Widdekind
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"hierarchy problem" of particle physics has saved spacetime ?
Widdekind replied to Widdekind's topic in Relativity
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 exponentialwell, for a mathematical model, of a strongforce 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 strongforce 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 wavefunction, w/o any feedback from the particle to the field (apparently). So, seemingly, MC of particle <> EM fields is distinguished, from MC of particle <> spacetime curvature, where the spacetime 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 implicitlypresent wavefunction 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(rx)[/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 "integrodifferential" equation approach  particles' wavefunctions (amplitudes) are evolved thru a differential timestep, knowing the fields & curvatures; and then the updated densities determine the updated fields & curvatures. Prima facie, such a twostep integrodifferential approach seems sound, w/o any obvious reason why such would not work well 
http://en.wikipedia.org/wiki/Hamiltonâ€“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 wavefunctions, 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 ?

[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, rearranging 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 oftomitted restmassenergy 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 wavefunctions (in the Schrodinger equation) to the antimatter wavefunctions (from the Dirac equation). So, seemingly, antimatter 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 antimatter implicitly present, in the Schrodinger equation, scales as the ratio of the particle's propagation vector, to its Comptonkvector (when the wavelength equals the Compton wavelength)  whereever the particle possesses momentum, there antiparticles 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 wavefunctions begin generating copious quantities of antielectrons ???

trying to understand Photons & Gluons (both "actual" & "virtual")
Widdekind replied to Widdekind's topic in Quantum Theory
http://en.wikipedia.org/wiki/Resonanceenhanced_multiphoton_ionization intuitively, for one electron, to simultaneously absorb two (or more) photons, would require the simultaneous overlap, of all of the involved wavefunctions (in a quantum equivalent, of a multibody collision). If so, then that would inform you of the necessary photon density (assuming photon wavefunctions are of order their wavelengths across). If so, then the multiple simultaneous overlap of several wavefunctions seems somewhat similar, to the "Rydberg blockade" effect, crucial to the formation of "photonic molecules", when (seemingly) several atomic wavefunctions are overlapping, so that excitation of one excludes excitation of overlapping nearby wavefunctions (??)  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 (??) 
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 antimatter exists, i.e. [math]\Psi_ = 0[/math], then matter must exist in a "superconducting state", to wit [math]\left( \vec{p}  q \vec{A} \right) = 0[/math]. So, oppositely, if matter does not propagate in such a "superconducting BCSsimilar state", then any residual [math]c\vec{\sigma}\circ\left(\hat{p}q\vec{A}\right)\Psi_+ \ne 0[/math] will begin to generate antimatter wavefunctions [math]\Psi_ \ne 0[/math]. Can such solutions, seemingly representing mattergenerated antimatter wavefunctions, be considered some sort of "friction", between the matter momentum [math]\vec{p}[/math], and the electromagnetic 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 "electromagnetic drag", between moving matter, and the vector potential field in which said matter is immersed, conceivably create antimatter ?? Note that the spinaxis [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 spinaxis derivative couples to the antimatter (may the nonspin axes derivatives be nonzero?).

Fermions are spin1, half of spin points "hyperspatially" ?
Widdekind replied to Widdekind's topic in Speculations
Table of Fundamental Fermions charge (col) vs. Hypercharge (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' charge4vectors 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 antiFermions Selection Rules for Fundamental Fermions: charge of spin axis [math]\left( \hat{z} \right)[/math] cannot oppose hypercharge of hyperspin axis [math]\left( \hat{w} \right)[/math] charges of nonspin axes [math]\left( \hat{x} \hat{y} \right)[/math] cannot oppose charge of spin axis [math]\left( \hat{z} \right)[/math] charges of nonspin 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 nonspin axes [math]\left( \hat{x} \hat{y} \right)[/math] must be less like hypercharge of hyperspin 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 antiFermions Higgs Boson could conceivably convert Fermions fromandto antiFermions (??) Fermions have hyperspin Sw = 1/2, and antiFermions have hyperspin Sw = +1/2 Higgs Boson has spatial spin Sxyz = 0 Higgs Boson may have hyperspatial spin Sw = 1 Higgs Boson may hyperspinflip Fermions fromandto antiFermions ?? 
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

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 forcefield 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

"hierarchy problem" of particle physics has saved spacetime ?
Widdekind replied to Widdekind's topic in Relativity
minimal coupling would seem appropriate, for single particle wavefunction 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 ? 
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/ nonEinsteinequationcompliant 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 3jet decays, in particle colliders, one of the jets derives from a spin1 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 ?

skipping subscripts, then, the hypercharge of normal antineutrinos is +1 (the opposite of normal neutrinos)... and normal pions are spinless unnethypercharged particles (+0). So, if backwards electrons still carried the same hypercharge, 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 hypercharge 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 misaccountings and errors would be best to be corrected quickly

"hierarchy problem" of particle physics has saved spacetime ?
Widdekind replied to Widdekind's topic in Relativity
what about the generalized KleinGordon 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" 
you would want to write, for righthanded antineutrinos [math]\bar{\nu}_L[/math] ?

can neutral Kaons decay by doubleW boson emission ?
Widdekind replied to Widdekind's topic in Modern and Theoretical Physics
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 quarkstates, emitted from W bosons (which would be "mixed" states, according to the appropriate mixing matrices, e.g. CKM) 
question about Strongforce potential due to gluons ?
Widdekind replied to Widdekind's topic in Speculations
if gluons are spin1 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 ? 
Fermions are spin1, half of spin points "hyperspatially" ?
Widdekind replied to Widdekind's topic in Speculations
"Charge Vector" model explains 3Jet events in Lepton collisions Nicholas Mee, Higgs Force: Recall the charge4vectors q4 = ( qx qy qz  qw) of electrons & antielectrons: [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 highenergy headon collisions, the electrons & antielectrons could conceivably "break up", a little like a headon collision, between fighterjets, at an airshow... the electron is (presumably) spinbackwards (i.e. lefthanded) and the antielectron is (presumably) spinforwards (i.e. righthanded)... 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 directionalpairs, producing a pair of partialcharged 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] "greenantigreen") [math]\left( d \bar{d} \right) = \pi^0[/math] (spin 0, in [math]z \bar{z} \leftrightarrow[/math] "blueantiblue") Conversely, twojet 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] "blueantiblue") [math]\left( d \bar{d} \right) = \pi^0[/math] (spin 0, in [math]z \bar{z} \leftrightarrow[/math] "blueantiblue") 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 charge4vector of a proton q4 = ( +1 +1 +1  3) is (spatiallyspeaking) the same as that of an (anti)electron, q4 = ( +1 +1 +1  +1). So, if the Strong force binds the separate charged subcomponents (quarks) of nucleons together; then perhaps the Strong force operates inside electrons, too, binding their three separate standardspatial chargeunitvectors together... normally, the Strong forcecarrier gluons are hidden inside the whole electron, like fuel is hidden inside fighterjets... only in headon collisions, does all the internal jetfuel spray far and wide, becoming apparent to outside observers. In some semiClassical sense, electrons and protons are the same femptometer size; and both are held together internally, by Strongforce 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 hypercharge). spin of quarks could account for Strongforce 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 Strongforce field... indeed, gluons could only then begin to emanate away, at (up to) the speedoflight... 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 nonzero probability, of pointing in directions orthogonal (note  not antiparallel, though) to the spin state, because the overlaps < Sx  Sy,z > are nonzero... 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}_{redantigreen}[/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 lowmass (few MeV), lacking a surrounding gluon field... which field, when generated, dramatically increases the mass of the quark (70350 MeV) in mesons & baryons... then perhaps electrons are "born bare", being lowmass (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 massenergy of electrons... such suggests, that the restmassenergy 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 electronrestmass worth of massenergy. Such suggests, that the effective restmassenergy of (say) electrons, is the simple sum: [math]E = q \left( \Phi_{internal} + \Phi_{external} \right)[/math] Similar logic leads to (in the lowenergy 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 nucleonsized charge... then all energymomentum, of charged particles, could be construed, and attributed, to external/internal EM forces / interactions (and selfinteractions). 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] 
can neutral Kaons decay by doubleW boson emission ?
Widdekind replied to Widdekind's topic in Modern and Theoretical Physics
so, that site states, that Wbosons are only involved, in Fermion generationchanging decays... if some Fermion (Hadron or Lepton) changes generation, thenandonlythen was the Weakforce, with its WF bosons, involved... tangentiallyrelated questions of a similar sort: 1) can photons generate quarks ?? 2) can gluons generate leptons ?? http://hyperphysics.phyastr.gsu.edu/hbase/particles/allfor.html [math]D^+ \longrightarrow K^ + \pi^+ + \pi^+[/math] the cquark must decay into an squark (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_Y3g.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 
Wikipedia's page provides, "The antineutrinos observed so far all have righthanded helicity (i.e. only one of the two possible spin states has ever been seen), while the neutrinos are lefthanded. "

question about Strongforce potential due to gluons ?
Widdekind replied to Widdekind's topic in Speculations
2quark mesons mass, orderofmagnitude, 100MeV = 102 3quark baryons mass, orderofmagnitude, 1000MeV = 103 Q1: does that not suggest, that the interaction energy, of the Strongforce, 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 threequark 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 cointeraction, with some single "center") ? an interaction strength, proportional to the product of the three quarks' individual strongforce 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 ~= quarkgluon plasma ? 
electrons + antielectrons <> Z0 evidently couple together, during decays of Higgs bosons (according to the book by Jim Baggott, Higgs)

how does KleinGordon equation admit "spontaneous transitions" ?
Widdekind replied to Widdekind's topic in Quantum Theory
do you not mean "sign of the frequency" (not "energy") ? antiFermions' phase frequencies are negative (not necessarily their energies). antiFermions also have opposite electric (and weak) charge... is there some sort of close connection, between charge <> energy / frequency ? 
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

"hierarchy problem" of particle physics has saved spacetime ?
Widdekind replied to Widdekind's topic in Relativity
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 deBroglie wavelengths were on Planck length long, and their energy density would only then begin to become "bothersome" to the fabric of spacetime i perceive that the rubbersheet analogy may mislead minds, into underestimating the enormous structural strength, of spacetime, as some sort of substance (w/ an ontological existence independent of massenergydensity)... crudely, you could construct a bridge a million km long from spacetime, and load it w/ a starmass of material, and see it sag in the center solely 3km 
all antiFermions are righthanded, yes ? and all antiFermions have opposite charge, and also opposite hypercharge, to Fermions, yes ? the hypercharge of (LH) Leptons (electrons & neutrinos) = 1 so the hypercharge of (RH) antiLeptons = +1 i thought that the existence of the antineutrino 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 "headon car crash", wherein the "pieces and parts" of the two colliding "cars" became confused and reconfigured, into new Fermions, which would be the surviving "Fermionic chassis" of the original two particles, but with their "charge loadouts" swapped toandfro forth and back between them vaguely like two trucks collide, and all the loads they were carrying get transferred around, before the newlyreloadedout Fermions emerge from the "crash scene" as reconfigured 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 loadouts" 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