Widdekind Posted June 14, 2010 Share Posted June 14, 2010 (edited) 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 The effect of colour combined with the Pauli Exclusion Principle is that any two quarks having the same flavour (two up , two down , two strange ) in the lowest energy state must spin parallel -- precisely the opposite of what happens in atoms & nuclei... If two flavours are identical, and the third differs ( e.g. , [math]\Delta^{++}[/math] ( uud ) or p (uud) ), then the identical pair must spin parallel, but the third is not constrained it can spin parallel (hence, total spin 3/2, as in the [math]\Delta[/math]) or antiparallel (hence, total spin 1/2, as in the proton). Frank Close. The New Cosmic Onion, pg. 95. 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]" It is analogous to quarks being held to one another by an elastic band that is slack. The quarks are apparently free, but after being struck, they recoil, and the elastic becomes tighter, preventing them from escaping. The elastic may become so stretched, that it snaps. The two new ends created have quarks on them, and so mesons are formed, but not free quarks... QCD predicts the following scenario, and experiment is consistent with it. Immediately after their production, the quark & antiquark move off in opposite directions. Initially, they feel no force, but as they separate, the energy in the force field between them grows, eventually becoming so great that its energy "E" exceeds the "mc2" needed to create further quarks & antiquarks. The initial quark & antiquark cluster together with these spontaneously created ones so quickly, forming mesons & baryons, that the original quark & antiquark are not detected; instead, two oppositely directed showers of hadrons emerge... QCD predicts that, at high energies, the quark & antiquark usually carry off most of the energy, with the glue collimated along their direction of motion, carrying little energy itself. In such circumstances, two distinct jets of particles emerge. Frank Close. The New Cosmic Onion, pp. 89-89,100-101. 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 mergedThe 'MIT Bag Model' [suggests that] free quarks of mass m, confined within a spherical shell of radius R, are found to have an effective mass [math]m_{eff} = \sqrt{m^2 + (\hbar x)/(R c)^2}[/math], where x is a dimensionless number around 2.5. Using the radius of the proton (say, 1.5 x 10-13 cm) for R, we obtain meff = 330 MeV / c2 for the up & down quarks. D.Griffiths. Intro. to Elementary Particles (2nd ed.), pg. 151. 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. Edited June 12, 2010 by Widdekind Consecutive posts merged. Link to comment Share on other sites More sharing options...
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