正八面体惑星の重力ポテンシャルの多重極展開

　正八面体惑星についても重力ポテンシャルを以下のように\(\rm{Legendre}\)多項式を用いて展開する。 \begin{align*} &\iiint_{V}\dfrac{-G\rho}{|\vec{r}-\vec{r}'|}dx'dy'dz'=\dfrac{-G\rho}{r}\iiint_{V}\sum_{l=0}^{\infty}\Bigl(\dfrac{r'}{r}\Bigr)^{l}P_{l}(cos \theta ')dx'dy'dz'\\ =& \sum_{l=0}^{\infty}\dfrac{1}{r^{2l+1}}\iiint_{V}(rr')^{l}P_{l}(cos \theta ')dx'dy'dz' = \sum_{l=0}^{\infty}\dfrac{1}{r^{2l+1}}\iiint_{V}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'dy'dz',\ (r' \lt r) \end{align*} 　上記の積分を行うため、以下の座標変換を行う。 \[u=\frac{1}{\sqrt{2}}(x+y),\ v=\frac{1}{\sqrt{2}}(y-x),\ u'=\frac{1}{\sqrt{2}}(x'+y'),\ v'=\frac{1}{\sqrt{2}}(y'-x')\] 　正八面体の\(6\)つの頂点\((\pm a,0,0),\ \)\((0,\pm a,0),\ \)\((0,0,\pm a)\)は上記の座標変換により、\((\pm a/\sqrt{2},\pm a/\sqrt{2},0),\ \)\((0,0,\pm a)\)に移るので、関数\(f(x,y,z)\)の正八面体内での積分は、 \begin{align*} & \iiint_{V}f(x',y',z')dx'dy'dz' \\ =& \int_{-a}^{0}\Big[\int_{-(1/\sqrt{2})(a+z)}^{(1/\sqrt{2})(a+z)}\Big[\int_{-(1/\sqrt{2})(a+z)}^{(1/\sqrt{2})(a+z)}f(u',v',z')du'\Big]dv'\Big]dz' +\int_{0}^{a}\Big[\int_{-(1/\sqrt{2})(a-z)}^{(1/\sqrt{2})(a-z)}\Big[\int_{-(1/\sqrt{2})(a-z)}^{(1/\sqrt{2})(a-z)}f(u',v',z')du'\Big]dv'\Big]dz' \end{align*} 　と表される。各\(l\)について積分を行うと、\(l=2n+1\)の場合、空間反転\(I\)に対し、 \[IY^{m}_{2n+1}(\theta,\phi)=Y^{m}_{2n+1}(\pi-\theta,\phi+\pi )=(-1)^{2n+1}Y^{m}_{2n+1}(\theta,\phi)=-Y^{m}_{2n+1}(\theta,\phi)\] であるので、 \[-\dfrac{G\rho}{r^{4n+3}}\iiint_{V}(rr')^{2n+1}P_{2n+1}(cos \theta ')dx'dy'dz'=0\] であり、その他の\(l\)について、 \[-\dfrac{G\rho}{r}\iiint_{V}P_{0}(cos \theta ')dx'dy'dz'=-\dfrac{4G\rho a^{3}}{3r}\] \[-\dfrac{G\rho}{r^{5}}\iiint_{V}(rr')^{2}P_{2}(cos \theta ')dx'dy'dz'= 0\] \[-\dfrac{G\rho}{r^{9}}\iiint_{V}(rr')^{4}P_{4}(cos \theta ')dx'dy'dz' =-\dfrac{G\rho a^{7}}{r^{9}}\cdot\dfrac{1}{30}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})]\] \[-\dfrac{G\rho}{r^{13}}\iiint_{V}(rr')^{6}P_{6}(cos \theta ')dx'dy'dz' =-\dfrac{G\rho a^{9}}{r^{13}}\cdot\dfrac{1}{504}[(x^{6}+y^{6}+z^{6})-\frac{15}{2}(x^{4}y^{2}+x^{2}y^{4}+y^{4}z^{2}+y^{2}z^{4}+z^{4}x^{2}+z^{2}x^{4})+90x^{2}y^{2}z^{2}]\] \[-\dfrac{G\rho}{r^{17}}\iiint_{V}(rr')^{8}P_{8}(cos \theta ')dx'dy'dz' =-\dfrac{G\rho a^{11}}{r^{17}}\cdot\dfrac{1}{160}[(x^{8}+y^{8}+z^{8})-14(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+35(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})]\] \begin{align*} -\dfrac{G\rho}{r^{21}}\iiint_{V}(rr')^{10}P_{10}(cos \theta ')dx'dy'dz' =& -\dfrac{G\rho a^{13}}{r^{21}}\cdot\dfrac{1}{1056}[(x^{10}+y^{10}+z^{10})-\frac{45}{2}(x^{8}y^{2}+x^{2}y^{8}+y^{8}z^{2}+y^{2}z^{8}+z^{8}x^{2}+z^{2}x^{8}) \\ &+21(x^{6}y^{4}+x^{4}y^{6}+y^{6}z^{4}+y^{4}z^{6}+z^{6}x^{4}+z^{4}x^{6})+504(x^{6}y^{2}z^{2}+y^{6}z^{2}x^{2}+z^{6}x^{2}y^{2})\\ &-630(x^{4}y^{4}z^{2}+y^{4}z^{4}x^{2}+z^{4}x^{4}y^{2})] \end{align*} \begin{align*} -\dfrac{G\rho}{r^{25}}\iiint_{V}(rr')^{12}P_{12}(cos \theta ')dx'dy'dz' =& -\dfrac{G\rho a^{15}}{r^{25}}\cdot\dfrac{731}{349440}[(x^{12}+y^{12}+z^{12})-33(x^{10}y^{2}+x^{2}y^{10}+y^{10}z^{2}+y^{2}z^{10}+z^{10}x^{2}+z^{2}x^{10}) \\ &+\frac{9745}{43}(x^{8}y^{4}+x^{4}y^{8}+y^{8}z^{4}+y^{4}z^{8}+z^{8}x^{4}+z^{4}x^{8})\\ &-\frac{17353}{43}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{5385}{43}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{12565}{43}(x^{6}y^{4}z^{2}+x^{6}y^{2}z^{4}+y^{6}z^{4}x^{2}+y^{6}z^{2}x^{4}+z^{6}x^{4}y^{2}+z^{6}x^{2}y^{4})+\frac{62825}{43}x^{4}y^{4}z^{4}] \end{align*} 　正八面体惑星の総質量は\(M=\frac{4}{3}\rho a^{3}\)なので、正四面体惑星のときに用いた球面調和関数を用いて、 \begin{align*} U(r,\theta,\varphi)=&-\dfrac{GM}{r}[1 +\big(\dfrac{a}{r}\big)^{4}\cdot\dfrac{1}{10}\sqrt{\dfrac{\pi}{21}}Te_{4}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{6}\cdot\dfrac{1}{168}\sqrt{\dfrac{\pi}{26}}Te_{6,o}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{8}\cdot\dfrac{3}{40}\sqrt{\dfrac{\pi}{561}}Te_{8}(\theta,\varphi)\\ &-\big(\dfrac{a}{r}\big)^{10}\cdot\dfrac{1}{2112}\sqrt{\dfrac{\pi}{910}}Te_{10,o}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{12}\cdot (\dfrac{6103}{44772000}\sqrt{\dfrac{41\pi}{11}}Te_{12,o8}(\theta,\varphi) + \dfrac{7429}{164000}\sqrt{\dfrac{246\pi}{676039}}Te_{12,o12}(\theta,\varphi)) +\cdots] \end{align*} と表される。最低次の非球対称の項は\(4\)次となる。各球面調和関数の前に現れる係数が力学的形状係数に相当するものであるが、それぞれ、 \[\dfrac{1}{10}\sqrt{\dfrac{\pi}{21}}=0.03867811\cdots, \ \dfrac{1}{168}\sqrt{\dfrac{\pi}{26}}=0.00206908\cdots, \ \dfrac{3}{40}\sqrt{\dfrac{\pi}{561}}=0.00561247\cdots,\] \[-\dfrac{1}{2112}\sqrt{\dfrac{\pi}{910}}=-0.00002782\cdots, \ \dfrac{6103}{44772000}\sqrt{\dfrac{41\pi}{11}}=0.00466452\cdots, \ \dfrac{7429}{164000}\sqrt{\dfrac{246\pi}{676039}}=0.00153159\cdots, \] となる。