Mr. Larson has worked. out the static relations between particles in the time region; specifically, he has calculated the equilibrium interatomic distances for all the elements and many compounds (see pages 2749 of The Structure of the Physlcal Universe). This paper will explore he dynamic relations between particles in the time region.
Consider a particle (say an alpha particle) moving directly towards a stationary atom (say a gold atom fixed in thin foil). Initially the particle has a velocity v_{o}. Once it enters the tlme region, that is, when its distance is less than one natural unit of space, two forces are encountered: the progression and gravitation. In the time region, the progression acts to bring particles closer together, whereas gravitation acts to repel particles — the reverse of gravitation in the timespace region. The progression is stronger until the equilibrium distance is reached, then the gravitational force becomes stronger. I believe that the equation of motion is
F_{p}  K_{G} / (x_{u} x)^{4 }= m d²_{x} / d_{t}² 
(1)

where
F_{p} = unit force of the progression
K_{G} = magnitude of the rotatianal motion of the partlcles
x_{u} = natural unit of space
m = mass of the moving particle
x = distancemeasured from start of time region
Dividing by m gives
F_{p} / m  K_{G} / m (x_{u } x^{4 }= d²x / dt² 

The right hand side reduces to
d²x/d_{t}² = d_{v}/d_{t} = d_{v}/d_{x }d_{x}/d_{t} = v d_{v}/d_{x} 
(2)

Thus
F_{p}/m  K_{G}/m (x_{u}  x)^{4 }= v d_{v}/d_{x} 
Separating variables and integrating, we have
ò

x_{f} 
ò

x_{f} 
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v_{f} 
F_{p/m dx}   K_{G} d_{x}/m (x_{u }x)^{4} =  vdv  
o  o  v_{o} 
or
F_{p}/m x_{f } K_{G} /3m [(x_{u } x_{f})^{3}  (x_{u})^{3}] = ½ (v_{f}²  v_{o}�) 
(3)

There are two cases of interest with this equation.
Case 1: Suppose we want to know the initial velocity required to bring the particles to a certain distance apart from each other. Equation (3) is solved for v_{o}, letting v_{f }be zero.
V_{o} = (2[K_{G}/3m{(x_{u}  x_{f})^{3}  (x_{u})^{3}}F_{p}/m x_{f}] )^{½} 
(4)

Case 2: Suppose we want to know the final separation between two particles, given v_{o}. Let
x_{sep} = x_{u}^{x} 
Equation (3) becomes
F_{p}/m (x_{u}  x_{sep})  K_{G}/3m x_{sep}^{3}  x_{u}^{3}] = ½ (v_{f}²  v_{o}²) 
With v_{f} = 0, and putting the terms irevolving x_{sep} on one side a_{f} the equation, we have
F_{p}/m x_{u} + K_{G} x_{u}^{3}/3m + 1/2 v_{o}² = F_{p}/m x_{sep} + KG/3m x_{sep}^{3} 
Define the following coefficients:
C_{2} = F_{p}/m x_{u} + K_{G} x_{u}^{3}/3m + ½ v_{o}² 
C_{2} = F_{p}/m 
C_{3} = K_{G}/3m 
C_{4} = C_{1}/C_{2} 
C_{5} = C_{3}/C_{2} 
With these coefficients, the result is a quartic equation:
x_{sep}^{4} + C_{4} x_{sep}^{3} + C_{5} = 0 
(5)

This equation can then be solved by the usual means.
Now, going back to equation (3) we can solve for v as a function of x:
v = d_{x}/d_{t} = {2[ F_{p}/m x  K_{G}/3m[(x_{u}  x)^{3} x_{u}^{3}]] + v_{o}²}^{½} 
Separating variables and integrating we have
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x  
t =  d_{x}/{2[F_{p}/m x K_{G}/3m((x_{u}  x)^{3} x_{u}^{3})] + v_{o}²}^{½}  
o 
The integral can be evaluated numerically by Romberg’s method.
Example
Consider an alpha particle moving directly towards a gold atom in a foil, at an initial velocity of 2.06x10^{7} meters/sec. What is the distance of closest approach? How long does it take to get there? What happens afterward.?
Here we have
v_{o} = 2.06 x 10^{7 }m/sec
m = 6.64 x 10^{27} kg
x_{u} = .455884 x 10^{7} m
F_{p} = 1.09699 x 10^{3} n
Now,
K_{G} = 1.09699 * 10^{3 }* (.455884 * 10^{7} )^{4} /(156.44) * l_{n}² t_{A} l_{n²} t_{B} 
For gold, t_{A} =4.5; for helium, t_{B} = 3. But helium has only one active dimension so the force is multiplied by 1/3. Thus
KG = 1.09699 * 10^{3} * (.455884 * 10^{7} )^{4} /(156.44)^{4 * }l_{n}� (4.5) l_{n}� (3) 
* 1/3 = 7.20006 * 10^{42} Nm^{4} 
(This assumes that since helium is inert, the electric displacements of gold have no bearing on the motion). The coefficients are next calculated:
C_{2} = 2.7438094 * 10^{15} 
C_{2} = 1.6520934 * 10^{23} 
C_{3} = 3.614488 * 10^{16} 
C_{4} = 4.6872709 * 10^{8} 
C_{5} = 2.197923 * 10^{39} 
The quartic equation is
x_{sep}^{4}  4.6972709 * 10^{8} x_{sep}^{3} + 2.187823 * 10^{39} = 0 
The only physical solution is
x_{sep} = 3.6014328 * 10^{11} m » .36 Å [3.6226287 * 10^{11} m with revised Fp] 
Note that his is cons:lderably greater than that predicted by use of classical atomic theory and Coulomb’s law: 2.581 x 10^{14} m.
Using equation (6) and Romberg’s method I find that
t = 6.3041312 * 10^{16} sec 
The average velocity of the particle to the point of closest approach is
4.552396 * 10^{8} /6.3041312 * 10^{16} = 7.2212742 * 10^{7} m/sec 
The initial velocity having been dissipated, the particle goes back to the equilibrium point. Of course, at room temperature, helium is a gas, and so the particle would not remain in the time region!
Situations in which the particle is not moving directly towards the atom will be treated in a future paper.