Direct interaction energy

In nuclei, the range of interaction is significantly smaller than the typical scale of the distance at which the local density varies. Therefore, we may expand the local densities $\rho(\bboxr_1)$ and $\rho(\bboxr_2)$ around their average position, and use this expansion to calculate the direct term in Eq. (3).

Denoting the standard total ( $\mbox{{\boldmath {$R$}}}$) and relative ( $\mbox{{\boldmath {$r$}}}$) coordinates and derivatives as

$$\mbox{{\boldmath {$R$}}}= {\textstyle{\frac{1}{2}}}(\bboxr_... ... \quad\quad \mbox{{\boldmath {$r$}}} = \bboxr_1 - \bboxr_2 ,$$ (4)

$$\bbox{\nabla} = \frac{\partial}{\partial\mbox{{\boldmath {$... ...artial\bboxr_1} - \frac{\partial}{\partial\bboxr_2}\right) ,$$ (5)

we have the expansion of local densities to second order,

$$\rho(\bboxr_1) = \rho(\mbox{{\boldmath {$R$}}}+ {\textstyle{\frac{1}{2}}}\mbox{{\boldmath {$r$}}}) = \rho(\mbox{{\boldmath {$R$}}})+ {\textstyle{\frac{1}{2}}}r_a\nabl... ...frac{1}{8}}} r_a r_b \nabla_a\nabla_b\rho(\mbox{{\boldmath {$R$}}}) + \ldots ,$$ (6)

$$\rho(\bboxr_2) = \rho(\mbox{{\boldmath {$R$}}}- {\textstyle{\frac{1}{2}}}\mbox{{\boldmath {$r$}}}) = \rho(\mbox{{\boldmath {$R$}}})- {\textstyle{\frac{1}{2}}}r_a\nabl... ...frac{1}{8}}} r_a r_b \nabla_a\nabla_b\rho(\mbox{{\boldmath {$R$}}}) + \ldots ,$$ (7)

and hence

$$\rho(\bboxr_1)\rho(\bboxr_2) = \rho^2(\mbox{{\boldmath {$R$... ...$}}})][\nabla_b\rho(\mbox{{\boldmath {$R$}}})]\Big) + \ldots,$$ (8)

where we implicitly assumed the summation over the repeated Cartesian indices $a$ and $b$.

Assuming that the local potential $V(\bboxr_1,\bboxr_2)$ depends only on the distance between the interacting particles, $V(\bboxr_1,\bboxr_2)$=$V(\vert\bboxr_1$$-$$\bboxr_2\vert)$=$V(r)$, the direct interaction energy is given by the integral of a local energy density ${\cal H}^{\text{int}}_{\text{dir}}(\mbox{{\boldmath {$R$}}})$,

$${\cal E}^{\text{int}}_{\text{dir}} = \int{\rm d}^3\mbox{{\... ...{\cal H}^{\text{int}}_{\text{dir}}(\mbox{{\boldmath {$R$}}}),$$ (9)

where up to second order,

$${\cal H}^{\text{int}}_{\text{dir}}(\mbox{{\boldmath {$R$}}}... ...ig(\rho\Delta\rho -(\bbox{\nabla}\rho)^2\Big)\Big] + \ldots,$$ (10)

where the coupling constants, $V_0$ and $V_2$, are given by the lowest two moments of the interaction,

$$V_n = \int{\rm d}^3\mbox{{\boldmath {$r$}}}\,r^n V(r) = 4\pi \int{\rm d}{r}\,r^{n+2} V(r) .$$ (11)

After integrating by parts, the two second-order terms in Eq. (10) are identical, so we can equally well use:

$${\cal H}^{\text{int}}_{\text{dir}}(\mbox{{\boldmath {$R$}}}... ...\textstyle{\frac{1}{6}}} V_2 \rho\Delta\rho \Big] + \ldots,$$ (12)

We see that the separation of scales between the range of interaction and the rate of change of the local density leads to a dramatic collapse of information that is transferred from the interaction potential to the interaction energy. Namely, the two constants, $V_0$ and $V_2$, completely characterize the interaction in the direct term, and the detailed form of the potential $V(r)$ becomes irrelevant. Moreover, it can be easily checked that in this approximation, the direct energy density is exactly equal to that corresponding to the contact force corrected by the second-order gradient pseudopotential,

$$\tilde{V}(\mbox{{\boldmath {$r$}}}) = V_0\delta(\mbox{{\bold... ...rtial} \cdot\delta(\mbox{{\boldmath {$r$}}})\bbox{\partial} .$$ (13)