Feynman Diagram Practice

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  • Опубліковано 7 бер 2021

КОМЕНТАРІ • 15

  • @SamuelDrummer
    @SamuelDrummer 2 роки тому +3

    For question 7b.ii) I guess that as you said all three types of interaction are possible, however the most probable Feynman diagram is the one with the gluon because in terms of probability:: Strong > EM > Weak

    • @danielm9463
      @danielm9463  2 роки тому +2

      That's a really great point--thanks for sharing your knowledge, Samuel!! I didn't mention that consideration anywhere in the video, so I'm pinning your comment so others can read it too.
      I'm looking at this again and realizing (for the first time?) that the situation of 7(b)(ii) is: an energetic up quark releases some of its energy which eventually leads to a quark/antiquark pair production. That feels like a classic example of how the strong interaction operates (energetic quarks producing more quarks). I don't know if it's possible for the Z boson to mediate that interaction. If so, I think you're spot on that the Z boson would be *way* less frequent than a gluon. This got me reading about weak neutral currents (interactions with the Z boson). All the examples on this page have the Z arising from annihilation. If you stumble across any examples where the Z is simply emitted (like in this case), please share--I'd love to learn more about it! There are lots of other rules (that I don't know and that go beyond the IB Physics curriculum) that further constrain the possible interactions.
      hyperphysics.phy-astr.gsu.edu/hbase/Particles/neucur.html

    • @SamuelDrummer
      @SamuelDrummer 2 роки тому +1

      ​@@danielm9463 To my knowledge, we can simply say that the Z boson respects the same rules as a photon would, it shares the same vertices coupling an electron/muon/tau with their associated anti-electron/muon/tau and a quarks with its associated anti-quark. However, contrary to the photon, the Z boson can also couple a neutrino with its associated anti-neutrino. Respecting the same rules as the photons, the Z can also be emitted from the desexcitation of a lepton, or a quark. (See slide 12 www.hep.phy.cam.ac.uk/~chpotter/particleandnuclearphysics/Lecture_10_Electroweak.pdf)

    • @danielm9463
      @danielm9463  2 роки тому +1

      @@SamuelDrummer That's a great resource. I hadn't ever studied electroweak unification but it's a really cool way to think about the similarities between photon couplings and Z0 couplings. I'll have to dig into it more--thanks for sharing!

  • @tmo314
    @tmo314 Рік тому +2

    Thanks! Also, what a legendary sense of humor for a particle physics video! 😂

  • @christopherjiang30
    @christopherjiang30 2 роки тому +3

    Very helpful thank you very much. I couldn't solve these problems and was struggling to find a solution

  • @retepredlef5212
    @retepredlef5212 3 роки тому +2

    This is very instructive!

  • @Marie-Gold
    @Marie-Gold 9 місяців тому

    This was so amazing. We want more of these !!

    • @danielm9463
      @danielm9463  8 місяців тому

      You all are physics juggernauts. I'm honestly in awe that you made it through such a long and tedious video!!

  • @okboomer9078
    @okboomer9078 5 місяців тому

    Watched it all, great video thank you learned a lot and I will forever remember that the strangeness is not conserved in weak interaction lol

    • @danielm9463
      @danielm9463  5 місяців тому

      I can't believe you watched the whole thing!! You all are physics champions.

  • @markus8030
    @markus8030 7 місяців тому

    Regarding question 12 a): I don't think they mean elementary particles but particles in general. So a viable answer would be the pi0

    • @danielm9463
      @danielm9463  7 місяців тому

      Agreed, that seems very plausible. I no longer have access to the mark scheme--do you? Would be interested in whether that was listed.

  • @imaginarynumber416
    @imaginarynumber416 8 місяців тому +1

    Hi, in example 12. b, how is it possible that antistrange decays into antidown? Is it not a violation of the principle that allows only diagonal quarks to decay into each other? Strange and down are not diagonal...

    • @danielm9463
      @danielm9463  8 місяців тому

      Hello, I'm not familiar with this rule and would be grateful for a source, if you have one. (I wasn't sure what diagonal means in this context.) I'm no expert, but here's some explanation that may be helpful.
      QUARK DECAYS IN GENERAL
      Footnote [a] on the Wikipedia page for the strange quark states: "The overwhelming majority of top quark decays produce a bottom quark, whose mass is closest to the top's. On very rare occasions it may decay into a strange quark; almost never a down quark." In other words, we can have t → b (most probable), t → s (rarely occurs), and t → d (almost never occurs). However, just because a process is extremely rare doesn't mean it can't occur, and it also doesn't mean we wouldn't draw a Feynman diagram for it. The Feynman diagrams can each represent a part of an equation, and when we calculate the future state, we sum over all diagrams weighted by their probability. (Here's a great video with some of these ideas: ua-cam.com/video/hk1cOffTgdk/v-deo.html)
      The *usual* path of quark decays is t → b → c → s → u ⟷ d, so maybe we could write these rules: (a) when quarks decay, their charge changes, and (b) they usually (but not always) decay into the quark with the most similar mass.
      (hyperphysics.phy-astr.gsu.edu/hbase/Particles/qrkdec.html)
      FLAVOR-CHANGING NEUTRAL CURRENTS:
      However, even those rules above may not be 100% accurate. Something called flavor-changing neutral decays are theoretically permitted, but they have not been observed. (This is my crude understanding.) Here's a nice physics SE answer that shows s → d via a flavor-changing neutral current: physics.stackexchange.com/a/294772
      Q12(b) IN PARTICULAR:
      This question showed up on a past IB exam, and the more I look into it, the more erroneous it appears. This seems to be the representation people usually show when they talk about a kaon decaying into two pions: hst-archive.web.cern.ch/archiv/HST2002/feynman/exampl8.gif I think this is closer to your understanding. It's very possible the IB test authors just made a mistake (this is not terribly uncommon). Or maybe they were trying to suggest a neutral flavor-changing current and they simplified the digram in a way that also might make it wrong.