P = 0 | {BBM} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, |
| | ${k}_{3}={k}_{4}^{* }={k}_{3r}+{\rm{i}}{k}_{3i},\quad {p}_{3}={p}_{4}^{* }={p}_{3r}+{\rm{i}}{p}_{3i}$, |
| | $\tfrac{{k}_{1r}}{{k}_{3r}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3r}\ {p}_{3r}-{k}_{3i}\ {p}_{3i}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3r}^{3}-3{k}_{3r}\ {k}_{3i}^{2}}$ |
P = 1 | {BBM, S} | the same as above |
P = 2 | {BBM, B} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, |
| | ${k}_{3}={k}_{4}^{* }={k}_{3r}+{\rm{i}}{k}_{3i},\quad {p}_{3}={p}_{4}^{* }={p}_{3r}+{\rm{i}}{p}_{3i}$, |
| | $\tfrac{{k}_{1r}}{{k}_{3r}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3r}\ {p}_{3r}-{k}_{3i}\ {p}_{3i}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3r}^{3}-3{k}_{3r}\ {k}_{3i}^{2}}$, |
| | ${k}_{5}={k}_{6}^{* }={k}_{5r}+{\rm{i}}{k}_{5i},\quad {p}_{5}={p}_{6}^{* }={p}_{5r}+{\rm{i}}{p}_{5i}$ |
P = 2 | {BBM, L} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, |
| | ${k}_{3}={k}_{4}^{* }={k}_{3r}+{\rm{i}}{k}_{3i},\quad {p}_{3}={p}_{4}^{* }={p}_{3r}+{\rm{i}}{p}_{3i}$, |
| | ${k}_{5}={k}_{6}^{* }={k}_{5r}+{\rm{i}}{k}_{5i},\quad {p}_{5}={p}_{6}^{* }={p}_{5r}+{\rm{i}}{p}_{5i},$ |
| | $\tfrac{{k}_{1r}}{{k}_{3r}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3r}\ {p}_{3r}-{k}_{3i}\ {p}_{3i}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3r}^{3}-3{k}_{3r}\ {k}_{3i}^{2}}$, |
| | ${k}_{5r}^{2}+{k}_{5i}^{2}\to 0$ |
P = 2 | {BBBM} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, |
| | ${k}_{3}={k}_{4}^{* }={k}_{3r}+{\rm{i}}{k}_{3i},\quad {p}_{3}={p}_{4}^{* }={p}_{3r}+{\rm{i}}{p}_{3i}$, |
| | ${k}_{5}={k}_{6}^{* }={k}_{5r}+{\rm{i}}{k}_{5i},\quad {p}_{5}={p}_{6}^{* }={p}_{5r}+{\rm{i}}{p}_{5i}$, |
| | $\tfrac{{k}_{1r}}{{k}_{3r}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3r}\ {p}_{3r}-{k}_{3i}\ {p}_{3i}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3r}^{3}-3{k}_{3r}\ {k}_{3i}^{2}}$, |
| | $\tfrac{{k}_{1r}}{{k}_{5r}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{5r}\ {p}_{5r}-{k}_{5i}\ {p}_{5i}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{5r}^{3}-3{k}_{5r}\ {k}_{5i}^{2}}$ |