TIME REGION PARTICLE DYNAMICS

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 27-49 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 time-space 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

Fp - KG / (xu- x)4 = m d²x / dt²

(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 = distance-measured from start of time region

Dividing by m gives

Fp / m - KG / m (xu - x4 = d²x / dt²

The right hand side reduces to

d²x/dt² = dv/dt = dv/dx dx/dt = v dv/dx

(2)

Thus

Fp/m - KG/m (xu - x)4 = v dv/dx

Separating variables and integrating, we have

ò	xf	ò	xf	ò	vf
	Fp/m dx -		KG dx/m (xu -x)4 =		vdv
	o		o		vo

or

Fp/m xf - KG /3m [(xu - xf)-3 - (xu)-3] = ½ (vf² - vo�)

(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.

Vo = (2[KG/3m{(xu - xf)-3 - (xu)-3}-Fp/m xf] )½

(4)

Case 2: Suppose we want to know the final separation between two particles, given v_o. Let

xsep = xu -x

Equation (3) becomes

Fp/m (xu - xsep) - KG/3m xsep-3 - xu-3] = ½ (vf² - vo²)

With v_f = 0, and putting the terms irevolving x_sep on one side a_f the equation, we have

Fp/m xu + KG xu-3/3m + 1/2 vo² = Fp/m xsep + KG/3m xsep-3

Define the following coefficients:

C2 = Fp/m xu + KG xu-3/3m + ½ vo²

C2 = Fp/m

C3 = KG/3m

C4 = -C1/C2

C5 = C3/C2

With these coefficients, the result is a quartic equation:

xsep4 + C4 xsep3 + C5 = 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 = dx/dt = {2[ Fp/m x - KG/3m[(xu - x)-3 -xu-3]] + vo²}½

Separating variables and integrating we have

	ò	x
t =		dx/{2[Fp/m x -KG/3m((xu - x)-3 -xu-3)] + vo²}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 2.06x10⁷ 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 10⁷ 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,

KG = 1.09699 * 10-3 * (.455884 * 10-7 )4 /(156.44) * ln² tA ln² tB

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 * ln� (4.5) ln� (3)

* 1/3 = 7.20006 * 10-42 N-m4

(This assumes that since helium is inert, the electric displacements of gold have no bearing on the motion). The coefficients are next calculated:

C2 = 2.7438094 * 1015

C2 = 1.6520934 * 1023

C3 = 3.614488 * 10-16

C4 = 4.6872709 * 10-8

C5 = 2.197923 * 10-39

The quartic equation is

xsep4 - 4.6972709 * 10-8 xsep3 + 2.187823 * 10-39 = 0

The only physical solution is

xsep- = 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 10¹⁴ 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 * 107 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.