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Question on liouville's theorem

  1. Feb 26, 2010 #1
    Liouville's theorem says that the phase space density moves as an incompressible fluid.

    In other words, if you follow a point in phase space, the number of points surrounding that point will always be the same.

    Does this contradict the concept of chaos, which says that points in phase space spread out with time?
  2. jcsd
  3. Feb 26, 2010 #2
  4. Feb 26, 2010 #3
    I figured out the answer, but is the paper that you linked to describing how to turn lead into gold?

    How does that work? If you accelerate a proton near a nucleus, then the beam proton changes into a neutron beam and emits a +1 pion, which is captured by a neutron in the nucleus, turning it into a proton?

    What's the energy of the beam proton? I assume less than .2 GeV because we're talking about hadrons and not free quarks and gluons? What happens to the neutron beam: how is it stopped (what particle process stops it)?

    Since lead has a greater atomic number than gold, then you would have to do the converse process - shoot neutron beams into lead. How do you accelerate a neutron beam?

    Wouldn't the resulting atom be imbalanced as far as proton/neutron ratio? Or can you send a beam of neutrons and the neutrons just stick to the nucleus without inducing a change that'll turn a proton into a neutron? Or do you send a beam of neutrons, and a neutron attaches to the nucleus, which will later beta decay (so a weak process instead of a strong process), giving you an extra proton?

    And a little less serious question: should I be worried about the commodity price of gold?
  5. Feb 26, 2010 #4
    Multi-MeV proton beams in particle accelerators do obey Liouvill's theorem phase space predictions during acceleration, and exhibit phase space growth with chaos. But unfortunately it is both difficult and expensive to convert platinum (Z=78) or mercury (Z=80) to gold (Z=79) using a particle beam.

    Bob S.
  6. Feb 26, 2010 #5
    The beams are collimated one proton at a time aren't they? Why would there be chaos in such a situation?

    Multi-MeV: thanks. Is that the typical range of protons under a Linac? Electrons would be the same range right, or would they brem even more than protons, so won't reach the multi-MeV range?

    Platinum is worth more than gold, so you wouldn't want to turn it into gold anyway. So maybe turn whatever is Z=77 into platinum would be good.

    On Wikipedia's transmutation page, there's the following quotation about converting gold into lead:

    "It would be easier to convert gold into lead via neutron capture and beta decay by leaving gold in a nuclear reactor for a long period of time."

    Is this better than shooting multi-MeV protons at the gold hoping to induce a transformation of a neutron into a proton via pion exchange? There's a big difference between shooting a neutron trying to induce neutron capture and shooting a proton trying to induce a transmutation of a neutron into a proton. I guess what's bothering me is if you shoot a neutron at the gold nucleus, who says neutron capture will happen instead of the neutron causing a proton in the nucleus to change into a neutron?
  7. Feb 26, 2010 #6
    In a 100-milliamp proton beam bunched at 400 MHz, there are roughly 1.5 billion protons confined by electric and magnetic fields into a bunch about 1 centimeter long. The electric and magnetic fields preserve phase space (six dimensional) as the protons are accelerated, but RF noise, magnet misalignment, and intra-beam scattering does cause the phase space to grow. A lot of opportunity for chaos.

    Bob S
    Last edited: Feb 26, 2010
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