Background

The measured mass of a composite nucleus is generally less the combined masses of the nucleons that make it up. This mass deficit is an enigma. Seemingly no one knows why it occurs. It occurs when nucleons are brought into close proximity. This the same situation involved in the loss of potential energy. The mass deficit is reversed when the nucleons are taken out of close proximity and gain potential energy. There thus may be a connection between the mass deficit of a nucleus and the loss in potential energy.

Binding energy is just the mass deficit expressed in energy units using the Einstein formula E=mc².

The Measurement of the
Inertial Mass of a Charged Particle

When a charged particle is injected into a magnetic field it executes a helicial trajectory. The radius R of the helix depends up the mass m, velocity v, and charge of the partcle q as well as field intensity B of the magnetic field.

R = mv/(qB)
and therefore
m = RqB/v

This is the standard formula for computing mass. Thus with the observed radius and the velocity of the particle its mass maybe determined. A smaller radius corresponds to a smaller mass.

Now suppose there is a composite charged particle that is rotating and generating a magnetic field of intensity b. That generated magnetic field may enhance the effect of the outside magnetic field. Let us say the effective magnetic field is enhanced by some ratio of b. That is to say the field inducing the helix trajectory is

B' = B + λb

The radius of the helix is then

R' = mv/(q(B + λb))

The computed mass m' using the standard formula is then

m' = R'qB/v = mv/(q(B + λb))qB/v
which reduces to
m' = m/(1 + λb/B)
and hence
m − m' = λbm/B/(1 + λb/B)

Thus the mass deficit Δm=m−m' is given approximately by

Δm ≅ λm(b/B)
and
BE = Δmc² ≅ λmc²(b/B)

The Magnetic Field Induced by a
Rotating Charge

The magnetic field intensity generated by a rotating charge anywhere in space is proportional to the product of its charge q, the radius of its orbit ρ and the rate of rotation ω; i.e., qρω. This is dimensionally a chargeq times a velocity v, in effect a current Therefore .

b = μqρω = μqv

where μ is a constant.

Potential Energy

Now consider the relations that prevail in a Bohr model of an atom. This consists of two particles each with charge q. One particle is very heavy compared to the other one. The lighter particle of mass m_q revolves around the heavier particle. The potential energy PE of the two particle system is given by

PE = −kq²/(2ρ)

where ρ is the radius of the orbit of the lighter particle and k is the Coulomb constant.

The tangential velocity of the revolving charge is given by a balance of centrifugal force with the Coulombic (electrostatic) attraction

m_qv² = kq²/ρ = −2PE = 2|PE|
and hence
v = (2|PE|/m_q)^½

Therefore b is given by

b = q(2|PE|/m_q)^½

But q can be expressed as

q² = 2ρ|PE|/k
and hence
b = (2|PE|)(ρ/m_q)^½

Similar relationships concwening potential energy and velocity prevail for positronium but positronium does not generate a magnetic field since it has effectively a net current of zero.

Binding Energy and
Potential Energy

Now the Bohr model is abandoned and the analysis reverts to

BE = Δmc² ≅ λmc²(b/B)

where m is the mass of the composite nucleus and equal to 2m_q.

This leads to

BE = 2λc²(2|PE|)(ρm_q)^½/B)

Equivalently

|PE| = γ(BE)

where γ is equal to [B/((4λc²(m_qρ)^½)].

In particular

Δ|PE| = γΔBE

Thus there is a one-to-one relationship between binding energy and the potential energy lost in the formation of a composite nucleus. In this analysis the mass deficit of a composite nucleus is apparent rather than real.