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Using Methods Derived from the Old Quantum Physics of Niels Bohr |
The components of a two-particle system of the nucleus would be held together by the nuclear force alone. This nuclear force is carried by the π meson. As with gravitation or the Coulomb force there is an inverse-square-of-the-distance dependence because force-carrying particles are spread over a spherical surface whose area is proportional to the square of the distance from the source.. However there is an essential difference for the nuclear force as compared to the electrostatic Coulomb force in that the particles carrying the electrostatic force, photons, do not decay whereas the π mesons of the nuclear force, a.k.a the strong force, do decay quite rapidly with distance. The proportion of the mesons which survive is an exponential function of distance. Therefore the formula for the nuclear force between a subsystem of n_{1} nucleons and a subsystem of n_{2} would be of the form
where d is the distance between the nucleon subsystems. H* and λ* are constants with dimensions of kg·m³/s² and m^{−1}, respectively. The material which follows is an attempt to work out the implications of this force formula for the quantum level phenomena of such a two-subsystem subsystem. The subsystems hereafter will be referred to as particles and their locations are just their centers of mass. The analysis applies only for the case in which the separation of the components of the the subsystems is insignificant compared to the separation between the subsystems. This analysis is a necessary preliminary to the analysis for subsystems with significant internal separation.
At this point it is convenient to switch notation. Let r stand for the radius of the orbit of a particle with respect to the center of mass of the particle pair. For the electron in an proton-electron pair r is essentially equal to the distance d between the particles. For the proton-neutron pair of a deuteron r is essentially equal to one half of the distance between the particles.
To keep things simple it will be presumed that all nucleons have the same mass, m_{p}. The proportional difference between the masses of the proton and neutron is only 0.1 of 1%. The relationship between the particle orbit radii, r_{1} and r_{2}, and the distance of their separation is derived as follows.
Let the center of mass be at the origin with particle one at −r_{1} and particle two at +r_{2}. Since the masses of the particles are m_{p}r_{1} and m_{p}r_{2} it must hold that
Since d=r_{1}+r_{2} this means that
This means the formula for the force experienced by Particle 1 as a function of the radius r_{1} of its orbit with respect to the center of mass of the system is
where H=H*/(1 + n_{1}/n_{2})² and λ=(1 + n_{1}/n_{2})λ*. It is conventient to express λ* as 1/d_{0}; i.e., d_{0} = 1/λ*. This means that
Hideki Yukawa established that the mass of the force-carrying particles is related to the λ parameter. From the mass of the π mesons it is known that d_{0} is about 1.5 fermi (1.5×10^{−15} meters).
In mechanics the force is conveniently represented as the negative of the derivative of a potential function V(r). This means that the potential function is given by the integration of the force function with with respect to distance. The potential at an infinite distance is taken to be zero.
For the Coulomb force of −α/r² this gives a potential function of V(r)=−α/r. Yukawa hypothesized a potential function which is just this potential function multiplied by an exponential factor; i.e., V(r)=(−α/r)e^{−λr}. However it is the force function that is multiplied by the exponential factor due to the decay of the force-carrying mesons; i.e.,
where d is the distance between the two nucleons and H* and λ* are constants characteristic of the nuclear force. Therefore the potential function for the nuclear force is really of the form
The standard procedure in quantum mechanics is to solve Schroedinger's equation for the particular potential function which applies. This works beautifully for the Coulombic potential and about a half dozen other special cases but gives no analytic results for the cases that deviate from those special cases. The Schroedinger equation cannot be solved even for the Yukawa potential function. There is no way to solve the Schroedinger equation analytically for even more complicated potential functions. Physicists then rely upon purely numerical methods of solution based upon perturbation theory.
What does work analytically is the analysis of Niels Bohr's Old Quantum Theory. Using
Bohr's analysis it can be established that the angular momentum
is quantized for any potential function in exactly the same way that it is for the
Coulombic force. This means that angular momentum must be equal to an integral multiple of
h, Planck's constant divided by 2π. This is also true whether the kinetic
energy is of the Newtonian form ½mv², where m is the mass of the particle and
v is its velocity or the relativistic form m_{0}c²[1/(1−β²)^{½} − 1 ]
where β is the velocity relative to the speed of light, v/c and m_{0}
is the rest mass of the particle.
In a circular orbit the attractive force balances the centrifugal force; i.e.,
This means that orbital velocity v is a function of orbit radius r; i.e.,
Angular momentum is defined as p_{θ}=mvr, which on the basis of the relation between v and r means
But previous analysis found angular momentum to be quantized. That is to say,
where l is an integer and h is Planck's constant divided by 2π.
This means that
Now let r/r_{0} be denoted as z.
The function ze^{−z} rises from 0 at z=0 to a maximum of e^{−1} at z=1 and falls asymptotically to 0 as z goes to +∞, as shown below.
For sufficiently small values of angular momentum there will be two values of r but there is a maximum angular momentum for which there are any solutions for r.
The levels shown in green above correspond to the
quantity
h²/[n_{1}²(Hn_{2}m_{p}r_{1})]
multiplied by the square of an integer l.
The value of this coefficient can be calculated. The value of
h² in SI units is 1.11×10^{−68}. The value of m_{p}
is 1.67×10^{−27} kg.
The potential energy function is
The first step in its evaluation is to express the integral in terms of the dimensionless variable q=s/r_{0}. This substitution gives
where p=r/r_{0}.
The integral can be, by integration by parts, put into the form
The integral in the above expression is the E_{1} function which is closely related to the exponential integral function, Ei(z). It has been evaluated and is available in standard tables. In particular E_{1}(1) = 0.2193839344.
Under the Special Theory of Relativity the kinetic energy is given by the expression
where β=v/c and m_{0} is the rest mass of the particle.
The first two terms of this kinetic energy function are
The balance of the attractive nuclear strong force with the centrifugal force under Special Relativity is
This can be reduced to
Multiplying by the denominator of the left-hand side produces a quadratic equation in β²; i.e.,
where ζ² = [Hn_{2}e^{−r/r0}/(m_{p}c²r)]². The solutions are
The negative solution must be discarded. Thus
For large values of ζ the solution is approximately β=1.
It is shown elsewhere that even in the relativistic case angular
momentum p_{θ}=mvr is quantized in increments of h so
where l is an integer.
For a circular orbit
Therefore
Since
p_{θ} = lh it follows that
where γ=Hn_{1}n_{2}/(hc). This is not an entirely satisfactory
expression of quantization because r on the RHS is quantized in some as yet undetermined
manner, yet the similarity with the non-relativistic case makes it of interest.
From the analysis of the previous section it is known that
where ζ = [Hn_{2}e^{−r/r0}/(m_{p}c²r)], which is
the same as γ(hc/(n_{1}m_{p}c²r).
The two expressions for β can be equated the result solved for r as a function of l. This provides the quantization of r and subsequently that of β and the rest of the characteristics of the system.
The equating of the two expressions for β gives
After squaring and rearranging the above equation reduces to
Since 1/ζ=m_{p}c²r/(Hn_{2}e^{−r/r0}) the previous equation is equivalent to
Previously the expression Hn_{1}n_{2}/(hc) was defined as γ. Utilizing this
term the previous equation can be put into the form
It is mathematically convenient to deal with z=2r/r_{0} as a variable so the above equation takes the form
To simplify matters still more for computation let (γ/(n_{1}l))² be denoted as ε and (m_{p}c²r_{0}/(2Hn_{2})) as μ. The equation to be solved for z is then
Although the above equation is transcendental and cannot be solved analytically, approximations to any degree of accuracy can be readily be obtained numerically, provided of course the equation has a solution at all. Since ε and μ² are positive there will always be a negative solution, but a negative solution is physically meaningless.
The potential envergy V(r) at r=r_{1} is given by
The equation e^{z}=ε+μ²ε²z² has solutions for some values of ε and μ and not for others. At the dividing point between the range for solutions and the range for no solution there would be a tangency solution. In other words there would be a value of z such that not only are the LHS and RHS equal but also the derivative of the two sides are equal. This means that
Equating the expressions for e^{z} gives a quadratic equation in z which has a solution that can be put into the form
Thus there will be a real valued solution for z if μ²ε≥1. The values of ε and μ reduces this condition to
Since r_{0}=d_{0}/(1+n_{1}/n_{2}) the above reduces to the condition
When numerical values are substituted into this expression
Thus when n_{1}/n_{2}=1, l ≤ 1.778 so l can only have the value 1 and hence there is only one angular momentum and energy state.
If n_{1}/n_{2}=1/2, l ≤ (2/3)(3.556)=2.3 so l can have the values 1 and 2 and hence there are two angular momentum and energy states.
Consider a proton-neutron pair interacting with a single neutron. There are three distance quantities:
The distance r is the radius of the orbit of the single neutron about the center of mass of the system. The angle θ is half of the angle spanned by the proton-neutron pair from the viewpoint of the single neutron. The relations which prevail among these distances are
The force on the single neutron is
Thus the force can be presented as
This means the previous analysis may be applied using these values for H and r_{0}.
(To be continued.)
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