tag:blogger.com,1999:blog-7307840825023135484.post7765569355931984891..comments2023-04-16T03:44:23.949+12:00Comments on Arcadian Pseudofunctor: Top Down NumbersKeahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-7307840825023135484.post-31920835838629061032010-05-09T20:39:31.323+12:002010-05-09T20:39:31.323+12:00"Nigel, you must stop putting big links on my..."Nigel, you must stop putting big links on my blog, or I will ban you from the blog"<br /><br />It was a small link. I will stop reading your blog now, so I won't make any more comments, go celebrate with more rudeness!<br /><br />NigeAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-16660836752778978512010-05-09T20:37:42.009+12:002010-05-09T20:37:42.009+12:00" I don't care what their content is"..." I don't care what their content is"<br /><br />I know you don't care a damn about simple mechanisms, that's why I'm asking where you claim to get Pi from, and if it is related to the factors of Pi in the link you deleted. If you cared about being clear rather than just rude, you'd give an answer.<br /><br />NigeAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-68549062454117799392010-05-09T12:20:21.280+12:002010-05-09T12:20:21.280+12:00OK, Nigel, so if the masses are responsible for th...OK, Nigel, so if the masses are responsible for the sign flip mentioned in the update (which has nothing at all to do with factors of $\pi$) then it makes the phased CKM no longer the sum of two circulants. That is, we end up looking at a matrix with the same norm squares, but unequal row sums. It could be relevant, because the Fourier matrix (unlike $R_n$) is like this.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-46447120636570265412010-05-09T08:43:33.633+12:002010-05-09T08:43:33.633+12:00Nigel, you must stop putting big links on my blog,...Nigel, you must stop putting big links on my blog, or I will ban you from the blog. I don't care what their content is. And of course I think the CKM parameters are related to masses ... duh. If you actually bothered reading anything I wrote you would understand that.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-56947103680577272132010-05-09T00:56:05.386+12:002010-05-09T00:56:05.386+12:00Just to clarify:
(1) the CKM parameters determine...Just to clarify:<br /><br />(1) the CKM parameters determine the relative weak force interaction strengths.<br /><br />(2) the weak force interaction rate depends on the relative mass of the weak gauge boson to the particle being considered. Neglecting electroweak theory for a moment, in Fermi's theory of beta decay, the weak force has a strength equal to the electromagnetic force strength multiplied by the impressive looking ratio ratio: (π^2)h(M^4)/[T(c^2)m^5)] (reference: Matthews <i>Quantum Mechanics</i> textbook), where T is the effective lifetime of the particle (i.e. halflife times the factor 1/ln2 ~ 1.44), m is the mass of the emitted fermion and M is the mass of the particle before it decays. <br /><br />(3) <i>therefore, there seems to be a link between the CKM parameters and the masses of fundamental particles.</i><br /><br />Do you agree? (Sorry for posting 2 comments in a row. Delete them if you like, I'll copy to my blog just in case.)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-28263758001227282512010-05-08T18:42:01.817+12:002010-05-08T18:42:01.817+12:00For those who don't remember: the current esti...For those who don't remember: the current estimates from CDF/D0 do not agree with this $0.04$ (although the SM must agree with this) but they do seem to agree with the supposedly <a href="http://kea-monad.blogspot.com/2009/10/lhc-era.html" rel="nofollow">bad</a> (Carl's matrix has not changed) number that we calculated last year. That is, the difference between (i) the 'true' phase $0.04$ and (ii) the phase one gets by ignoring a bunch of signs is EXACTLY the discrepancy between (i) the SM fit and (ii) the current experimental range.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-31895214464170827422010-05-08T18:26:58.870+12:002010-05-08T18:26:58.870+12:00P.S. I am not upset that we get the SM value, but ...P.S. I am not upset that we get the SM value, but this means that those experimentalists are going to have to get off their asses and sort this number out ...Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.comtag:blogger.com,1999:blog-7307840825023135484.post-12794644887227458002010-05-08T18:17:45.325+12:002010-05-08T18:17:45.325+12:00OK, I think Maxima is doing this right, and $0.04$...OK, I think Maxima is doing this right, and $0.04$ is the right answer. If you look at it by hand, the quotient of phases comes down to a number close to $\pi$. So it matters that we keep track of the $\pi$s when summing all the phases, and this is probably what we did wrong when we were doing it by hand.Keahttps://www.blogger.com/profile/05652514294703722285noreply@blogger.com