Jump to content


Senior Members
  • Posts

  • Joined

  • Last visited


  • Atom

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

Widdekind's Achievements


Organism (8/13)



  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)
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.