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to Estimate the Parameters of the Nuclear Force |

This material develops a theory of the nuclear force. It arose out of an examination of Hideki Yukawa's theory of the nuclear force being carried by π mesons. Yukawa applied an exponential factor to the Coulomb potential but the exponential factor should be applied to the force because the exponential factor arises from the decay of the force-carrying mesons with distance. The formula for the nuclear force is then

where d is the distance between the nucleon centers, H* and λ* are constants with dimensions
of kg·m³/s² and m^{−1}, respectively. It is more convenient
to represent λ* as 1/d_{0} so the force formula is

The value of d_{0} can be determined from the mass of the particles, the π mesons,
which carries the nuclear force.

In previous work an estimate was obtained for the value of H* based upon the energy required to disassociate the proton and neuteron of the deuteron. However in that work approximations and simplifications were used. This material uses the most accurate data available and takes into account the slight difference in the masses of the proton and neutron. Those masses are

Mass of neutron: m

and thus the mass ratio is

m

Let r_{p} be the radius of the proton orbit in a deuteron. The ratio
of that radius to the separation distance between the two nucleons is then

If the nuclear force formula in terms of the separation distance d between the two nucleons is

where H* and d_{0} are constants with dimensions of kg·m³/s² and m,
respectively, then in terms of the radius of the proton orbit r_{p} the
force formula is

or

−He

where

and

r

=0.7617 fermi.

For convenience let r_{p}/r_{0} be denoted as ρ.
Previous work derived a condition which ρ must satisfy for
a given level of H. Also the previous work derived the relationship between the energy
V required to disassociate the deuteron and the values of H and ρ.
The two conditions to be satisfied are

where E(ρ)=(∫

or, in logarithmic form

ln(V) = ln(H) − ln(r

and

ρ = ln(γ) + ½ln(1 + ν²ρ²)

where γ=H/(

where s

a natural unit of length.

The value of V is 2.22457 MeV, which in joules is 3.5642×10^{−13}.

For numerical solution it is more convenient to express the two conditions as

g(ρ, H) = ρ − ln(γ) − ½ln(1 + ν²ρ²) = 0

The numeric procedure is based upon computing f_{0} and g_{0} for trial values of ρ and H and then finding adjustments
to the trial values that bring f and g closer to zero.

g

where the partial derivatives are computed at the trial values of ρ and H.

Let the trial values of f and g be expressed as a column vector F_{0} and likewise the adjustments
are expressed as a column vector dR. Then the adjustments dR to the trial solutions are found as
the solutions to the matrix equation

where M is the matrix

of the partial derivatives.

That is to say

dR = −M

The elements of the matrix M are given by

(∂f/∂H) = −1/H

(∂g/∂ρ) = 1 −(ν²ρ)/(1+ν²ρ²)

(∂g/∂H) = −1/H

The above system converges to the solution ρ=2.711535 and H=4.820872×10^{−26}.
This means that the radius of the orbit of the proton in the deuteron is

Since H=H*/3.99451 it means that H*=1.92570×10^{−25} kg·m³/s².

The tangential
velocity of the nucleons may be obtained from the expression for β; i.e., β =
γe^{−r/r0}/*l*. The parameter *l* is the angular
momentum number for the proton and, as explained in previous work,
must necessarily be equal to unity. The parmeter γ is
is equal to H/(~~h~~c)=1.5249. Since γ/*l*=1.5249/1 and ρ=2.711535,
β=(1.5249)(0.066435)=0.1013. The results indicate that for the deuteron that the
velocity of the nucleons in the lowest energy state is 10 percent of the speed
of light. Their individual kinetic energies as a ratio of their rest mass energy
are given by [(1−β²)^{−½}−1]. This means that in the deuteron the kinetic
energy of the nucleons about 0.5 of 1 percent of their rest mass energy.

Since H*=1.92570×10^{−25} and the constant for the electrostatic force is
2.31×10^{−27} the nuclear force is greater than the
electrostatic force up to the point where e^{ρ}=19.257/0.23102 or ρ=
ln(83.36)=4.42. This corresponds to a distance 4.42d_{0} or d_{0}=6.73 fermi.
The ratio H*/(~~h~~c), which represents a fine structure
constant for the nuclear force, is equal to 6.09. Since this value is greater than unity the analysis
of the nuclear force cannot be achieved by a series approximation in the way
that the electrostatic force can be where the fine structure constant is
1/137.036. This does not mean that the nuclear force cannot be analyzed; it just
means that it has to be done by a different scheme than what is used for the electrostatic
force.

The results imply that the formula for nuclear force expressed with respect to particle separation d is:

where

H* = 1.92570×10

and

d

The radius of the single feasible orbit for the proton is
2.0654 fermis. The radius of the neutron orbit is a slightly smaller 2.0626 fermi. The separation distance d of the
nucleons is then
4.129 fermi. (The accepted value for the diameter of the deuteron is 4.2 fermi.)
The potential energy for the nucleons in the deuteron is then 2.22457 MeV.
The tangential velocity of the component particles of the deuteron relative
to the speed of light is 0.1013. The angular velocity of the nucleons is 1.470×10^{22} radians per second or
2.34×10^{21} revolutions per second.

The angular momentum for the proton is

The ratio of this angular momentum to ~~h~~ = 1.0545715×10^{−34} kg m²/s is 0.99576. This is
amazingly
consistent with the quantization of angular momentum in terms of units of ~~h~~. There is no where in the above model any
element of quantum theory, yet the result is consistent with quantum theory.

The ratio of the angular momentum of the neutron to that of the proton is the product of the square of the ratio of the
orbit radii and the ratio of masses. Thus the angular momentum of the neutron is (0.9973)(1.001375)(0.99576)=0.99444~~h~~.
All in all this is a great vindication of the model.

computed angular momentum of neutron in a deuteron = 0.99444

The above analysis was based upon the assumption that the mass deficit of the deuteron is equal to the energy of the gamma ray emitted upon its formation. If the mass deficit is taken equal to the decrease in potential energy upon its formation and that this energy is divided between the increase in kinetic energy and the energy of the gamma ray then a different value of H* is found.

and

d

This estimate is based upon the diameter of the deuteron being 4.2 fermi and the diameters of the proton and neutron both being 1 fermi. The separation distance of the proton and neutron in the deuteron is 4.2-0.5-0.5=3.2 fermi.

Further details are given in the Spectrum of Two-Particle Systems although there H* is denoted as H.

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