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The Trajectories of Particles in a Deuteronic
System Under a Central Force

In atomic and subatomic systems the lowest energy state is usually one of zero angular momentum. This means from a classical physics perspective that a particle appears to pass through another particle. This is not impossible even on a macroscopic level. All that is necessary is for at least one of the particles to have an empty center of mass, such as when there are subparticles distributed about that center of mass. That exists for example in a vortex-ring.

In order to investigate quantization in such deuteronic systems using the Wilson-Sommerfeld condition the details of the dynamics of the system must be determined.

Consider first the simple case of two particles of equal mass m subject to a force proportional to the inverse of the square of their separation distance s. The center of mass is at the midpoint of their separation and thus their distances from the center of mass r is equal to one half their separation distance. The being considered first is that of zero angular momentum.

The equation of motion for both particles is

m(d²r/dt²) = −k/s²

where k is a constant.

Since r=s/2 this equation reduces to

m(d²r/dt²) = −4k/r²

The determination of the trajectory starts at time t=0 with r=r0 and v=dr/dt=0. Although there is an analytic solution to this problem let us procede with a difference scheme approximation to that solution.

At any point in time the acceleration a(t)=(d²r/dt²) can be computed as

d²r/dt² = −(4k/m)/r²

From the acceleration the velocity at the next time step can be computed as

v(t+Δt) = v(t) + (d²r/dt²)Δt

The position at the next time step is then given by

r(t+Δt) = r(t) + v(t)Δt

The sequence may then be repeated. It would be desirable to repeat the computation until the particle returns to its initial position. That would be a complete cycle. However for this case the process cannot go past the singularity at r=0.

Below are the results of the computation when r0=−5, (4k/m)=1 and Δt=0.1. The particle reaches r=0 at about t=12.5. This means that the cycle period is about 60.

Particles with Empty Centers

A model can be constructed that eliminates the singularity at r=0. Suppose the subparticles of one particle are located at a distance b from their center of mass. Below is an illustration of one such possible arrangement.

Although the other particle would not have to have a smaller scale for it to pass through the first particle such a case is easy to visualize, as for example below.

The reason this case was illustrated with three subparticles is of course because the nucleons are believed to be composed of three subparticles called quarks.

The square of the separation distance of the subparticles of one particle to the center of mass of the other particle is then (r²+b²).

The radial component of the force is the force multiplied by the cosine of the angle between the vector to the subparticles and the radial vector. This cosine is equal to r/(r²+b²)½. Thus the acceleration on the particle is given by

a(t) = d²r/dt² = −(4k/m)r/(r²+b²)3/2

The trajectory for a particle in such a model is given below along with the velocity and acceleration for the case (4k/m)=1 and b=1.

The particle goes from −10 to +10 but not quite back to −10. This is an artifact of the particular computation scheme rather than the physics of the problem. The cycle period is about 30. The limits of the trajectory depend upon the parameters of the problem and the starting point. The maximum and minimum r is where the total energy of the system is in its potential energy and none is in its kinetic energy. Thus if the particle starts at S with a velocity of 0 then ±S is the limits of its trajectory.

This procedure can be extended to forces with other formulas; such as F=−H*exp(−|r|/r0)/r². This will be referred to as the Exponentially Weighted Force. Below are shown the graph of r(t) for the same parameter values as above with r0=10.

There is little perceptible qualitative difference between the case of the strictly inverse distance squared force and the exponentially weighted version of that force. One could say that the curve for the strictly inverse squared law is more peaked than the one with exponential weighting.

The value of r(t) does not quite return to −10 before reversing direction, but this probably an artifac of the computational scheme rather than a physical phenomenon. The cycle time for the exponentially weighted case is much longer, nearly 200, compared to the 60 for the unweighted case. This is because the exponential factor reduces the strength of the force. If the force constant for the exponentially weighted force were higher this would reduce the cycle time.

(To be continued.)

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