A NEW LOOK AT GRAVITY

There is a new way of looking at gravity that may be “peculiar” but it does give instructive results. The fundamental Newtonian equation for gravity is:

F = m a = G m₁ m₂ / r² eq 76

where r is the radius to the center of the mass m₂ and G is the gravitational constant 6.6732 x 10^-8 dyne cm² / gm² Then taking a man for example mass, m₁ , standing on a large planet such as earth which is mass, m₂, the acceleration on the man will be

F = m₁ a = m₁ ds 2 /dt² = G m₁ m₂ / r² eq 77

and dropping the m₁ from the equation, and letting the mass of the planet with radius s be its volume (4/3 p s³) times its density, r , the equation now becomes:

ds ² /dt² = G r 4/3 p s³ / s²= 4/3 p r G s eq 78

At this point equation 74 is identical to equation 78 if we realize that the "constant" terms are equal, and thus:

H² = 4/3 p r G eq 79

and the expansion / contraction of the universe equation 69 can be rewritten with H² being substituted for 4/3 p r G. Thus gravity is nothing more or less than the expansion of the universe, gravity is just an acceleration, no different from any other acceleration, and not a "special force" at all. The need for a gravitron etc. disappears.

The key to thinking about gravity is to realize that the surface of the earth is NOT and inertial reference system at all, in fact I can feel the acceleration as I sit here, and thus is not a proper frame of reference. There is a zero G reference in the center of the earth, and it perhaps could be used, but the surface of the earth is definitely an accelerated system. Taking the central point the surface is actually accelerating very slightly away from that point. That implies that the whole earth is very like a big balloon being blown up slowly, it is expanding.

Then why do we not measure this expansion? The very yard stick which we propose to use also is expanding and since it is made of the exact same materials as the earth it is expanding at the same rate. Thus the acceleration we feel is the only major evidence of the expansion of the universe locally. There is another analogy to this "gravitational" term. The so called Corriolis forces which had to be invented to account for the actual rotation of the earth under a projectile fired either north or south. This virtual force was required because we persisted in using an improper assumption that the earth was a stationary reference plane. When we account for the rotation then the need for the "Corriolis force" disappears.

Similarly when we account for the real acceleration in the expansion of all mass in the entire universe the need for a separate "gravitational" force also disappears, and this explains why no "gravitron" has so far been found. There is no need for a gravitron to carry a fictious force.

The whole universe is expanding, and thus the "real" motions of planets includes their own more or less spherical expansion combined with what would appear to be an increasing spiral orbit if we had an absolute meter stick to measure the distances. Since we and our measures are part of the universe, we can not do this, and must account for the changing distance by inventing an apparent force. If we account for the expansion of the universe we no longer need the force. Thus gravitation is no more or no less than the expansion of the universe which also is no more than time passing as new spasons are created to provide the change needed to allow time to pass.

STRONG and WEAK NUCLEAR FORCES

The strong and weak nuclear forces are in nature different from the two body forces, which now include Gravity linked into electromagnetic forces as they all are simply two body photon interactions. The strong nuclear force will be assumed as an induction to be a THREE body interaction and the weak nuclear force will be assumed to be a FOUR body interaction. All the two body interactions will be found to be proportional to 1/r² while all of the three body will be proportional to 1/r⁶ and four body interactions proportional to 1/r¹². To complete this list even though they are not being developed at this time, five body interactions would be proportional to 1/r²⁰.

The two body forces are interactions of two things taken one at a time, i.e. 2 factorial divided by 0 factorial 2!/0!=2. The three body forces are three things taken two at a time, but exclude two things one at a time already counted, which is 3 factorial divided by 1 factorial = 3!/1!=2x3= 6. The four body forces are four things taken three at a time, excluding the lower interactions and they are 3x4 = 12 = 4 factorial divided by 2 factorial 4!/2!){five body would be 5!/3! = 5 x 4 = 20}.

F_{g or em} = k/r²

F_s = B₁/r⁶ eq 80

F_w = B₂/r¹² eq 81

and since by definition E º dF/dr, differentiating, and letting the constant be subsumed into the proportionality constants:

E_{g or em} = k’/r

E_s = B₁’/r⁵ eq 82

E_w = B₂’/r¹¹ eq 83

These force equations should describe the electrical charge density, Ec, of nuclear isotopes. R. Hofstader has reported this data for numerous isotopes, and using that it is possible to evaluate B₁ and B₂, taken in the form:

E_c = B₁/r⁶ + B₂/r¹² eq 84

The charge density is related only to the proton atomic number, A, and the neutrons do not contribute to charge. However, the strong and weak forces operate on both neutrons and protons. Thus we need to relate neutron count to proton count or both to the total atomic weight of the nucleus. This then requires one other relationship. Accounting for neutron to proton ratios the stable isotopes¹³⁶ are related to proton count such that:

E_c = 1.53 - 0.7667 A^1/3 eq 85

Equation 84 was placed into a least square error fit program which varied B₁ and B₂ to evaluate B₁ and B₂ and then the values found placed back into the same program but with those values to find the curve that best fit the data. Also remember that the strong nuclear force term is positive inside the nucleus and repulsive outside the nucleus, so this requires two equations to account for this. This program is shown below:

FORTRAN Computer program for Fs and Fw

DIMENSION R(20), E(20), NAME(20)

1 READ (5,100,END=900)Z,A,NAME

100 FORMAT (2F10.0,20A1)

RHALF=1.05*A**(1./3.)

RC=RHALF-3.0287

BONE=920.

BTWO=328460.

EHALF=0.6407

EC=1.53-0.7667*(A**0.33333)

DO 10 J=1,20

R(J)=0.5*J

IF(R(J).LT.RHALF GO TO 11

150 EA=BONE/((R(J)-RHALF+3.00287)**6-BTWO/((R(J)-RHALF+3.0287)**12)

E(J)=0.5*EC*EA/EHALF

500 GO TO 12

11 S=2.*RHALF-R(J)-RC

151 E(J)=EC+EC*((-BONE/S**6)+BTWO/(S**12))/EHALF)*0.5

12 CONTINUE

10 CONTINUE

WRITE(6,200) Z,A,NAME,EC,RHALF

WRITE(6,300) R

WRITE(6,300) E

200 FORMAT(20(1X,F5,3))

WRITE(6,444)

444 FORMAT

GO TO 1

900 STOP

END

The actual measured data and the calculated curves are shown below. The values shown in the program are the best fit with B₁ being 920 and B₂ being 328460 when Ec is given in units of 10¹⁹ coulombs per cubic centimeter and the radius is given in Fermis i.e. Fermis to the 6th power or Fermis to the 12th power for the respective constants. There is less than 1% total deviation from the reported data for the 12 isotopes calculated. (He-4, C-12, O-16, Mg-24, Ca-40, V-51, Co-59, Sr-88, In-115, Sb-122, Au-197, and Bi-209). Because of use of equation 85, the isotopes below about C-12 should be run on an individual basis, not with the general stability zone equation. The fact that a simplistic relationship like equation 84 would fit the whole family of isotopes strongly supports the hypothesis of strong and weak nuclear forces.

Fig 4 Charge Density of selected isotopes from computer program