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

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

E = mc2 = I@

E = mc2 is a rest energy equation where mass has 0 LINEAR velocity, therefore 0 LINEAR kinetic energy. Therefore, mc2 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 mr2 which for a sphere is ~2/5 mr2 . Since mc2 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 3rd dimension z at this time. This means a line with radius r becomes a circle when

x2 + y2 = r2, AND all possible considerations of x and y are made with one condition, r is a constant.

[x2 + y2 + z2 = r2 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 = Ic2 /r2 and r =~1/c then E = Ic4 and using Newton's laws of motion, it is possible to show rotational acceleration @ as c4 rotations/second/second such that Ic4 = I@ the torque of one sphere at rotational velocity c2 rotations per second.

E = mc2 = I@ Rest energy or torque of a sphere, size, radius ~ wavelength 1/c rotating at set velocity c2 rotations/second

I = mr2 And m = I/r2 = I/(1/c2 ) = Ic2

And again from Newton's laws of motion c2 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 c2 rotations/second (Note TH = THETA)

E = m@/v = m(TH/t2) / (TH/t) = m/t where time = 1/c2 seconds/rotation as the # of seconds of one rotation of a CDM sphere.

E = m/t = mc2 Rest energy as mass with moment of inertia mr2 rotating at c2 rot/second.

Note also that for this sphere,v2t, is equivalent to the v2 /c2 such that mf(final) = mo(rest)/SQRT(1-v2/c2) then (1-v2/c2) becomes 1-v2 t and since v = c2 and t = 1/c2, then v2t = c2 and (1-v2/c2) becomes ~ -c2 and mass becomes mo/c. (Velocities are absolute so - sign is irrelevant, in other words SQRT(1-v2/c2) is actually SQRT(absolute |1-v2/c2|). But if one uses linear velocity, this is the wrong equation and since above it was stated mc2 = Energy with NO LINEAR velocity so the above (1-v2/c2) becomes (1-0/c2) or simply 1 and rest mass = final mass with 0 linear velocity. Using w for wavelength, de Broglie showed m = h/wv but this is ROTATIONAL velocity and from above m = h/(1/c)c2 = h/c

E = mc2 = (h/c)c2 = 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/c2, such that Et = mc and then E(1/c)=mc or E =mc2 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) = c2 rot/sec = c2/1/c2 rot-periods/sec2 and this is the rotational acceleration @.of CDM as c4 . 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/c2 and v =c2 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 = c2 rotations per second or 1 rotation every 1/c2 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) = mv2/r = ma and rotational force, torque/radius (L)/r = F = I@/r

ma = I@/r

mar = (2/5) mr2@

@ = (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 mr2.........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 mr2 , which is the value for a thin shell hollow sphere. A hollow sphere moment of inertia would appear to be between 2/5 mr2 and mr2 . I believe 2/5 mr2 is correct.