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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @InertialObservr @SamuelGWalters

      So how do we predict the spectrum? .. well for the lighter mesons we can .. QCD has an approximate global SU(3)_L x SU(3)_R family symmetry which is also spontaneously broken to SU(3)_V by the quark condensate <qq> .. (cont)

      1 reply 0 retweets 4 likes
    2. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @InertialObservr @SamuelGWalters

      So by Goldstone's theorem we will have 3^2 - 1 = 8 goldstone bosons .. these are the light pseudoscalar mesons K^\pm π^\pm etc Since the symmetry is Global this means that they should be massless, though which is not the case .. the small masses are treated as a pertbtn .. cont

      2 replies 0 retweets 3 likes
    3. Sam Walters  ☕️‏ @SamuelGWalters 15 Oct 2019
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      Replying to @InertialObservr

      The Goldstones always come out massless, so could approximately represent mesons (as I understand). Would symmetry breaking not give them mass thereafter?

      1 reply 0 retweets 1 like
    4. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @SamuelGWalters

      breaking the symmetry doesn't give them mass, since the symmetry isn't broken by the Higgs mechanism the vev <qq> is a so called 'non perturbative' effect, it's role strictly breaks the global SU(3)xSU(3) chiral symmetry .. we can't expand a strongly coupled theory abt a vev

      1 reply 0 retweets 1 like
    5. Sam Walters  ☕️‏ @SamuelGWalters 15 Oct 2019
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      Replying to @InertialObservr

      I knew (or read) that the symmetry breaking doesn't give mass to gluons since they're massless, but wondered about mesons. How QCD explains their having nonzero mass (?). The non-perturbative aspect of QCD is something rather intriguing and which one day I'll have to study.

      1 reply 0 retweets 1 like
    6. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @SamuelGWalters

      I agree 100% about the non-pert aspect of QCD .. I will admit i'm not an expert on it, though would love to learn more about it Well the mass term you're referring to is just the mass they obtain via the SM higgs mechanism m_q ~ v*y_q ! Then ..(cont)

      1 reply 0 retweets 1 like
    7. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @InertialObservr @SamuelGWalters

      the pions will have a mass far too small!! That is m_π \neq m_{quark content} Once the quarks confine and become pions their masses are given by m_π ~ \sqrt{ Λ_{QCD}^3 m_{quark content} }/f_π where f_π is determined *experimentially*.. it's somewhat of a fudge factor, as it were

      1 reply 0 retweets 1 like
    8. Sam Walters  ☕️‏ @SamuelGWalters 15 Oct 2019
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      Replying to @InertialObservr

      I think f_π is related to the decay rate constant for the neutral pion decaying into 2 photons. Your expression reminded me of a similar one in Weinberg's (Vol II, p. 368) where the pion's decay rate is related to f_π and m_π. The two would seem compatible.

      1 reply 0 retweets 1 like
    9. 〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @SamuelGWalters

      you're right that's precisely what it is .. the pion decay constant .. however the pion decay can't be predicted from QCD and is determined experimentially

      1 reply 0 retweets 4 likes
    10. Sam Walters  ☕️‏ @SamuelGWalters 15 Oct 2019
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      Replying to @InertialObservr

      Weinberg says that a calculation by Steinberger (in 1949) can give a theoretical expression for the decay rate (for π0 to 2 photons) that is much closer to the actual measured value. (Section 22.1, Vol. II.) Which, incidentally, was a beginning of those anomalous diagrams.

      1 reply 0 retweets 7 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 15 Oct 2019
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      Replying to @SamuelGWalters

      interesting, i'll have to look into it .. My guess is that it's mainly from symmetry & dimensional analysis to get an estimate .. but even so the point is that this cannot be done in general for other couplings/masses

      9:42 PM - 15 Oct 2019
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