Lepton Masses

There are interesting relations between the lepton masses (see Koide formula).

Also, we can see that the 3 masses come approximately from the equation:

x3 - 41 x2 + 90 x -1 = 0

Where the answers (|x0|, |x1|, |x2|) * 45.76MeV

Is this significant? Probably not. Most equations seem to have integer values.

Comments

Post a Comment

Popular posts from this blog

Vortex Solution to Navier Stokes Equations

Image
This is a solution I found to the Navier Stokes equations . It uses the flow lines of the Hopf Fibration (which is basically just a way of covering the entire 3D space in circles). The flow lines can be seen in this image below: You can see my paper here. Solution to Navier-Stokes equation . Unfortunately I won't get a million pounds for winning the Clay Millennium prize because this is a compressible not incompressible solution. :( Oh, well. Nice try! I think it's pretty nice anyway. :) Also it uses the nice identity: which paramaterizes the Pythagorean quadruples!
Read more

Conjugate of Lie Algebra

Conjugate of Lie Algebra A little article noting the similarities between momentum conservation (continuous) and charge conservation (discrete). We all know that the conjugate coordinate to position is momentum . Thus a term such as  ∫O(-k)O(-q)O(k+q)dk 4 dq 4 = ∫O'(x)O'(x)O'(x)dx 4 But a lie group given by it's root vectors (charges) also has a similar form which looks a lot like momentum: O {-A} O {-B} O {A+B} + (A n )O {A} O {-A} O n So we should be able to find conjugate variables to the root vectors. So we can write the above as: O' C O' C O' C The O' C  being linear combinations of the Os. In the case of E 8 , the {A} are 8 dimensional unit vectors. One difficulty are the neutral particles, O n  , which seemingly have no equivalent in the momentum picture. Such a term would look like:  ∫O(k)O(-k)kμAμ(0)dk 4 =  ∫O(x)Bμ∂μO(x)dx 4   which is not Lorenz invariant.... [ unless O(x) were spinors and Bμ were Dirac matrices?? B
Read more