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Koide and Quarks

  1. Mar 29, 2011 #1
    Hi everyone, I'm new to PhysicsForums. I don't really know if what I have to offer is anything new to anyone or if it is simply meaningless numerological garbage, but it just looked too beautiful for me to ignore, and if this is common knowledge, then I humbly apologize. I did a search on this site and on google to see if this had already been noticed and didn't find anything. Hopefully I'm not mistaken. If I am, please forgive me and point me in the right direction.

    I don't really have the physics background everyone that I've seen posting seems to have, so bear with me. I went to RIT as an undergraduate in electrical engineering a few years ago, but I dropped out halfway for various personal reasons. I know basic things about relativity and quantum theory from some courses I took, as well as books I have read like The Elegant Universe and The Fabric of the Cosmos by Brian Greene. So, if you respond to this post with something overly technical, I may or may not understand it. However, I will most definitely try my best.

    I recently became very interested in Koide's Formula. It is fascinating to me. From what I've read so far on the forums, it seems to be pretty well known, but just in case you don't know what it is, it can be found on wikipedia here (http://en.wikipedia.org/wiki/Koide_formula). To reiterate what the website says,

    [tex] \frac{m_e + m_\mu + m_\tau}{(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2}\approx\frac{2}{3} [/tex]

    The article suggested a geometric interpretation to this equation. It has to do with the cosine of the angle between the [itex] <1,1,1> [/itex] vector and [itex] <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau> [/itex]. So, I figured that quarks may have something very similar. I grouped the quarks' masses up in terms of charge and made vectors out of them similar to the vector previously stated. Then, just for kicks, I took their magnitudes and compared the down-quark vector's magnitude to the up-quark vector and got a ratio of about 4.99825, which is very close to 5! Now, I rechecked my work and realized I had mistakenly used GeV units in the up-quark vector versus MeV units in the down-quark, but had I not, I would never have seen such a simple ratio. After correcting the mistake, the ratio was determined as [itex] 1/(2\sqrt 10) [/itex], which I would have never seen if I didn't make that initial mistake. This theoretical value is about 99.9651 percent accurate to the calculated real value. I'm terrible with sig figs, so anyone please feel free to correct me on any and all percentages I give.

    To make more clear what I did, let,

    [tex] \vec a_1 = <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau>\approx<\sqrt{0.511}, \sqrt{105.7}, \sqrt{1777}> [/tex]
    [tex] \vec a_2 = <\sqrt m_u, \sqrt m_c, \sqrt m_t>\approx<\sqrt{2.4}, \sqrt{1270}, \sqrt{171200}> [/tex]
    [tex] \vec a_3 = <\sqrt m_d, \sqrt m_s, \sqrt m_b>\approx<\sqrt{4.8}, \sqrt{104}, \sqrt{4200}> [/tex]

    All vectors have units of MeV.

    [tex] \frac{|\vec a_3|}{|\vec a_2|}\approx\frac{1}{2\sqrt{10}} [/tex]

    The values I got for the masses were from the wikipedia article on elementary particles (http://en.wikipedia.org/wiki/Elementary_particle). I verified that wikipedia's numbers were accurate using this other site just to make sure (http://pdg.web.cern.ch/pdg/2010/tables/rpp2010-sum-quarks.pdf0.

    I did the same thing for the third and first vectors and found a nice little ratio involving four numbers. If you want it, PM me. I don't want to post it because it might be taken as overly numerological, even though it is 99.9166 percent accurate.

    Getting back to Koide's Formula, I plugged the wikipedia numbers in and found that this well known ratio was about 0.6666323, which is about 99.9949 percent of 2/3. I used this as a baseline to determine whether what I found next was at all valid, and this is what blew me off my feet. I believe I must have checked and rechecked the numbers I'm about to give more than thirty times. Can someone please verify this to make sure that I'm not making a simple mistake over and over? The first equation's theoretical value is 99.9859 percent accurate and the second equation 99.9218. So, here it goes...
    [tex] \varphi=\frac{\sqrt 5 + 1}{2} [/tex]
    This is the golden ratio. And here are the two equations I discovered.

    [tex] \frac{m_u + m_c + m_t}{(\sqrt{m_u}+\sqrt{m_c}+\sqrt{m_t})^2}\approx\frac{2}{3}\ast\sqrt{\varphi} [/tex]
    [tex] \frac{m_d + m_s + m_b}{(\sqrt{m_d}+\sqrt{m_s}+\sqrt{m_b})^2}\approx\frac{1}{\sqrt{5}}\ast\varphi [/tex]


    I have a gut feeling this is important, but I simply don't have the means to verify it scientifically. If I algebraically manipulate Koide's Formula using the [itex] <1,1,1> [/itex] vector, the cosine of the angle of Koide's Formula becomes [itex] \sqrt{2}/2 [/itex], which has an angle of [itex] \pi/4 [/itex], which is known. Doing the same with the above two equations is a bit more difficult to obtain a simple angle. Instead, all I got was a ratio of cosines for the second equation and nothing for the first. Anyone who has access to Matlab would probably have better luck than me for finding simple angles. The ratio I got could be seen as either [itex] \cos((3\pi)/10)/\cos(\pi/6) [/itex] or [itex] \sin(\pi/5)/\sin(\pi/3) [/itex]. They are the same thing. I'm going to refrain from hypothesizing what this means. Instead, I'd like to hear what all of you have to say. I don't know if I'm seeing something that's not there or if this significant. Can someone please help me?

    Thanks,

    J Rivera
     
  2. jcsd
  3. Mar 29, 2011 #2

    fzero

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    The masses reported by the PDG are valid at different scales. Because of quantum effects, the mass of a particle is dependent on the scale at which it is being measured; this is called mass renormalization. For example, the reported mass of the charm quark is the value at the charm quark mass itself, 1.27 GeV. This scale is the relevant one when considering production of charmed particles at the threshold for production, that is the minimum energy needed to produce them. Similarly the top quark mass is the one measured at 172 GeV. It makes no sense to compare quark masses at different scales. One must use the renormalization group equations to convert them to their values at the same scale. An older paper that actually shows these results is one by Koide http://arxiv.org/abs/hep-ph/9410270 (see Table VI).

    If you actually do this properly and then check the Koide formula, you will find that the ratios change by several percent, drastically reducing the possibility that these purported relationships are anything but numerology.
     
  4. Mar 29, 2011 #3
    Got it. Thanks for clearing that up.

    -J
     
  5. Mar 29, 2011 #4

    phyzguy

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    This paper:

    http://arxiv.org/abs/hep-ph/0601031v2

    argues that the Koide formula is general, and applies to the up quark family and the down quark family across a wide range of energy scales, although the constant of 2/3 is slightly different (being almost exactly 2/3 for leptons, about (2/3) * 1.06 for the down quark family and about (2/3) * 1.33 for the up quark family). It looks to me like more than numerology, but I don't claim to be able to explain it.
     
  6. Mar 29, 2011 #5

    fzero

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    The thing is, to that level of precision we might as well note that [tex]m_c/m_t[/tex] is small and [tex]m_u/m_t[/tex] is practically zero (especially when we properly evaluate [tex]m_u[/tex] at the top mass), so we should just expand

    [tex]
    \frac{m_u + m_c + m_t}{(\sqrt{m_u}+\sqrt{m_c}+\sqrt{m_t})^2}\approx\ 1 - 2\sqrt{\frac{m_c}{m_t}} + 4\frac{m_c}{m_t} + \cdots
    [/tex]

    [tex]m_\mu/m_\tau[/tex], [tex]m_c/m_t[/tex], and [tex]m_s/m_b[/tex] have nothing much to do with one another, other than they are all relatively small, so that we get approximately the same answer when we subtract powers of them from 1.
     
  7. Nov 11, 2011 #6

    arivero

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    Has anybody puzzled about the rare triplet charm, bottom, top? I am sure something has, but I do not remember who. It doesn't seems a bad fit.

    EDIT: arxiv:1101.5525v1 mentions it. But I would expect earlier mentions, either in the arxiv or in PF
     
    Last edited: Nov 11, 2011
  8. Nov 11, 2011 #7

    arivero

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    To be precise, with the quark raw values of 0601031v2
    1.302 GeV, 4.339 GeV, 170.8 GeV
    the quotient is 1.5046, but it is due to a very low top mass. For 174.1, it is 0.99986, tenfold better.

    If we input this bottom and charm into Koide and solve, we get as solutions 173.9 GeV and 111.6 MeV. In some sense, it seems that Koide can be use to predict both the mass of the top and the mass of the strange quark, but at the price of a very peculiar triplets: scb and cbt.
     
  9. Nov 12, 2011 #8

    arivero

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    I had missed http://arxiv.org/abs/1111.0480 which proposes Koide relationships for cbt and uds, as well as a total one for udscbt!

    They fail to find the relationship for scb because they forget the possibility of minus sign. For instance, for

    s=.086 c=1.29 b=4.19

    we have

    (-sqrt(s)+sqrt(c)+sqrt(b))^2/(s+c+b)=1.500...
     
  10. Nov 13, 2011 #9

    arivero

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    Ok, I tink I have the answer. The right sequence should be t,b,c,s,u,d. With this sequence, the worst fit is with the up quark.

    Take for instance the central values t=172,9 GeV and b=4,19 GeV. Then

    Koide(t,b,c) implies c=1.356 GeV
    Koide(b,c,s) implies s= 92 MeV
    Koide(c,s,u) implies u= 36 KeV (the bad prediction, but then...)
    Koide(s,u,d) implies d= 5.3 MeV (... miracle, just in target again).

    The ladder is built by iterating the rule
    [tex]m_3=\left((\sqrt m_1+\sqrt m_2)*(2-\sqrt 3 \sqrt {1+2 {\sqrt{m_1*m_2} \over (\sqrt{m_1}+ \sqrt{m_2})^2} } \right)^2
    [/tex]
    which comes just from solving Koide's.

    It seems that d and u either get some extra contributions or are sensible to the scale. It is reported elsewhere that slights runs of the mass can improve the match.
     
    Last edited: Nov 13, 2011
  11. Nov 14, 2011 #10

    arivero

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    Even funnier. Using t=172.9 and b=4.19 to predict c=1.356 and then
    this value to predict s=(-.3033)^2=.092, I note that

    1/9 * (-sqrt(s)+sqrt(c)+sqrt(b))^2 = 939.68 MeV, almost exactly the triple of the lepton fundamental mass
    1/9* ((sqrt(0.511)+sqrt(105)+sqrt(1777))^2)= 313.5 MeV

    It could be good if someone takes some time to check this, as well as the phases going into the cosines. Now that I see it in this way, I wonder if we had already check this scb and just forgot about it.
     
  12. Nov 15, 2011 #11

    arivero

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    to conclude, up to now we have found that
    Code (Text):

    pi=4*a(1)
    313.83*(1+sqrt(2)*c(0*pi/3+0.222222))^2
    1776.81957798694484607752
    313.83*(1+sqrt(2)*c(2*pi/3+0.222222))^2
    .51095720602758346680
    313.83*(1+sqrt(2)*c(4*pi/3+0.222222))^2
    105.64946480702757041487
    3*313.83*(1+sqrt(2)*c(0*pi/3+3*0.222222))^2
    4197.22350443870492221734
    3*313.83*(1+sqrt(2)*c(2*pi/3+3*0.222222))^2
    92.26695681377014578294
    3*313.83*(1+sqrt(2)*c(4*pi/3+3*0.222222))^2
    1359.44953874752493178316
     
    and also that

    Code (Text):

    charm=3*313.83*(1+sqrt(2)*c(4*pi/3+3*0.222222))^2
    b=3*313.83*(1+sqrt(2)*c(0*pi/3+3*0.222222))^2
    (sqrt(b)+sqrt(charm)+sqrt(172900))^2/(b+charm+172900)
    1.50050048925970773022
    (sqrt(b)+sqrt(charm)+sqrt(173249))^2/(b+charm+173249)
    1.50000053244468838037
     
    And there are some good expectations for neutrinos. Other groups have reported a koide relationship for uds but it depends a lot of the running of the mass.
     
  13. Nov 16, 2011 #12

    arivero

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    Hmm, perhaps it is also important to tell that with the negative value for the sqrt of the mass of the strange quark,
    strange=3*313.83*(1+sqrt(2)*c(2*pi/3+3*0.222222))^2
    the angle between electron,muon,tau and bottom, charm, strange is a lot near to the orthogonal than in the case with the positive sqrt. Perhaps it is the most orthogonal possible angle.
     
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