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CarlB in Hawai!

  1. Oct 29, 2006 #1

    arivero

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    I do not know if he is going, but Carl happens to have a 25 min talk allocated next Wednesday, before Koide's talk:

    http://www.phys.hawaii.edu/indico/sessionDisplay.py?sessionId=116&confId=3

    It is a middle-to-long talk, but too early morning, fortunately Koide will follow so people has two opportunities :smile:

    the slides for Brannen and powerpoint for Koide are online there. Carl has gone "only eV" in terms of units, which is better for a presentation. But 41 slides is one every 30 seconds, I guess he will be planning the shortcut now...
     
    Last edited: Oct 29, 2006
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  3. Oct 29, 2006 #2

    Kea

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    Hee, hee. From what he said he is planning to get in as much as he can! Oh...I just looked at the schedule on the link and noticed that the guy before Carl is also talking about a similar business. Wish I was in Hawaii.

    We're smiling for you, CarlB
    :smile:
     
    Last edited: Oct 29, 2006
  4. Oct 29, 2006 #3

    CarlB

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    I really don't know what I should do with the talk.

    I figure I have to tell about the incredible coincidence of two 5-digit accurate small fractions in the charged lepton masses. But after that, I've become so adrift from standard physics that it is very difficult for me to talk at all. I can either spout what I think is bullcrap, or I can spout what the audience thinks is bullcrap. Hell of a choice.

    A few slides in I show that the -1 you get in spinors when you rotate a spin-1/2 around an axis by 2 pi appears, in the density theory, to show up only because you failed to rotate the "vacuum" that defined the spinor (i.e. complex numbers are geometric objects so you can't go around rotating the states without rotating the complex numbers too). This point is the same thing as the usual assumption that swapping two fermions causes the wave function to get negated.

    But none of that -1 stuff happens in the density theory, and you can move from the density theory back to the spinor theory, so are they a part of reality or not? To me, the fact that a simple version exists (without -1s) implies that the usual way of doing QM, the spinor theory, is a mathematical trick; the density operator theory is closer to reality.

    Unfortunately, the same sort of thinking has spread through the rest of my understanding of physics, some before, some afterwards. I think that the Wick rotated calculations correspond to the real world, the Minkowski geometry is a bad geometric trick. I think that the geometry should define the symmetry, but physics is defined the opposite way. I believe relativity follows from a simple geometry (not Minkowski). I made a list once of how far I had split from the standard view and found that there were 20 basic assumptions I have that are "cranky".

    I'm sort of inclined to stop the lecture there, and ask the audience. "Okay, along this line, would the graduate students in the audience like to know a bunch of facts about physics that will piss off your professors?"

    Also, I've got notes for the apparent evidence in cosmic rays for particles traveling sqrt(3) c. I wonder if I should show those too. You can't take the idea of "velocity as eigenstate of Dirac equation" and "electron made out of three/six particles with perpendicular velocities" without concluding that the maximum electron speed is the speed of the constituents/sqrt(3). But I doubt anyone wants to hear it.

    By the way, I don't know about pub hopping. I'm not much of a late night drinker. Oh oh, low battery. I'd better go get on a plane.

    Thanks for the good vibes.

    Carl
     
    Last edited: Oct 29, 2006
  5. Oct 29, 2006 #4

    Kea

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    Carl, there will be some people in the audience who have taken on board this lesson, although maybe not in your Clifford algebra way.

    Er, you might be surprised. There are a lot of people talking about [itex]c[/itex] change now....at least as of the last few weeks! So go for it.

    All the best
    Kea
    :smile:
     
  6. Oct 29, 2006 #5

    CarlB

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    Kea & Arivero,

    I had the most useful plane flight from Seattle.

    The most controversial thing in the neutrino mass predictions is the supposition that the masses scale like
    [tex]\mu_1/\mu_0 = 3^{12} / 3^1 = 3^{11}[/tex]

    And this was supposed to be associated with the assumption that the neutrino mixing matrix is a 12th root of unity (after you correct it by making it translate from mass eigenstates [tex](\nu_1,\nu_2,\nu_3)[/tex] to mass eigenstates [tex](e,\mu,\tau)[/tex]which you do by multiplying it by the matrix of eigenvectors of a circulant matrix). I hadn't even included the neutrino mixing angle stuff in the slides.

    Now the justification for all this was that the mass conversion L->R->L is correct for the charged leptons, but is off by an angle of 2 pi/12 for the neutrinos. However, if that angle were taken literally, you'd end up with a complex mass, not a mass that was taken to the 12th power.

    On the plane ride over, I realized that the two circulant matrices that represent the charged leptons and the neutrinos were simply:
    [tex]
    \begin{array}{rcl}
    M_1 &=& \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right),\\
    M_0 &=& \left(\begin{array}{ccc}0&i&0\\0&0&i\\i&0&0\end{array}\right),
    \end{array}[/tex]

    where the charged lepton transformation acts like [tex]L_e -> R_e -> M_1L_e[/tex] and the neutrino transformation acts like [tex]L_\nu -> R_\nu ->M_0L_\nu[/tex]

    The point is that the M_0 matrix can act like a true 12th (or is it 6th) root of unity in a way that does not overuse (burn up) any degrees of freedom the way that trying to make it be a complex 12th root of unity does.

    Now the place where all this gets tied back into the theory is this: I now assume that the mass interaction is a change in the sign of [tex]\gamma_0[/tex]. But [tex]\gamma_0[/tex] is not one of the commuting roots of unity that defines the primitive idempotents associated with the elementary particles. So what does nature do? The elementary particles are mass eigenstates and therefore have to be eigenstates of parity.

    Nature does the same thing as happens when you want to make a spin-1/2 in the +x direction spinor out of the two +/-z spinors (1,0) and (0,1). She makes a complex linear combination. The result is that one of the linear combinations ends up as M_1, the other as M_0. But instead of the complex linear combination being built out of nice simple complex numbers, instead nature has to use those nasty 3x3 matrices. The result is that the mass interaction is far less probable with the neutrinos than the charged leptons. Neutrinos have to go through the 12 step program in order to achieve sobriety.

    Also, I realized that I didn't explain how one got the Koide formula from the 3x3 solutions to the idempotent equation. One does this by adding together the right handed and left handed matrices. When you do this, all the vectors cancel and all that is left is the scalars, which have the form of the Koide mass formula [tex]1+\sqrt{2}\cos(\delta+2n\pi/3)[/tex]. The only problem is that the angle delta is wrong.

    The reason that all the non scalar parts cancelled is because the potential energy was assumed to be order Planck energy for every part of a multivector except the scalar part (which corresponds to order of magnitude of muon mass, I guess). The potential energy is a sum of squares of coefficients, and so the mass is the square of the scalar coefficient, which explains why Koide's formula works for square roots of masses instead of masses.

    It is possible to derive that the observed particles are what you get when you demand that all but the scalar parts sum to zero but it isn't easy. It is easy, however, to show that this happens with the observed leptons. And the quarks show that of the three (non scalar) roots of unity that define the ``fermion cube'', one of those roots has a potential energy that, while being very high compared to the scalar energy, its low enough that we can sort of see it a little.

    As a last addition, I thought I need to put in a slide showing how all this came about, particularly the PhysicsForums contribution, which I sem to recall was Alejandro's doing.

    None of this has taken place yet. I got in 7 hours ago, and since I was in my relatively inexpensive hotel by 3PM local time, I engaged a very helpful cab driver to drive me all over Honolulu looking for what we eventually figured out was the Punchbowl Memorial. My father lost a brother, missing in action, in the Pacific in 1944. He had always imagined that some day his children would play with his big brother's children but that was not to be. Instead, as an inadequate substitute, he named his first born son (me) after his brother. In the US, not returning from wars gets your name listed on a memorial, in this case at the Punchbowl Memorial, so of course I went to visit it. I had been thinking that I would do this on the day I flew out, but this worked out much better. It was a beautiful day to be sad. The sky was 40% covered with those sharp edged, very white clouds that make the blue look more intense. Like this:
    http://en.wikipedia.org/wiki/Cumulus_cloud

    Carl
     
  7. Oct 30, 2006 #6

    arivero

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    It is going to be an intimate audience anyway, having six (seven?) paralell sessions and an early start. I am amazed your talk is not the first in the morning; this implies you hava some advantage scoring, perhaps because of the mention on 's slides, perhaps because you and Koide happen to represent the meaning of the meeting: collaboration across the Pacific.

    graduate students do not have professors, they have advisors.

    It seems 6th to me.

    Seems a good point to isolate in a single slide.

    A point to discuss, if you leave time for that, is to wonder if a square root of the masses is an square root of the Dirac operator, and thus a hint for low energy supersymmetry.

    I hope you hat got slide pens. These hotels happen not to have blackboards. Besides, blackboard is slower, count your minutes.

    Lat me mention it because it was travel-related, as this plane jump you are doing. I was worried Hans guesswork was to be lost, at a time were the forum structure for speculation was not defined. I had been following it during a travel from Spain to Greece; the travel combined cargo ships with trains and then a lot of stations and some spare time to think and to follow physicsforums. I remember to crawl the pubs of Igumenitsa looking for a internet terminal. When I returned I decided to start that long long thread. No need for an slide, but if you bump against someone having good quality guesses, address them to the thread!
     
  8. Oct 30, 2006 #7

    CarlB

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    Arivero, Kea,

    I've got a new cut of slides. Way too long for 20/25 minutes, but I think the slides should be something people can look at who aren't there:
    http://www.brannenworks.com/jpp06.pdf

    Those are only half modified, but you can see that the mass comes from an assumption that is based on thinking of a potential energy between two particles that, in the spinor language is:
    [tex]V = |<A|B>|^2 = <A|B><B|A>[/tex]
    but with two modifications. First, it is made into the square of a sum of density operators, and second, the scalar part is interpreted as related to gravity and is interpreted as the particle mass. The non scalar parts of the sum are interpreted to have natural masses around the Planck mass, and so they amount to infinities that have to be arranged to cancel. Then the elementary particles are the things with scalar sums, and the scalar sum, when squared, is the measured low energy mass.

    Note that the above is 4th order in spinors but only 2nd order in density operators. So we can solve the problem in the density operator language but it makes a more difficult problem in spinors.

    Here it is 6:30AM local time. I had 5 hours of sleep and can hardly wait to start the day. A mild touch of hypomania perhaps.

    One cool thing about Hawaii is the high degree of foreign influence here. For me, this means that I can buy unusual treats. Last night I picked up a bag of something made from rice, soy sauce and sea weed that turned out to be quite tasty. $2.49 + tax for 4oz. I took the photos to prepare for blogging the event.

    Yes, I will include a slide mentioning Physics Forums and your contributions. PF is a remarkable resource.

    Carl
     
  9. Oct 30, 2006 #8

    arivero

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    I think that somewhere between slide 7 and 11 you must stress "up to here the empirical data, now lets search for some model". Because Koide will present other models, and it must be clear that, while Koide original formula comes from a model, you both now take the reversal way: given this surprising empirical input, look for a model.
     
  10. Oct 30, 2006 #9

    CarlB

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    Yes, I set it up as a transition from observation to theory, but I need to mention that there are alternatives. Also, the lecture preceding mine is about tribimaximal neutrino mixing and I really need to tie in.

    This is the first time I've attended the opening remarks at a physics meeting. Perhaps coincidentally, this is the farthest west I've attended one of these things.

    I haven't seen Dr. Koide. Reading between the lines of one of his emails, I suspect that he will not arrive until Wednesday. He is well known among the Japanese physicists I have talked with so far. Our presentations are particularly natural at a joint meeting of the US and Japanese particle physicists.

    First session starts in 15 minutes. I will attend the session "Low energy tests of the standard model", as this seems germane to me:
    http://www.phys.hawaii.edu/~dpf06/post/lowenergy/ms1.pdf

    The other two parallel sessions I'm planning on are beyond the standard model:
    http://www.phys.hawaii.edu/~dpf06/post/bsm/ms2.pdf
    http://www.phys.hawaii.edu/~dpf06/post/bsm/ms3.pdf

    I can't imagine paying $400 to attend one of these without attending all the sessions. Since I'm not connected with academia, these are quite a rush.

    Carl
     
  11. Oct 30, 2006 #10

    Kea

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    Hi Carl

    Goodness, you managed a lot on such a short flight! We're still working on recovering the idempotents the M-theory way (which will take a while at the pace I go) but we've been distracted by gluon tree amplitudes, which isn't a bad thing because it would (a) indicate a rigorous QFT framework and (b) be useful at the LHC.

    Enjoy Hawaii!

    :smile:
     
  12. Oct 30, 2006 #11

    CarlB

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    Kea,

    It turns out that I love taking photographs and Hawaii is going to provide me with some entertainment. However, as a fat old bald guy, with a tendency towards skin cancer, the island does not provide a whole lot of attractions for me. Now Iceland, that I think I would really enjoy.

    Which reminds me. I am at a cheap hotel and have to walk to the Waikiki Sheraton where the meetings are. Because of limitations on the human capacity of fat old balding guys to tolerate friction, I picked up a small bottle of baby powder downstairs at the colorful hotel I'm staying at. It has worked well.

    The meetings started at a convenient hour for me but there are signs of some very sleepy Japanese here. I like to sit in a single session for the full 2 hours. There were four talks this past session, on the subject of low energy tests of the standard model. At the start of the second talk, a young man came in and sat next to me. Immediately he began dozing off, his head bobbing. Each time he fell over against me, he woke upright. At the end of the talk he vigorously ran out of the room, presumably to attend the next meeting in a different session. So many capabilities are wasted on youth.

    Density operators are all about keeping the initial and final states together in a single package. Now in normal density matrix stuff the initial and final states are not necessarily very close together. But in what I'm doing, they are literally at the same point in space time. Is there some relationship between that concept and string theory where the point particles are replaced by pairs, also very close together?

    Polchinski is here and will give a talk. There are also a series of sessions devoted to string theory but I wasn't planning on visiting any. There are no printed programs, everyone is complaining. What I'd like to attend would be a meeting titled "string theory for idiots".

    Carl
     
  13. Oct 30, 2006 #12

    Kea

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    Yes, personally I'm an Iceland type of person myself. Maybe if you get a chance to gain some altitude...go on an astronomy excursion?

    Er...the way we see it with operads, string theory is just a moduli space technique for doing rigorous QFT (sorry guys). But yes, the similarities are more than superficial. The important point is that the interpretation of a point is permitted in a (let's call it) measurement geometry but not in any other arbitrary geometry, such as interpreting the spacetime literally as a universal background.

    Oh, I wonder if he'll come to your talk...:smile:
     
  14. Oct 30, 2006 #13

    Hurkyl

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    M_0 looks like a twelfth-root to me; if you cube it, you get (-i)I, and you need to square it twice more to get the identity.
     
  15. Oct 30, 2006 #14

    CarlB

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    Hurkyl, Kea, and all,

    you're right, but in the context of a mass interaction, the minus stops the process at 6 stages. What I need is an object that has to be raised to the 12th power before it gives unity, and it doesn't repeat any degrees of freedom along the way. Among the reals, the largest Nth such power is 1, but with 3x3 matrices of Clifford algebras you can do something much more interesting. For example,
    [tex]\left(\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right)[/tex]
    is a 3rd root of unity that doesn't repeat any degrees of freedom. If I replace the "1" with a 4th root of unity that doesn't repeat any degrees of freedom then I'm golden.

    I know way too much about square roots of unity in Clifford algebras but I don't know much about 4th roots of unity. I kind of doubt that they exist.

    Oh, and Kea, I'm suddenly having doubts about my construction. The problem is with the sentence "When you do this, all the vectors cancel and all that is left is the scalars". While listening to a lecture just now I convinced myself that the bivectors didn't cancel. I had assumed that they didn't appear because of stuff I did a long time ago when I was using a different potential energy, but now I have my doubts.

    This old man is getting tired. To be young again and able to continuously party for 6 days running. Oh, and to not make as many mistakes.

    I made a rookie tactical mistake. I looked up the slides for two lectures I was interested in before attending them. And last night I realized that "Babe in the Universe", who has the speed of light cosmology theory, lives somewhere around here so I sent her an email. I am still thinking that there has to be a symmetry breaking for spacetime that chooses a preferred parity. To get this in Clifford algebra, one replaces c with a multivector that has a parity term in it. The spinor Dirac equation is preserved under this sort of transformation, but the particles end up distinguished from one another. The transformation amounts to a change to the potential energy.

    Carl
     
  16. Oct 30, 2006 #15

    Kea

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    Don't worry. I have no doubts whatsoever that the silent army is busily working it all out, and they should soon put us old(ish) folks out of our misery.

    It would be wonderful if you could meet Louise Riofrio. Make sure you get some photographs!
    :smile:
     
  17. Oct 30, 2006 #16

    CarlB

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    I send a lot of emails to academic types and don't get replies. My assumption has been that they have already read enough emails from people on yahoo, they don't need anymore, and they've set their filters to eliminate everything but .edu.

    If anyone ever complains that I didn't send them an email with notification of something important back in 2004, I'm going to tell them, "but I did send it to you. I was very excited by the results and I knew you would be too. I remember it clearly. It was titled "Better than sex!!!", so that I was sure it would get your attention.

    I went to a neutrino session. It was well attended in a fairly large room. Mostly experimentalists talking about cuts.

    I'm thinking maybe that factor 2/9 is going to explain itself this week. I used to think it was associated with statistical mechanics. The logic goes something like this: QM probabilities are proportional to the density matrix, [tex]\rho[/tex]. In statistical mechanics, probabilities look something like [tex]\exp(-H/kT)[/tex]. Equating these two, it shouldn't be a surprise when you find that your probabilities end up with terms in them like [tex]\exp(i m/n)[/tex]. And mass is, of course, a probability.

    Carl
     
  18. Oct 30, 2006 #17

    Kea

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    LOL. :smile: But Louise is really friendly. Why not leave a note on her blog?
     
  19. Oct 31, 2006 #18

    Hans de Vries

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    Congratulations with your talk Carl! I think you did a poster session before
    but this is the first real talk. :smile:

    Regards, Hans
     
  20. Oct 31, 2006 #19

    CarlB

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    Kea,

    she got back to me and we may meet. But I had forgotten how tiring these things were for me. I've just got back from the "reception". I had pleasant (well, for me), conversation with a group of youngsters working on the D0 at Cern. So long as I don't talk about physics I can carry on fairly intelligent conversations with physicists, LOL.

    And I know a little more about 4th roots of unity that touch 4 degrees of freedom:
    [tex]\left(\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{array}\right)[/tex]

    The above is a matrix example, of course, so needs to be converted into geometric form in order to understand it. A simple conversion to "Dirac bilinears" on the left and idempotents, on the right, is:

    [tex]=\left(\begin{array}{cccc}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{array}\right)
    \left(\begin{array}{cccc}0&&&\\&1&&\\&&0&\\&&&1\end{array}\right)[/tex]

    [tex]+\left(\begin{array}{cccc}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{array}\right)
    \left(\begin{array}{cccc}1&&&\\&0&&\\&&1&\\&&&0\end{array}\right)[/tex]

    Geometrically, the above is defined by three roots of unity, [tex]a,b,c[/tex] The three roots each anticommute with the other (i.e. they could be three vectors from a Clifford algebra)). Then the 4th root of unity should be:

    [tex]0.5(a(1-c) + b(1+c))[/tex]

    but alas my Clifford algebra calculator says I made a mistake. I need to add a function to the calculator that allows you to input a matrix and get back the geometric equivalent. Or maybe I should try this again tomorrow.

    Hans, this may not be my first talk, but I think it's the first talk I've ever given where someone who had an academic position was present who actually wanted to listen to it.

    Carl
     
    Last edited: Oct 31, 2006
  21. Oct 31, 2006 #20

    arivero

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    Cof, true. :uhh: Somehow I modded -1 out.
     
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