computer Posted November 1, 2022 Report Share Posted November 1, 2022 (edited) Field objects moving at the speed of light Let us consider what field "blobs" can be, moving at the speed of light in a certain direction while maintaining their shape. That is, compact formations capable of traveling long distances compared to their size without significant changes in structure. Unlike dipole radiation, which propagates spherically in all directions. Perhaps such structure have emissions of atoms during the transitions of electron clouds to less energetic levels. Discussion of how justified use of the term "photon" in relation to such objects is beyond the scope of this article. Let us take as basis the equations, existence of which in the real world is justified in the topic on dipole radiation: The following symbols are used: Scalar potential = a Vector potential = A Electrical field = E Speed of light in vacuum = c Time derivatives are denoted by singlequote ' a' = - c2 · div A A' = - E - grad a E' = c2 · rot rot A The formulas are given in cylindrical coordinate system (ρ,φ,z), associated with the point of space where the geometric center of field blob is located at the time of observation. Let us put r2 = ρ2 + z2 Motion occurs along z-axis at the speed of light and structure of field object remains unchanged, that is, ∂/∂t = - c · ∂/∂z for all physical quantities. Also, integral of internal energy throughout all the space must be finite, density of which is expressed by the law: u = ε0/2 · E2 + μ0/2 · H2 where E2 = Eρ2 + Eφ2 + Ez2, H2 = Hρ2 + Hφ2 + Hz2 H = 1/μ0 · rot A, B = rot A = μ0 · H Let us put J = rot B = rot rot A Let us start with the mathematically simplest descriptions possible from the point of view of field laws mentioned above. In cylindrically symmetric case, when ∂/∂φ = 0 for all physical quantities. Basic equations are divided into two independent systems: 1. With circular electric field. Aφ' = - c · ∂Aφ/∂z = - Eφ → Eφ = c · ∂Aφ/∂z → ∂Eφ/∂z = c · ∂2Aφ/∂z2 Eφ' = - c · ∂Eφ/∂z = c2 · Jφ = c2 · (- ∂2Aφ/∂z2 - ∂2Aφ/∂ρ2 - ∂Aφ/∂ρ / ρ + Aφ / ρ2) → ∂Eφ/∂z = c · (∂2Aφ/∂z2 + ∂2Aφ/∂ρ2 + ∂Aφ/∂ρ / ρ - Aφ / ρ2) Equating ∂Eφ/∂z from two equations, we get ∂2Aφ/∂ρ2 + ∂Aφ/∂ρ / ρ - Aφ / ρ2 = 0 → ∂/∂ρ (∂Aφ/∂ρ + Aφ / ρ) = 0 If Aφ is not zero in all the space, so ∂Aφ/∂ρ + Aφ / ρ = 0, and Aφ is proportional to 1 / ρ, that gives infinite energy integral. Hence, such non-zero components of compact radiations can not exist. After artificial creation or computer modeling such structures will diverge in waves in all directions, instead of moving in one direction at the speed of light. 2. With circular magnetic field. a' = - c · ∂a/∂z = - c2 · (∂Aρ/∂ρ + Aρ / ρ + ∂Az/∂z) → ∂a/∂z = c · (∂Aρ/∂ρ + Aρ / ρ + ∂Az/∂z) Aρ' = - c · ∂Aρ/∂z = - Eρ - ∂a/∂ρ → Eρ = c · ∂Aρ/∂z - ∂a/∂ρ ∂Eρ/∂z = c · ∂2Aρ/∂z2 - ∂2a/∂ρ/∂z Az' = - c · ∂Az/∂z = - Ez - ∂a/∂z → Ez = c · ∂Az/∂z - ∂a/∂z ∂Ez/∂z = c · ∂2Az/∂z2 - ∂2a/∂z2 Eρ' = - c · ∂Eρ/∂z = c2 · Jρ → ∂Eρ/∂z = c · (∂2Aρ/∂z2 - ∂2Az/∂ρ/∂z) Ez' = - c · ∂Ez/∂z = c2 · Jz → ∂Ez/∂z = c · (∂2Az/∂ρ2 - ∂2Aρ/∂ρ/∂z - ∂Aρ/∂z / ρ + ∂Az/∂ρ / ρ) Equating ∂Eρ/∂z from the equations for Aρ' и Eρ', we get c · ∂2Aρ/∂z2 - ∂2a/∂ρ/∂z = c · (∂2Aρ/∂z2 - ∂2Az/∂ρ/∂z) and conclude that a = c · Az if we are talking about quantities decreasing to zero with distance from the center goes to infinity. From the equation for a' then follows ∂Aρ/∂ρ + Aρ / ρ = 0, which means Aρ = 0 if Aρ is not proportional to 1 / ρ with infinite energy integral. From the equation for Az' follows Ez = 0 at a = c · Az The following equations remain valid: Eρ = - ∂a/∂ρ = - c · ∂Az/∂ρ whereas from ∂Ez/∂z = c · (∂2Az/∂ρ2 + ∂Az/∂ρ / ρ) = 0 it follows that with non-zero Az must be Az proportional to ln(ρ) and energy integral is infinite. Thus, no valid expressions for field formations were found. The situation changes if we assume that div E ≠ 0 (non-zero charge density) and introduce additional terms into formulas for E' using the velocity field: E′ = c2 · J - grad (E · V) - V · div E where div E = ∂Eρ/∂ρ + Eρ / ρ + ∂Ez/∂z in case of circular magnetic field, whereas case of circular electric field remains within previous calculations, since there div E = 0 Assuming that Vz = c is in the entire space around isolated field object, whereas Vρ = 0 and Vφ = 0, and since E · V = Ez · c, we get Eρ' = - c · ∂Eρ/∂z = c2 · Jρ - c · ∂Ez/∂ρ - 0 · div E → ∂Eρ/∂z = ∂Ez/∂ρ - c · Jρ → ∂Eρ/∂z = ∂Ez/∂ρ - c · (∂2Az/∂ρ/∂z - ∂2Aρ/∂z2) Ez' = - c · ∂Ez/∂z = c2 · Jz - c · ∂Ez/∂z - c · div E → ∂Ez/∂z = - c · Jz + ∂Ez/∂z + div E → c · Jz = div E → c · (∂2Aρ/∂ρ/∂z - ∂2Az/∂ρ2 + ∂Aρ/∂z / ρ - ∂Az/∂ρ / ρ) = div E The following equations remain true ∂a/∂z = c · (∂Aρ/∂ρ + Aρ / ρ + ∂Az/∂z) Eρ = c · ∂Aρ/∂z - ∂a/∂ρ Ez = c · ∂Az/∂z - ∂a/∂z From the expression for Ez' after substitutions it follows: c · (∂2Aρ/∂ρ/∂z - ∂2Az/∂ρ2 + ∂Aρ/∂z / ρ - ∂Az/∂ρ / ρ) = ∂Eρ/∂ρ + Eρ / ρ + ∂Ez/∂z = c · ∂2Aρ/∂ρ/∂z - ∂2a/∂ρ2 + c · ∂Aρ/∂z / ρ - ∂a/∂ρ / ρ + c · ∂2Az/∂z2 - ∂2a/∂z2 → ∂2a/∂ρ2 + ∂a/∂ρ / ρ + ∂2a/∂z2 = c · (∂2Az/∂ρ2 + ∂Az/∂ρ / ρ + ∂2Az/∂z2) Which leads to the conclusion a = c · Az Then Ez = 0, also ∂Aρ/∂ρ + Aρ / ρ = 0, hence Aρ = 0 to avoid infinity of energy integral. As result we get: a = c · Az, Aρ = 0, Ez = 0 Eρ = - ∂a/∂ρ = - c · ∂Az/∂ρ Which corresponds to the equation derived earlier from Eρ' ∂Eρ/∂z = ∂Ez/∂ρ - c · (∂2Az/∂ρ/∂z - ∂2Aρ/∂z2) Herewith Bφ = - ∂Az/∂ρ = Eρ/c Charge, spin and polarization If one looks in the direction of movement of field object, it is easy to notice that in the above version with annular magnetic field it is possible to orient this field clockwise or counterclockwise. Accordingly, radial electric field will be directed from z-axis outward or inward to this axis. To one type of field formations can be attributed conditional positive "spin", to the second negative. Let us try to find out how intensity of fields can decrease at distance from the geometric center of object. Let a = A0 / s, где A0 = amplitude constant, and s2 = R2 + ρ2 + z2, where R = object's scaling constant, possibly having an indirect relation to conditional "wavelength" in experiments. Note that ∂s/∂ρ = ρ / s, ∂s/∂z = z / s Then Az = A0 / c / s, Aρ = 0, Eρ = A0 · ρ / s3, Ez = 0 div E = ∂Eρ/∂ρ + Eρ / ρ = A0 · (2 / s3 - 3 · ρ2 / s5) The integral of charge density (divided by dielectric constant) over the entire space will be equal to ∫-∞+∞∫02·π∫0∞ (2 / s3 - 3 · ρ2 / s5) · ρ ∂ρ ∂φ ∂z = 0 That is, although charge density is not locally zero, the object as a whole is charged neutrally. This is natural, for example, for radiation arising from atoms and molecules, taking into account laws of conservation, since the particles located there will not give up part of their charge. In general, when E = Eρ = - ∂a/∂ρ, the subintegral expression ρ · div E = ρ · (∂Eρ/∂ρ + Eρ / ρ) = ρ · (- ∂2a/∂ρ2 - ∂a/∂ρ / ρ) = - ρ · ∂2a/∂ρ2 - ∂a/∂ρ = ∂/∂ρ (- ρ · ∂a/∂ρ) Computing the integral ∫0∞ ρ · div E ∂ρ we get for ρ = 0 the function - ρ · ∂a/∂ρ = 0, for ρ = ∞ the function - ρ · ∂a/∂ρ = 0 if ∂a/∂ρ decreases by absolute value with a distance faster than 1 / s Further computation of integrals by φ and z will not change zero result. The author of this article tested using MathCAD zero equality of the triple integral for a = A0 · ρ2 / s3 with Eρ = A0 · (2 · ρ / s3 - 3 · ρ3 / s5), also for a = A0 · ρ4 / s5 with Eρ = A0 · (4 · ρ3 / s5 - 5 · ρ5 / s7), for a = A0 · ρ / s2, a = A0 · z / s2, a = A0 / s2 Very wide range of such objects is neutrally charged in general, although it is likely that field formations are statistically inclined to take simplest geometric shapes, with minimum number of spatial extrema. It should be noted that when a = A0 / s2 or s appears with even higher degrees, field formation receives significantly greater ability to penetrate matter than with a = A0 / s or a = A0 · ρ2 / s3 Accordingly, the probability of registration of field object by measuring instruments is reduced. Which may be similar to the behavior of neutrinos in experiments. Polarized field object can be described as follows: s2 = R2 + X · x2 + Y · y2 + Z · z2 where R, X, Y, Z are scaling constants ∂s/∂x = X · x / s, ∂s/∂y = Y · y / s, ∂s/∂z = Z · z / s If a = A0 / s, where A0 is amplitude Az = A0 / c / s, Ax = 0, Ay = 0 Ex = A0 · X · x / s3, Ey = A0 · Y · y / s3, Ez = 0 Bx = - A0 / c · Y · y / s3, By = A0 / c · X · x / s3, Bz = 0 div E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = A0 · (X / s3 - 3 · X · x2 / s5 + Y / s3 - 3 · Y · y2 / s5) At the same time, all the above formulas for case of circular magnetic field remain true, E′ = c2 · J - grad (E · V) - V · div E Ex' = c2 · (∂Bz/∂y - ∂By/∂z) - 0 - 0 = 3 · A0 · c · X · Z · x · z / s5 Ey' = c2 · (∂Bx/∂z - ∂Bz/∂x) - 0 - 0 = 3 · A0 · c · Y · Z · y · z / s5 Ez' = c2 · (∂By/∂x - ∂Bx/∂y) - 0 - c · div E = 0 That is, there may be no cylindrical symmetry, with different X and Y, the field object will be stretched or extended along x- axis or y-axis. Compression or extension along z-axis is determined by multiplier Z. With significant differences between coordinate multipliers, structures arise with predominant orientation of fields in one direction (and the opposite also) in areas with high field energy density. This topic can be seen as a preface: This topic can be considered as a continuation: Edited November 11, 2022 by computer Quote Link to comment Share on other sites More sharing options...
computer Posted November 2, 2022 Author Report Share Posted November 2, 2022 15 hours ago, JeffreysTubes8 said: Not even light Of course, absolute vacuum does not exist, even far from Solar system. But the main goal of article was other: how many fundamental fields are required to describe photon-like objects. Quote Link to comment Share on other sites More sharing options...
computer Posted November 3, 2022 Author Report Share Posted November 3, 2022 Gravitational constant is too small for nano-world of photons. Cosmic masses are required to make gravity really acting. Quote Link to comment Share on other sites More sharing options...
computer Posted November 4, 2022 Author Report Share Posted November 4, 2022 (edited) 10 hours ago, JeffreysTubes8 said: If that were true how would we detect neutrinos? If a particle-like object is charged neutrally as a whole thing, it may contain locally areas with non-zero charge density. Neutrinos can be very small, but it does not mean they are not detectable. Neutrons also. Maybe additional "weak interaction", promoted in official physics theories, not only electromagnetic. Edited November 4, 2022 by computer Quote Link to comment Share on other sites More sharing options...
computer Posted November 6, 2022 Author Report Share Posted November 6, 2022 On 11/4/2022 at 6:42 PM, JeffreysTubes8 said: There's also this fact that a neutron is heavier than a proton Neutron is much heavier than photon or electron but is more difficult to detect. It seems many factors have to be taken into account: charged or neutral particle is, its "dimensions" or "wavelength" (photon is big, neutrino is small), velocity, and mass is not first in the list of priorities. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.