electrons can emit gluons ??

[Widdekind]

http://en.wikipedia.org/wiki/Weak_hypercharge

 

the Weak-hyper-charge, of right-handed Fermions, is one-charge-unit larger (in magnitude), than their regular left-handed cousins...

 

looking at the table provided on the page, the only way of conserving hyper-charge, as well as charge & spin, in transitions, from L <----> R...

 

is to emit a spin-less pion = L-quark + R-antiquark (or vice versa)...

 

only L/R quark/antiquark pairs, have no spin, and no charge, but a unit of hyper-charge, to carry away the unit of hyper-charge "lost", when (say) some R-electrons ----> L-electron

 

so, can electrons expel pions, in transitioning from R ----> L handedness ??

 

 

 

to conserve charge, hyper-charge, momentum, angular momentum...

 

a right-propagating, right-handed electron, would have to recoil backwards, whilst preserving its spin... and emitting a neutral pion (of zero spin) in its original forward direction...

 

and originally have had enormous energy, to account for the 140MeV rest-mass of the pion

 

pion emission resembles gluon emission, and so seems to smack of E/W + S unification... can electrons actually emit gluons / pions, in the process of reversing R-to-L handedness ??

[swansont]

Where would the energy come from in emitting a pion?

[Widdekind]

http://hyperphysics.phy-astr.gsu.edu/hbase/particles/hadron.html

 

The symmetry which suppresses the electron pathway is that of angular momentum, as described by Griffiths. Since the negative pion has spin zero, the electron and antineutrino must be emitted with opposite spins to preserve net zero spin. But the antineutrino is always right-handed, so this implies that the electron must be emitted with spin in the direction of its linear momentum (i.e., also right-handed). But if the electron were massless, it would (like the neutrino) only exist as a left-handed particle, and the electron pathway would be completely prohibited. So the suppression of the electron pathway is attributed to the fact that the electron's small mass greatly favors the left-handed symmetry, thus inhibiting the decay.

 

 

[math]\pi^- \longrightarrow \mu^- + \bar{\nu_{\mu}}[/math]

[math]\gg [/math]

[math] \longrightarrow e^- + \bar{\nu_e}[/math]

 

however, when the decay does occur, the emitted electron is [math]e^-_R[/math]

http://en.wikipedia.org/wiki/Pion#Neutral_pion_decays

 

[math]\pi^0 \longrightarrow \gamma + e^+ + e^-[/math] (1%)

 

so, if the reaction can occur one way, then it can occur the other way...

 

leptons (electrons + neutrinos) can convert, to hadrons (quarks)...

 

(which seems a strong step towards Strong / Weak unification)

[swansont]

Where would the energy come from in an electron emitting a pion?

[Widdekind]

Where would the energy come from in an electron emitting a pion?

 

KE ? (the same place the pion's energy is deposited, during decay)

 

i think i managed to muddle matter-antimatter X right-handed/wrong-handed (for want of worthier words)

 

in the decay [math]e^-_R \rightarrow e^-_L[/math], the hyper-charge changes by (plus) one unit, whilst the (regular) charge is conserved...

 

to balance hyper-charge, without affecting charge, requires an electrically-neutrally, hyper-charged, particle...

 

potentially a neutral pion...

 

but antiquarks are intrinsically opposite-handed from quarks...

 

so q + !q = 0 net charge...

 

but only -1 hyper-charge, if one of the quarks is "wrong" handed, and the other "right"...

which means both must be spinning in the same sense

 

e.g. down-right (Y_W = -2/3) + anti-down-right (Y_W = -1 x 1/3)...

or up-left (Y_W = 1/3) + anti-up-left (Y_W = -1 x 4/3)...

 

BUT if both quarks are spinning in the same sense...

then the electrically-neutral pion, would have a net spin of one, yes??

 

the hyper-physics site states, that the decay

 

[math]\pi^- \longrightarrow e^-_R + \bar{\nu}_R[/math]

 

can occur...

 

but the net total hyper-charge on the RHS = (-2) + (-1 x -1) = -1

 

so 'tis the same on the other (LHS) side also...

 

but the only way to have a negative pion w/ net negative-one hyper-charge, is

 

down-right (-2/3) + antidown-right (-1 x 1/3)

 

if both quarks are bound into a common pion particle, and so are traveling together, then their linear momenta point parallel...

so if they're spinning in the same sense also, then their spins are pointing parallel...

and the pion would have to have a strong spin

 

the decay does occur, commonly enough to note

but Wikipedia purports that pions are spinless scalars

(?!)

[swansont]

KE ? (the same place the pion's energy is deposited, during decay)

That won't work, because the decay always has to be possible in all frames, including the rest frame of the particle. No KE available.

 

No sense in looking for new physics in a proposed decay that violates conservation of energy

[Widdekind]

what about the exact inverse of the hyper-physics site:

 

[math]\bar{\nu}_R + e_R^- \rightarrow \pi^-[/math]

 

?

 

are you then stating, that right-handed fermions, cannot flip over to left-handed... w/o interacting, w/ another fermion... only collisions can spin-flip, spin-reverse, RH'd fermions ?

[swansont]

I'm saying that your proposed decay can't happen, because it won't conserve energy. Mass always has to decrease for a spontaneous reaction. Induced reactions, i.e. with more than one particle, can happen if the particles have sufficient energy.

[Widdekind]

the combo, of an electron + neutrino, suggests the intermediary to be the W^- boson, i.e. for the reaction to involve the Weak force, not the Strong force

still... under certain circumstances, an electron can collide with a neutrino, and they both "break up" into two new Fermions, i.e. quarks (which, after the collision and "break ups", then begin emitting gluons, but not before, if i understand correctly)...

 

that seems sensible, and potentially important... electrons can "break up", losing some of their electrical charge, to a neutrino, in intensely energetic interactions... then, afterwards, subsequently, the "broken" particles begin emitting gluons, in "trying to knit themselves back together", which is my understanding of "color confinement"... the electron ----> d-quark, !v ----> !u-quark, so some of the electron's charge is "chipped off" and "lodges" in the neutrino... then, in their strange broken configuration, they then afterwards (only) begin giving off gluons...

 

only particles possessing partial electrical charges, emit gluons, and interact Strong-ly, yes ?

 

so i guess that the electron and anti-neutrino interact Weak-ly... re-configure into quarks... and only then begin generating "glue"

 

no spontaneous spin / chiral transitions occur ?

 

so, then, the energy of RH = LH = invariant w.r.t. spin-polarization-to-momentum-direction ? cp. photons, E = hv either forward-or-backward spinning ? simplistically stated, particles "don't care" whether they spin one-way or another... only Weak-interactions "care" about chirality ?

 

if i understand correctly, then there are no natural LH <----> RH transitions, at all... no RH electron can spin-flip, into a normal LH electron, via any known reaction... RH electrons can be produced, or be destroyed, by interactions... but no electron can "flip over on their own", or even "be flipped over" w/o also being destroyed, as an electron, in intense (Weak) interactions, which reconfigure them into quarks...

 

still, something seems amiss... in the reaction(s)

 

[math]e_R^- + \bar{nu}_R \leftrightarrow \pi^-[/math]

 

what is the hyper-charge, on the RHS ?? is the hyper-charge not (-1) on the LHS ?? so it cannot be (0) on the RHS ??

[swansont]

 

that seems sensible, and potentially important... electrons can "break up", losing some of their electrical charge, to a neutrino, in intensely energetic interactions...

 

Not sure what makes you think this is correct.

[Widdekind]

 

Not sure what makes you think this is correct.

 

is this correct :

 

[math]e_R^- + \bar{\nu}_R \longrightarrow W^- \longrightarrow \pi^-[/math] ?

 

and, can i ask, how to balance the so-called "hyper-charge", in the above equation ? For instance, is not the (net) hyper-charge, on the RHS, = 0 ? (And, is not the hyper-charge, of [math]\bar{\nu}_R = +1[/math] ?

[swansont]

Are you just going to ignore my comment?

is this correct :

[math]e_R^- + \bar{\nu}_R \longrightarrow W^- \longrightarrow \pi^-[/math] ?

I'm not a particle physics person. AFAIK right-handed antineutrinos don't exist.

But this is not an example of an electron "breaking up" and losing some of its charge to a neutrino.

and, can i ask, how to balance the so-called "hyper-charge", in the above equation ? For instance, is not the (net) hyper-charge, on the RHS, = 0 ? (And, is not the hyper-charge, of [math]\bar{\nu}_R = +1[/math] ?

Wikipedia doesn't list it. But since the weak hypercharge on the e_R is -2, why would you assume the antineutrino (if it exists) is +1?

[Widdekind]

Are you just going to ignore my comment?

 

 

I'm not a particle physics person. AFAIK right-handed antineutrinos don't exist.

 

But this is not an example of an electron "breaking up" and losing some of its charge to a neutrino.

 

 

Wikipedia doesn't list it. But since the weak hypercharge on the e_R is -2, why would you assume the antineutrino (if it exists) is +1?

 

all anti-Fermions are right-handed, yes ?

 

and all anti-Fermions have opposite charge, and also opposite hyper-charge, to Fermions, yes ?

 

the hyper-charge of (LH) Leptons (electrons & neutrinos) = -1

 

so the hyper-charge of (RH) anti-Leptons = +1

 

i thought that the existence of the anti-neutrino was well established

 

i want to know, whether one could consider the collision [math]e^- + \bar{nu} \rightarrow W^- \rightarrow \pi^-[/math] as a kind of "head-on car crash", wherein the "pieces and parts" of the two colliding "cars" became con-fused and re-configured, into new Fermions, which would be the surviving "Fermionic chassis" of the original two particles, but with their "charge load-outs" swapped to-and-fro forth and back between them

 

vaguely like two trucks collide, and all the loads they were carrying get transferred around, before the newly-re-loaded-out Fermions emerge from the "crash scene" as re-configured Fermions

 

i personally perceive, that it may be helpful to view such collisions, that way, such that the "underlying Fermion chassis" survive through the collision, during which their "charge load-outs" are swapped around...

conversely, [math]e^+ e^- \rightarrow \gamma \gamma[/math] would then be construed, as "more destructive", such that the underlying "Fermion chassis" did not survive the collision...

 

i want a way to visualize the "moment by moments" of the collisions... seems as if "charge" somehow has an existence of its own, (somewhat) independent of the existence of the "Fermion chassis" that "carry" the charge, vaguely like truck carrying loads

[swansont]

From what I can tell from the wikipedia page, they are listing right-handed neutrinos [math]{\nu}_R[/math], and the entry is blank since these don't exist. And when you write [math]\bar{\nu}_R[/math], you are talking about the antiparticle of something that doesn't exist.

[Widdekind]

Wikipedia's page provides, "The antineutrinos observed so far all have right-handed helicity (i.e. only one of the two possible spin states has ever been seen), while the neutrinos are left-handed. "

[swansont]

Wikipedia's page provides, "The antineutrinos observed so far all have right-handed helicity (i.e. only one of the two possible spin states has ever been seen), while the neutrinos are left-handed. "

But these are the antiparticles of a left-handed neutrino. That's why they have right-handed helicity.

[Widdekind]

you would want to write, for right-handed anti-neutrinos

 

[math]\bar{\nu}_L[/math]

 

?

[swansont]

you would want to write, for right-handed anti-neutrinos

 

[math]\bar{\nu}_L[/math]

 

?

 

No, I would simply write [math]\bar{\nu}[/math], since there's no reason to distinguish between a right-handed and left-handed family of particles. We know neutrinos are left-handed, and about antiparticles.

 

The only reason to designate the handedness is for a new, hypothesized particle whose only distinction is that it has opposite helicity, since saying "left-handed neutrino" is redundant for the physics we have.

[Widdekind]

skipping subscripts, then, the hyper-charge of normal anti-neutrinos is +1 (the opposite of normal neutrinos)... and normal pions are spin-less un-net-hyper-charged particles (+0). So, if backwards electrons still carried the same hyper-charge, as normal electrons... then the following decay could be quickly accounted for:

 

[math]e_R^- + \bar{\nu} \longrightarrow \pi^-[/math]

If the Standard Model is completely correct, and hyper-charge is an invariant quantity under Parity, then the above decay poses no problems, and the appropriate page on Wikipedia would be better if duly updated. i personally perceive, that accurately accounting for Hypercharge is crucial, because Hypercharge is a crucial quantity, affecting E/W and also Higgs interactions... so any mis-accountings and errors would be best to be corrected quickly