Re: Lorentz violation of the Standard Model

> beg your pardon if you find
>my writing inappropriate and please understand my poor English skills.

No, Your English is excellent.

>I am not sure about your notations, but it looks like one of the two
>states you have in the rest frame is usually what we have saved
>for the "antiparticle" ones.

No what i did was to orient the spins in a different direction so as to
to show explicitly that you can have a phase difference between the
left and right states, which keeps track of the spin as boost to rest,
and
then to the opposite direction.

If you use the Halzen and Martin conventions from David Griffith,
introduction to elementary particles with p_x=p_y=0, then you have

u1 = ( 1 0 c.p_z/(E+mc^2) 0)

u2 = ( 0 1 0 -c.p_z/(E+mc^2)

Which hides the phase difference.

But the general solution, any spin, no antiparticles) is

u = A.u1 + B.u2 , with A^2 + B^2 = 1, A and B complex,

so the phase difference between A and B doesn't go away.

>I agree that we can pick our spinor in that way
>and I just picked one simple case, but I don't think it changes or fails
>any of my argument.

Well the center of your argument, seems to be, boosting from

(1, 1, 0 , 0 ) to ( sqrt(E+p^3 , sqrt(E-p^3), ?, ?)

[Three means third component here, i take it, and c=1)

then you take the limit p=~E, to get (sqrt(2E), 0, 0,0),
i.e. dropping a very small but present right handed component,

You then boost back to rest, and find you don't have (1,1,0,0) any
more.
Whats happened is you've to a frame where one component look
very small, done a small rotation, boosted back to rest, your small
rotation now looks very big. Thats because during the boost the right
handed component gets very small, but that doesn't mean you
can ignore it. If you used Halzen and Martin's convention,
then, at high speed

c.p_z / E+mc^2 ->1, so

u1 = (1 0, 0, 0)/sqrt(mc)
-> sqrt(c/E)(1, 0, 1, 0)

while u2 = (0,1,0,0) /sqrt(mc)
-> sqrt(c/E) (0,1,0,-1)

You'd find that you couldn't make the same approximation

>In my article, I have followed Peskin & Schoeder
>and showed one state we can relate to the boosted frame solutions.
>Are you saying what Peskin & Schoeder has shown in their book is wrong?

I've never read Peskin, but i've heard its a common omission. Griffiths
states that
using the Bjorken an Drell normalization
convection, u^dagger u = |E|/mc^2, leads to spurious
difficulties when m->0

>Also, I don't see how zero part of your states of field can be smoothly
>connected to nonzero part of the massive ones.

Hopefully its clear from the above description.

Barry

.

Relevant Pages

  • Re: Lorentz violation of the Standard Model
    ... >my writing inappropriate and please understand my poor English skills. ... i.e. dropping a very small but present right handed component, ... You then boost back to rest, and find you don't have any ... >Are you saying what Peskin & Schoeder has shown in their book is wrong? ...
    (sci.physics.research)
  • Re: Lorentz violation of the Standard Model
    ... > so the phase difference between A and B doesn't go away. ... > i.e. dropping a very small but present right handed component, ... > rotation now looks very big. ... Thats because during the boost the right ...
    (sci.physics.research)