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Natural Science Forum / Physics / Particle Physics / August 2006Dirac Spinors and Degrees of Freedom
Hello to all: I have over the past few months been studying Zee's excellent book "QFT in a Nutshell." I had some questions regarding the degrees of freedom for a Dirac spinor. Just as the Klein-Gordon equation for a non-zero mass projects one unphysical degree of freedom out of a four-vector potential leaving it with three degrees of freedom (two transverse, one longitudinal), the Dirac equation for a non-zero mass projects two degrees of freedom out of a four-spinor wavefunction leaving it with two degrees of freedom (spins up and down). We also know that to give mass to a massless vector boson, one starts with a four-vector potential with two transverse degrees of freedom, and adds a third longitudinal degree freedom by "swallowing" a Nambu-Goldstone scalar. Importantly, such a mechanism is also *predictive* of the mass. Questions: 1) Is it proper, linguistically, to say that a Weyl spinor (R and L projections using 1 +/ gamma^5) which solves the Dirac equation for a zero-mass Fermion has a single degree of freedom, namely, spin up, or spin down, because such a massless spinor, which travels at the speed of light and can never be overtaken by a Lorentz boost, thereby yields a precise correspondence between helicity and chirality? 2) In rough analogy to how boson mass is "revealed," would it make sense, in principle, to consider the possibility that a *predictive* mechanism for Fermion masses might start with two decoupled massless Weyl spinors, one L and one R, each with one degree of freedom, and have these two Weyl spinors "swallow" up the degrees of freedom from one another (yes, I know that sounds obscene ;-) ) yielding a massive Fermion (four-component Dirac spinor) with two degrees of freedom? Of course, one would want to ultimately show such a mechanism in detail and arrive at the right masses. But, my question is whether this line of inquiry makes basic theoretical sense at the outset? Thanks, Jay. _____________________________ Jay R. Yablon Email: jyablon@nycap.rr.com > Hello to all: > [quoted text clipped - 42 lines] > Jay R. Yablon > Email: jyablon@nycap.rr.com I just replied to you at sci.physics.research, but I may as well repost since it will take some time to moderate there. Klein-Gordon doesn't project out a degree of freedom associated with a component, it ensures the momentum is on the mass shell. Any component of any field satisfies that equation, including a spinor field. The 'projection' for a massive 4-vector field is given by something else called the Proca equations. 1) yes 2) In the standard model you do start with massless spinors, but you have the mechanism wrong. Read further in Zee. Hope that helps, Dan hidden-irony.blogspot.com >> Questions: >> [quoted text clipped - 24 lines] >> >> Jay. . . . > Klein-Gordon doesn't project out a degree of freedom associated with a > component, it ensures the momentum is on the mass shell. Any [quoted text clipped - 11 lines] > Dan > hidden-irony.blogspot.com Dan, thanks for your reply. Perhaps I need to clarify my question, because the emphasis I am suggesting is on "a *predictive* mechanism for Fermion masses " When a vector boson mass is revealed because a Nambu / Goldstone scalar is swallowed by a vector potential, that mass is *predicted" because it is a function *only* of strength of known dimensionless running couplings -- in the case of the W and Z bosons of electroweak theory, the electromagnetic and weak isospin running couplings. I am familiar with the standard model for giving mass to fermions, but the mass values arrived at are *arbitrary* because the (Yukawa type, I believe?) couplings do not relate to any known couplings like the electromagnetic and/or weak and/or strong (though one may suspect that they should, we just don't know how). This is a well-known limitation of the standard model, because the couplings are made proportional "by hand" to the experimentally-observed fermion mass values but the fermion mass values are not in any way predicted as are the W and Z boson masses. Not to mention that we have no clue as to why nature replicates fermions into three mass-distinguished generations and so far can do nothing more than play games with masses and Cabibbo-type mixing angles. One would have to at least suspect that we will never have a true answer as to why the fermions have the masses they do until we understand why nature has dealt us the Fermion generation redundancy for which nobody has ever uncovered a theoretical necessity. So, what I am asking is whether the mechanism I am suggesting with a mutual swallowing / sharing of degrees of freedom between a pure R and a pure L Weyl spinor might make some sense in thinking about a *predictive* mechanism for revealing fermion masses which, by definition as something predictive, would go beyond the standard model to new territory. I am not asking if this is "right" for sure because we can't know that without the details. Just asking if this seems to be a sensible way to think about this particle physics research problem (or at least part of this problem) for which nobody today has any sure answers. Jay. > >> Questions: > >> [quoted text clipped - 80 lines] > > Jay. Well, it's very sensible in a certain way, because mass (at least at the field equation level) can be thought of as a coupling between the pure left and right fields. What you're vaguely saying sounds very similar to what actually occurs in the standard model. But look at the chapter on the Higgs mechanism again (the "swallowing"), it depends crucially on the properties of the gauge field. We need those extra parameters which aren't predictive to give the fermions mass too, we can't get it 'for free' with the Higgs mechanism. But if you're suggesting something completely different and radical, and I don't think so, you should get someone else to answer. Dan hidden-irony.blogspot.com $$ Virtually between w & z bosons at-a-distance. $$ Question: $$ What is ..THEORETiCALLY, *in BETWEEN* ADjACENT w or z bozons? $$ $$ Answer: $$ A *ViRTUAL SPACE* is in BETWEEN adjacent ViRTUAL *PARTicles". $$ $$ Distinguish a "degree-of-freedom", from "NOthing-at-a-distance". Jay R. Yablon wrote: > > >> Questions: > >> > >> 1) Is it proper, > >> > >> 2) In rough analogy to how boson mass is "revealed," > >> these two Weyl spinors "swallow" > >> Jay. > . . . > > > > Klein-Gordon doesn't project out a degree of freedom > > Hope that helps, > > > > Dan > When a vector boson mass is revealed because a Nambu / Goldstone scalar > is swallowed > One would have to at least suspect that we will never have a true answer > as to why the fermions have the masses they do until we understand why [quoted text clipped - 11 lines] > at least part of this problem) for which nobody today has any sure > answers. $$ What exactly are you asking? <snicker> > Jay. Re: Dirac Spinors and Degrees of Freedom. END of POST. > Hello to all: > [quoted text clipped - 42 lines] > Jay R. Yablon > Email: jyablon@nycap.rr.com I just replied to you at sci.physics.research, but I may as well repost since it will take some time to moderate there. Klein-Gordon doesn't project out a degree of freedom associated with a component, it ensures the momentum is on the mass shell. Any component of any field satisfies that equation, including a spinor field. The 'projection' for a massive 4-vector field is given by something else called the Proca equations. 1) yes 2) In the standard model you do start with massless spinors, but you have the mechanism wrong. Read further in Zee. Hope that helps, Dan hidden-irony.blogspot.com > Hello to all: > [quoted text clipped - 42 lines] > Jay R. Yablon > Email: jyablon@nycap.rr.com I'm sorry if this is double posted. Google isn't treating me right. It's the Proca equations, not the Klein-Gordon, that eliminates one of the degrees of freedom of a massive vector field. Klein-Gordon just puts the momentum on the mass shell, and it's satisfied by any component of any field. 1) yes 2) In the standard model you do start with massless fermions, but you have the mechanism wrong. Read further in Zee. Hope this helps, Dan http://hidden-irony.blogspot.com > Hello to all: > [quoted text clipped - 37 lines] > Thanks, > Jay. I'm having moderate success constructing an electron (e-) or an (e+) using 3 charges (-) (-) (+), or (+) (+) (-), respectively. >From that, construct a proton using (e+) (e+) (e-). Each of those particles decay when combined with their anti-particle. Aside from antiparticle induced decay I think those particles are stable, i.e. there is no evidence of spontaneous decay. I get fair results by defining the charge above as being a finite vector orthogonal to space & time, which is suggestive of a 5D orthogonality, where the 5th is finite and possess only +/-1, sticking out of an orthogonal spacetime, along the lines of Kaluza's 5D theories. However, I find it's more realistic to use asymmetrical metrics in 4D for practical applications, albiet somewhat unconventionally. I'll use "4" for time following classics here. For example: Using F_14 for the electric field, and setting in a rest frame g_14 = F_14, we will obtain classically charge "q" by q = E * r^2, where E = q/r^2 = Electric Field, (Charge "q" is experimentally determined to be quantized), but in GR to include that we need to fracture the metric into symmetric and asymmetric g_uv = s_uv + a_uv, and for brevity I'll set a_uv = q*F_uv. Following the classical demo above, we can do a product in the spirit of q = E*r^2, written in terms of GR as, (A and B are charges, A^1 and B^1 are spatial distances relativity to charges A and B, A^4 and B^4 are the time difference due to the differneces in spatial location), 2AB = a_uv A^u B^v = A*F_14 A^1 B^4 + A*F_41 A^4 B^1. >From this we can use F_14 = - F_41= E r = A^1 = - B^1, because A and B are in relatively opposite directions, but A^4 = B^4 =r because the're in the same direction relatively temporally and get, 2AB = A* E * r^2 + A*(-E) *r*(-r) = A*E*r^2 , where E = B/r^2. Clearly a spinor like a_uv = -a_vu was employed to provide the relativity of charges A and B. By dividing by "r" the mass/energy, "m" of a pair of charges is, mc^2 = A*B/r , classically. Yup, Jay's ideas make sense. Regards Ken S. Tucker kst |
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