find the potential seen by the negative muon for r < R . If it were sitting exactly at the center of the nucleus, it would (by symmetry) feel no force. That is, there is no term in the potential linear in r , as the force at the origin is − dV nuc /dr ]$_{r}$ =0 = 0. This suggests that we can take V nuc ( r ) = A + Br $^{2}$. The coefficient B can be fixed from the requirement that the slopes dV nuc /dr and dV Coul /dr match at r = R . The coefficient A in turn is fixed by the requirement of continuity at r = R . Thus a potential reasonably appropriate for such a muonic atom is V$_{v}$ ( r ) =    αZ ( r 2 − 3 R $^{2}$) / 2 R 3 for r ≤ R , − αZ/r for r > R. (47)  This potential no longer has the 1 /r singularity at the origin, so the BCs for the outgoing integration for the 1 s state are more like those in Eqs. (13) and (14) for l = 0: bearing in mind that we here are making the equations dimensionless by setting m$_{µ}$ = 1 and that the eigenenergy of the bound state is less than m$_{µ}$ . Unlike the pure Coulomb case, here f ( r ) will have curvature near r = 0, where it will vanish. Running the numerical integrations with potential V$_{v}$ ( r ) for $^{40}$Ca gives a binding energy B = 1 . 07006, some 60 keV smaller than the pure Coulomb result for this nucleus. Figure 17 displays the difference between the ψ$_{a}$ ( x )’s and ψ$_{b}$ ( x )’s for the case with this nuclear charge distribution and the pure Coulomb case. We now consider the solution of the radial Dirac equations, Eqs. (7) and (8), when both potentials, the Lorentz scalar V$_{s}$ ( x ) and Lorentz vector V$_{v}$ ( x ), come into play. We discuss two different situations. As our first case, we take the scalar potential V$_{s}$ ( r ) to be the same as in Eq. (9) and the vector potential as V$_{v}$ ( x ) = − α$_{s}$/x . Here α$_{s}$ is the (strong) gluon-quark coupling constant of quantum chromodynamics (QCD), having a value of O (1) at low energies and becoming small at high energies (asymptotic freedom). 20 A. When V$_{v}$ ( x ) Is Coulombic IV. COMBINING THE TWO POTENTIALS g ( ϵ ) = a 0 ϵ , and f ( ϵ ) = a 0 (1 − ˜ E ) ϵ $^{2}$/ 3 , (48)