earth), then if the particle is emitted at 0.89c (relative to Earth), it goes off in the opposite direction at about -0.05c (relative to the ship) and never reaches the detector. If emitted at 0.91c (relative to Earth) then the particle is moving slowly at roughly 0.055c (relative to the ship) towards the front and eventually reaches the front detector.
     
    Last edited: May 14, 2012
  24. May 14, 2012 #23

    Mentz114

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    If a detector rings in one frame, it will ring in all frames. This is because the detection event is an intersection of two worldlines, which can never be transformed away.
     
  25. May 14, 2012 #24
    I may have got the wrong picture of the experiment in that case. I assumed that since the addition of velocities was being discussed, the emitter was on the ship. Picture the ship moving from left to right away from the Earth, the detector stationary (relative to the Earth) ahead of the ship and the particle fired from right to left from the ship. What I meant was:

    It could only "not reach counter A" if it were emitted in the opposite direction. In that case, if the ship is moving at 0.9c (relative to Earth), the particle reaches the detector slowly (at about 0.050c relative to the Earth) if it is emitted at 0.89c (relative to the ship), it goes off in the opposite direction (at about -0.055c relative to the Earth) and never reaches the detector if emitted at 0.91c relative to the ship, and just stays at a fixed distance from the detector if emitted at exactly 0.9c, the two speeds cancelling.​
     
  26. May 14, 2012 #25

    Mentz114

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    Does this mean that some observers will see a different count of particles from that in the rest frame of the apparatus ? I don't think that's possible, although the detection events may be in the distant future of some obervers.
     
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