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I Why would interacting K0L produce K+ more often than K-?

  1. Aug 20, 2016 #1
    I am simulating lots of [itex]K^0_{L}[/itex] going into a volume of rock, and they sometimes produce a [itex]K^{+}[/itex] or a [itex]K^{-}[/itex]. However, this does not happen in equal amounts - I am seeing roughly 3 times as many [itex]K^{+}[/itex] as I am [itex]K^{-}[/itex]. What is the explanation for this?
     
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  3. Aug 20, 2016 #2

    Vanadium 50

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    Are the K0-nucleon and K0bar-nucleon cross-sections identical?
     
  4. Aug 21, 2016 #3
    I'm not sure where to look to check this. If they are different then that would be a simple explanation I think, but what if they are identical?

    Also, can you tell me if the following diagrams are the correct interpretation of what's happening?
    WP_20160821_14_49_34_Pro.jpg
     
  5. Aug 21, 2016 #4

    Vanadium 50

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    OK, take one step back. Are the K+ and K- cross-sections identical? Based on that, what can you say about K0 and K0bar?
     
  6. Aug 21, 2016 #5

    mfb

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    The weak interaction is only relevant if there is no strong process doing the same.
     
  7. Aug 21, 2016 #6
    The K+ and K- must not be identical if I'm seeing a disparity between them, so the K0, K0bar must differ too?

    Is that as simple as replacing the W bosons in those diagrams with charged pions?
     
  8. Aug 21, 2016 #7

    mfb

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    There is an easier process, but the relative contribution might depend on the kaon energy.
     
  9. Aug 21, 2016 #8
    Gluon-mediated?
     
  10. Aug 21, 2016 #9

    mfb

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    Yes.
     
  11. Aug 21, 2016 #10

    Garlic

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    Is this how the reaction diagrams would look like?
    IMAG0335_1.jpg

    Note: I would argue that the neutral Kaon is instead the one with the strange quark, and the anti neutral Kaon is the one with the strange antiquark. Is that right?
     
  12. Aug 21, 2016 #11

    Vanadium 50

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    Feynman diagrams are the wrong tool to solve this problem, especially if they use the wrong interaction or in Garlic's case don't conserve charge.

    Are the K+ and K- cross-sections identical? Based on that, what can you say about K0 and K0bar? Is there some symmetry that relates them?
     
  13. Aug 21, 2016 #12
    Not sure if you missed my reply earlier but I said the K+, K- cross-sections mustn't be identical seeing as different amounts are produced?

    And regarding symmetries... do you perhaps mean that K0 and K0bar are both odd-parity states? [itex] P | K^0 \rangle = - | K^0 \rangle \hspace{3 mm}[/itex], [itex] P | \overline{K}^0 \rangle = - | \overline{K}^0 \rangle[/itex] ? Or that K0 and K0bar are related by charge conjugation [itex] C | K^0 \rangle = | \overline{K}^0 \rangle[/itex] ...?
     
  14. Aug 21, 2016 #13

    Vanadium 50

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    Hint 1: I am not talking about production cross-sections for K+ and K-; I am talking about the K+-nucleon and K- -nucleon cross-sections.
    Hint 2: Isospin
     
  15. Aug 21, 2016 #14
    Hmm, ok, so there are two [itex]\pm\tfrac{1}{2}[/itex] isospin doublets related to kaons: [itex](K^-, \overline{K}^0) [/itex] and [itex](K^+, K^0) [/itex], and the [itex]\pm\tfrac{1}{2}[/itex] isospin doublet for the nucleons [itex] (p, n) [/itex]... and if isospin is conserved in strong interactions... maybe there are fewer ways for an isospin-conserving K+ - nucleon interaction to occur than there are for K- - nucleon... so more survive? I feel very lost here.
     
  16. Aug 22, 2016 #15

    Garlic

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    Where did I make the mistake?
    In the left diagram, a neutral particle (kaon) emits a negative charged particle (pion) and becomes positive. A positive charged particle (proton) captures a negative charged particle and becomes neutral.
    Is the mistake in the right diagram?
     
  17. Aug 22, 2016 #16

    Vanadium 50

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    Garlic, the right diagram doesn't conserve charge, and everywhere you have the K0 and K0bar switched. I don't think posting wrong things will help Anchovy.

    Anchovy, what does isospin tell you about the relation between K+ n reactions and K0 p reactions? And what does it tell you about the relation between K- p reactions and K0bar n reactions?
     
  18. Aug 22, 2016 #17
    The K+ has isospin +1/2, neutron has -1/2, so a total of zero in the initial state so the final state should be zero overall too?
    The K0 has -1/2, proton has +1/2, so final state should be zero?

    The K- has +1/2, proton has +1/2, so final state should be +1 overall.
    The K0bar has -1/2, neutron has -1/2, so final state should be -1 overall.

    The K+ n and K0 p reactions involve two objects with opposite isospin, and the K- p and K0bar n reactions involve two objects with the same isospin... I'm still confused.
     
  19. Aug 22, 2016 #18

    Garlic

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    I understand, you are right. But I was only asking if they were right, I wasn't claiming anything.
     
  20. Aug 22, 2016 #19

    Vanadium 50

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    Let's go back a bit. Calculating isospin is not the important part. What's important is that there is an isospin symmetry. Let's consider two reactions,
    p+ + S -> anything and n + S --> anything. Here S is an isospin scalar, like 4He. What does isospin tell you about their rates?

    Now consider:
    K+ + S -> anything and K0 + S --> anything. What does isospin tell you about their rates?

    Now consider:
    K0bar + S -> anything and K- + S --> anything. What does isospin tell you about their rates?

    Finally, does isospin tell you anything about
    K0+ S -> anything and K-bar + S --> anything?
     
  21. Aug 22, 2016 #20
    Is it that their rates have to differ because p and n have different isospin 3rd components?
     
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