The strong nuclear force acts between any pair of nucleons which are sufficiently close. It is noteworthy that a neutron and a proton can form a bound state, whereas two neutrons or two protons cannot. In this Chapter we explain why this is the case. We examine the degree of fine tuning of the strong nuclear force required to bring about this state of affairs. How much weaker would the nuclear force need to be to fail to bind the deuteron? How much stronger would the nuclear force have to be to result in stable dineutrons and diprotons?
To address these questions we use an elementary approach to the inter-nucleon nuclear force. It is represented as a potential. We then apply the Schrodinger equation, which is particularly well suited to bound-state problems. The analytic solution of the Schrodinger equation for a 'square well' potential of depth V0 and radial range 'a' shows that there is exactly one bound state if M.V0.a^2/h^2 lies between 1.234 and 4.935 where M is the reduced mass. Empirical values for V0 and 'a' for the deuteron, a spin-one (triplet) state, give M.V0.a^2/h^2 = 1.779, consistent with the deuteron being bound (and, incidentally, having no excited states). We find that decreasing the strength of the strong nuclear coupling constant (gs) by about 15% would result in the deuteron being unbound.
In contrast, the nuclear force between nucleons with anti-aligned spins (i.e. a spin-zero, or singlet, state) is weaker than the triplet force. Specifically M.V0.a^2/h^2 = 1.113 for the singlet state. Since this is less than 1.234 it follows that there is no singlet spin state which is bound. Now nn and pp, being composed of identical particles, must have a ground state which is a spin singlet, i.e. their spins must be opposite as a consequence of the Pauli Exclusion Principle. It follows that the di-neutron and di-proton, nn and pp, are not bound in this universe. We find that an increase in gs of ~10% would be sufficient to bind the diproton, and an increase in gs of only ~5% is sufficient to stabilise the dineutron.
The non-existence of diprotons and dineutrons in this universe is therefore a consequence of the relative weakness of the nuclear force for spin-singlet states. It is not, as some authors have implied, a consequence of electrostatic repulsion. Whilst this adds to the instability of the diproton, it would not explain the absence of a bound dineutron.
If gs were reduced by more than ~15%, and hence the deuteron did not exist, the implications for the universe would be catastrophic. The lack of deuterium would remove the pathway for the formation of heavier elements. Thus, whilst heavier nuclei might still be stable, there would be no means of producing them. Thus, the lower bound on gs to ensure that deuterium is bound is more restrictive than the lower bound on gs derived in Chapter 9 to ensure stability of the heavier nuclei (at least if the electrostatic force is unchanged).
The implications of increasing gs by more than 10%, resulting in a bound diproton, are discussed fully in Chapter 9C. The reason for this special attention is that the implications of diproton stability have been repeatedly misrepresented in the literature. It has been claimed that diproton stability would result either in an all-helium primordial universe, or in a universe in which stars would explode. An all-helium primordial universe, being without hydrogen, and hence without water, hydrocarbons, proteins, hydrogen bonds, etc, would indeed be catastrophic. Similarly, the absence of stable stars would be catastrophic to elemental synthesis. However, diproton stability results in neither an all-helium universe nor in explosive stars, as will be demonstrated in detail in Chapter 9C. Anticipating the conclusions of Chapter 9C, the existence of the diproton would probably not be catastrophic for the evolution of a complex, chemically rich, universe.
Hence, we conclude that, as regards di-nucleon stability, gs exhibits a single-sided fine tuning of Type C1, namely that gs is required to exceed 85% of gs in this universe. The degree of fine tuning in gs is comparable to that found in Chapter 9 regarding the stability of the heavier elements (carbon to calcium) if similar variations in the electrostatic charge, e, are also considered.
We have taken the opportunity in this Chapter to derive an expression for how the binding energy of a nucleus might scale as gs is varied. Our formula is given in Equ.(16) of this Chapter. This formula is devised so that the binding energy reduces to zero at the non-zero value of gs when the nucleus becomes unbound - in the case of the deuteron at gs = 0.85 x gs(actual).
This formula is believed to be appropriate for nuclear forces which approximate to finite range potentials, e.g. square-well potentials or Yukawa potentials. The characteristic of finite range potentials is that they have no bound states if the magnitude of the potential is below some non-zero lower bound. Our Equ.(16) for the nuclear binding energy in an alternative universe stands in contrast to the scaling which would apply for an inverse square force law, i.e. for V proportional to gs/r. Such an infinite range potential has bound states however small we make gs. The binding energy for this potential is proportional to gs^4. Equ.(16) is generally more sensitive to changes in gs than is gs^4.
