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Constraints for the vector-isoscalar channel (fourth and sixth orders)

At fourth order, we again found two identical sets of linear combinations of the isoscalar and isovector coupling constants, in which the scalar and vector coupling constants are in same case connected to one another, that is,

$\displaystyle C_{00,4000}^{0000,t}$ $\textstyle =$ $\displaystyle \frac{3 C_{00,2202}^{2202,t}}{2 \sqrt{5}},$ (113)
$\displaystyle C_{00,2000}^{2000,t}$ $\textstyle =$ $\displaystyle \frac{1}{2} \sqrt{5} C_{00,2202}^{2202,t},$ (114)
$\displaystyle C_{00,3101}^{1101,t}$ $\textstyle =$ $\displaystyle -\frac{6 C_{00,2212}^{2212,t}}{\sqrt{5}},$ (115)
$\displaystyle C_{00,3110}^{1110,t}$ $\textstyle =$ $\displaystyle -2 \sqrt{\frac{3}{5}} C_{00,2202}^{2202,t},$ (116)
$\displaystyle C_{00,3111}^{1111,t}$ $\textstyle =$ $\displaystyle -\frac{6 C_{00,2202}^{2202,t}}{\sqrt{5}},$ (117)
$\displaystyle C_{00,3112}^{1112,t}$ $\textstyle =$ $\displaystyle -2 \sqrt{3} C_{00,2202}^{2202,t},$ (118)
$\displaystyle C_{00,4011}^{0011,t}$ $\textstyle =$ $\displaystyle \frac{3}{2} \sqrt{\frac{3}{5}} C_{00,2212}^{2212,t},$ (119)
$\displaystyle C_{00,2011}^{2011,t}$ $\textstyle =$ $\displaystyle \frac{1}{2} \sqrt{15} C_{00,2212}^{2212,t},$ (120)
$\displaystyle C_{00,2211}^{2211,t}$ $\textstyle =$ $\displaystyle \sqrt{\frac{3}{5}} C_{00,2212}^{2212,t},$ (121)
$\displaystyle C_{00,2213}^{2213,t}$ $\textstyle =$ $\displaystyle \sqrt{\frac{7}{5}} C_{00,2212}^{2212,t},$ (122)
$\displaystyle C_{00,2211}^{2011,t}$ $\textstyle =$ $\displaystyle C_{00,3312}^{1112,t}=C_{00,4211}^{0011,t}=0,$ (123)

with the two coupling constants $C_{40,0000}^{0000,t}$ left unrestricted. Apart from these 6 free and 20 dependent coupling constants, the vector-isoscalar channel of the CE requires that all the remaining fourth-order coupling constants are forced to be equal to zero. In particular in the Eq. (123) we showed the vanishing coupling constants, belonging to the set of ones with indices $m=0$ and $I=0$, which were found to be non-vanishing in the scalar-isoscalar channel.

At sixth order the pattern of the results is the same, and we have,

$\displaystyle C_{00,3303}^{3303,t}$ $\textstyle =$ $\displaystyle -\frac{1}{3} \sqrt{\frac{7}{15}} C_{00,4212}^{2212,t},$ (124)
$\displaystyle C_{00,6000}^{0000,t}$ $\textstyle =$ $\displaystyle -\frac{C_{00,5112}^{1112,t}}{2 \sqrt{15}},$ (125)
$\displaystyle C_{00,4000}^{2000,t}$ $\textstyle =$ $\displaystyle -\frac{7 C_{00,5112}^{1112,t}}{2 \sqrt{15}},$ (126)
$\displaystyle C_{00,4202}^{2202,t}$ $\textstyle =$ $\displaystyle -\frac{2 C_{00,5112}^{1112,t}}{\sqrt{3}},$ (127)
$\displaystyle C_{00,5101}^{1101,t}$ $\textstyle =$ $\displaystyle -\frac{3 C_{00,4212}^{2212,t}}{2 \sqrt{5}},$ (128)
$\displaystyle C_{00,3101}^{3101,t}$ $\textstyle =$ $\displaystyle -\frac{21 C_{00,4212}^{2212,t}}{10 \sqrt{5}},$ (129)
$\displaystyle C_{00,5110}^{1110,t}$ $\textstyle =$ $\displaystyle \frac{C_{00,5112}^{1112,t}}{\sqrt{5}},$ (130)
$\displaystyle C_{00,5111}^{1111,t}$ $\textstyle =$ $\displaystyle \sqrt{\frac{3}{5}} C_{00,5112}^{1112,t},$ (131)
$\displaystyle C_{00,3110}^{3110,t}$ $\textstyle =$ $\displaystyle \frac{7 C_{00,5112}^{1112,t}}{5 \sqrt{5}},$ (132)
$\displaystyle C_{00,3111}^{3111,t}$ $\textstyle =$ $\displaystyle \frac{7}{5} \sqrt{\frac{3}{5}} C_{00,5112}^{1112,t},$ (133)
$\displaystyle C_{00,3112}^{3112,t}$ $\textstyle =$ $\displaystyle \frac{7}{5} C_{00,5112}^{1112,t},$ (134)
$\displaystyle C_{00,3312}^{3312,t}$ $\textstyle =$ $\displaystyle \frac{2}{9} C_{00,5112}^{1112,t},$ (135)
$\displaystyle C_{00,3313}^{3313,t}$ $\textstyle =$ $\displaystyle \frac{2}{9} \sqrt{\frac{7}{5}} C_{00,5112}^{1112,t},$ (136)
$\displaystyle C_{00,3314}^{3314,t}$ $\textstyle =$ $\displaystyle \frac{2 C_{00,5112}^{1112,t}}{3 \sqrt{5}},$ (137)
$\displaystyle C_{00,6011}^{0011,t}$ $\textstyle =$ $\displaystyle \frac{1}{4} \sqrt{\frac{3}{5}} C_{00,4212}^{2212,t},$ (138)
$\displaystyle C_{00,4011}^{2011,t}$ $\textstyle =$ $\displaystyle \frac{7}{4} \sqrt{\frac{3}{5}} C_{00,4212}^{2212,t},$ (139)
$\displaystyle C_{00,4211}^{2211,t}$ $\textstyle =$ $\displaystyle \sqrt{\frac{3}{5}} C_{00,4212}^{2212,t},$ (140)
$\displaystyle C_{00,4213}^{2213,t}$ $\textstyle =$ $\displaystyle \sqrt{\frac{7}{5}} C_{00,4212}^{2212,t},$ (141)
$\displaystyle C_{00,4011}^{2211,t}$ $\textstyle =$ $\displaystyle C_{00,4413}^{2213,t}=C_{00,4211}^{2011,t}= 0,$ (142)
$\displaystyle C_{00,3312}^{3112,t}$ $\textstyle =$ $\displaystyle C_{00,5312}^{1112,t}=C_{00,6211}^{0011,t}=0 ,$ (143)

with the two coupling constants $C_{60,0000}^{0000,t}$ left unrestricted. Apart from these 6 free and 36 dependent coupling constants, the vector-isoscalar channel of the CE requires that all the remaining sixth-order coupling constants are forced to be equal to zero. As before, we listed explicitely the vanishing coupling constants (see Eqs. (142)-(143)), which were found to be non-vanishing in the scalar-isoscalar channel.

The general rule that we have specified at the end of the Appendix A, can be applied now to explain the results of this section, where at all the orders we found constraints that are nondiagonal in the spin space. Here, the reason is the possibility of having pairs of secondary densities that are coupled to rank $v=1$. These pairs can appear at both scalar and vector coupling constants, which results in relating them to one another.

next up previous
Next: Constraints for the vector-isovector Up: Continuity equation and local Previous: Constraints for the scalar-isovector
Jacek Dobaczewski 2011-11-11