Gravitational Red Shift and the Speed of the Fast Solar Wind
If you refer back to Figures 5 and 6 in Article 6, “How Backward-Through-Time Particles Explain Gravity, Inertia and Angular Momentum,” and the accompanying explanations about how Universal Field Cosmology says gravity works, you will recall that a “backward through time particle” (BTTP), that would otherwise be on a straight-on collision course with a “basic particle” (BP) within an object, is slightly deflected by the presence of other matter nearby in a direction away from that other matter, which then creates a gravitational force by its repulsion of the BPs in the object it is near, pushing it toward the other matter.
It follows from this deflection of BTTPs that if the two particles are to collide in a gravitational field then the BTTP would need to be, after its deflection, arriving at the collision at a slight angle to the direction in which the piece of matter is moving forward through time. Since the deflection of the BTTP is proportional to the mass of the other nearby matter, and inversely proportional to its distance, the larger the gravitational field the piece of matter is in, the larger the angle of approach of the BTTP will be at the point of collision.
When this non-straight-on collision occurs, though, the net velocity of the particles after colliding cannot be zero, as is required for them to become EMR in the Universal Energy Field. What could happen, though, and this theory predicts does happen, is that some of the particles remain as matter and carry away the net kinetic energy, and others effectively collide head on and become a part of the Universal Energy Field.
Let’s consider a case where a collision, when straight on, would involve 100 BPs directly colliding with 100 BTTPs, producing a photon of EMR with energy Es. If the collision were at a slight angle, such that one percent of the total energy needed to be carried off as kinetic energy, then one BP and one BTTP could carry off this energy as kinetic energy, leaving 99 BPs to collide head on with 99 BTTPs and make the quantum jump to the Energy Field. The quantum of EMR produced would then have 99/100 as much energy as Es. Its energy, Ea, would be: Ea = 0.99Es. Since the energy carried by EMR is proportional to its frequency, fa = 0.99fs. In other words, in this gravitational field, there would be a 1% red shift in radiation emitted, and this 1% of lost energy would be carried away by the particles as kinetic energy, half in our Universe by the piece of matter containing the BP and half by the BTTP in the backward through time domain it occupies. This can be illustrated as follows:
The reverse situation, with the same red shift result, occurs when photons are absorbed (creating absorption spectra lines) in a gravitational field. Since the BTTPs and the BPs matter that are created need to end up traveling in exactly opposite directions through time to satisfy the conservation of momentum, but the gravity of the sun will deflect the BTTP, the particles have to launch off at an angle to each other rather than in directly opposite directions. This is achieved with the help of an incoming particle (perhaps attracted by the gravitational field) which imparts the required KE and mass to the particles (say 1%), so a photon of only 99% the energy normally required need be absorbed. This creates a red-shifted absorption line in the spectrum.
Note from Figure 9 that the velocity of the piece of matter carrying away the KE, when a red-shifted photon is created, is always directly away from the center of mass causing the gravitational field. It is also clear that these pieces of matter could be carrying substantial amounts of kinetic energy (KE), as they could be traveling at very high speeds. If the piece of matter consisted of just the “basic particle” that carried away this energy, then it would be traveling at the speed of light, and have kinetic energy E = ½mbc².
If a piece of matter, such as a proton, collides with a backward through time particle of the same mass at the surface of the sun, producing slightly red-shifted photons of light and other EMR, then the net energy lost in the red-shift is carried off as kinetic energy by another nearby particle, most likely to be the same type of particle, in this case a proton. Since the red shift at the surface of the Sun is Z = 2.12239 x 10-6, and energy is proportional to frequency, then the total energy lost as kinetic energy in the forward through time domain is Z x ½mc2. Since it is carried away by a particle close to the same mass as the one brought to rest by the collision, and its energy is Z times as high, it will be traveling much more slowly, and its speed can be calculated by equating the energies:
½mwv2 = Z x ½mcc2
where mc = the mass of the colliding particle(s) and mw is the mass of the solar wind particle. So,
v2 = Zc2mc/mw