trying to understand W-bosons ?

[Widdekind]

By analogy to Strong interactions, wherein quarks "extrude" gluons that "carry off" the quarks' color charges, perhaps in Weak interactions, fermions "extrude" W-bosons, that "carry off" the fermions' electric charges ??

 

And, by analogy, to gluons consisting of anticolors & colors, and "breaking" between them, into antiquarks & quarks, perhaps W-bosons consist of antineutrinos & neutrinos, and "break" between them, into typical W-boson decay products ??

 

If so, then would the "anti-screening effect", of electrons "extruding" their electric charges, "out to a Weak-arms' length away", i.e. ~0.01fm, help with "renormalization", of all the infinities arising, from viewing electrons as point particles ? I.e. if electrons actually "emanated" their electric charges, out to 10^-2fm, would that "soften" the EM self-energy, of electrons, per the short-range "asymptotic freedom", of non-Abelian gauge fields, like the Weak & Strong interactions ?

 

---

 

If Weak bosons have rest-masses [math]E_0 = m_0 c^2[/math], then is not the range, of the Weak interaction, calculated according to:

 

[math]\Delta t \approx \frac{\hbar}{\Delta E}[/math]

 

[math]\Delta E = \gamma E_0[/math]

 

[math]r = v t = c \Delta t \left( 1 - \left( \frac{E_0}{\Delta E} \right)^2 \right)^{1/2}[/math]

 

[math]= \frac{\hbar c}{\Delta E} \left( 1 - \left( \frac{E_0}{\Delta E} \right)^2 \right)^{1/2}[/math]

 

[math]= \frac{\hbar c}{\Delta E^2} \left( \Delta E^2 - E_0^2 \right)^{1/2}[/math]

 

[math]= \frac{\hbar c}{E_0} \frac{ \sqrt{x^2 - 1}}{x^2}[/math]

 

[math]\therefore r = \frac{\hbar c}{E_0} \frac{ \sqrt{x^2 - 1}}{x^2} [/math]

That function can be plotted, e.g. at the Wolfram Alpha website. The range r(x) is maximized, for [math]x = \sqrt{2}[/math], when r(x) = r₀/2, and declines slowly with increasing x thereafter. If the rest-mass energy, of Weak bosons, is ~80 GeV, then why isn't the Weak interaction most strongly observed, in particle colliders, "tuned" to ~120 GeV ??

Edited January 10, 2012 by Widdekind

[Widdekind]

I understand, that after an interaction, between quanta, via a 'Fundamental Force', e.g. Weak & Strong, that the quanta, emerging from said interaction, have been 'conformed' into eigenstates, of said force. For example, after a down quark 'emanates' a W^- boson, that quark emerges from that Weak interaction, in a Weak eigenstate, e.g. [math]u' \approx 0.97 u + 0.23 c[/math]. Then, only after a subsequent Strong interaction, e.g. with neighboring quarks in the same nucleon, would that quark's wave-function "collapse", into a Strong eigenstate, i.e. [math]0.96 \rightarrow u, 0.04 \rightarrow c[/math]. How does the 'energy barrier' eliminate the second possibility, i.e. "collapse" into a charm quark ?

 

Note, neutrinos only interact Weakly. Er go, once neutrinos emerge, from a Weak interaction, in a Weak eigenstate, neutrinos remain in those Weak eigenstates, for-ever-more, until & unless their next Weak interaction. Meanwhile, [math]\nu_e, \nu_{\mu}, \nu_{\tau}[/math] are Weak eigenstates, and so are combinations, of the "true" neutrino Gravity mass eigenstates, e.g. [math]\nu_e \approx 0.9 \nu_1 + 0.5 \nu_2[/math].

 

Thus, emerging from the Weak interaction, depicted in the figure above, are a "Weak up" quark u', a "Weak electron" e^-, and a "Weak electron neutrino" v_e:

 

[math]u' \approx 0.97 u + 0.23 c[/math]

 

[math]e^{-}[/math]

 

[math]\nu_e \approx 0.9 \nu_1 + 0.5 \nu_2[/math]

Then, only subsequent interactions, with other quanta, via other forces, can "collapse" those emergent wave-functions, into eigenstates, of those other forces. Presumably, in the case of quarks, which readily & rapidly interact Strongly, such "collapses" occur "quickly". Is a "Weak electron", emerging from a Weak interaction, in the same quantum state, as a "normal", i.e. "EM electron" ?

 

Note, neutrinos may interact Gravitationally. And, if neutrinos interacted via the Gravity force, then they would "collapse" into Gravity mass eigenstates, e.g. [math]\nu_e \rightarrow \nu_1[/math]. And, if neutrinos "collapsed" into a mass eigenstate, then they would have a fixed "flavor", i.e. they would not undergo "flavor oscillations". Er go, if Solar neutrinos undergo flavor oscillations, then they have not interacted Gravitationally, en route to earth, from the center of our sun ?

 

Ipso facto, "free fall" trajectories, along geodesics, through curved space-time, do not constitute Gravity interactions. Presumably, any interaction, via any force, must impute some change in momentum, i.e. scattering. Perhaps a "quantum of curvature", in the fabric of space-time, i.e. a Graviton, must scatter other quanta, e.g. neutrinos, in order to have "interacted" Gravitationally ? Perhaps Gravitons, in the fabric of space-time, are like Phonons, in crystal lattices ? Perhaps, in regions of intense & rapidly varying gravity, e.g. cores of collapsing stars undergoing Super-Novae, Gravitons are generated, in the fastly flexing fabric of space-time, and scatter neutrinos, into "true" mass eigenstates ?

 

In EM, EM radiation is attributed to EM quanta, i.e. photons. By analogy, perhaps G radiation is attributable, to G quanta, i.e. gravitons ? Perhaps gravitons obey E = h f ? If so, then gravitons with wave-lengths of km's would have energies of neV's, naively equivalent to temperatures of 10^-5K. Perhaps super-cooled quantum detectors could absorb such low-frequency, low-energy gravitons ? The quantum coupling constant [math]\alpha_G \equiv G m_p^2 / \hbar c \approx 0.5 \times 10^{-38}[/math] is miniscule.

 

 

---

I understand, that Weak bosons can decay, into both quarks & leptons. In the following figure, the EM charges carried, by charged Weak bosons, "enters into" one of the quanta, emerging from the decay, of the boson, e.g. [math]\left( -1 \right) + \nu_e \rightarrow e^{-}, \left( +1 \right) + \bar{u} \rightarrow \bar{d}[/math]. From the following figure, neutral Weak bosons would resemble one, or the other, of the charged bosons, except without the electric charge, and without the implied charge "entering into" any of the fermions:

 

Edited January 10, 2012 by Widdekind

[Widdekind]

energy is "pluri-potent" ?

 

All force-carrying bosons seem to be "pluri-potent", able to generate both matter & anti-matter, e.g.

 

[math]\gamma \rightarrow \bar{e}e[/math]

 

[math]W^{+} \rightarrow \bar{e}\nu_e[/math]

 

[math]g \rightarrow \bar{q}q[/math]

Now, photons & gluons are mass-less, er go "energy" can generate both "matter & anti-matter" ?

[Widdekind]

I understand, that after an interaction, between quanta, via a 'Fundamental Force', e.g. Weak & Strong, that the quanta, emerging from said interaction, have been 'conformed' into eigenstates, of said force. For example, after a down quark 'emanates' a W^- boson, that quark emerges from that Weak interaction, in a Weak eigenstate, e.g. [math]u' \approx 0.97 u + 0.23 c[/math]. Then, only after a subsequent Strong interaction, e.g. with neighboring quarks in the same nucleon, would that quark's wave-function "collapse", into a Strong eigenstate, i.e. [math]0.96 \rightarrow u, 0.04 \rightarrow c[/math]. How does the 'energy barrier' eliminate the second possibility, i.e. "collapse" into a charm quark ?

 

 

I understand, that wave-function "collapse" occurs, after each & every force interaction I, between "particle" quanta & "force" quanta, i.e. between fermions & bosons; and that that "collapse" conforms the "particle" quanta, emerging from the interaction, into eigenstates [math]|\phi_I>[/math], of the interaction force:

 

In particular, I understand, that fermion "particle" quanta, which emit boson "force" quanta, need not "wait around", before "collapsing", whilst the emitted boson propagates away, towards some potential future interaction, with some other "particle", at some other place & time:

 

[Widdekind]

Z⁰ bosons

 

Z⁰ bosons convey momentum, energy, & spin:

 

The Z boson is its own antiparticle. Thus, all of its flavour quantum numbers and charges are zero. The exchange of a Z boson between particles, called a neutral current interaction, therefore leaves the interacting particles unaffected, except for a transfer of momentum...

 

The neutral Z boson obviously cannot change the electric charge of any particle, nor can it change any other of the so-called "charges" (such as strangeness, baryon number, charm, etc.). The emission or absorption of a Z boson can only change the spin, momentum, and energy of the other particle

Z⁰ bosons are generated, with appreciable probability, only at high energies, comparable to the rest-mass-energy of the boson, i.e. 100GeV = 10¹⁵K:

 

The Z boson can couple to any Standard Model particle, except gluons and photons [or other Weak bosons ??]. However, any interaction between two charged particles that can occur via the exchange of a virtual Z boson, can also occur via the exchange of a virtual photon. Unless the interacting particles have energies on the order of the Z boson mass (91 GeV) or higher, the virtual Z boson exchange has an effect of a tiny correction [math]\propto \left( E/M_Z \right)^2[/math], to the amplitude of the electromagnetic process... Z-bosons behave in somewhat the same manner as photons, but do not become important, until the energy of the interaction is comparable with the relatively huge rest mass of the Z boson.

 

Z boson interactions involving neutrinos have distinctive signatures: They provide the only known mechanism for elastic scattering of neutrinos in matter; neutrinos are almost as likely to scatter elastically (via Z boson exchange) as inelastically (via W boson exchange).

From whence arises that energy-dependence, of the interaction cross-section? It resembles the Rayleigh-Jeans limit, i.e. [math]\frac{E}{M_Z} \rightarrow \frac{k_B T}{M_Z c^2}[/math], but would not the super-massive W-bosons correspond to the "ultra-violet limit" ??

 

 

 

 

"running" range of Weak interaction ?

 

If Weak bosons have rest-masses [math]E_0 = m_0 c^2[/math]; and if a 'particle' experiment is performed at some energy [math]E[/math]; then is not the range, of the Weak interaction, calculated according to:

 

[math]\Delta t \approx \frac{\hbar}{\Delta E}[/math]

 

[math]\gamma E_0 = \Delta E + E[/math]

("loan + cash-on-hand")

 

[math]r = t v = \frac{\hbar c}{\Delta E} \left( 1 - \left( \frac{E_0}{\Delta E + E} \right)^2 \right)^{1/2}[/math]

 

[math]= r_0 \frac{1}{x} \sqrt{1 - (x+f)^{-2}}[/math]

where:

 

[math]r_0 \equiv \frac{\hbar c}{E_0}[/math]

 

[math]x \equiv \frac{\Delta E}{E_0}[/math]

("loan")

 

 

[math]f \equiv \frac{E}{E_0}[/math]

("cash-on-hand")

For [math]f \ll 1[/math], the function [math]r(x,f)[/math] is maximized at [math]x_{max} \approx \sqrt{2} \left( 1 - f \right)[/math], whereat [math]r_{max} \approx \frac{r_0}{2} \left(1+\frac{f}{\sqrt{2}} \right)[/math]. For [math]f \geq 1[/math], [math]x_{max} \rightarrow 0[/math], and [math]r_{max} \rightarrow \infty[/math], i.e. once 'particles' can "pay their own way", the potential range of interaction is no longer limited, by a need "to pay back the Heisenberg bank". Perhaps that explains the large 'resonances', in the interaction cross-sections, when the 'particle' experiment is 'tuned' to the bosons' rest-mass-energy, i.e. at [math]f = 1, E = E_0[/math] ? Note, the above approximations seem to remain accurate, if imprecise, for the entire range of [math]0 \leq f < 1[/math].

 

 

PROOF

 

Let

 

[math]u = x + f[/math]

("loan + cash-on-hand")

then

 

[math]r = r_0 \frac{1}{u-f} \sqrt{1 - u^{-2}}[/math]

which is maximized at

 

[math]0 = \frac{\partial r}{\partial u} = r \times \left( -\frac{1}{u-f} + \frac{1}{1-u^{-2}} \frac{1}{u^3} \right)[/math]

 

[math]u-f = u^3 - u[/math]

 

[math]0 = u^3 - 2u + f[/math]

Assume

 

[math]u \equiv \sqrt{2} + \alpha f[/math]

Then, to first order in [math]f[/math]

 

[math]0 \approx \left( 2^{3/2} + 3 \sqrt{2} \alpha f + ... \right) - \left( 2^{3/2} + 2 \alpha f \right) + f[/math]

 

[math]0 \approx f \left( 3 \sqrt{2} \alpha - 2 \alpha + 1 \right)[/math]

 

[math]\alpha \approx \frac{1}{2 - 3 \sqrt{2}}[/math]

 

[math] \therefore x \approx \sqrt{2} - \frac{ 3 \sqrt{2} - 1 }{ 3 \sqrt{2} - 2 }f \approx \sqrt{2} \left(1 - f \right)[/math]

Now

 

[math]r = r_0 \frac{1}{u-f} \sqrt{1 - u^{-2}}[/math]

 

[math]\approx r_0 \frac{1}{\sqrt{2} \left( 1 + \frac{\alpha-1}{\sqrt{2}}f \right)} \sqrt{1 - \frac{1}{2 + 2 \sqrt{2} \alpha f + ...}}[/math]

 

[math]\approx \frac{r_0}{\sqrt{2}} \left( 1 - \frac{\alpha-1}{\sqrt{2}}f \right) \sqrt{ 1 - \frac{1}{2 \left( 1 + \sqrt{2} \alpha f \right) } }[/math]

 

[math]\approx \frac{r_0}{\sqrt{2}} \left( 1 - \frac{\alpha-1}{\sqrt{2}}f \right) \sqrt{ 1 - \frac{1}{2} \left( 1 - \sqrt{2} \alpha f \right) }[/math]

 

[math]\approx \frac{r_0}{\sqrt{2}} \left( 1 - \frac{\alpha-1}{\sqrt{2}}f \right) \sqrt{ \frac{1}{2} \left( 1 + \sqrt{2} \alpha f \right) }[/math]

 

[math]\approx \frac{r_0}{2} \left( 1 - \frac{\alpha-1}{\sqrt{2}}f \right) \left( 1 + \frac{\sqrt{2} \alpha f}{2} \right) [/math]

 

[math]\approx \frac{r_0}{2} \left( 1 - \frac{\alpha-1}{\sqrt{2}}f + \frac{ \alpha }{\sqrt{2}}f \right) [/math]

 

[math]\approx \frac{r_0}{2} \left( 1 + \frac{f}{\sqrt{2}} \right) [/math]

Q.E.D.

[Widdekind]

W⁰ would be a 'Flavor-Changing Neutral Current' ??

 

FCNCs are generically predicted by theories that attempt to go beyond the Standard Model, such as the models of supersymmetry or technicolor. Their suppression is necessary for an agreement with observations, making FCNCs important in model-building

If both the [math]W^{\pm}[/math] bosons can change 'flavor', then would not the already-hypothesized [math]W^0[/math] boson change 'flavor' too? I.e. would not the [math]W^0[/math] be a FCNC ?

 

Somehow, the "mixing" of the already-hypothesized [math]W^0, B[/math] bosons, of the unified EW force, into the [math]Z^0, \gamma[/math] bosons, of the separated EM & W forces, "suppresses" FCNCs, by "balancing" the effects, of the [math]W^0, B[/math] ??

 

I understand, that the eigenstates, of both the [math]Z^0, \gamma[/math], are the "canonical", "mass" eigenstates, of EM & S interactions, i.e. emission / absorption, of either [math]Z^0, \gamma[/math] "conforms" the wave-functions of 'particles', into "collapsed" eigenstates, of the non-Weak forces. If so, then the [math]Z^0[/math] is "merely heavy light".

[Widdekind]

calculating masses, for Weak-eigenstate quarks ??

 

If

 

[math]\left( \begin{array}{c}

d_W \\

s_W \\ \end{array} \right) = \left( \begin{array}{cc}

cos(\theta_W) & sin(\theta_W) \\

-sin(\theta_W) & cos(\theta_W) \\ \end{array} \right) \left( \begin{array}{c}

d \\

s \\ \end{array} \right)[/math]

then is not the 'expectation value', of the masses, of the Weak-eigenstate quarks (in MeV), assuming that the Weak mixing-angle is [math]\theta_W \approx 30^{\circ}[/math]

 

[math]\left( \begin{array}{c}

<m_{d_W}> \\

<m_{s_W}> \\ \end{array} \right) = \left( \begin{array}{cc}

cos^2(\theta_W) & sin^2(\theta_W) \\

sin^2(\theta_W) & cos^2(\theta_W) \\ \end{array} \right) \left( \begin{array}{c}

<m_d> \\

<m_s> \\ \end{array} \right) \approx \frac{1}{4} \left( \begin{array}{cc}

3 & 1 \\

1 & 3 \\ \end{array} \right) \left( \begin{array}{c}

300 \\

500 \\ \end{array} \right) \approx \left( \begin{array}{c}

350 \\

450 \\ \end{array} \right) [/math]

and

 

[math]\left( \begin{array}{c}

<m_{u_W}> \\

<m_{c_W}> \\ \end{array} \right) \approx \frac{1}{4} \left( \begin{array}{cc}

3 & 1 \\

1 & 3 \\ \end{array} \right) \left( \begin{array}{c}

300 \\

1500 \\ \end{array} \right) \approx \left( \begin{array}{c}

600 \\

1200 \\ \end{array} \right) [/math]

If so, then the Weak-eigenstate quarks are intermediate in mass, between their closest-corresponding "canonical" quarks. (Note that the inverse matrix involves negative numbers??) How would this affect the "exo-thermy" or "endo-thermy" of quark Weak decays ? For example, the Weak kaon decay [math]K^0 \rightarrow \bar{\mu}\mu[/math] begins with the "collapses", of the kaon's [math]s, \bar{d}[/math] quarks, in "mutually compatible" Weak eigenstates. Naively, the "lower flavor" decay pathway "ought" to be energetically favored

 

I understand, that the Weak-boson generating interactions are:

 

[math]\bar{u}_W d_W \longleftrightarrow W^{-}[/math]

 

[math]\bar{c}_W s_W \longleftrightarrow W^{-}[/math]

et vice versa. For some reason, no 'flavor-changing neutral currents (FCNC)' occur, as if "the strangeness goes with the (electric) charge".

 

 

 

---

Please ponder the "Weak mixing", of the hypothesized [math]W^0, B[/math] unified-EW bosons, into [math]Z^0, \gamma[/math]. Given that [math]m_Z > m_W[/math], whilst [math]m_{\gamma} \rightarrow 0 \ll m_W[/math], perhaps, via that mixing, "the [math]Z^0[/math] gains mass, whilst the [math]\gamma[/math] loses mass" ?? And, if "the [math]Z^0[/math] has something more than the [math]W^0[/math]", and if "the [math]\gamma[/math] has something less than the [math]B[/math]", then perhaps:

 

[math]W^{+} = W_0 + \{+1\}[/math]

("loaded with positive electric charge")

 

[math]W^{-} = W_0 + \{-1\}[/math]

("loaded with negative electric charge")

 

[math]W^0 = W_0 + \{ \}[/math]

("unloaded")

 

[math]Z^0 = W_0 + \{+1\} + \{-1\}[/math]

("double loaded")

 

[math]\gamma = B - \{+1\} - \{-1\}[/math]

("double unloaded" ??)

What might a hypothetical "non-specified (mass &) electric charge packet", i.e. [math]\{+1\}, \{-1\}[/math], represent ?? Seemingly, they are associated with mass, and with electric charge, but not with Weak charge, or Strong "color" charge.

 

An intimate connection, between mass & electric charge, seems suggested, e.g. in flavor-changing charged currents, both mass & charge change. And, saying that [math]B = \gamma + \{+1\} + \{-1\}[/math] seems possible, i.e. "a [math]B[/math] would be a doubly-charge-loaded [math]\gamma[/math]". Somehow, photons "forsake all charges, both positive & negative", and so "opt out of mass-inducing interactions" ?? Could that, qualitatively, characterize the hypothesized Higgs interaction ?? And, why would not such hypothetical "massive charge packets" not be able to exist alone, i.e. what would they "lack w.r.t. particles", i.e. "charge must be confined within particles" -- could there be some sort of similarity, to color-confinement, in the Strong interaction ??

 

Stated another way, without Weak mixing:

 

[math]W^{+} = W_0 + \{+1\}[/math]

("loaded with positive electric charge")

 

[math]W^{-} = W_0 + \{-1\}[/math]

("loaded with negative electric charge")

 

[math]W^0 = W_0 + \{ \}[/math]

("unloaded")

 

[math]B = \gamma + \{+1\} + \{-1\}[/math]

("double loaded" ??)

but with Weak mixing:

 

[math]Z^0 = W_0 + \{+1\} + \{-1\}[/math]

("double loaded")

 

[math]\gamma = B - \{+1\} - \{-1\}[/math]

("fully unloaded" ??)

Pictorially, the "double charge load" has been transferred, from the [math]B[/math], to the [math]W^0[/math], rendering the former a [math]\gamma[/math], and the latter a [math]Z^0[/math]. Naively, if [math]m_Z \approx m_W + 10 GeV[/math], then perhaps the "massive charge packets" mass [math]m_{\{\pm1\}} \approx 10 GeV[/math] ?? Naively, perhaps then [math]m_B \approx 20 GeV[/math] ??

Edited January 12, 2012 by Widdekind

[Widdekind]

Is the following chart accurate, especially regarding the Weak charges Q_W ?

 

   mass    q    Qw   C
u   4    +2/3  +1/3  +1
d   5    -1/3  +1/3  +1

v   >0     0    -1   0
e   0.5   -1    -1   0

[Widdekind]

According to Wikipedia, Z⁰ bosons decay, to fermion-antifermion pairs, in "branching ratios", proportional, to those fermions' "effective Weak hyper-charges", i.e. not their Weak charges Y_W = 2(Q-T₃), but a "mixed" factor, involving the Weak mixing angle, Y_W,Z⁰ = 2(xQ-T₃), where [math]x = sin^2(\theta_W)[/math]. By contrast, W⁺,W^- bosons decay to fermions, without such "bias".

 

I understand, that if "there was no Weak mixing", i.e. x = 0, that the Z⁰ would "unmix", back into the W⁰; and that, therefore, the W⁰ would couple to fermions, only proportional to their iso-spin T₃, whose magnitude (1/2) is the same for all fermions.

 

I understand, that if "there was full Weak mixing", i.e. x = 1, that the Z⁰ would "become" the B; and that therefore the B must couple to fermions, in proportion, to their canonical Weak hyper-charge Y_W.

 

I understand, that if "Weak mixing was reversed", i.e. [math]\theta_W \rightarrow \theta_W'[/math], such that [math]sin^2(\theta_W) \rightarrow cos^2(\theta)[/math], that the Z⁰ would "become" the photon; and that, therefore, the photon must couple to fermions, in proportion to another "biased" Weak hyper-charge derivative, i.e. [math]Y_{W,\gamma} = 2 \left( cos^2(\theta_W) Q - T_3 \right)[/math].

 

Accordingly, assuming [math]\theta_W \approx 30^{\circ}[/math], w.h.t.:

 

   mass    q     Qb    Qg    Qz    Qw    C
u   4    +2/3  +1/3    0    -2/3   -1    +1
d   5    -1/3  +1/3  +1/2   +5/6   +1    +1

v   >0     0    -1    -1     -1    -1    0
e   0.5   -1    -1   -1/2   +1/2   +1    0

So, up-type quarks barely couple, to photons, i.e. [math]Q_{\gamma,u} \approx 0[/math] ??

Edited January 15, 2012 by Widdekind

[Widdekind]

According to this presentation,

 

[math]\sigma_{W^+ \rightarrow \bar{e}\nu} \approx 3 nb[/math]

[math]\sigma_{W^- \rightarrow e\bar{\nu}} \approx 2 nb[/math]

[math]\sigma_{W^+ \rightarrow \bar{l}l} \approx 0.5 nb[/math]

where all of the above figures are, subsequently, divided by an "acceptance factor" of [math]\approx 0.5[/math], to derive "total" cross sections. Naively, perhaps the square-roots, of those figures, represent estimates, of the linear range, of the Weak interaction, via those pathways ? If so, then those cross-sections represent ranges, of approximately 0.1-1 10^-18m = 0.1-1 m-fm.

Edited January 18, 2012 by Widdekind