THE LIGHT SPEED HYSTERESIS LOOP- Chapter 30 Physics of the Universe
Copyright (c) 1998 by Gerald Grushow BSEE
There are many possible solutions to the universe. Throughout this
book the constant light speed solution has been chosen as the most likely.
This solution involves the expansion of space and time while the light
speed remains fairly constant. At present we measure a light speed of
2.99792458E8 meters per second. This light speed will vary slightly as
we move from the minimum radius to the maximum radius of the universe.
The variation of light speed will be very small but sufficient
to cause the universe to oscillate. In this chapter we will explore the
exponential e^x solution and then apply the results to the e^sinx solution.
In the e^x solution, the universe oscillates from a minimum radius
to a maximum radius accordeing to the e^x function. The most likelly
solution for the universe is that e^x varies from e^-pi to e^+pi. Thus:
Ru = Roe^x (30-1)
Tu = Toe^x (30-2)
Cu = (Ro/To) (30-3)
Let us assume that we exist at the zero point where (x=0)
Ru = Ro (30-4)
Tu = To (30-5)
Cu = Co (30-6)
Since the universe is following an exponential function, you can
always make today, the zero time. The past will be minus time and the
future will be positive time. The mass and energy of the universe can also
be expressed as an exponential. Thus:
Mu = Moe^-x (30-7)
Eu = (MoCo^2)e^-x (30-8)
As the universe expands the mass of the universe drops. The energy drops
as well.The function (x) varies as follows:
-pi pi (30-9)
Let us now call the light speed of the universe at minimum radius Ca,
and the mass of the Universe at minimum radius Ma. Let us also call
our mass Mo and the mass at maximum radius Mb. The energy of the
universe at minimum radius and maximum radius are:
Ea = Ma Ca^2 (30-10)
Eb = Mb Cb^2 (30-11)
However Mb = Ma e^-2pi = 0.0018674 Ma (30-12)
The energy of the universe at maximum radius is:
Eb = Ea e^-2pi = 0.0018674 Ea (30-12)
We see that the energy of the universe at maximum radius is very
small compared to the energy at minimum radius. In order for the
universe to oscillate, this small energy must be identical with the
energy lost due to a drop in light speed.
Delta Energy = Ma (Ca^2 - Cb^2) = Ma e^-2pi (30-13)
Solving for Cb we get:
(Cb/Ca)^2 = 1- e^-2pi = .998132557 (30-14)
Therefore :
Cb/Ca = 0.999065842 (30-15)
Our light speed is the geometric mean of the maximum light speed
and the minimum light speed. Thus:
Co = (Ca Cb)^0.5 = Ca(0.999065842)^0.5 (30-16)
Co = 0.999532812 Ca (30-17)
Solving for Ca we get:
Ca= 2.999325829E8 (30-18)
Solving for Cb we get:
Cb =2.996523985E8 (30-19)
The difference in light speed is:
Ca- Cb = 2.80184E5 (30-20)
This is a light speed change of:
% change in light speed = 0.0935 % (30-21)
Thus it only takes a 0.09 percent change in light speed for the
universe to oscillate in an e^x function. This is a light speed
hysteresis loop. Likewise both time and distance have a hysteresis
loop.
For an exponential sinusoidal solution, this could be
considered the bottom light speed. This solution has a light speed
which falls from the minimum radius to the present, and then rises toward
the maximum radius. On the reverse cycle, the light speed rises even
further and then falls back to the original. Assuming that the
difference in light speeds are the same, we can calculate the various
light speeds as follows.
Minimum Light speed = 2.99792E8 (30-22)
High Light Speed = 2.99792 .(2.99932/2.996524) (30-23)
High Light Speed = 3.0007173E8 (30-24)
The geometric mean is the speed at the minimum radius and also
the speed at the maximum radius for the e^sinx solution. Thus:
Geometric Mean Light speed = 2.999318E8 (30-25)
The e^x solution hits the boundaries hard. This causes
a big bang and a little bang. The e^sinx solution is rather smooth.
It will precipitate the neutrons at the minimum radius and break them
apart at the maximum radius. However, both solutions will produce
essentially the same results.
The cycle time of the universe for the e^x solution depends
upon the function (x). For the sinusoidal solution it depends upon
the inductive capacitive oscillator. That will produce a pure sine
wave as the driving function. There are several solutions possible
for the cycle time. These will be covered in a later chapter.