Widdekind

The calculation of entropy, from "counting microstates", weights every such state, by its probability, yes? If so, the entropy of a|A> + b|B> is constant, for all a,b s.t. a+b = constant. Now, when Wave Functions (WFs), of quantum objects, interact "weakly" (when it's "business as usual" under "normal & ordinary" kinds of conditions, such that the SWE still applies, without WF collapse), their WFs become "entangled". So, strongly seemingly, you're quite correct -- [math]\Psi = c_1 (v_1 \otimes u_1) + c_2 (v_2 \otimes u_2) \rightarrow v_1 \otimes u_1[/math] (or, [math]v_2 \otimes u_2[/math]) conserves entropy. What if particle number is not conserved, in the interaction, which causes WFC ? Please ponder a standard double-slit experiment, in which a detector screen (D) is irradiated by an incident electron beam (e-), in what amounts to a "reverse photoelectric effect": [math]e^{-} + D \rightarrow e^{-} : D + \gamma[/math] where the ejected photon is rapidly reabsorbed, within the detector (D), as thermal radiation [the incident electron beam heats up the detector screen]. At deepest root, this rather resembles the WF "localization", of when a formerly free electron becomes bound to a bare proton during Recombination in the ISM or IGM. The SWE is completely "probability preserving", so the irreversible act of particle creation / destruction, in which a whole new probability distribution is "born", or an old probability distribution "vanishes", cannot mathematically be described with said SWE. This is quite akin to WFC, which also represents a "discontinuous quantum jump", beyond the scope or purview of the SWE. So, could not the "quantum measurement problem", of what causes WFC, have something to do, with particle number non-conservation ? And, would not a particle number increase, in an interaction, increase the ensemble's entropy (a whole new particle, with new states available to it, in addition to those available to the previous particles...) ?

(Thanks for the responses.) Typically, interacting Wave Functions do not "collapse", but become "entangled", entering a single state for all of the entangled particles. "Measurement" destroys this singular entangled state, causes all of the entangled Wave Functions to collapse, and, thereby, puts all of the particles into individually separate states. Would not that increase the system's entropy (making more states available to the ensemble) ?

I found these references, seemingly suggesting, that nucleons "clique" together, inside nuclei, w/ pions apparently representing "thermal collisional" excitations (my words), which I would guess come from "stressing & stretching" the "glue" linking & locking quark-triplets together, inside nucleons: EDIT: These last two sources seemingly suggest, that pions represent a radial, quasi-periodic, "breather mode", expansion of core [constituent / valence / structural] quarks: [math](d u u) \rightarrow (d u) u[/math] [up quark in proton gets "quantum kick" (of some sort)...] [math]\rightarrow (d u) + g + u[/math] [up quark "puffs out" radially, stretching gluon bonds, which thusly "ooze more glue"] [math]\rightarrow (d u) + (d \bar{d}) + u[/math] [up quark "swells" further still, stretching gluon bonds until they break, "ripping" into quark-antiquark pairs] [math]\rightarrow ((d u) + d) + (\bar{d} u)[/math] [up quark's expansion "stalls out", as it Hadronizes with the antidown antiquark, whilst down quark, left behind at smaller radii, Hadronizes with "spectator" structural quarks, to form a neutron] [math]\rightarrow (d u) + (d \bar{d}) + u[/math] [up quark begins being brought back inwards, in "infall"...] [math]\rightarrow (d u) + g + u[/math] [up quark continues "deflating", quark-antiquark recombine into gluon bond] [math]\rightarrow (d u) u[/math] [quasi-stable proton has reformed] This "Cepheid Variable pulsating breather mode" model, of virtual pion production in protons, seems consistent w/ claims, that the proton spends ~80% of its time being "bare", and ~20% of its time "dissociated" into neutron + pion [looking for the book every which where]. EDIT to EDIT: According to I.S.Hughes' Elementary Particles (3rd. ed.), p.51, " part of the time the proton exists as a neutron plus a [math]\pi^{+} meson[/math], [math]p \rightleftharpoons n + \pi^{+}[/math] ", which orbits the central nucleon in an L=1 p-state. Given that gluons have S=1, the emission of a gluon, by a quark (S=1/2), Conservation of Angular Momentum must imply that the quark spin-flips; and, that, when the gluon bond breaks, the resulting quark-antiquark pair produced must be "born" spin-parallel: [math]p^{+} = (d \downarrow u \uparrow u \uparrow) \rightarrow (d \downarrow u \uparrow) \; u \uparrow[/math] [math]\rightarrow (d \downarrow u \uparrow) + g \uparrow + \; u \downarrow[/math] [math]\rightarrow (d \downarrow u \uparrow) + (d \uparrow \bar{d} \uparrow) + u \downarrow[/math] [math]\rightarrow \left( (d \downarrow u \uparrow) + d \uparrow \right) + (\bar{d} \uparrow u \downarrow)[/math] [math]\rightarrow \left( d \downarrow u \uparrow d \downarrow \right) + \pi^{+}_{l=1}[/math] [the spin-up down quark, from the torn tendon of glue, Hadronizes with spectator quarks in the core, but must spin-flip to spin-down, to align with its new-found fellow down quark. That spin-flip can account for the "Angular Momentum boost" (L=0 to L=1) given to the newly Hadronized pion.] [math]= n + \pi^{+}_p[/math] [math]\rightarrow \left( (d \downarrow u \uparrow) + d \uparrow \right) + (\bar{d} \uparrow u \downarrow)[/math] [math]\rightarrow (d \downarrow u \uparrow) + (d \uparrow \bar{d} \uparrow) + u \downarrow[/math] [math]\rightarrow (d \downarrow u \uparrow) + g \uparrow + \; u \downarrow[/math] [math]\rightarrow (d \downarrow u \uparrow u \uparrow) = p^{+}[/math] In-so-far as any kind of crude comparison can be construed, between the L=1 p-state pion, and the corresponding Hydrogenic Wave Function (L=1, m=1 [?]), then the Wave Function of the "spun up" pion perhaps takes on a toroidal, donut-like shape, rotating around the neutron at the center. According to Hughes (ibid.), this simple model predicts that protons spend a time-fraction x=0.7 in the "bare" proton state, and the rest of the time in the "dressed" neutron + pion state. Repeating the process for neutrons ([math]n \leftrightharpoons p + \pi^{-}[/math]), yields a bare-state fraction x=0.76. If meaningful, this might imply, that the lighter, and more electrostatically repulsive, up quarks, in protons, are more likely to "poof out", into the "pion breather" mode, than the lighter & less repulsive down quarks, in neutrons. Note that the "splitting" of gluons into quark-antiquark pairs seems superficially similar to the "splitting" of photons into electron-positron pairs.

Are there any known examples, of interactions involving Wave Function Collapse, which conserve (interacting) particle number (or, is there always an "accompanying" annihilation / creation event, of a whole quanta of Wave Function) ?

If quantum mechanical "measurement" represents an irreversible alteration, to the Wave Function, of a quantum system (causing its "collapse" into a singular state)... then doesn't QM measurement resemble Entropy, which irreversibly increases with time? What other physical processes, or quantities, are also "irreversible" ?? (Thanks in advance)

I understand, that you cannot send meaningful information on individual photons, since their behavior is random. But, what about the patterns produced by ensembles of photons ("fringes" vs. "heaps"), passing through the slits ? Measurement / observation causes Wave Function Collapse, which changes the statistical distribution of detector-screen hits to change, from "fringes mode" to "heaps mode". If so, using a conventional "carrier beam signal" (of suitably entangled photons), and assuming that the lengths of the "legs" were (sufficiently) the same, each user could cause, or cause not, the collapse of the photons' Wave Functions, switching the system from "fringes" to "heaps", which could encode information (if the carrier beam was continuously maintained, and by pre-arranged agreement). What am I missing ?

(Thanks for pointing that out.) Virtual pions are "off-mass shell" ? Actual pions would not help bind nuclei together, for fusion effects ?

How big are Collider Beams, used in High Energy collisions ? For example, in Deep Inelastic Scattering experiments, beaming Leptons (electrons, neutrinos) at nucleons, how physically big & broad are the Leptonic Wave Functions ?

"Quarks never appear in isolation. This process of Hadronization occurs before quarks formed in high-energy collisions can interact in any other way. The only exception is the top quark, which may decay before it Hadronizes" (ISBN 9781155255507, pg. 50). Thanks for all the responses

A pion is a Meson, a "middle-weight" combo of a quark & antiquark, whose mass-energy equivalent is ~140 MeV (~1/7th of nucleon mass-energies). They flit between nucleons, in a nucleus, and bind them all together, thereby accounting for the Strong Force, to my understanding. Their Wave Functions are apparently more extended, distended, or delocalized, than those of the nucleons themselves, which are typically spread out over ~1 fm. Thus, the "pion cloud" has a "tail" which extends out well beyond the "surface" of the nucleons to which they are bound.

According to the book Biocentrism (pp. 78ff), by Robert Lanza, by means of a continuous beam of Quantum Entangled, separated photons (emerging from a BBO crystal), the presence or absence of Interference Fringes, observed on one side of the set-up, can be instantaneously determined, by inserting / withdrawing a scrambler, from the beam on the second side of the set-up. In essence, the person on the second side plays "handsies in front of the projector screen" with their half of the split beam. Assuming that (1) such a split beam was sent to two remote locations (S & P) conventionally, at the speed of light; and, (2) that said split beam was continuously maintained, from the central source; then, (3) could not binary bits of information, be instantaneously sent, from P to S, based upon the presence / absence, of interference fringes, observed at S ??? Such a set up would allow a "central HQ" to feed a "laser umbilical" out to "soldiers" in the field, who would then be able to "cross talk", between themselves, instantaneously (in binary).

Nucleons are apparently immersed in a "pion cloud", of extended, delocalized, diffuse pions. And, pions apparently produce the Yukawa Potential, to mediate the Strong Nuclear Force. What role do pions play, in nuclear fusion processes ? Could you create some sort of "pion catalyzed fusion", along the lines of [math]p^{+} + p^{+} + \pi \rightarrow He^{2+}:\pi \rightarrow He^{2+} + \pi[/math] ? Could you "pre-treat" fusion fuels, by bombarding the protons in order to "stress" their gluon fields, thereby (electro- or neutrino-)producing quark/antiquark pairs (as gluon bonds break), which could "hadronize" into pions, and make the protons "pion rich" and "extra (Strong Force) sticky" ??? Anthony W. Thomas, Wolfram Weise. The Structure of the Nucleon:

Asymptotic Freedom -- Quarks are like Tether-Balls, "feeling free off the fist", but bound back to the pole after a few feet (???) According to the Bag Model, inside Hadrons, quarks are kept confined by a bag-like "skin" of "sticky slime". That "glue" is concentrated at the surface "skin" of the Hadron (r ~ 10-15m). When high-energy Leptons are beamed at, and into, Hadrons, they stream through that "sticky slime skin", probably (???) w/o interacting with those gluons, since Leptons are "color blind", and do not interact via the Strong (Color) Force. Once inside the Hadron, the Leptons interact with the constituent 'structural' quarks, causing their Wave Functions to collapse, or localize, into little (apparently) point-particles (r < 10-18m). At first, before those collapsed constituent quarks can careen out into the "sticky slime skin" of the gluon field, they scatter the incident Leptons like free particles. Now, at low Lepton energies, when gluon bonds begin to break, perhaps (???) a little like taughtened taffy thinning in the middle, those gluons can beget more gluons, to help hold together the tendon-like bond. In this "slow stretch" mode, the Lepton's incident energy can be absorbed into the "glue", perhaps (???) a little like kevlar bringing a bullet to a soft stop. But, at high Lepton energies, quarks are thrust through the "sticky slime" far too fast for that "re-forming effect", of the gluons' broken bonds, to take place. Thus, the threads of "sticky slime" simply "snap", before any incident Kinetic Energy can be absorbed into those bonds -- perhaps (???) a little like a swiftly speeding sniper bullet penetrating a kevlar coat. Thereby, harder & harder hit quarks act as if they are "entrained" with less & less glue, and so look like lighter & lighter (point) particles. To resolve the interior structures of nucleons (r ~ 10-15m), Leptons must have high energy (l < 10-15m, E > 1 GeV). Thus, nucleons (m c2 ~ 1 GeV) can disintegrate under such circumstances, shattering into a "scree spray" of naked quarks, antiquarks, & gluons, each carrying only a tiny fraction of the original nucleon's momentum. However, such collisions are often quasi-elastic, so that the incident Lepton only transfers a tiny fraction, of its incident energy, into the nucleon. In some sense, the Lepton's Wave Function, then, "flows through", and only "ruffles", the nucleonic Wave Functions, of the three structural quark cores, and all of their entrained gluons. References: G. Aubrecht. Quarks, Quasars, & Quandries, pp. 68-72. Fig. 3 -- Apparent relation, of Gluon density, to gradient of Quark density, seemingly suggests (???) some sort of "self-consistent" solution, and/or that Quark momentum is associated w/ Gluon emission. This picture apparently purports, that Nucleons are a little like "glue bubbles". The "sticky slime" slathers the surfaces of quarks, "constricting" them radially inwards, with powerful implosive pressure. Fig. 5 -- By Camouflage, quarks expel their color charge, thereby "blanching" and becoming color neutral (White). In crude conception, quarks "blanch" by "oozing" a shrouding "Camouflaging cocoon" cladding of "sticky slime" glue, composed of a gluon triplet. Each gluon in the triplet carries, quantum mechanically, 1/3rd of the emanating quark's color charge, combined with 1 of the 3 anticolors. Thus, each gluon triplet, in sum, carries 1 color, whilst being anticolor antineutral (Black). Then, each "blanched" color neutral quark (White) attracts the "blackened" anticolor antineutral gluon triplet (Black) that it emanated (White <---> Black). Meanwhile, those 3 gluon triplets, each carrying, quantum mechanically, the monochromatic color charge, of their parent quarks, are themselves mutually attracted & bound, by the Strong (Color) Force, thereby becoming, in combination, color neutral (White). Thus, in Lepto-production collisions, the incident Leptons interact Electromagnetically with the Camouflaged constituent quark cores, which interact Chromodynamically with their own Camouflaging "cocoon cladding" of colored glue, which "sticky slime shrouding shells" mutually interact, again Chromodynamically: Lepton <---EM---> quarks <---CD---> own gluons <---CD---> others' gluons This "loose linkage" may act as a kind of "shock absorber", complicating the dynamics of Lepto-production collisional interactions. Andrew Watson. The Quantum Quark, pp. 285-287. Books LLC. Quarks: Quark, Strange Quark, Top Quark, Isospin, Down Quark, Up Quark, Bottom Quark, Hypercharge, Charm Quark, Baryon Number.

"Best of Both Worlds" Resolution (???) -- Unperturbed Quark Wave Functions are extended [~10-15m], but collapse into quasi-structureless points [< 10-18m] upon Perturbation (1) Quark Wave Functions collapse upon perturbation by impinging Electrons / Neutrinos According to Robert Lanza's Biocentrism: (2) Immediately after perturbation impact, Nucleons are comprised of 3 point-like [Valence] Quarks in a "sea" of Gluons & Quark/Antiquark pairs [produced upon Gluon bonds breaking] According to F.E.Close's An Introduction to Quarks & Partons:

Have scientists seen "bare" exotic quarks (s, c, b, t), or do they only appear in Mesons & Baryons ?

According to The Teaching Company [Great Courses] Impossible: Physics beyond the Edge (lecture 21), the Quantum No-Cloning Theorem denies the possibility of "perfect" copying of quantum states. So, what is the "best" quantum copying which can be created ? Could you "quasi-clone" a quantum system with, say, 90% fidelity ?

According to The Teaching Company's [Great Courses] Impossible: Physics beyond the Edge (lecture 22), Electro-Magnetic Fields "enforce" Electric Charge conservation; and, Gravitational Forces "enforce" Mass-Energy conservation. Does this mean, that Color Charge conservation is "enforced" by the Strong Force ???

From analyzing the aforecited figure, it looks like, compared to those of up quarks, the Wave Functions, of down quarks, are more "squeezed", into intermediate radii. To wit, the positive charge distributions of up quarks dominate, even relative to their greater electrical charges, in the inner core, and outer limbs, of the nucleons. Uploaded with ImageShack.us

Not to my knowledge, which was why I'm confused. Frank Close's The New Cosmic Onion explicitly states that the charge of the proton is concentrated inside 3 distinct quarks, embedded in "electrically neutral material", thought to be the gluon field "glue". Merged post follows: Consecutive posts mergedIn D.Griffiths' Intro. to Elementary Particles (2nd ed.), quark Wave Functions are often modeled using standard, electronic, Hydrogen Atom Wave Functions -- the same also "recycled" for modeling Positronium. Thus, maybe quarks are "small" (10-18 m) in the same sense as electrons are always said to be "small" (10-18 m) (as seen here), even tho their WFs are much larger (10-18 m for quarks, 10-10 m for electrons).

The "(pseudo)scalar" (S=0) mesons, [math]\pi^0 = (u \bar{u} - d \bar{d})/ \sqrt{2}[/math] and [math]\eta = (u \bar{u} + d \bar{d} - 2 s \bar{s})/ \sqrt{6}, \eta^{'} = (u \bar{u} + d \bar{d} + s \bar{s})/ \sqrt{3}[/math] contain contributions from d,u,s type quarks, in a Spin anti-parallel configuration. Thus, with their Spins separated by 1 unit, the individual quarks comprising those mesons could (hypothetically) emit, absorb, & exchange W-Bosons. Is it possible, that the reason why those mesons are not "pure" [math]d \bar{d}, u \bar{u}, s \bar{s}[/math] states, but rather admixtures of them, is b/c the individual quarks comprising those mesons are (actually) undergoing Weak Force interactions, passing W-Bosons back & forth, and Flavour Oscillating ([math]d (W^{-} W^{+}) \bar{d} \to u (W^{+} W^{-}) \bar{u} \to s (W^{-} W^{+}) \bar{s} \to u (W^{+} W^{-}) \bar{u}[/math])... ???

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(Virtual) photons mediate all Electromagnetic phenomena, including the "Electrostatic Coulomb force". That's what confuses me -- how can two, static, point charges, each carrying opposite "signs" of electrical charge, attract each other, by exchanging (virtual) photons ?? Explaining how two point charges, each carrying the same "sign" of electrical charge, could repel each other, by exchanging (virtual) photons, is easy -- as Swansont pointed out, photons carry momentum, so that the two positive point charges, could "shoot" photons between themselves, and each time they "fired" a photon, they'd be "kicked backwards" (a little like a rocket), and each time they absorbed the other charge's photons, they'd be "kicked backwards" again. In such a way, two same-signed charges would repel each other. But, how can you use photons, to explain Electrostatic attraction ?

According to Fundamentals in nuclear physics: from nuclear structure to cosmology, by Jean-Louis Basdevant, James Rich, Michel Spiro (pg. 156), the Electric Charge distributions, of nucleons, are essentially spherically symmetric: Uploaded with ImageShack.us Where is the evidence of 3 quarks, in the aforegoing figures ?

Simple Model of Quarks inside Nucleons Neglects rapidly rising Strong Force Potential at "rim" of nucleon, as well as Spin (& Orbital ???) angular momentum. Electrostatic Binding Energy of neutron is predicted to be about -1/3 MeV more negative than that for proton. Uploaded with ImageShack.us Nucleon Clocks ??? Immediately after a Down quark decays into an Up, the supposed spatial arrangement of the quarks (dud --> duu) is "out of balance", compared to the stable state for protons (duu --> udu). Perhaps it takes some time for the "newborn" perturbed proton to "settle down" into said stable state ?? Could detailed scattering experiments observe this "settling down" process, which perhaps proceeds according to some sort of "decay half-life", as the "slightly excited" proton "leaks" away a little more energy (p* --> p) ??? Merged post follows: Consecutive posts mergedMagnetic effects 25x less Uploaded with ImageShack.us Merged post follows: Consecutive posts mergedHigh-energy Electrons, scattered off of nucleons, react according to how many Quarks they encounter (??) Frank Close. The New Cosmic Onion, pg. 88. Frank Close. The New Cosmic Onion, pg. 104. [math]\lambda_e = \frac{\hbar c}{\sqrt{E^2 - (m c^2)^2}} \approx \frac{\lambda_c}{\gamma}[/math] (from Klein-Gordon Equation) 1 GeV Electrons face a "phalanx" of Quarks, seeing a "shield wall" of Quarks "wading" or "swimming" thru Gluons. But, higher-energy Electrons can "hammer" on individual Quarks, isolated from their "phalanx formation". Lone Quarks are increasingly easy to "push around". Merged post follows: Consecutive posts mergedNucleons are "3 quarks swimming in slime [= glue]" David Griffiths. Introduction to Elementary Particles, pg. 72. Snapped "slime strings" bloop back into balls of slimy glue, imparting momentum (w/o carrying any momentum away) Merged post follows: Consecutive posts mergedHooke's Potential (1/2 k x2) approximation [math]1 GeV \approx \frac{1}{2} k r_0^2[/math] [math]k \approx 10^{20} N/m[/math] When a quark strays 1 fm from its fellows, the force from this formula is 105 N -- to wit, 10 tons of force, acting across 1 fm, on 1 quark's mass. This amounts to more than a billion billion trillion G's. Quarks are rather rugged. Merged post follows: Consecutive posts merged"Slime ball" model = "Bag Model" From Hyperphysics: The quarks of a proton are free to move within the proton volume If you try to pull one of the quarks out, the energy required is on the order of 1 GeV per fermi, like stretching an elastic bag. The energy required to produce a separation far exceeds the pair production energy of a quark-antiquark pair, so instead of pulling out an isolated quark, you produce mesons as the produced quark-antiquark pairs combine. http://hyperphysics.phy-astr.gsu.edu/hbase/particles/quark.html#c6 http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/scatele.html#c1 Merged post follows: Consecutive posts mergedPerhaps quarks in "glue" could be modeled as point masses in Visco-elastic fluid ?? Merged post follows: Consecutive posts mergedMagnetic moment effects are PHENOMENAL Assuming Quark masses, of 2.4 Mev (Up) & 4.8 Mev (Down), this simple model suggests that the Magnetic binding energies, of nucleons, are over 1 GeV, with the neutron being about 370 MeV heavier (less bound). A characteristic Magnetic Binding Energy: [math]U_0 \equiv \frac{\mu_0 \mu_q^2}{4 \pi r_0^3} \approx 390 MeV[/math] compares favorably, with the mass difference (~300 MeV), between protons (ground state: all 3 quarks' magnetic moments aligned), and the [math]\Delta^{+}[/math] (excited spin-state: all 3 quarks' spins aligned). Thus, although not particularly precise, this simple model strongly suggests, that magnetic moment effects can account for the mass differences between baryons. Further, from the "snapping slime-string" model mentioned above (PPs), that the [math]\Delta^{+}[/math] decays by emitting pions might mean, that gluon bonds break when "twisted a half-turn" (180 degrees), as the "middle" quark's spin flips back to anti-parallel (i.e., given its opposite charge, as its magnetic moment flips back to parallel). If so, "slime" might not be the best mental model of the "consistency" of gluon glue -- it might be more like Earth's semi-plastic Mantle magma. Finally, the best-fit models, for the Potential Wells confining quarks, predict forces of order 900 MeV fm-1, or about 16 tons of force, at 1 fm, on each quark (D.Griffiths. Intro. Elementary Particles (2nd ed.), pg. 173) ! And, the magnetic moment binding energies, suggested by this simple model, are also right around 1 GeV fm-1. Thus, perhaps a "Magnetic Well" (~r-3) might poignantly approximate the potential keeping quarks confined ?? Merged post follows: Consecutive posts mergedThe Neutral Rho Meson ([math]\rho^0[/math]) is an excited spin-state of the Neutral Pi-Meson / Pion ([math]\pi^0[/math]), where both quarks' mechanical Spins are parallel (making their Magnetic Moments anti-parallel, b/c of the opposite charge on the anti-quark). This "magnetic mis-alignment" makes the mass of the particle increase from ~135 MeV to 775 MeV, or by 640 MeV. This is of the same order of magnitude, as the ~300 MeV mass increase caused by the spin-excitation of a proton into a [math]\Delta^{+}[/math]. This would seem to suggest, that even this simple model, accurately captures "the gist" of the mighty Magnetic Moment interactions amongst quarks, in hadrons. If anything, the numbers might mean, that the quarks in mesons orbit a bit closer together, by an amount (given the r-3 dependency of the Magnetic Moment energy) of order (640 MeV / 300 MeV)1/3 [math]\approx[/math] 1.3. It does not seem implausible, that 2-quark mesons might be somewhat smaller than 3-quark baryons. And, if mesons were somewhat smaller spatially, that could account for their being allot less "glue" in mesons vs. baryons. Merged post follows: Consecutive posts merged Since, even for mesons, meff >> m, we find from that formula that [math]R \approx \frac{\hbar x}{c \, m_{eff}}[/math]. And, since, in mesons, meff is roughly 5x less, Rmeson is roughly 5x more -- to wit, ~7.5 fm. Note, too, that, since gluons carry Color Charge, and so can interact with other gluons, one could conceivably claim that baryons have 6x more "glue" (see following figure) than mesons -- which would account for the fact that nucleons have ~6x more "glue mass" than pions (~930 MeV : ~130 MeV). Uploaded with ImageShack.us Merged post follows: Consecutive posts mergedOpposite electric charges means that mechanical Spins parallel produce Magnetic Moments anti-parallel, et vice versa. Thus, for the ground states of nucleons (S = 1/2, n/p), that the oppositely charged quarks also Spin in opposite directions, means that all 3 quarks' Magnetic Moments are always co-aligned. Thus, this simple picture predicts, that the Magnetic Moments, of protons vs. neutrons, should stand in the ratio of: [math]\mu_p : \mu_n = \frac{+1 + \frac{1}{4} +1}{- \frac{1}{4} - 1 - \frac{1}{4} } = \frac{9/4}{-6/4} = -\frac{3}{2}[/math] in close agreement with the actual value of the ratio (-1.46). Moreover, the neutron magnetic moment = -1.91 [math]\mu_N[/math], where [math]\mu_N[/math] is the Nuclear Magneton, [math]\mu_N = \frac{e \hbar}{2 m_p} \approx 5.05 \; 10^{-27} J T^{-1}[/math] Thus, incorporating the constants from my model, w.h.t.: [math]-1.91 \mu_N = - \frac{6}{4} \mu_u[/math] so that: [math]\mu_u \equiv \frac{2/3 e \hbar}{2 m_u} = (2/3) (-1.91) \mu_N[/math] from which it becomes quite clear, that the numbers work out exactly, were one to set mu = mp / 1.91 = 490 MeV, which (for such a simple model) is surprisingly close to the actual mu,eff = 340 MeV. Of course, if one must use the effective mass, in all of the Magnetic Moment formulae, then magnetic effects are, indeed, "25x less" than the electrostatic effects, as originally claimed. But, again against that, is the fact that the excited spin-states of protons ([math]\Delta^{+}[/math]) and neutral pions ([math]\rho^0[/math]) are 300-600 MeV more massive, a "phenomenal" effect. Merged post follows: Consecutive posts mergedQuark effective masses reduce Magnetic Moments Based upon Wikipedia's List of Baryons & List of Mesons, I compiled the following figures, for (1) the Effective Masses, of the various flavors of quarks, when bound in Baryons & Mesons; and (2) the energy cost, for flipping the spins of quarks, in Baryons & Mesons. All such values are in MeV. Flavor Bare Mass Effective Mass (baryons) (mesons) Spin Flip Cost (baryons) (mesons) u 2 312 68 295 640 d 5 315 71 295 640 s 95 485 425 220 400 c 1300 1660 1800 65 140 b 4200 4990 5210 20 45 t 174000 Method: For example, u+d quarks, in nucleons, must mass about 1/3 of that of their nucleon (~940 MeV), implying an meff of ~310 MeV per quark. Likewise, u+d quarks, in pions, must mass about 1/2 of that of their pion (~140 MeV), implying an meff of ~70 MeV per quark. Then, effective masses for the further flavors were estimated, successively, by "subtracting off" those of the flavors estimated already. Spin-flip energy costs were calculated, in turn, by comparing excited spin states (S=3/2) with their ground-states (S=1/2). For instance, the [math]\Delta^{+}[/math] and [math]\Delta^0[/math] are the excited spin-states of the proton & neutron, each massing about 295 MeV more than their nucleon counterparts. Likewise, the [math]\rho^{+}[/math] and [math]\rho^0[/math] are the excited spin-states of the pions, each massing about 640 MeV more than their pion counterparts. Care, for the more massive quarks, was taken to try to select excited spin-states, where it was clear, which quark was flipped. Results: The product: [math]m_{eff} \times SpinFlipCost \approx constant[/math] for baryons (but not mesons): Flavor Meff * SFC (baryons) (mesons) u 92040 43520 d 92925 45440 s 106700 170000 c 107900 252000 b 99800 234450 t This makes perfect sense. For, the energy cost, of flipping a Magnetic Moment, in a magnetic field, is [math]\Delta U = \mu B[/math], and quarks' Magnetic Moments are reduced by their effective masses: [math]\mu_q = \frac{\hbar e_q}{2 m_{q,eff}}[/math] Merged post follows: Consecutive posts mergedApproximating Nucleons as uniformly Magnetized w/ uniform internal Magnetic Field From the formula for Magnetic Dipoles, w.h.t.: [math]B(0) = \frac{2 \mu_0}{3} m \delta^3(\vec{r})[/math] [math]m = k_{p,n} \mu_N[/math] [math]k_p = +2.82, k_n = -1.91[/math] [math]\delta^3(\vec{r}) \to \left( \frac{4 \pi r_0^3}{3} \right)^{-1}[/math] [math]r_0 \approx 1.4 \; fm[/math] [math]B(0) = k_{p,n} \frac{2 \mu_0}{4 \pi r_0^3} \mu_N \approx k_{p,n} \; 368 \; GT[/math] [math]\mu_q \equiv \kappa_{u,d} \frac{\hbar e/3}{2 m_p/3} = \kappa_{u,d} \mu_N[/math] [math]\kappa_u = 2, \kappa_d = -1[/math] [math]\Delta U = 2 \times \mu_q B(0) = 2 \kappa_{u,d} k_{p,n} \frac{2 \mu_0}{4 \pi r_0^3} \mu_N^2[/math] To flip the Spin of the d in a p+ ([math]\to \Delta^{+}[/math]) therefore requires: [math]2 \times 1 \times 2.82 \frac{2 \mu_0}{4 \pi r_0^3} \mu_N^2 \approx 65 KeV[/math] or 4 MeV were one to use the Bare Mass of the d in the determining its Magnetic Moment. And to flip the Spin of the u in a n0 ([math]\to \Delta^0[/math]) therefore requires: [math]2 \times 2 \times 1.91 \frac{2 \mu_0}{4 \pi r_0^3} \mu_N^2 \approx 88 KeV[/math] or 14 MeV were one to use the Bare Mass of the u in determining its Magnetic Moment. Merged post follows: Consecutive posts mergedThe Magnetic Moments, of Deuterium & Tritium, appear to be (basically) simple sums, of the Magnetic Moments of protons (+2.82 [math]\mu_N[/math]) and neutrons (-1.91 [math]\mu_N[/math]): particle Magnetic Moment (10[sup]-27[/sup] J/T) proton +14.106067 neutron -9.66236 deuteron +4.3307346 (= p+n) triton +15.046094 (= p+n-n) This seemingly says, that in Deuterium (S=1), the proton & neutron are Spin-aligned (so that their Magnetic Moments add); and, in Tritium (S=1/2), the two neutrons are Spin-anti-aligned (so that their Magnetic Moments cancel out). INTERPRETATION: The Magnetic Moments, of nuclear particles, can be constructed, as simple sums, of the Magnetic Moments of their 'sub-particle' constituents. APPLICATION: The Magnetic Moments, of protons & neutrons, can be constructed, as simple sums, of the Magnetic Moments of their up & down quarks: [math]\mu_p = + 2 \mu_u + \mu_d = +2.82 \mu_N[/math] [math]\mu_d = - \mu_u - 2 \mu_d = -1.91 \mu_N[/math] Note that, in the ground states (S=1/2), of protons & neutrons, the mechanical Spins, of the oppositely charged quarks, are oppositely aligned ([math]p (u \uparrow u \uparrow d \downarrow), n (u \uparrow d \downarrow d \downarrow)[/math]), so that their Magnetic Moments are actually all aligned. The solution, to this simple system of equations, is: [math]\left( \stackrel{2}{1} \stackrel{1}{2} \right) \left( \stackrel{\mu_u}{\mu_d} \right) = \mu_N \left( \stackrel{+2.82}{+1.91} \right)[/math] [math]\left( \stackrel{\mu_u}{\mu_d} \right) = \frac{\mu_N}{3} \left( \stackrel{2}{-1} \stackrel{-1}{2} \right) \approx \mu_N \left( \stackrel{5/4}{1/3} \right)[/math] Assuming that [math]\mu_q = \hbar e_q / 2 m_{q,eff} \times S[/math], with S=1/2, this strongly suggests, that the Effective Masses, of the up & down quarks, are roughly mp/4 & mp/2. Merged post follows: Consecutive posts mergedThe Magnetic Moments, of Deuterium & Tritium, appear to be (basically) simple sums, of the Magnetic Moments of protons (+2.82 [math]\mu_N[/math]) and neutrons (-1.91 [math]\mu_N[/math]): particle Magnetic Moment (10[sup]-27[/sup] J/T) proton +14.106067 neutron -9.66236 deuteron +4.3307346 (= p+n) triton +15.046094 (= p+n-n) This seemingly says, that in Deuterium (S=1), the proton & neutron are Spin-aligned (so that their Magnetic Moments add); and, in Tritium (S=1/2), the two neutrons are Spin-anti-aligned (so that their Magnetic Moments cancel out). INTERPRETATION: The Magnetic Moments, of nuclear particles, can be constructed, as simple sums, of the Magnetic Moments of their 'sub-particle' constituents. APPLICATION: The Magnetic Moments, of protons & neutrons, can be constructed, as simple sums, of the Magnetic Moments of their up & down quarks: [math]\mu_p = + 2 \mu_u + \mu_d = +2.82 \mu_N[/math] [math]\mu_d = - \mu_u - 2 \mu_d = -1.91 \mu_N[/math] Note that, in the ground states (S=1/2), of protons & neutrons, the mechanical Spins, of the oppositely charged quarks, are oppositely aligned ([math]p (u \uparrow u \uparrow d \downarrow), n (u \uparrow d \downarrow d \downarrow)[/math]), so that their Magnetic Moments are actually all aligned. The solution, to this simple system of equations, is: [math]\left( \stackrel{2}{1} \stackrel{1}{2} \right) \left( \stackrel{\mu_u}{\mu_d} \right) = \mu_N \left( \stackrel{+2.82}{+1.91} \right)[/math] [math]\left( \stackrel{\mu_u}{\mu_d} \right) = \frac{\mu_N}{3} \left( \stackrel{2}{-1} \stackrel{-1}{2} \right) \approx \mu_N \left( \stackrel{5/4}{1/3} \right)[/math] Assuming that [math]\mu_q = \hbar e_q / 2 m_{q,eff} \times S[/math], with S=1/2, this strongly suggests, that the Effective Masses, of the up & down quarks, are roughly mp/4 & mp/2.