On Monday, June 30, 2008 at 8:58:19 PM UTC-4, Jay R. Yablon wrote:

"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:6csk1lF3i22a3U1@mid.individual.net...
. . .

Is there a reason why a velocity measurement would need to be so
precise as to cause infinite energy to be required? IMHO, this is not
a good reason to reject a velocity operator. Is there perhaps a
different reason? What am I missing here?

In the Standard Model, elementary fermions are massless without an interaction with the Higgs field.

Best,

Fred Diether

Fred,

I did a calculation of what happens to the mass matrix during the transformation from the Dirac-Pauli representation to the Newton-Wigner representation via Foldy-Wouthuysen. This is shown in:

http://jayryablon.wordpress.com/files/2008/06/foldy-wouthuysen.pdf

Not sure where to go from there, but I'll be away the rest of the week
on vacation, so I'll take another look when I return.

Best,

Jay.

Hi Jay

From :

https://jayryablon.files.wordpress.com/2008/06/foldy-wouthuysen.pdf

You said:

In the third line we 1 2 b = , and also b@ = -@b which flips the sign in the b term. In the final
line, we use 1 2 @ = and p p = ( + + ) + + = p 2 2 2 2 2 2 2 / / x y

This did not copy and paste properly so I will translate

we use a^2=1 and p^2/p = (px^2 + py^2 + pz^2)/(px^2 + py^2 + pz^2)^.5 = p

Allow me to sprinkle a little Koide magic on this.

(px + py + pz )/( px^2 + py^2 + pz^2 )^.5 = p

We know ahead of time if p = 2/3 then x y z will render the lepton line.

Is there a way to hammer a round peg into a square hole to use the Koide magic?

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