|San José State University|
& Tornado Alley
Using Methods Derived from
the Old Quantum Physics
of Niels Bohr
The deuteron, the nucleus of heavy hydrogen, is a very interesting, important special case. Its two nucleons, the proton and neutron, are 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 would be of the form
where d is the distance between the nucleons, 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 the deuteron.
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. This means the formula for the force experienced by each nucleon as a function of the radius r of their orbits with respect to their center of mass is
where H=H*/4 and λ=2λ*. It is conventient to express λ* as 1/d0; i.e., d0 = 1/λ*. This means that r0 = 1/(2λ*) = ½d0
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 d0 is about 1.5 fermi (1.5×10−15 meters). Thus r0 is about 0.75 fermi.
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. 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 m0c²[1/(1−β²)½ − 1 ]
where β is the velocity relative to the speed of light, v/c.
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 pθ=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/r0 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
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
is 1.67×10−27 kg. As
indicated above the value of r0 is 7.5×10−16 m.
A ball-park estimate of H can be obtained by noting that for small separation d the electrostatic force is less than the nuclear force but for large d the opposite is the case. At some point the two forces would be equal. If that point of equality is at 6d0, where d0 equals 1.5 fermi then the nuclear force constant would be e6=403.4 times the constant for the electrostatic force. The effective force constant for force expressed in terms of the orbital radius rather than particle separation is one fourth of that value.
Thus an estimate of H based upon this assumption
of where the nuclear force equals the electrostatic force
(403.4/4)(2.3×10−28)=2.32×10−26 and since
The maximum value that ze−z can attain is e−1=0.3679 and this is achieved at z=1. Thus strictly speaking no value of r/r0 can attain the figure of 0.3828. Thus H would have to have a value at least 1.041 times its above estimated value for there to be a solution.
Assume at the lowest level (l=1) of angular momentum r/r0=1 and H=2.32×10−26.
Since in a circular orbit v = [He−r/r0/(mr)]1/2 this means that the lowest level corresponds to a tangential velocity of a nucleon of 6.825×107 m/s, which is 22.75 percent of the speed of light. Thus relativistic effects should be considered.
The kinetic energy is ½mv²=2.7×10−10 joules.
The potential energy function is
The first step in its evaluation is to express the integral in terms of the dimensionless variable q=s/r0. This substitution gives
The integral can be, by integration by parts, put into the form
The integral in the above expression is the E1 function which is closely related to the exponential integral function, Ei(z). It has been evaluated and is available in standard tables. In particular E1(1) = 0.2193839344. This means that
It has been known since 1935 that if a deuteron is hit by a sufficiently high energy photon the deuteron disintegrates into a free proton and a free neutron. The energy level required for this photodisintergration of the deuteron is 2.23 Mev. This is 3.57×10−13 joules. This is reasonably identified as the potential energy of each of the l=1 level nucleons. A photon that knocks one nucleon out of its orbit disassociates the deuteron.
An estimate of the magnitude of H may be obtained as
This is roughly the same order of magnitude as the value previously used, 2.32×10−26
(Well, O.K. it is a one order of magnitude difference.)
If the lowest order of potential energy occurrs at r/r0=2 then the calculation would involve (E1(2.0)−e−2/2)=0.0489−0.0677=0.0188 and the value of H would be 1.4242×10−26, the same order of magnitude as the previously used estimate of H. However for r/r0 to be equal to 2, H would have to be such that
This is on the same order of magnitude although one value is 2.2 times the other.
The value of r/r0=2 is a significant figure. The accepted diameter of the deuteron is 4.2 fermis. The diameters of the proton and neutron are both about 1 fermi. Deducting 1 fermi, for the combined radii of the proton and neutron, from 4.2 fermis gives 3.2 fermis as the distance between the centers of mass of the two nucleons. This represents orbit radii of 1.6 fermis for both of the two nucleons. This figure divided by r0=0.75 fermi is 2.13.
For H to have a value of 3.24×10−26 the point of equality between the nuclear and electrostatic force would have to be at 6.33d0 rather than 6.25d0. This is obviously possible.
Before resolving the discrepancy about the estimated value of H it is advisable to consider the relativistic adjustment which should be made.
Under the Special Theory of Relativity the kinetic energy is given by the expression
where β=v/c and m0 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 on the left produces a quadratic equation in β²; i.e.,
where ζ² = [He−r/r0/(m0c²r)]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 pθ=mvr is quantized in increments of
where l is an integer.
For a circular orbit
pθ = l
h it follows that
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 = [He−r/r0/(m0c²r0)], which is
the same as γ(
In principle 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=m0c²r/(He−r/r0) the previous equation is equivalent to
Previously the expression H/(
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/r0 as a variable so the above equation takes the form
To simplify matters still more for computation let (γ/l)² be denoted as ε and (m0c²r0/(2H)) 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 equation ez=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 ez 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
Thus l can only have the value 1 and hence there is only one stable angular momentum and energy state for the deuteron.
A ball-park estimate of H was obtained previously by noting that for small separation d the electrostatic force is
less than the nuclear force but for large d the opposite is the case. At some point the
two forces would be equal. Previously the point of equality was taken at 6d0,
but that appeared to give a slightly too low value of H for the equations to have a solution.
Here the point of equality will be taken at 6.246d0,
equals 1.5 fermi. Under this assumption the nuclear force constant would be e6.246=518
times the constant
for the electrostatic force; i.e., H*=516(2.304×10−28)=1.1935×10−25.
The effective force constant is one fourth of that value; i.e.,
H=2.972×10−26. The value of γ is H/(
Thus for l=1 the value of ε is (0.7333)²=0.5378.
The value of μ is the ratio of the rest mass energy of a proton times r0 to twice the value of H. The rest mass energy of a proton is 938.72 Mev or 1.5×10−10 joules. Thus the value of μ is (1.5×10−10)(7.5×10−16)/(2*2.972×10−26)) or (1.893). Therefore μ²=3.582. The coefficient of z² which is μ²ε² is then 1.0358
The equation to be solved is then
The above transcendental equation does not have a postive solution.
Using the value of H corresponding to r1=1.5 fermi, the values of γ=3.24×10−26/(3.17×10−26)=1.025. Thus ε=1.051. The value of μ is then (1.5×10−10)(7.5×10−16)/(2*3.24×10−26)) or (1.736). The square of μ is then 3.014 and εμ²=3.089. The equation to be solved is then
Its solution is 3.929 which, since z=2r/r0, means that r1=3.929×0.75/2=1.473 fermis. This is not far different from the value of r=1.5 fermi upon which the value of H was based.
The results imply that for the deuteron that the velocity of the nucleons in the lowest energy state is 13.88 percent of the speed of light. Their individual kinetic energies are then about 1 percent of their rest mass energy. Their potential energy is
Although this is off by a factor of 2.27 from the experimental value of 2.23 Mev found for the photodisintergration of deuterons, it seems clear the model has a measure of validity.
One of the implications of the model is that there is only one of quantum state for the deuteron.
If r1/r0=2.746 then H has to be equal to 5.03776×10−26. The computed value of V(r1) is then
Since H=5.03776×10−26 then H*=2.0151×10−25. This means that the nuclear force is
greater than the electrostatic force up to 6.775d0 or
10.16 fermi. The ratio H*/(
hc), which corresponds to
the fine structure constant, is equal to 636. 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 for the electrostatic force.
The results imply that the formula for nuclear force expressed with respect to particle separation d is:
The radius of the single feasible orbit for the nucleons is (2.746)(0.75)=2.06 fermis. The separation distance d of the nucleons is then 4.12 fermi. The potential energy for a nucleon in the deuteron is then 2.23 Mev. The tangential velocity of the component particles of the deuteron relative to the speed of light is 0.28. The angular velocity is then 2.04×1023 radians per second. This corresponds to a frequency of 3.245×1022 per second.
The value of ε is 2.53566 and μ equals 1.11656. For these values the solution to the transcendental equation is 5.50207 which is 2r1/r0 so r1/r0=2.751, very close to the value derived from the photodisintegration energy of the deuteron. The two values are within 0.2 of 1% of each other.
Thus the model of the nuclear force presented here is logically consistent and empirically compatible with the data for the deuteron.
The above formula is for the force between two single nucleons. The general formula for the force between one collection of n1 nucleons and another of n2 nucleons is;
(To be continued.)
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