tag:blogger.com,1999:blog-7307840825023135484.post7892368045896725354..comments2012-02-07T15:48:38.662+13:00Comments on Arcadian Pseudofunctor: M Theory Lesson 321Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-7307840825023135484.post-579236313437509892010-05-03T12:03:38.564+12:002010-05-03T12:03:38.564+12:00Ah, with the $1,0,-1$ row at the end there too ......Ah, with the $1,0,-1$ row at the end there too ...Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-66864384838819974022010-05-03T12:03:00.509+12:002010-05-03T12:03:00.509+12:00Actually, a bit better ... again subtracting the $...Actually, a bit better ... again subtracting the $(24^2)$ from the diagonal, and splitting it up into a $1/24$ piece and a $1/144$ piece (the (ub) correction) we get:<br /><br />$(1/24)*$<br />$-681$, $705$, $0$;<br />$704$, $-704$, $24$;<br />$1$, $23$, $0$ plus<br /><br />$(1/144)*$<br />$-1$, $0$, $1$;<br />$0$, $0$, $0$.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-84068523817277180322010-05-03T11:39:36.485+12:002010-05-03T11:39:36.485+12:00Mitchell, as Nigel points out, the square roots at...Mitchell, as Nigel points out, the square roots at least are not numerology ... they are the amplitudes in any parameterisation of the CKM. Anyway, I wanted to find a (and this probably is numerological) representation with a denominator of $(24^2 + 1)$. That is, find numerators so that the squares of the CKM entries have a simple form. The best I found, after subtracting $24^2$ times the identity matrix, was $1/144$ times:<br /><br />$-4087$, $4230$, $1$;<br />$4225$, $-4224$, $143$;<br />$6$; $138$; 0<br /><br />which has the advantage that one can easily check that the row sums are all exactly $1$. Of course the zero comes from the $0.999133$. These numbers (with the $144$) are combinations of things like $1/24$ and $143/144$.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-30837920170594982762010-05-03T01:16:10.924+12:002010-05-03T01:16:10.924+12:00You have to square wavefunctions to get relative p...You have to square wavefunctions to get relative probabilities of finding particle within a given volume, or the effective particle density. Born originally discovered that in the 1920s, from <a href="http://en.wikipedia.org/wiki/Wave_power#Wave_energy_and_wave_energy_flux" rel="nofollow">an analogy with transverse water waves where the energy contained by the wave is proportional to square of the amplitude (height) of the wave.</a> Also, in longitudinal sound waves, the energy per unit area is proportional to the square of the amplitude (pressure). The same holds for electromagnetic energy, where the energy per unit volume of a field is proportional to the square of the field strength.<br /><br />Vice versa, you take square roots of probabilities to get wavefunctions, etc. For a unit mass, kinetic energy is proportional to the square of the velocity. In Balmer and Rydberg's empirical equations for line spectra, you square the reciprocals of a series of integers and combine that with subtraction to get model the line positions. This led to Bohr's quantum model of electron energy levels in Rutherford's atom. It's lucky Bohr didn't ignore Rydberg's empirical formula as mere numerology.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-68714094322344433022010-05-03T00:12:49.081+12:002010-05-03T00:12:49.081+12:00whoops, the LaTeX interpreter got me. That last se...whoops, the LaTeX interpreter got me. That last sentence reads: I'll bet 1000 dollars this latest epicycle means nothing.Mitchellhttps://www.blogger.com/profile/10768655514143252049noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-57684007509483264422010-05-03T00:11:16.449+12:002010-05-03T00:11:16.449+12:00Careful with the numerology. If you allow reciproc...Careful with the numerology. If you allow reciprocals, square roots, powers, along with addition and multiplication, you can approximate *anything*, even if you start with just a few "magic numbers". The value of such schemes has to be judged by their complexity in bits rather than just the number of parameters. The business about circulants and MUBs might have something going for it, because those concepts have other connections. But I'll bet $1000 this latest epicycle means nothing.Mitchellhttps://www.blogger.com/profile/10768655514143252049noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-80502768957892850962010-05-02T20:59:28.056+12:002010-05-02T20:59:28.056+12:00Now that you've pointed it out, I can see it v...Now that you've pointed it out, I can see it very clearly.<br /><br />:-)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-41217846455147212352010-05-02T20:57:12.373+12:002010-05-02T20:57:12.373+12:00And on that note ... the block of $0.97334$, $0.04...And on that note ... the block of $0.97334$, $0.0416666$, $0.04075$ and $0.999133$ (and hence the whole matrix) is easily described by the square roots of (resp.):<br /><br />$18/19$, $1/(24^2)$, $23/24 \cdot 1/(24^2 + 1)$ and $24^2/((24^2 + 1))$<br /><br />How many parameters does that count as?Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-88581114953937059732010-05-02T18:39:52.589+12:002010-05-02T18:39:52.589+12:00P.S. Oh, and thanks for the history, Nigel. I gues...P.S. Oh, and thanks for the history, Nigel. I guess you spotted that my 0.999133 is the square root of $24^2/(24^2 + 1)$.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-91425037465373761682010-05-02T13:07:29.356+12:002010-05-02T13:07:29.356+12:00Nigel, the stringers will just point out that I ha...Nigel, the stringers will just point out that I have used 4 parameters to fit a 4 parameter matrix. They will not be impressed.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-20083982346766823502010-05-02T11:03:07.569+12:002010-05-02T11:03:07.569+12:00Congratulations to you and Carl on being able to f...Congratulations to you and Carl on being able to fit the model so accurately to the CKM parameters!<br /><br />This is an exciting area to me, because I loved the story told of how Cabibbo discovered the first one of these amazing Standard Model parameters:<br /><br />"Here we see Nature giving a clear message that quarks and leptons are somehow profoundly related to one another. ... the Italian physicist Nicola Cabibbo in 1964 [discovered that] the strength of the weak force when acting within either one of the quark generations is (to within 1 part in 25) identical to that when acting on the leptons: e to v; however, its strength is only about 1/25 as powerful when leaking between one pair and the other, c to d and u to s. ... What Cabibbo had done was to ... assume that the true strength between pairs of the same [quark] generation is therefore essentially 24/25 relative to that of the leptons. This inspired him to the following insight into the nature of the weak interaction acting on quarks and leptons. It is as if a lepton has only one way to decay, whereas a quark can choose one of two paths, with relative chances of A^2 = 1/25 and 1-(A^2) = 24/25, the sum of the two paths being the same as that for the lepton. Today we know that this is true to better than one part in a thousand. This one part in a thousand is itself a real deviation from Cabibboâ€™s original theory, and is due to the effects of a third generation, which was utterly unknown in 1964."<br /><br />-Professor Frank Close, <i>The New Cosmic Onion,</i> Taylor and Francis, N.Y., 2007, pp. 154-8.<br /><br />I personally hope you manage to write a paper explaining your model physically in terms of preons (without an excess of mysterious stringy epicycles), but if you can get any research money for "stringy phenomenology", at least the string theorists will turn you into a media megastar.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-28185354450676070782010-05-02T09:05:35.764+12:002010-05-02T09:05:35.764+12:00Good idea, Mitchell! Um, but there is a limit to w...Good idea, Mitchell! Um, but there is a limit to what I can do in my cold, isolated flat with an old laptop.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-1309466195298357582010-05-01T17:39:35.647+12:002010-05-01T17:39:35.647+12:00It's not your intention, but it would be very ...It's not your intention, but it would be very interesting to hybridize this with the <a href="http://motls.blogspot.com/2008/12/ckm-matrix-from-f-theory.html" rel="nofollow">mainstream model-building in string theory</a>. Even if they can't be brought into alignment, it would be instructive for everyone to understand why not.Mitchellhttps://www.blogger.com/profile/10768655514143252049noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-75157171432571464962010-05-01T17:37:52.675+12:002010-05-01T17:37:52.675+12:00Oops, that's (td) ...Oops, that's (td) ...Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-9827929133058344822010-05-01T16:25:01.500+12:002010-05-01T16:25:01.500+12:00OK, so the best I can get for (ts) is 0.00987, wit...OK, so the best I can get for (ts) is 0.00987, with similar parameters. Not under 0.009, but close.<br /><br />Mitchell, if you look at this matrix carefully, you can permute signs around and invert parameters (duality) and the mixing matrix basically cycles around the flavours. So 1/24 works too. It's a magic flavour mixing matrix from quantum information theory. Just a simple thing. <br /><br />Note also that it follows the tribimaximal form discussed on a recent post, only instead of two R2 factors there are three R2 factors. I think of this as representing three 'state sets' for quarks, as opposed to two for weak/mass for the neutrinos. The small $c$ factor therefore acts as a 'colour' correction to what is a simple weak/mass mixing rule, although the $(a,b)$ are different for the quarks. (The 24 probably comes from the dimension of the Leech lattice somehow ... hmmm).Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-27431165073474648162010-05-01T16:13:42.902+12:002010-05-01T16:13:42.902+12:00Hmmmm. I have to teach tomorrow very early and so ...Hmmmm. I have to teach tomorrow very early and so can't look at this right now.<br /><br />Meanwhile, I gave up looking for the method of deriving the MU(3) matrix from the probabilities. It seems that I have to make a unitary matrix from the probabilities, convert it to magic, and then look up the parameters.CarlBrannenhttps://www.blogger.com/profile/17180079098492232258noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-24013394677486853222010-05-01T15:28:40.402+12:002010-05-01T15:28:40.402+12:00OK, I'm trying to understand what you just did...OK, I'm trying to understand what you just did. The elements of the CKM matrix are basically flavor-changing amplitudes for quarks. You approximate them using a "triple product of ... scaled R2 factors". Is there a physical conception behind that, or at this point is it still just a generalization of Koide's formula?Mitchellhttps://www.blogger.com/profile/10768655514143252049noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-37946403684156356802010-05-01T13:13:04.143+12:002010-05-01T13:13:04.143+12:00That's only THREE real parameters, and a very ...That's only THREE real parameters, and a very nice matrix. Check it out.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.com