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The Deuteron, the Virial Theorem and
the Allocation of Potential Energy Loss
Between Increased Kinetic Energy
and Energy of Photon Emission

This is an effort to estimate the parameters of a nuclear force formula using the physical characteristics of deuterons. The force between two unit charges or two unit masses has the form

F = ±K/s²

where s is the separation distance between the particles and K is a constant. The inverse distance squared relations comes from the forces being carried by particles; photons in the case of the charges and gravitons in the case of the masses. These particles last forever and at a distance s they are spread over a spherical area proportional to s². The nuclear force is carried by π mesons and they decay over time. The decay over times results in the number surviving being an exponential function of distance. The mesons that survive to a distance s are also spread over a spherical area proportional to s². This means that the formula for the force between two nucleons should be

F = −H*e-s/s0/s²

H and s0 are constants; i.e., parameters of the nuclear force.

While this formula does not incorporate the spin dependent effects it captures the overall dependence of nuclear force on distance in the same way that the inverse distance square law captures the nature of the force between charge particles even though it does allow for the effects of particles' magnetic moments.

The Virial Theorem

The Virial Theorem says the time-averaged kinetic energy K for a system of particles satisfies the following relation

K + ½ΣFiri = 0

where Fi is the force vector on the i-th particle and ri is its position vector. The summation is over the particles of the system.

Suppose the system is a proton and neutron forming a deuteron. Furthermore suppose the strong nuclear force binding the two nucleons together is given by the formula

F = −H*e-s/s0/s²

where s is the separation distance of the centers of the nucleons and H and s0 are constants.

Assume the origin of the coordinate system is at the center of one of the nucleons, say the neutron. Then ri is equal to zero for the neutron and equal in magnitude to the separation distance for the proton. Thus, according to the Virial Theorem

2K −H*e-s/s0/s = 0
or, equivalently
2K = H*e-s/s0/s

The term on the RHS is just the Yukawa potential. The Yukawa potential is not the true potential but it is clearly relevant.

The True Potential Energy Function

The true potential V is

V(s) = −∫s[H*e-z/s0/z²]dz

By changing the variable of integration from s to z=s/s0 the potential energy can be expressed as

V(z0) = −(H/s0)∫z0[e-z/z²]dz

where z0=s/s0.

The technique of Integration by Parts, ∫AdB=AB−∫BdA, may be applied to ∫z0[e-z/z²]dz with A=e-z and dB=dz/z² and thus B=−1/z. The result is

z0[e-z/z²]dz = [−e-z/z]z0 − ∫z0[−(e-z/z)]dz
which reduces to
z0[e-z/z²]dz = [e-z0/z0] + ∫z0[(e-z/z)]dz

The substitution of the above formula into the expression for V results in

V(z0) = −(H/s0)[e-z0/z0 + ∫z0[e-z/z]dz]
or, equivalently
V(z0) = −(H/s0)[e-z0/z0 + (H/s0) ∫z0[e-z/z]dz
which reduces to
V(z0) = −(H/s0)(e-z0/z0) − (H/s0) ∫z0[e-z/z]dz

The above relation would also apply in a time-averaged form. Because of the cumbersomeness of including an over-line to denote a time-averaged expression this notation will be dropped. From this point forward an unadorned variable is the average over a cycle.

By virtue of the Virial Theorem the first term on the right hand side reduces to

−(H/s0)(e-z0/z0) = −2Z

Therefore

V(z0) = −2K − (H/s0) ∫z0[e-z/z]dz

When the separation distance goes from +∞ where V=0 and K=0 to s where V has a negative value and K a positive value there is a discrepancy in the change in K compared to the change in V. In the following the changes, denoted by Δ, are for the change from infinite separation to finite separation. Conservation of energy is maintained by the emission of a gamma ray. The energy γ has to be such that

ΔV + ΔK + γ = 0
or, equivalently
ΔV = −ΔK − γ = − (ΔK +γ)

The previous expression for V can be written as

ΔV = −ΔK −(ΔK + (H/s0) ∫z0[e-z/z]dz)
which means that
γ = ΔK + (H/s0) ∫z0[e-z/z]dz)

The ratio of the energy of the emitted gamma ray γ to the increase in kinetic energy ΔK can be computed; i.e.,

γ/ΔK = 1 + 2(∫z0(e-z/z)dz)/(e-z0/z0)

where use was made of the fact that K=½(e-z0/z0). This result indicates that the allocation of the loss of potential energy between increased kinetic energy and the energy of the emitted gamma ray is independent of the value of the constant H in the nuclear force formula.

The integral ∫x(e-z/z)dz is called the exponential integral and is denoted as E1(x). Let the ratio E1(z0)/(e-z0/z0) be denoted as r.

Since γ is a measured quantity equal to 2.224575 MeV the ratio desired is

ΔK/γ=1/(1+2r).

Hence ΔK is 2.22475/(1+2r).

The values of the quantities involved in the above formula are tabulated below for z over the range of 1.0 to 2.0. The quantity z is the ratio of the separation distance between the centers of the nucleons in a deuteron and the scale factor s0 in the nuclear force formula.

 z        E1(z)          exp(-z)/z        Ratio r           1+2r         1/(1+2r)        K (MeV)
1.0	0.219384	0.367879441	0.596347541	2.192695081	0.456059763	1.014539148
1.1	0.185991	0.302610076	0.614622627	2.229245255	0.448582316	0.997905006
1.2	0.158408	0.250995177	0.631119698	2.262239396	0.442039866	0.983350836
1.3	0.135451	0.209639841	0.646112874	2.292225748	0.436257206	0.970486874
1.4	0.116219	0.176140689	0.659807799	2.319615598	0.431105913	0.959027436
1.5     0.100020	0.14875344	0.672387811	2.344775622	0.426480039	0.948736834
1.6	0.086308	0.126185324	0.683978116	2.367956232	0.422305103	0.939449374
1.7	0.074655	0.107460897	0.694717822	2.389435645	0.418508865	0.931004359
1.8	0.064713	0.091832716	0.704683505	2.409367011	0.415046772	0.923302672
1.9	0.056204	0.078720326	0.713970621	2.427941243	0.411871582	0.916239224
2.0	0.048901	0.067667642	0.722664465	2.445328929	0.408942939	0.909724239

The graph of ΔK versus z is shown below.

What the table and the graph indicate is that over the range of z0 from 1.0 to 2.0 the value of the kinetic energy only varies from 1 MeV to 0.9 MeV. This means that while there might be some uncertainty about the value of z0 the value of ΔK is going to be significant, on the order of 1 MeV.

The Dimensions of the Deuteron

The best estimate of the separation distance of the nucleons in the deuteron is 2.252 fermi. The best estimate of s0 is 1.522 fermi, based upon the Yukawa relation concerning the mass of the π meson.

Thus the value of z0 is the pure number 1.4796.

Evaluation of the Energies for a Deuteron

For z0 equal to 1.4796 the value of e-z0/z0 is 0.153904. The value of the integral E1 for z0=1.4796 is 0.1031064. Thus

γ/ΔK = 1 + 2(0.1031064/0.153904)
= 1 + 1.2966 = 2.2966

Since γ is measured as 2.224575 MeV, this means that K=2.224575/2.2966 = 0.98638 MeV. This makes ΔV = -(2.224575+0.98638) = -3.19322 MeV. Since 1 MeV is equal to 1.60217646×10-13 joules, ΔV = −5.1161×10-13 joules.

Previously it was noted that

V(z0) = −(H/s0)[e-z0/z0 + ∫z0[e-z/z]dz]
therefore
−5.1161×10-13 = −(H/s0)[0.153904+0.1031064]
hence
(H/s0) = 1.99062×10-12 kg m²/sec²

Since s0 =1.522×10-15 meters this means that

H = (1.99062×10-12)(1.522×10-15)
= 3.0297×10-27 kg m³/sec².

The constants of force formulas can be represented as a multiple of the product of h, Planck's constant divided by 2π, and the speed of light, c. Since hc is equal to 3.1615×10-26 kg m³/sec², this means that H is equal to 0.0958hc.

The potential energy function V can be expressed in the form

V(s/s0) = (H/s0)W(s/s0)

where

W(z) = ∫z(exp(−x)/x²)dx

The value of (H/s0) is 3.0297×10-27/1.522×10-15=1.9906×10-12 joules. In units of millions of electron volts (MeV) this is 12.2424 MeV.

Implications

The analysis indicates that the mass deficit of the deuteron is 3.19322=(2.224575+0.98638) MeV rather than 2.224575 MeV as conventionally presumed. This means that the estimate of the mass of the neutron is 0.98638 MeV too low. This makes the mass of the neutron 940.5519 MeV instead of 939.56556 MeV. The mass of the neutron is thus about four and a half electron masses greater than that of the proton instead of two and a half. This is an increase of only 0.1 of 1 percent in the mass of the neutron, but it signficantly alters the values of the binding energies of nuclides. For example, the binding energy of the Berylium isotope Be5 is conventionally −0.75 Mev. This is an anomaly because the binding energies of all of the other nuclides are positive. With the adjustment suggested by the above analysis the binding energy of Be5 is a positive 0.2364 MeV.

(To be continued.)


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