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Producing K+ from K0L via 'charge exchange'?

  1. Nov 16, 2015 #1
    I'm looking for ways that a [itex]K^{0}_{L}[/itex] can generate a [itex]K^{+}[/itex] and I'm aware that a so-called 'charge exchange' process can allow this. However, I can't access the well-cited publication that appears when googling, ("Kaon inelastic scattering and charge exchange on nuclei - Phys. Rev. C 19 1393 - CB Dover, 1979"), nor can I find a full downloadble version of any textbook that googling turns up, or even any powerpoint showing what happens.

    Can anyone explain what such a process involves?
     
  2. jcsd
  3. Nov 16, 2015 #2

    ChrisVer

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    I guess it means that the Kaon interacts with a nucleon and changes...
    for example in the paper you are asking they have:
    [itex] ^{30}Si (K^- \bar{K}^0) ^{30}Al^{*}[/itex]
    charge exchange..
    So the Si's proton interacted with the negative charged Kaon and became a neutron for Al and a neutral Kaon....
     
  4. Nov 16, 2015 #3
    I'm not quite understanding... not sure why there's a negative kaon being mentioned now... is it illegal or against forum rules for you to post the pdf if you've been able to access it?
     
  5. Nov 16, 2015 #4

    fzero

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    I don't have access to the paper right now, but I can discuss generalities. First of all, ##K^0+X \rightarrow K^\pm +Y## implies that we are exchanging a ##d## or ##\bar{d}## for a ##u## or ##\bar{u}## quark, respectively. For the ##s\bar{d}## component, this is kind of easy, since we have ##d##s directly in ##p## or ##n## so an annihilation process can produce a ##u\bar{u}## and pair the ##\bar{u}## with the ##s## to get a ##K^-##. There are other processes available at higher orders.

    From the abstract of the paper, they are talking about incident momenta of order ##300-800~\text{MeV}##, which is still of order the kaon masses, so one could imagine that an adequate description would be in terms of pion-intermediated strong interactions with the nucleus. This gives plenty of opportunities to match up ##u## and ##d## quark exchanges with the kaons.
     
  6. Nov 24, 2015 #5
    OK, many thanks.
     
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