正二十面体惑星の重力ポテンシャルの多重極展開

　正二十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、正二十面体の各面を底面とし、中心を頂点とする正三角錐について積分を行い、\(20\)個の正三角錐について総和をとる。頂点\((0,0,a)\), \((\tfrac{2}{\sqrt{5}}a\cos (4\pi /5),\tfrac{2}{\sqrt{5}}a\sin (4\pi /5),\tfrac{1}{\sqrt{5}}a)\), \((\tfrac{2}{\sqrt{5}}a\cos (4\pi /5),-\tfrac{2}{\sqrt{5}}a\sin (4\pi /5),\tfrac{1}{\sqrt{5}}a)\)の正三角形を底面とし、 \((0,0,0)\)を頂点とする正三角錐について積分を行うため、以下の座標変換を行う。 \[u=\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}x+\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}z, \ w=-\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}x+\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}z,\] \[u'=\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}x'+\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}z', \ w'=-\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}x'+\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}z',\] 頂点\((0,0,a),\ \) \((\tfrac{2}{\sqrt{5}}a\cos (4\pi /5),\pm \tfrac{2}{\sqrt{5}}a\sin (4\pi /5),\tfrac{1}{\sqrt{5}}a)\)は上記の座標変換により、 \((\tfrac{1}{30}\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}a,0,\tfrac{1}{60}\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}a),\ \)\((-\tfrac{1}{60}\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}a,\pm \tfrac{2}{\sqrt{5}}a\sin (4\pi /5),\tfrac{1}{60}\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}a)\)に移るので、関数\(f(x,y,z)\)の正三角錐\(V_{\Delta}\)内での積分は、 \[\iiint_{V_{\Delta}}f(x',y',z')dx'dy'dz' =\int_{0}^{w_{max}}\Big[\int_{-u_{max}/2}^{u_{max}}\Big[\int_{-y_{max}}^{y_{max}}f(u',y',w')dy'\Big]du'\Big]dw' \] \[w_{max}=\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}a,\ u_{max}=(3-\sqrt{5})w',\ y_{max}=\frac{1}{\sqrt{3}}(u_{max}-u')\] 　と表される。そして、 \[F_{l}(x,y,z)=\iiint_{V_{\Delta}}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'dy'dz'\] とすると、 \[\begin{pmatrix} x_{i} \\ y_{i} \\ z_{i} \end{pmatrix}=Z^{-i}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ \begin{pmatrix} x_{i+5} \\ y_{i+5} \\ z_{i+5} \end{pmatrix}=(Z^{i}Y)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\] \[\begin{pmatrix} x_{i+10} \\ y_{i+10} \\ z_{i+10} \end{pmatrix}=(Z^{i}X)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ \begin{pmatrix} x_{i+15} \\ y_{i+15} \\ z_{i+15} \end{pmatrix}=(Z^{i}YX)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ (i=0,1,2,3,4)\] \[Z=\begin{pmatrix} \cos (2\pi /5) & \sin (2\pi /5) & 0 \\ -\sin (2\pi /5) & \cos (2\pi /5) & 0 \\ 0 & 0 & 1 \end{pmatrix},\ Y=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix},\ X=\begin{pmatrix} 1/\sqrt{5} & 0 & -2/\sqrt{5} \\ 0 & -1 & 0 \\ -2/\sqrt{5} & 0 & -1/\sqrt{5} \end{pmatrix}\] として、 \[\iiint_{V}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'dy'dz=\sum_{i=0}^{19}F_{l}(x_{i}, y_{i}, z_{i}, a)\] である。
　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、 \[IY^{m}_{2n+1}(\theta,\varphi)=Y^{m}_{2n+1}(\pi-\theta,\varphi+\pi )=(-1)^{2n+1}Y^{m}_{2n+1}(\theta,\varphi)=-Y^{m}_{2n+1}(\theta,\varphi)\] 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、 \[-\dfrac{G\rho}{r}\iiint_{V}P_{0}(cos \theta ')dx'dy'dz'=-\dfrac{2G\rho\sqrt{10+2\sqrt{5}}a^{3}}{3r}\] \[-\dfrac{G\rho}{r^{5}}\iiint_{V}(rr')^{2}P_{2}(cos \theta ')dx'dy'dz'= 0\] \[-\dfrac{G\rho}{r^{9}}\iiint_{V}(rr')^{4}P_{4}(cos \theta ')dx'dy'dz'= 0\] \begin{align*} -\dfrac{G\rho}{r^{13}}\iiint_{V}(rr')^{6}P_{6}(cos \theta ')dx'dy'dz' =&G\rho\dfrac{11(8+\sqrt{5})\sqrt{10+2\sqrt{5}}a^{9}}{50400r^{13}} [x^{6} - \frac{42}{5}x^{5}z + 3x^{4}y^{2} - 18x^{4}z^{2} + 84x^{3}y^{2}z \\ &+ 3x^{2}y^{4} - 36x^{2}y^{2}z^{2} + 24x^{2}z^{4} - 42xy^{4}z + y^{6} - 18y^{4}z^{2} + 24y^{2}z^{4} - \frac{16}{5}z^{6}] \end{align*} \[-\dfrac{G\rho}{r^{17}}\iiint_{V}(rr')^{8}P_{8}(cos \theta ')dx'dy'dz'= 0\] \begin{align*} -\dfrac{G\rho}{r^{21}}\iiint_{V}(rr')^{10}P_{10}(cos \theta ')dx'dy'dz' =&-G\rho\dfrac{589\sqrt{10+2\sqrt{5}}a^{13}}{3300000r^{21}} [x^{10} - \dfrac{495}{31}x^{9}z - \dfrac{4365}{62}x^{8}y^{2} + \dfrac{1575}{62}x^{8}z^{2} + \dfrac{3960}{31}x^{7}y^{2}z \\ &+ \dfrac{4620}{31}x^{7}z^{3} + \dfrac{9660}{31}x^{6}y^{4} + \dfrac{3150}{31}x^{6}y^{2}z^{2} - \dfrac{4200}{31}x^{6}z^{4} + \dfrac{6930}{31}x^{5}y^{4}z - \dfrac{41580}{31}x^{5}y^{2}z^{3} \\ &- \dfrac{5544}{31}x^{5}z^{5} - \dfrac{9975}{31}x^{4}y^{6} + \dfrac{4725}{31}x^{4}y^{4}z^{2} - \dfrac{12600}{31}x^{4}y^{2}z^{4} + \dfrac{5040}{31}x^{4}z^{6} - \dfrac{23100}{31}x^{3}y^{4}z^{3} \\ &+ \dfrac{55440}{31}x^{3}y^{2}z^{5} + \dfrac{2025}{31}x^{2}y^{8} + \dfrac{3150}{31}x^{2}y^{6}z^{2} - \dfrac{12600}{31}x^{2}y^{4}z^{4} + \dfrac{10080}{31}x^{2}y^{2}z^{6} \\ &- \dfrac{1440}{31}x^{2}z^{8} - \dfrac{2475}{31}xy^{8}z + \dfrac{23100}{31}xy^{6}z^{3} - \dfrac{27720}{31}xy^{4}z^{5} - \dfrac{125}{62}y^{10} + \dfrac{1575}{62}y^{8}z^{2} \\ &- \dfrac{4200}{31}y^{6}z^{4} + \dfrac{5040}{31}y^{4}z^{6} - \dfrac{1440}{31}y^{2}z^{8} + \dfrac{64}{31}z^{10}] \end{align*} \begin{align*} -\dfrac{G\rho}{r^{25}}\iiint_{V}(rr')^{12}P_{12}(cos \theta ')dx'dy'dz' =&-G\rho\dfrac{4471(15+2\sqrt{5})\sqrt{10+2\sqrt{5}}a^{15}}{312000000r^{25}} [x^{12} + \dfrac{2860}{263}x^{11}z - \dfrac{23122}{263}x^{10}y^{2} + \dfrac{5764}{263}x^{10}z^{2} \\ &- \dfrac{20020}{263}x^{9}y^{2}z - \dfrac{45760}{263}x^{9}z^{3} + \dfrac{78045}{263}x^{8}y^{4} + \dfrac{572220}{263}x^{8}y^{2}z^{2} - \dfrac{138600}{263}x^{8}z^{4} - \dfrac{62920}{263}x^{7}y^{4}z \\ &+ \dfrac{366080}{263}x^{7}y^{2}z^{3} + \dfrac{128128}{263}x^{7}z^{5} - \dfrac{4620}{263}x^{6}y^{6} - \dfrac{2115960}{263}x^{6}y^{4}z^{2} - \dfrac{554400}{263}x^{6}y^{2}z^{4} \\ &+ \dfrac{295680}{263}x^{6}z^{6} - \dfrac{40040}{263}x^{5}y^{6}z + \dfrac{640640}{263}x^{5}y^{4}z^{3} - \dfrac{1153152}{263}x^{5}y^{2}z^{5} - \dfrac{73216}{263}x^{5}z^{7} \\ &- \dfrac{84975}{263}x^{4}y^{8} + \dfrac{2448600}{263}x^{4}y^{6}z^{2} - \dfrac{831600}{263}x^{4}y^{4}z^{4} + \dfrac{887040}{263}x^{4}y^{2}z^{6} - \dfrac{190080}{263}x^{4}z^{8} \\ &+ \dfrac{14300}{263}x^{3}y^{8}z - \dfrac{640640}{263}x^{3}y^{4}z^{5} + \dfrac{732160}{263}x^{3}y^{2}z^{7} + \dfrac{20350}{263}x^{2}y^{10} - \dfrac{405900}{263}x^{2}y^{8}z^{2} \\ &- \dfrac{554400}{263}x^{2}y^{6}z^{4} + \dfrac{887040}{263}x^{2}y^{4}z^{6} - \dfrac{380160}{263}x^{2}y^{2}z^{8} + \dfrac{33792}{263}x^{2}z^{10} + \dfrac{14300}{263}xy^{10}z \\ &- \dfrac{228800}{263}xy^{8}z^{3} + \dfrac{640640}{263}xy^{6}z^{5} - \dfrac{366080}{263}xy^{4}z^{7} - \dfrac{725}{263}y^{12} + \dfrac{27500}{263}y^{10}z^{2} \\ &- \dfrac{138600}{263}y^{8}z^{4} + \dfrac{295680}{263}y^{6}z^{6} - \dfrac{190080}{263}y^{4}z^{8} + \dfrac{33792}{263}y^{2}z^{10} - \dfrac{1024}{263}z^{12}] \end{align*} となる。一方、正二十面体の対称性をもつ球面調和関数は以下のとおりである。 \begin{align*} Ic_{6}(\theta,\varphi)=&\dfrac{\sqrt{11}}{5}Y_{6}^{0}(\theta,\varphi) +\dfrac{\sqrt{7}}{5}(-Y_{6}^{5}(\theta,\varphi)+Y_{6}^{-5}(\theta,\varphi)))\\ =&\dfrac{1}{160}\sqrt{\dfrac{11\cdot 13}{\pi}}(231\cos^{6}\theta-315\cos^{4}\theta+105\cos^{2}\theta-5) +\dfrac{21}{160}\sqrt{\dfrac{11\cdot 13}{\pi}}\sin^{5}\theta\cos\theta\cdot[e^{5i\varphi}+e^{-5i\varphi}]\\ =&\dfrac{1}{160}\sqrt{\dfrac{143}{\pi}}r^{-6}(231z^{6}-315z^{4}r^{2}+105z^{2}r^{4}-5r^{6}) +\dfrac{21}{160}\sqrt{\dfrac{143}{\pi}}r^{-6}z[(x+iy)^{5}+(x-iy)^{5}]\\ =&-\dfrac{1}{32}\sqrt{\dfrac{143}{\pi}}r^{-6} [x^{6} - \frac{42}{5}x^{5}z + 3x^{4}y^{2} - 18x^{4}z^{2} + 84x^{3}y^{2}z + 3x^{2}y^{4} - 36x^{2}y^{2}z^{2} + 24x^{2}z^{4} - 42xy^{4}z + y^{6} - 18y^{4}z^{2} + 24y^{2}z^{4} - \frac{16}{5}z^{6}] \end{align*} \begin{align*} Ic_{10}(\theta,\varphi)=&\dfrac{1}{5}\sqrt{\dfrac{13\cdot 19}{3}}Y_{10}^{0}(\theta,\varphi) +\dfrac{1}{5}\sqrt{11\cdot 19}(Y_{10}^{5}(\theta,\varphi)-Y_{10}^{-5}(\theta,\varphi)) +\dfrac{1}{5}\sqrt{\dfrac{11\cdot 17}{3}}(Y_{10}^{10}(\theta,\varphi)+Y_{10}^{-10}(\theta,\varphi))\\ =&\dfrac{1}{2560}\sqrt{\dfrac{7\cdot 13\cdot 19}{\pi}}(46189\cos^{10}\theta - 109395\cos^{8}\theta + 90090\cos^{6}\theta - 30030\cos^{4}\theta + 3465\cos^{2}\theta - 63)\\ &+\dfrac{33}{1280}\sqrt{\dfrac{7\cdot 13\cdot 19}{\pi}}\sin^{5}\theta(323\cos^{5}\theta - 170\cos^{3}\theta + 15\cos\theta)\cdot[e^{5i\varphi}+e^{-5i\varphi}] +\dfrac{187}{5120}\sqrt{\dfrac{7\cdot 13\cdot 19}{\pi}}\sin^{10}\theta\cdot[e^{10i\varphi}+e^{-10i\varphi}]\\ =&\dfrac{1}{2560}\sqrt{\dfrac{1729}{\pi}}r^{-10}(46189z^{10} - 109395z^{8}r^{2} + 90090z^{6}r^{4} - 30030z^{4}r^{6} + 3465z^{2}r^{8} - 63r^{10})\\ &+\dfrac{33}{1280}\sqrt{\dfrac{1729}{\pi}}r^{-10}(323z^{5} - 170z^{3}r^{2} + 15zr^{4})[(x+iy)^{5}+(x-iy)^{5}] +\dfrac{187}{5120}\sqrt{\dfrac{1729}{\pi}}r^{-10}[(x+iy)^{10}+(x-iy)^{10}]\\ =&\dfrac{31}{640}\sqrt{\dfrac{1729}{\pi}}r^{-10} [x^{10} - \dfrac{495}{31}x^{9}z - \dfrac{4365}{62}x^{8}y^{2} + \dfrac{1575}{62}x^{8}z^{2} + \dfrac{3960}{31}x^{7}y^{2}z + \dfrac{4620}{31}x^{7}z^{3} + \dfrac{9660}{31}x^{6}y^{4} + \dfrac{3150}{31}x^{6}y^{2}z^{2} \\ &- \dfrac{4200}{31}x^{6}z^{4} + \dfrac{6930}{31}x^{5}y^{4}z - \dfrac{41580}{31}x^{5}y^{2}z^{3} - \dfrac{5544}{31}x^{5}z^{5} - \dfrac{9975}{31}x^{4}y^{6} + \dfrac{4725}{31}x^{4}y^{4}z^{2} - \dfrac{12600}{31}x^{4}y^{2}z^{4} + \dfrac{5040}{31}x^{4}z^{6} \\ &- \dfrac{23100}{31}x^{3}y^{4}z^{3} + \dfrac{55440}{31}x^{3}y^{2}z^{5} + \dfrac{2025}{31}x^{2}y^{8} + \dfrac{3150}{31}x^{2}y^{6}z^{2} - \dfrac{12600}{31}x^{2}y^{4}z^{4} + \dfrac{10080}{31}x^{2}y^{2}z^{6} - \dfrac{1440}{31}x^{2}z^{8} \\ &- \dfrac{2475}{31}xy^{8}z + \dfrac{23100}{31}xy^{6}z^{3} - \dfrac{27720}{31}xy^{4}z^{5} - \dfrac{125}{62}y^{10} + \dfrac{1575}{62}y^{8}z^{2} - \dfrac{4200}{31}y^{6}z^{4} + \dfrac{5040}{31}y^{4}z^{6} - \dfrac{1440}{31}y^{2}z^{8} + \dfrac{64}{31}z^{10}] \end{align*} \begin{align*} Ic_{12}(\theta,\varphi)=&\dfrac{3}{25}\sqrt{\dfrac{7\cdot 17}{5}}Y_{12}^{0}(\theta,\varphi) +\dfrac{1}{25}\sqrt{\dfrac{2\cdot 11\cdot 13}{5}}(-Y_{12}^{5}(\theta,\varphi)+Y_{12}^{-5}(\theta,\varphi)) +\dfrac{1}{25}\sqrt{\dfrac{3\cdot 13\cdot 19}{5}}(Y_{12}^{10}(\theta,\varphi)+Y_{12}^{-10}(\theta,\varphi))\\ =&\dfrac{3}{51200}\sqrt{\dfrac{7\cdot 17}{5\pi}}(3380195\cos^{12}\theta - 9699690\cos^{10}\theta + 10392525\cos^{8}\theta - 5105100\cos^{6}\theta + 1126125\cos^{4}\theta - 90090\cos^{2}\theta + 1155)\\ &-\dfrac{143}{25600}\sqrt{\dfrac{7\cdot 17}{5\pi}}\sin^{5}\theta(-6555\cos^{7}\theta + 5985\cos^{5}\theta - 1425\cos^{3}\theta + 75\cos\theta)\cdot[e^{5i\varphi}+e^{-5i\varphi}]\\ &+\dfrac{741}{51200}\sqrt{\dfrac{7\cdot 17}{5\pi}}\sin^{10}\theta(115\cos^2\theta - 5)\cdot[e^{10i\varphi}+e^{-10i\varphi}]\\ =&\dfrac{3}{10240}\sqrt{\dfrac{119}{5\pi}}r^{-12}(676039z^{12} - 1939938z^{10}r^{2} + 2078505z^{8}r^{4} - 1021020z^{6}r^{6} + 225225z^{4}r^{8} - 18018z^{2}r^{10} + 231r^{12})\\ &-\dfrac{143}{5120}\sqrt{\dfrac{119}{5\pi}}r^{-12}(-1311z^{7} + 1197z^{5}r^{2} - 285z^{3}r^{4} + 15zr^{6})((x+iy)^{5}+(x-iy)^{5})\\ &+\dfrac{741}{10240}\sqrt{\dfrac{119}{5\pi}}r^{-12}(23z^{2} - r^{2})((x+iy)^{10}+(x-iy)^{10})\\ =&-\dfrac{789}{10240}\sqrt{\dfrac{119}{5\pi}}r^{-12}[x^{12} + \dfrac{2860}{263}x^{11}z - \dfrac{23122}{263}x^{10}y^{2} + \dfrac{5764}{263}x^{10}z^{2} - \dfrac{20020}{263}x^{9}y^{2}z - \dfrac{45760}{263}x^{9}z^{3} + \dfrac{78045}{263}x^{8}y^{4} + \dfrac{572220}{263}x^{8}y^{2}z^{2} \\ &- \dfrac{138600}{263}x^{8}z^{4} - \dfrac{62920}{263}x^{7}y^{4}z + \dfrac{366080}{263}x^{7}y^{2}z^{3} + \dfrac{128128}{263}x^{7}z^{5} - \dfrac{4620}{263}x^{6}y^{6} - \dfrac{2115960}{263}x^{6}y^{4}z^{2} - \dfrac{554400}{263}x^{6}y^{2}z^{4} \\ &+ \dfrac{295680}{263}x^{6}z^{6} - \dfrac{40040}{263}x^{5}y^{6}z + \dfrac{640640}{263}x^{5}y^{4}z^{3} - \dfrac{1153152}{263}x^{5}y^{2}z^{5} - \dfrac{73216}{263}x^{5}z^{7} - \dfrac{84975}{263}x^{4}y^{8} + \dfrac{2448600}{263}x^{4}y^{6}z^{2} - \dfrac{831600}{263}x^{4}y^{4}z^{4} \\ &+ \dfrac{887040}{263}x^{4}y^{2}z^{6} - \dfrac{190080}{263}x^{4}z^{8} + \dfrac{14300}{263}x^{3}y^{8}z - \dfrac{640640}{263}x^{3}y^{4}z^{5} + \dfrac{732160}{263}x^{3}y^{2}z^{7} + \dfrac{20350}{263}x^{2}y^{10} - \dfrac{405900}{263}x^{2}y^{8}z^{2} \\ &- \dfrac{554400}{263}x^{2}y^{6}z^{4} + \dfrac{887040}{263}x^{2}y^{4}z^{6} - \dfrac{380160}{263}x^{2}y^{2}z^{8} + \dfrac{33792}{263}x^{2}z^{10} + \dfrac{14300}{263}xy^{10}z - \dfrac{228800}{263}xy^{8}z^{3} + \dfrac{640640}{263}xy^{6}z^{5} \\ &- \dfrac{366080}{263}xy^{4}z^{7} - \dfrac{725}{263}y^{12} + \dfrac{27500}{263}y^{10}z^{2} - \dfrac{138600}{263}y^{8}z^{4} + \dfrac{295680}{263}y^{6}z^{6} - \dfrac{190080}{263}y^{4}z^{8} + \dfrac{33792}{263}y^{2}z^{10} - \dfrac{1024}{263}z^{12}] \end{align*} 正二十面体惑星の総質量は\(M=\frac{2\sqrt{10+2\sqrt{5}}\rho a^{3}}{3}\)なので、重力ポテンシャルは、 \begin{align*} U(r,\theta,\varphi)=&-\dfrac{GM}{r}[1 +\Bigl(\dfrac{a}{r}\Bigr)^{6}\cdot \dfrac{11(8+\sqrt{5})}{1050}\sqrt{\dfrac{\pi}{143}}Ic_{6}(\theta,\varphi) +\Bigl(\dfrac{a}{r}\Bigr)^{10}\cdot \dfrac{19}{34375}\sqrt{\dfrac{\pi}{1729}}Ic_{10}(\theta,\varphi) -\Bigl(\dfrac{a}{r}\Bigr)^{12}\cdot \dfrac{34(15+2\sqrt{5})}{121875}\sqrt{\dfrac{5\pi}{119}}Ic_{12}(\theta,\varphi) +\cdots] \end{align*} と表される。最低次の非球対称の項は\(6\)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、 \[\dfrac{11(8+\sqrt{5})}{1050}\sqrt{\dfrac{\pi}{143}}=0.01589437\cdots, \ \dfrac{19}{34375}\sqrt{\dfrac{\pi}{1729}}=0.00002356\cdots, \ -\dfrac{34(15+2\sqrt{5})}{121875}\sqrt{\dfrac{5\pi}{119}}=-0.00197362\cdots\] となる。