Emc2.text

"Well, I heard the Owl calling, softly as the night was falling,

With a question and I replied, but he's gone across the
borderline."^{1}

^{1}Kate Wolf, 1980, Across the Great Divide, from
Gold In California - A Retrospective of Recordings 1975-1985
-RHINO Records Inc.

E = mc^{2} = I@

E = mc^{2} is a rest energy equation where mass
has 0 LINEAR velocity, therefore 0 LINEAR kinetic energy.
Therefore, mc^{2} must represent a point mass energy
which can be shown as the ROTATIONAL energy of a fixed position
sphere. Mass is ~ to I, moment of inertia as mr^{2} which
for a sphere is ~2/5 mr^{2} . Since mc^{2} is a
fixed position mass, then it's position can be defined by x,y,z
but it appears that its actual size can be defined x,y with a
radius r. Since there is no linear movement, it is not necessary
to consider the 3^{rd} dimension z at this time. This
means a line with radius r becomes a circle when

x^{2} + y^{2} = r^{2}, AND all
possible considerations of x and y are made with one condition, r
is a constant.

[x^{2} + y^{2} + z^{2} = r^{2}
AND all possible considerations of x, y, z are made with one
condition, r is a constant. The result will be a sphere.]

Based on the constant c as the speed of light which is movement of energy without linear movement of mass "position" (at least not Cartesian - linear but Gaussian - rotational), then the smallest wavelength becomes the wavelength of ONE rotation, f = 1. This number can be calculated as 1/c and from body-centered (supersymmetry) structure, radius can be geometrically shown to be ~1/4c (actually probably slightly smaller but for demonstration here, one can use 1/4c). Also, since it appears to me most of our measurements are beyond 1/c cm, I shall use 1/c as distance knowing it must be modified once the correct size of CDM (cold dark matter) is determined.

E = Ic^{2 }/r^{2} and r =~1/c then E =
Ic^{4} and using Newton's laws of motion, it is possible
to show rotational acceleration @ as c^{4 }rotations/second/second
such that Ic^{4} = I@ the torque of one sphere at
rotational velocity c^{2 }rotations per second.

E = mc^{2} = I@ Rest energy or torque of a
sphere, size, radius ~ wavelength 1/c rotating at set velocity c^{2}
rotations/second

I = mr^{2} And m = I/r^{2 }= I/(1/c^{2 })
= Ic^{2 }

And again from Newton's laws of motion c^{2} can be
shown to = angular velocity (*v*), such that m = I^{ }or
Angular impulse of CDM sphere with radius ~ wavelength 1/c
rotating at set velocity of c^{2} rotations/second (Note *TH*
= THETA)

E = m@/*v* = m(*TH*/t^{2}) / (*TH*/t)
= m/t where time = 1/c^{2} seconds/rotation as the # of
seconds of one rotation of a CDM sphere.

E = m/t = mc^{2} Rest energy as mass with moment of
inertia mr^{2} rotating at c^{2} rot/second.

Note also that for this sphere,*v*^{2}t, is
equivalent to the v^{2} /c^{2} such that m_{f}(final)
= m_{o}(rest)/SQRT(1-v^{2}/c^{2}) then
(1-v^{2}/c^{2}) becomes 1-*v*^{2}
t and since *v* = c^{2} and t = 1/c^{2},
then v^{2}t = c^{2} and (1-v^{2}/c^{2})
becomes ~ -c^{2} and mass becomes m_{o}/c.
(Velocities are absolute so - sign is irrelevant, in other words
SQRT(1-v^{2}/c^{2}) is actually SQRT(absolute
|1-v^{2}/c^{2}|). But if one uses linear
velocity, this is the wrong equation and since above it was
stated mc^{2} = Energy with NO LINEAR velocity so the
above (1-v^{2}/c^{2}) becomes (1-0/c^{2})
or simply 1 and rest mass = final mass with 0 linear velocity.
Using w for wavelength, de Broglie showed m = h/w*v* but
this is ROTATIONAL velocity and from above m = h/(1/c)c^{2}
= h/c

E = mc^{2} = (h/c)c^{2} = hc Since LINEAR
velocity of light is c, then this formula also appears correct as
Et = h but only if time is set based on the speed of light and
not the rotational velocity of the CDM at 1/c^{2}, such
that Et = mc and then E(1/c)=mc or E =mc^{2} when LINEAR
velocity c is required to follow wave equation f (rot/sec)*
T(sec/period) = 1 rot/period so freq(vel) c must have period of T
of 1/c which means rotations(c) / time T (1/c) = c^{2}
rot/sec = c^{2}/1/c^{2} rot-periods/sec^{2}
and this is the rotational acceleration @.of CDM as c^{4}
. IF c were the rotational velocity of CDM, then the radius would
be 1/4 centimeter and one could directly observe this size.

But for Linear vel c, the wavelength is 1 cm (*wv* = c
cm/sec) when velocity is c. But velocity c is 3e10 rotations a
second, then the wavelength = distance/freq as 1 cm/3e10 rot and
1/c cm for ONE rotation. From linear vT =*w* . T becomes *w*/v
or 1/c/c or 1/c^{2} and *v* =c^{2} so the
universal almost constants based on earth velocity c =

Linear vel v of movement of light or angular impulse on CDM = c cm/sec

Rotational vel* v* of CDM = c^{2} rotations per
second or 1 rotation every 1/c^{2} second.

From here it is possible to calculate size, volume and surface
area of CDM as well as energy equations using ROTATIONAL laws and
noting the angular acceleration @ = 10a linear acceleration based
on Centrifugal force (F) = mv^{2}/r = ma and rotational
force, torque/radius (L)/r = F = I@/r

ma = I@/r

mar = (2/5) mr^{2}@

@ = (5/2)a/r and from above r ~ 1/4c then

@ = (5/2) a4c = 10ca. The missing power of 10 in Linear to Angular Energy conversions. (At least missing for me.)

@ = 10ca ..................a = @/10c...........and for sphere
with moment of inertia 2/5 mr^{2}.........m = 40 Ic2

@/a = 10 c rot/cm ( I believe those are the correct labels)
NOTE this number is dependent on use of moment of inertia as 2/5
mr^{2} , which is the value for a thin shell hollow
sphere. A hollow sphere moment of inertia would appear to be
between 2/5 mr^{2} and mr^{2} . I believe 2/5 mr^{2}
is correct.

Copyright Except words from Across the Great Divide by Kate Wolf ........Return to Homepage or Return to Cold Dark

Copyright September 17, 1998 - Dewey Birkhofer