Why is c quark heavier than s while u is lighter than d?

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In summary: is that c,b,t are all leptons, while u,d,s are made of up/down quarks and their antimatter partners.
  • #36
Vanadium 50 said:
There is not a shred of evidence that the e goes with (u,d), the mu goes with (c,s) or the tau goes with (t,b). They are usually placed in tables this way, but not for any reason other than mass.

Mean life time hierarchy also favors the current alignment of u-d-e, c-s-mu, t-b-tau. Lepton flavor conservation favors the alignment of the respective neutrino types with their respective lepton types. No second or third generation fundamental particles, and no composite particles containing second or third generation fundamental particles are stable. It also makes sense that t is the 3rd rather than the 2nd generation because its extremely short lifetime makes it an outlier as the non-hadronizing quark which fit an extreme rather than intermediate generation.
 
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  • #37
ohwilleke said:
Mean life time hierarchy also favors the current alignment of u-d-e, c-s-mu, t-b-tau.

That's the same as the mass hierarchy, since weak decays go as m^5.
 
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  • #38
Vanadium 50 said:
Like the Earth's polar diameter being half a billion inches?

Error of more than 0.1%

Or that there are pi x 10^7 seconds in a year?

Error of more than 0.3%

Koide formila says that Q = 0.666666666..., measured value is 0.666659(10), the difference is 0.000007... which is slightly more than 0.001% (and importantly, the difference is smaller than measurement uncertainty)

Of course, it can be spurious. There is no way to prove that it is not.
 
  • #39
nikkkom said:
slightly more than 0.001%
I think that he refers to the current context. The koide tuple for (s,u,d) with an up mass equal to zero fails a bit,
it is Q(95,0,4.8)=.70031 well far away from 0.666... The koide tuple for (c,s,u) does a bit better,
Q(1235,95,0)=.66003, so a error of 1%. EDIT: We travel better with the last pdg value for charm mass
Q(1275,95,0)=.66309, but still not as good as for leptons.

Of course if we do not need a mass exactly 0, we can invoke a waterfall to produce masses for the first generation: use c,s to solve for u, and then s,u to solve for d. I am not very easy about the last step (also (b,s,d) is a suitable tuple if you look at the S4 symmetry argument) but at least for the question in case in effectively predicts a very small u mass. Namely,

Code:
mc=1275
ms=95
((sqrt(mc)+sqrt(ms))*(2-sqrt(3)*sqrt(1+2*sqrt(mc*ms)/(sqrt(mc)+sqrt(ms))^2)))^2
.01478

To be, the result, exactly zero, the proportion mc::ms should be 13.92, while experimentally it is 13.42 or 11.73 depending of how you measure it. In any case too low.
 
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