Gravitational Red Shift and the Speed of the Fast Solar Wind
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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 Stationary Energy Theory 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 creates a repulsive force pushing this piece of matter toward the other matter, which is the attractive force of gravity.
It follows from this deflection of BTTPs that if the two
particles are to collide in a gravitational field then the
BTTP would 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
away from the center of mass causing the gravitational field, rather than toward it, though it may or may not be directly away from it, depending on exactly where the collision takes place with respect to the 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
consists 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
v = cZ½(mc/mw)½