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Everything posted by Widdekind

  1. first, if i understand the Lagrangian formalism, then any potential, of any form or origin, would wind up incorporated into the time component, of the (generalized) derivative... e.g. a hypothetical exponential-well, for a mathematical model, of a strong-force potential... second, i perceive a potential problem, in properly interpreting the symbols, since [math]|| d^4x \sqrt{det(g)} ||[/math] = length4 so, seemingly, for a Schwarzschild metric, [math]det(g) = r^4 sin^2(\theta)[/math], so that [math]d^4x \sqrt{det(g)} = c \; dt \; dr \; d\theta \; d\phi \; r^2 \; sin(\theta)[/math] so, similarly, for the KGE, paying particular attention to units, the inverse metric is appropriate, for the "inverse differentials" of the derivatives -- the determinant is of the metric matrix, the derivatives multiply the inverse metric matrix (??) looks like the Laplacian is reconstituted from the formulas, with the exception of the Schwarzschild metric terms, in the time and radius double derivatives moreover, looks like a hypothetical strong-force exponential scalar potential, to try to model quarks in nucleons, would fold into the time derivative component, exactly as an EM scalar potential i want to ask about "minimal coupling" First, MC usually refers to the generalization, of the derivative operators, to include EM scalar & vector potentials, "imposed" upon the particle's wave-function, w/o any feedback from the particle to the field (apparently). So, seemingly, MC of particle <----> EM fields is distinguished, from MC of particle <----> space-time curvature, where the space-time curvature is "imposed", w/o any feedback from the particle to the SET <----> Ricci tensor. Second, starting from the Schrodinger equation, for the H atom, for simplicity's sake, and observing that the proton's implicitly-present wave-function is basically [math]\Psi_p^* \Psi_p \approx \delta^3(0)[/math], the generalization seems sound: [math]V( r ) \rightarrow \int d^3x \Psi_p^*(x) \Psi_p(x) V(r-x)[/math] Thus, turning probability amplitudes into probability (and mass, charge) densities, seemingly allows one to calculate the fields (and SE / curvatures) caused by particles. That approach would be an "integro-differential" equation approach -- particles' wave-functions (amplitudes) are evolved thru a differential time-step, knowing the fields & curvatures; and then the updated densities determine the updated fields & curvatures. Prima facie, such a two-step integro-differential approach seems sound, w/o any obvious reason why such would not work well
  2. http://en.wikipedia.org/wiki/Hamilton–Jacobi_equation The "Classical" HJ equation [math]H + \frac{\partial S}{\partial t} = 0[/math] where the action [math]S = \int L dt[/math] gives rise to [math]\Psi = \Psi_0 e^{i \frac{S}{\hbar}}[/math] so the phase of wave-functions, at some particular point, is equal to the time integral, of the action, at that point, from [math]t=-\infty[/math] to the current time ?
  3. [math] \left( E - q \Phi \right) \Psi_+ - c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_- = m c^2 \Psi_+[/math] [math] - \left( E - q \Phi \right) \Psi_- + c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_+ = m c^2 \Psi_-[/math] If, for each component, of the matter spinor, [math]\Psi_+ \rightarrow \Psi_+' e^{- \imath \frac{m c^2}{\hbar}t}[/math], then [math]\hat{E} \Psi_+ = e^{-\imath \frac{m c^2}{\hbar}t} \hat{E} \Psi_+' + m c^2 \Psi_+[/math]. So, seemingly, the Dirac equation is (nearly) reducible to that of a ("pseudo-")massless particle. Next, re-arranging terms, for the matter spinor, w/o any vector potential: [math]\hat{E} \Psi_+ = c \vec{p} \circ \vec{\sigma} \Psi_- + m c^2 \Psi_+ + q \Phi \Psi_+[/math] comparison to the equivalent Schrodinger equation (including the oft-omitted rest-mass-energy term): [math]\hat{E} \Psi_+ = \frac{1}{2m} \vec{p} \circ \vec{p} \Psi_+ + m c^2 \Psi_+ + q \Phi \Psi_+[/math] so seemingly suggests [math]\vec{p} \Psi_+ = 2 m c \vec{\sigma} \Psi_-[/math] [math]\begin{bmatrix} \hat{p}_x \Psi_u \\ \hat{p}_x \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} -\imath \Psi_{-,d} \\ \imath \Psi_{-,u} \end{bmatrix}[/math] [math]\begin{bmatrix} \hat{p}_y \Psi_u \\ \hat{p}_y \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} \Psi_{-,d} \\ \Psi_{-,u} \end{bmatrix}[/math] [math]\begin{bmatrix} \hat{p}_z \Psi_u \\ \hat{p}_z \Psi_d \end{bmatrix} = 2 m c \begin{bmatrix} \Psi_{-,u} \\ - \Psi_{-,d} \end{bmatrix}[/math] linking the gradient of the matter wave-functions (in the Schrodinger equation) to the anti-matter wave-functions (from the Dirac equation). So, seemingly, anti-matter is implicitly present, in the Schrodinger equation (??). Order of magnitude: [math]\hbar \vec{k} \approx 2 m c \Psi_-[/math] [math]\frac{k}{k_C} \approx \Psi_-[/math] Thus, the amount of anti-matter implicitly present, in the Schrodinger equation, scales as the ratio of the particle's propagation vector, to its Compton-k-vector (when the wavelength equals the Compton wavelength) -- where-ever the particle possesses momentum, there anti-particles are implicitly present, apparently (w/ some mixing between spin up vs. down components). Would that affect degenerate matter, within White Dwarves ?? Would the squished electron wave-functions begin generating copious quantities of anti-electrons ???
  4. http://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization intuitively, for one electron, to simultaneously absorb two (or more) photons, would require the simultaneous overlap, of all of the involved wave-functions (in a quantum equivalent, of a multi-body collision). If so, then that would inform you of the necessary photon density (assuming photon wave-functions are of order their wavelengths across). If so, then the multiple simultaneous overlap of several wave-functions seems somewhat similar, to the "Rydberg blockade" effect, crucial to the formation of "photonic molecules", when (seemingly) several atomic wave-functions are overlapping, so that excitation of one excludes excitation of overlapping nearby wave-functions (??) -------------------- a quick Google'ing for particle decay rates and "mass squared" seemingly suggests that decay rates are proportional to mass squared (not inversely so) -- do not more massive particles decay more quickly, e.g. top quark as an extreme example (??)
  5. http://en.wikipedia.org/wiki/Dirac_equation#Comparison_with_the_Pauli_theory [math] \left( E - q \Phi \right) \Psi_+ - c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_- = m c^2 \Psi_+[/math] [math] - \left( E - q \Phi \right) \Psi_- + c \vec{\sigma}\circ\left(\hat{p}-q \vec{A}\right) \Psi_+ = m c^2 \Psi_-[/math] If no anti-matter exists, i.e. [math]\Psi_- = 0[/math], then matter must exist in a "super-conducting state", to wit [math]\left( \vec{p} - q \vec{A} \right) = 0[/math]. So, oppositely, if matter does not propagate in such a "super-conducting BCS-similar state", then any residual [math]c\vec{\sigma}\circ\left(\hat{p}-q\vec{A}\right)\Psi_+ \ne 0[/math] will begin to generate anti-matter wave-functions [math]\Psi_- \ne 0[/math]. Can such solutions, seemingly representing matter-generated anti-matter wave-functions, be considered some sort of "friction", between the matter momentum [math]\vec{p}[/math], and the electro-magnetic momentum [math]q\vec{A}[/math] that the matter "should" have ? Could i make an analogy, to matter entrained in a flowing fluid, such that if the matter moves with some speed different from the bulk fluid flow, then "friction" and "vortices" and "eddy currents" and such are generated around it ? If so, then could some sort of "electro-magnetic drag", between moving matter, and the vector potential field in which said matter is immersed, conceivably create anti-matter ?? Note that the spin-axis [math]\left( \hat{z} \right)[/math] seems "special", since, if [math]D_{\mu} \equiv \hat{p}_{\mu} - q \vec{A}_{\mu}[/math] [math]\begin{bmatrix} c D_z & c \left( D_x - \imath D_y \right) \\ c \left( D_x + \imath D_y \right) & - c D_z \end{bmatrix} \Psi_+ = \begin{bmatrix} E + mc^2 - q \Phi & 0 \\ 0 & E + mc^2 - q \Phi \end{bmatrix} \Psi_-[/math] so that only the spin-axis derivative couples to the anti-matter (may the non-spin axes derivatives be non-zero?).
  6. Table of Fundamental Fermions charge (col) vs. Hyper-charge (row) [math]\bordermatrix{ ~ & -1 & -\frac{2}{3} & -\frac{1}{3} & 0 & +\frac{1}{3} & +\frac{2}{3} & +1 \cr +1 & ~ & \bar{u} & ~ & \bar{\nu} & \bar{d} & ~ & \bar{e} \cr -1 & e & ~ & d & \nu & ~ & u & ~ \cr}[/math] for Fermions spinning in [math]\langle +\hat{z} \rangle[/math], Fermions' charge-4-vectors are: [math]\bordermatrix{ ~ & \hat{x} & \hat{y} & \langle \hat{z} \rangle & \hat{w} \cr e & -1 & -1 & \langle -1 \rangle & -1 \cr d & 0 & 0 & \langle -1 \rangle & -1 \cr \nu & 0 & 0 & \langle 0 \rangle & -1 \cr u & +1 & +1 & \langle 0 \rangle & -1 \cr}[/math] et vice versa for anti-Fermions Selection Rules for Fundamental Fermions: charge of spin axis [math]\left( \hat{z} \right)[/math] cannot oppose hyper-charge of hyper-spin axis [math]\left( \hat{w} \right)[/math] charges of non-spin axes [math]\left( \hat{x} \hat{y} \right)[/math] cannot oppose charge of spin axis [math]\left( \hat{z} \right)[/math] charges of non-spin axes [math]\left( \hat{x} \hat{y} \right)[/math] cannot differ (i.e. are indistinguishable, i.e. only the spin axis is "special") charges of non-spin axes [math]\left( \hat{x} \hat{y} \right)[/math] must be less like hyper-charge of hyper-spin axis [math]\left( \hat{w} \right)[/math], than the charge of spin axis [math]\left( \hat{z} \right)[/math] For Fermions: [math]q_{x,y} \ge q_z \ge q_w[/math] et vice versa for anti-Fermions Higgs Boson could conceivably convert Fermions from-and-to anti-Fermions (??) Fermions have hyper-spin Sw = -1/2, and anti-Fermions have hyper-spin Sw = +1/2 Higgs Boson has spatial spin Sxyz = 0 Higgs Boson may have hyper-spatial spin Sw = 1 Higgs Boson may hyper-spin-flip Fermions from-and-to anti-Fermions ??
  7. http://en.wikipedia.org/wiki/Quantum_electrodynamics#Equations_of_motion When you take the variational derivative, of the QED Lagrangian, with respect to the wave function [math]\Psi[/math]... why doesn't the derivative include terms, due to the conjugate transpose of the wave function [math]\bar{\Psi}[/math] ? In Classical analogy, for a Lagrangian with the KE term [math]\left( \frac{1}{2m} \vec{p}^T \circ \vec{p} \right)[/math], derivatives with respect to momentum would include (one) terms, from the transpose of momentum, which is essentially the same mathematical object
  8. is the following derivation correct (ignoring for simplicity's sake a few factors of c): for a single charged particle, instantaneously located at [math]\vec{r}[/math], in the Classical limit, the fields from said particle, at point [math]\vec{r}'[/math]: [math]\Phi(r') \propto \frac{q}{r}[/math] [math]r = \sqrt{ \left( \vec{r}' - \vec{r} \right) \circ \left( \vec{r}' - \vec{r} \right) }[/math] [math]\vec{A}(r') = \vec{v} \Phi(r')[/math] the potential generate the force fields: [math]\vec{E} = -\nabla \Phi - \frac{\partial \vec{A}}{\partial t} = -\nabla \Phi - \vec{a} \Phi - \vec{v} \frac{\partial \Phi}{\partial r} \frac{\partial r}{\partial t}[/math] [math]\frac{\partial r}{\partial t} = \frac{1}{2 r} \left( 2 \left( x' - x \right) \left( - \frac{\partial x}{\partial t} \right) + \cdots \right) = - \hat{r} \circ \vec{v}[/math] where [math]\hat{r}[/math] points away from the force-field generating particle, towards the other point. But [math]\hat{r} \frac{\partial \Phi}{\partial r} = \nabla \Phi[/math]. so [math]\vec{E}=-\nabla\Phi-\vec{a}\Phi+\vec{v}\left(\vec{v}\circ\nabla\Phi\right)[/math] [math]\vec{B} = \nabla \times \vec{A} = - \vec{v} \times \nabla \Phi[/math] [math]\vec{v}' \times \vec{B} = \nabla \Phi \left( \vec{v}' \circ \vec{v} \right) - \vec{v} \left( \vec{v}' \circ \nabla \Phi \right)[/math] the force felt by a test charge [math]q'[/math] at [math]\vec{r}'[/math]: [math]\vec{F}' = q' \left( \vec{E} + \vec{v}' \times \vec{B}\right)[/math] [math]\frac{\vec{F}'}{q'} = -\nabla \Phi \left( 1 - \left( \vec{v}' \circ \vec{v} \right) \right) - \vec{a} \Phi - \vec{v} \left( \left( \vec{v}' - \vec{v} \right) \circ \nabla \Phi \right)[/math] i think the above is equivalent to http://en.wikipedia.org/wiki/Lorentz_force#Lorentz_force_in_terms_of_potentials
  9. working backwards, from the Klein-Gordon equation... is the generalized mass-energy relation, for curved spacetime, full of EM fields: [math]\left(P - e A\right)^{\mu} g_{\mu \nu} \left( P - e A \right)^{\nu} = m^2[/math]
  10. minimal coupling would seem appropriate, for single particle wave-function solutions, given the "low energy" limit ? if the KG generalizes, in the presence of EM fields, to [math]D^{\mu} \rightarrow D^{\mu} - e A^{\mu}[/math]... then would any energy potential, be treated, similarly, inserted into the time component of the generalized differential operator ?
  11. Q1: trying to denote spin, w/ subscripts, what prevents the "trifurcation" of photons: [math]\gamma_{+\hbar} \longrightarrow \gamma_{+\hbar} + \gamma_{-\hbar} + \gamma_{+\hbar}[/math] The above hypothetical photon decay could conserve all quantum numbers, as well as spin, and energy + momentum. So, what prevents (presumably?) such processes ? Q2: If photons are spin=1... then why don't electrons, interacting w/ protons in atoms, constantly spin flip, each time they emit or absorb virtual photons ? Q2': If virtual bosons can exist "off mass shell", w/ non-Einstein-equation-compliant combinations of momentum + energy... then can they also exist w/o the canonically required spin ? Q2'': If actual photons carry oscillating EM fields... then do virtual photons, mediating and generating the scalar potential of particles, carry scalar potential ? Are they "scalar potential" photons, representing quantized amounts of scalar potential (as opposed to actual photons, which carry quantized amounts of vector potential, whose plane of oscillation is their plane of polarization, etc.) ? Q3: If gluons are spin = 1... then how could gluons "bifurcate" (as depicted in some Feynman Diagrams), [math]g_{+\hbar} \longrightarrow g_{+\hbar} + g_{\pm\hbar}[/math] w/o violating spin conservation ? Q3': Or, do virtual gluons not need to conserve spin ? Q3'': In 3-jet decays, in particle colliders, one of the jets derives from a spin-1 gluon... is that gluon an "actual" and "promoted" gluon (to quote Gary Zukov's Dancing Wu Li Masters), as opposed to "virtual" gluons mediating the strong / color force w/in nucleons ?
  12. 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
  13. what about the generalized Klein-Gordon approach [math]D^{\mu}D_{\mu} \Psi = m^2 \Psi[/math] [math]\longrightarrow[/math] [math]D^{\mu} \eta_{\mu \nu} D^{\nu} \Psi = m^2 \Psi[/math] (inserting Minkowski metric matrix) [math]\longrightarrow[/math] [math]D^{\mu} g_{\mu \nu} D^{\nu} \Psi = m^2 \Psi[/math] (swapping out Minkowski, inserting GR metric matrix) ? The statements of AJB seem completely correct, only somewhat understated... at an age of ~1 TPL, energy densities w/in the universe were ~1 EPL, so since 1 planck time, the whole history of the complete cosmos has been "low energy"
  14. you would want to write, for right-handed anti-neutrinos [math]\bar{\nu}_L[/math] ?
  15. perhaps important, is the fact that the eigenstates of the Weak interaction, are not the eigenstates of mass (or other Force interactions?). So, quarks emitted from gluons, would not be the same quark-states, emitted from W bosons (which would be "mixed" states, according to the appropriate mixing matrices, e.g. CKM)
  16. if gluons are spin-1 bosons... then how could one gluon, "bifurcate" into two ? would not needing to conserve spin, require "trifurcation" into three gluons ? If gluons do generate other gluons, why would one gluon be able to "duplicate" into two... wouldn't they always "triplicate" into three ?
  17. "Charge Vector" model explains 3-Jet events in Lepton collisions Nicholas Mee, Higgs Force: Recall the charge-4-vectors q4 = ( qx qy qz | qw) of electrons & anti-electrons: [math]\tilde{q}_{e^-} = \begin{bmatrix} -1 & -1 & -1 & | -1 \end{bmatrix}[/math] [math]\tilde{q}_{e^+} = \begin{bmatrix} +1 & +1 & +1 & | +1 \end{bmatrix}[/math] In high-energy head-on collisions, the electrons & anti-electrons could conceivably "break up", a little like a head-on collision, between fighter-jets, at an airshow... the electron is (presumably) spin-backwards (i.e. left-handed) and the anti-electron is (presumably) spin-forwards (i.e. right-handed)... without loss of generality, the collision occurs in the [math]\hat{x}[/math] direction... then, the internal charge components of the colliding pair of particles, could conceivably combine in directional-pairs, producing a pair of partial-charged quarks, and a gluon (along the collision axis): [math]\longrightarrow[/math] [math]\begin{bmatrix} \pm & 0 & 0 & | 0 \end{bmatrix} \times \hat{S}_{\hat{x}}[/math] [math]\begin{bmatrix} 0 & \pm & 0 & | 0 \end{bmatrix}[/math] [math]\begin{bmatrix} 0 & 0 & \pm & | 0 \end{bmatrix}[/math] [math]=[/math] [math]\tilde{g}_{xx}[/math] (spin 1) [math]\left( d \bar{d} \right) = \pi^0[/math] (spin 0, in [math]y \bar{y} \leftrightarrow[/math] "green-antigreen") [math]\left( d \bar{d} \right) = \pi^0[/math] (spin 0, in [math]z \bar{z} \leftrightarrow[/math] "blue-antiblue") Conversely, two-jet collisions could conceivably result, from fragmentation, into a pair of particles, e.g.: [math]\longrightarrow[/math] [math]\begin{bmatrix} \pm & \pm & 0 & | -1 \end{bmatrix}[/math] [math]\begin{bmatrix} 0 & 0 & \pm & | +1 \end{bmatrix}[/math] [math]=[/math] [math]\left( u \bar{u} \right) = \pi^0[/math] (spin 0, in [math]z \bar{z} \leftrightarrow[/math] "blue-antiblue") [math]\left( d \bar{d} \right) = \pi^0[/math] (spin 0, in [math]z \bar{z} \leftrightarrow[/math] "blue-antiblue") internal structure of Leptons could unify EW <---> S forces The Classical Electron Radius ~1fm, essentially the same size as normal nucleons. And, in net, the charge-4-vector of a proton q4 = ( +1 +1 +1 | -3) is (spatially-speaking) the same as that of an (anti-)electron, q4 = ( +1 +1 +1 | +1). So, if the Strong force binds the separate charged sub-components (quarks) of nucleons together; then perhaps the Strong force operates inside electrons, too, binding their three separate standard-spatial charge-unit-vectors together... normally, the Strong force-carrier gluons are hidden inside the whole electron, like fuel is hidden inside fighter-jets... only in head-on collisions, does all the internal jet-fuel spray far and wide, becoming apparent to outside observers. In some semi-Classical sense, electrons and protons are the same fempto-meter size; and both are held together internally, by Strong-force carrying gluons... which glue is allot less stressed, inside a single electron, as compared to the combination of quarks comprising protons (note the difference in "w" directed hyper-charge). spin of quarks could account for Strong-force field "gluon gob" "Bare" quarks have been observed, e.g. ephemeral top quarks, which decay before generating gluons. So, the exact instant that a quark is created in a collider, the quark is created "bare", "born" without any surrounding Strong-force field... indeed, gluons could only then begin to emanate away, at (up to) the speed-of-light... so that some time would be required, for the quark's gluon field to "inflate" around the particle... Now, quarks have (spatial) spin, which, due to quantum uncertainty, always has some non-zero probability, of pointing in directions orthogonal (note -- not anti-parallel, though) to the spin state, because the overlaps < Sx | Sy,z > are non-zero... and, gluons generate rotations, of quarks' spin axes, e.g. [math]d_{red} = d_x = \begin{bmatrix} -1 & 0 & 0 & | -1 \end{bmatrix}[/math] [math]\longrightarrow \begin{bmatrix} 0 & -1 & 0 & | -1 \end{bmatrix} + \begin{bmatrix} -1 & +1 & 0 & | 0 \end{bmatrix}[/math] [math] = d_y + \tilde{g}_{x\bar{y}}[/math] [math] = \left( d_{green} \right) + \tilde{g}_{red-antigreen}[/math] So, since the spin of quarks is constantly rotating them around; and if quark rotation requires gluon emissions; then the spin of quarks would immediately begin emanating a surrounding "cloud" of gluons, as the quark spun itself, into a superposition of spin states. "bare" electrons briefly resemble neutrinos ? If quarks are "born" "bare", being low-mass (few MeV), lacking a surrounding gluon field... which field, when generated, dramatically increases the mass of the quark (70-350 MeV) in mesons & baryons... then perhaps electrons are "born bare", being low-mass (few eV), lacking a surrounding photon field... which field, when emanated, increases the mass of the electron (half MeV) ? all energy / momentum is EM ? In the presence of (external) EM fields, the generalized Einstein relation is: [math]\left( E - q \Phi \right)^2 - \left( c \vec{p} - q \vec{A} \right)^2 = \left( m c^2\right)^2[/math] Thus, at rest, an electron (say) has an energy equal to [math]E = mc^2 + q \Phi[/math] where q = -e for an electron. So, external scalar potential (Voltage) fields affect the mass-energy of electrons... such suggests, that the rest-mass-energy of electrons could come, from the internal scalar potential field of the electron: [math]m c^2 \equiv q_e \Phi_e \approx \frac{e^2}{4 \pi \epsilon_0 R_e}[/math] assuming that the electron is a spherical ball of uniform charge density. The above equation is that, of the Classical Electron Radius ~1fm. So, if electrons where small balls, about the same size as normal nucleons... then the Voltage their own charge distribution generated, withinside themselves, would give their charge, an electron-rest-mass worth of mass-energy. Such suggests, that the effective rest-mass-energy of (say) electrons, is the simple sum: [math]E = q \left( \Phi_{internal} + \Phi_{external} \right)[/math] Similar logic leads to (in the low-energy limit): [math]c^2 \vec{p} = q \vec{v} \left( \Phi_{internal} + \Phi_{external} \right)[/math] since [math]c^2 \vec{A} \equiv \vec{v} \Phi[/math] for particles. If some force "glued" electrons tightly together, into small balls of nucleon-sized charge... then all energy-momentum, of charged particles, could be construed, and attributed, to external/internal EM forces / interactions (and self-interactions). Indeed, in the absence of external fields, the internal fields of (say) electrons would appear to increase, by the appropriate gamma factor, correctly accounting for the equal increase in the apparent mass, of moving electrons: [math]\Phi_{int} \rightarrow \gamma \Phi_{int}[/math] [math]E \rightarrow \gamma E[/math]
  18. so, that site states, that W-bosons are only involved, in Fermion generation-changing decays... if some Fermion (Hadron or Lepton) changes generation, then-and-only-then was the Weak-force, with its WF bosons, involved... tangentially-related questions of a similar sort: 1) can photons generate quarks ?? 2) can gluons generate leptons ?? http://hyperphysics.phy-astr.gsu.edu/hbase/particles/allfor.html [math]D^+ \longrightarrow K^- + \pi^+ + \pi^+[/math] the c-quark must decay into an s-quark (emitting a positive W which decays into one of the positive pions)... so accounting for the resulting negative Kaon (and the above positive pion)... but the other positive pion cannot conceivably come from another positive W, since no quark exists to have emitted the same... so hypothetical "double W" emissions seem incompatible, with some known reactions; and unnecessary to assume, for any reaction (if gluons can be attributed the other new particles produced) http://en.wikipedia.org/wiki/Gluon http://en.wikipedia.org/wiki/File:Feynman_Diagram_Y-3g.PNG [math]e^+ + e^- \longrightarrow \tilde{g}\tilde{g}\tilde{g}[/math] can occur... ipso facto, the reverse process is presumably possible, i.e. triple gluon fusion into charged lepton antilepton particle pairs
  19. 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. "
  20. 2-quark mesons mass, order-of-magnitude, 100MeV = 102 3-quark baryons mass, order-of-magnitude, 1000MeV = 103 Q1: does that not suggest, that the interaction energy, of the Strong-force, scales as ~(10 MeV)Nq ? Q2: what is the implication ? Q2A: if the Strong interaction, between three quarks, was the sum of three independent pairwise interactions... then the wouldn't the energies scale as ~3 x (10 MeV)2 ? Q2B: if the Strong interaction, between three quarks, was the product, of three dependent pairwise interactions... where the strength of the interaction depended on how much "red" was interacting with "green", and simultaneously how much "green" was interacting with "blue", and simultaneously how much "red" was interacting with "blue"... then the wouldn't the energies scale as ~((10 MeV)2)3 ? Q2C: so, does not a scaling relation of ~ (10 MeV)3 imply, that the Strong interaction is a three-quark interaction, where each quark is interacting with some "center" of their nucleon... so that the strength of the interaction is how much "red" is interacting with the "center", times how much "green" is interacting w/ said "center", times how much "blue" is interacting w/ said "center" (because all three quarks must be simultaneously involved, in a single co-interaction, with some single "center") ? an interaction strength, proportional to the product of the three quarks' individual strong-force color charges, ~(10 MeV)3, suggests, not that each quark is interacting w/ each other (which would be a sum, or product, of pairwise interactions), but that all three are interacting, as if "to" some single "center", of their nucleon ? -------------------- if a nucleon is composed of quarks, and gluons... then is there some crude resemblance, to ions and photons, in a plasma ? i.e. nucleons ~= quark-gluon plasma ?
  21. electrons + antielectrons <----> Z0 evidently couple together, during decays of Higgs bosons (according to the book by Jim Baggott, Higgs)
  22. do you not mean "sign of the frequency" (not "energy") ? anti-Fermions' phase frequencies are negative (not necessarily their energies). anti-Fermions also have opposite electric (and weak) charge... is there some sort of close connection, between charge <----> energy / frequency ?
  23. sometimes neutral Kaons decay into three pions... does that imply, that the strange quark in the neutral Kaon, can occasionally decay, by emitting, not one, but two W bosons (which each then decay into separate pions) ? that is, [math]K^0 \rightarrow \pi^0 + W + W[/math] [math] \rightarrow \pi^0 + \pi + \pi[/math] ? (All meson decays seem to proceed through the Weak force, e.g. link) http://en.wikipedia.org/wiki/Kaon
  24. is not the statement stronger still ? For, all energy densities are "low energy density" (<<Planck)... you could compress quanta, squishing them in size until their de-Broglie wavelengths were on Planck length long, and their energy density would only then begin to become "bothersome" to the fabric of space-time i perceive that the rubber-sheet analogy may mislead minds, into underestimating the enormous structural strength, of spacetime, as some sort of substance (w/ an ontological existence independent of mass-energy-density)... crudely, you could construct a bridge a million km long from space-time, and load it w/ a starmass of material, and see it sag in the center solely 3km
  25. 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
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