正十二面体惑星の重力ポテンシャルの多重極展開
正十二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、正十二面体の各面を底面とし、中心を頂点とする正五角錐について積分を行い、\(12\)個の正五角錐について総和をとる。底面が\((b\cos (2k\pi /5),b\sin (2k\pi /5),h)\),\((k=0,1,2,3,4)\)を頂点とし、頂点が\((0,0,0)\)の正五角錐\(V_{\Delta}\)内での積分
\[F_{l}(x,y,z)=\iiint_{V_{\Delta}}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'dy'dz'\]
を用いると、
\[\begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix}=\begin{pmatrix} x \\ y \\ z \end{pmatrix},\
\begin{pmatrix} x_{11} \\ y_{11} \\ z_{11} \end{pmatrix}=-\begin{pmatrix} x \\ y \\ z \end{pmatrix},\
\begin{pmatrix} x_{i} \\ y_{i} \\ z_{i} \end{pmatrix}=(Z^{i-1}X)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\
\begin{pmatrix} x_{i+5} \\ y_{i+5} \\ z_{i+5} \end{pmatrix}=-(Z^{i-1}X)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ (i=1,2,3,4,5)\]
\[Z=\begin{pmatrix} \cos (2\pi /5) & \sin (2\pi /5) & 0 \\ -\sin (2\pi /5) & \cos (2\pi /5) & 0 \\ 0 & 0 & 1 \end{pmatrix},\ X=\begin{pmatrix} 1/\sqrt{5} & 0 & 2/\sqrt{5} \\ 0 & -1 & 0 \\ 2/\sqrt{5} & 0 & -1/\sqrt{5} \end{pmatrix}\]
として、
\[\iiint_{V}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'dy'dz=\sum_{i=0}^{11}F_{l}(x_{i}, y_{i}, z_{i}, a)\]
である。また、正五角錐\(V_{\Delta}\)を、頂点が\((0,0,0),\ \)\((0,0,h),\ \)\((b\cos (2k\pi /5),b\sin (2k\pi /5),h),\ \)\((b\cos (2(k+1)\pi /5),b\sin (2(k+1)\pi /5),h),\ \)\((k=0,1,2,3,4)\)の\(5\)つの三角錐に分けると、積分\(F_{l}(x,y,z)\)は、
\[F_{l}(x_{i}, y_{i}, z_{i}, a)=\sum_{j=0}^{4}f_{l}(x_{j}, y_{j}, z_{j}, a),\
\begin{pmatrix} x_{j} \\ y_{j} \\ z_{j} \end{pmatrix}=(Z^{j})^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\]
\[f_{l}(x_{j}, y_{j}, z_{j}, a)=\int_{0}^{h}\Big[\int_{-b\cos(4\pi /5)}^{0}\Big[\int_{-x'\tan(\pi /5)}^{x'\tan(\pi /5)}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dy'\Big]dx'\Big]dz'\]
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、
\[IY^{m}_{2n+1}(\theta,\varphi)=Y^{m}_{2n+1}(\pi-\theta,\varphi+\pi )=(-1)^{2n+1}Y^{m}_{2n+1}(\theta,\varphi)=-Y^{m}_{2n+1}(\theta,\varphi)\]
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、
\[-\dfrac{G\rho}{r}\iiint_{V}P_{0}(cos \theta ')dx'dy'dz'=-\dfrac{G\rho(10+2\sqrt{5})\sqrt{3}a^{3}}{9r}\]
\[-\dfrac{G\rho}{r^{5}}\iiint_{V}(rr')^{2}P_{2}(cos \theta ')dx'dy'dz'= 0\]
\[-\dfrac{G\rho}{r^{9}}\iiint_{V}(rr')^{4}P_{4}(cos \theta ')dx'dy'dz'= 0\]
\begin{align*}
-\dfrac{G\rho}{r^{13}}\iiint_{V}(rr')^{6}P_{6}(cos \theta ')dx'dy'dz'
=-&G\rho\dfrac{11(51+19\sqrt{5})\sqrt{3}a^{9}}{163296r^{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(19+6\sqrt{5})\sqrt{3}a^{13}}{14434200r^{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(317+145\sqrt{5})\sqrt{3}a^{15}}{5458752000r^{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。\(z\)軸周りの\(1/2\)回転\((x,y,z) \to (-x,-y,z)\)に対して、\((r,\theta,\varphi) \to (r,\theta,\varphi+\pi)\)なので、球面調和関数は、
\[Y_{l}^{m}(\theta,\varphi) \to Y_{l}^{m}(\theta,\varphi+\pi)=(-1)^{m}Y_{l}^{m}(\theta,\varphi)\]
と変換するので、正二十面体惑星のときに用いた球面調和関数を用いて、
\begin{align*}
Ic_{6}(\theta,\varphi+\pi)=&\dfrac{\sqrt{11}}{5}Y_{6}^{0}(\theta,\varphi+\pi)
+\dfrac{\sqrt{7}}{5}(-Y_{6}^{5}(\theta,\varphi+\pi)+Y_{6}^{-5}(\theta,\varphi+\pi)))\\
=&\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}{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+\pi)=&\dfrac{1}{5}\sqrt{\dfrac{13\cdot 19}{3}}Y_{10}^{0}(\theta,\varphi+pi)
+\dfrac{1}{5}\sqrt{11\cdot 19}(Y_{10}^{5}(\theta,\varphi+\pi)-Y_{10}^{-5}(\theta,\varphi+\pi))
+\dfrac{1}{5}\sqrt{\dfrac{11\cdot 17}{3}}(Y_{10}^{10}(\theta,\varphi+\pi)+Y_{10}^{-10}(\theta,\varphi+\pi))\\
=&\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{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+\pi)=&\dfrac{3}{25}\sqrt{\dfrac{7\cdot 17}{5}}Y_{12}^{0}(\theta,\varphi+\pi)
+\dfrac{1}{25}\sqrt{\dfrac{2\cdot 11\cdot 13}{5}}(-Y_{12}^{5}(\theta,\varphi+\pi)+Y_{12}^{-5}(\theta,\varphi+\pi))
+\dfrac{1}{25}\sqrt{\dfrac{3\cdot 13\cdot 19}{5}}(Y_{12}^{10}(\theta,\varphi+\pi)+Y_{12}^{-10}(\theta,\varphi+\pi))\\
=&\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{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*}
となる。正二十面体惑星の総質量は\(M=\frac{\sqrt{3}}{9}(10+2\sqrt{5})\rho a^{3}\)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(40+11\sqrt{5})}{5670}\sqrt{\dfrac{\pi}{143}}Ic_{6}(\theta,\varphi+\pi)
+\Bigl(\dfrac{a}{r}\Bigr)^{10}\cdot \dfrac{38(65+11\sqrt{5})}{200475}\sqrt{\dfrac{\pi}{1729}}Ic_{10}(\theta,\varphi+\pi)
-\Bigl(\dfrac{a}{r}\Bigr)^{12}\cdot \dfrac{68(215+102\sqrt{5})}{710775}\sqrt{\dfrac{5\pi}{119}}Ic_{12}(\theta,\varphi+\pi)
+\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、
\[-\dfrac{11(40+11\sqrt{5})}{5670}\sqrt{\dfrac{\pi}{143}}=-0.01857493\cdots,
\ \dfrac{38(65+11\sqrt{5})}{200475}\sqrt{\dfrac{\pi}{1729}}=0.00072392\cdots,
\ -\dfrac{68(215+102\sqrt{5})}{710775}\sqrt{\dfrac{5\pi}{119}}=-0.0154008\cdots\]
となる。
内部ポテンシャルについての\(\rm{Taylor}\)展開
\(r = \sqrt{x^{2} + y^{2} + z^{2}} \ll a\)のとき、\(\xi =x/r,\
\)\(\eta =y/r,\ \)\(\zeta =z/r,\ \)\(t = r/a\)として、\(t\)について\(\rm{Taylor}\)展開をする。
\begin{align*}
u_{L}(x, y, z, a) =& \dfrac{h-z}{2}
[(-\dfrac{1+\sqrt{5}}{4}x+\dfrac{(1-\sqrt{5})\sqrt{10+2\sqrt{5}}}{8}y+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{12}\\
&+(\dfrac{\sqrt{5}-1}{4}x-\dfrac{\sqrt{10+2\sqrt{5}}}{4}y+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{23}\\
&+(x+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{34}\\
&+(\dfrac{\sqrt{5}-1}{4}x+\dfrac{\sqrt{10+2\sqrt{5}}}{4}y+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{45}\\
&+(-\dfrac{1+\sqrt{5}}{4}x+\dfrac{(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}}{8}y+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{51}]\\
\end{align*}
\[
u_{A}(x, y, z, a) = -\dfrac{(h-z)^{2}}{2}[S_{1}+S_{2}+S_{3}+S_{4}+S_{5}-3\pi\mathrm{sgn}(h-z)]
\]
として、まず、\(u_{L}(x, y, z, a)\)については、
\begin{align*}
r_{1}=&\sqrt{(x-b)^{2}+y^{2}+(z-h)^{2}}=a\sqrt{(t\xi-\tfrac{1}{30}\sqrt{30+6\sqrt{5}}(5-\sqrt{5}))^{2}+(t\eta)^{2}+(t\zeta-\tfrac{1}{60}\sqrt{30+6\sqrt{5}}(5+\sqrt{5}))^{2}} \\
=&a(1-\sqrt{30+6\sqrt{5}}(\dfrac{5-\sqrt{5}}{30}\xi+\dfrac{5+\sqrt{5}}{60}\zeta)t
+(\dfrac{\sqrt{5}-5}{15}\xi^{2}-\dfrac{\sqrt{5}+5}{15}\xi\zeta-\dfrac{2\sqrt{5}+5}{30}\zeta^{2}+\dfrac{1}{2})t^{2}+\cdots ),
\end{align*}
\begin{align*}
r_{2}=&\sqrt{(x- b\cos(2\pi /5))^{2}+(y-b\sin(2\pi/5))^{2}+(z-h)^{2}} \\
=&a\sqrt{(t\xi-\tfrac{1}{30}\sqrt{30+6\sqrt{5}}(5-\sqrt{5})\cos(2\pi /5))^{2}+(t\eta-\tfrac{1}{30}\sqrt{30+6\sqrt{5}}(5-\sqrt{5})\sin(2\pi /5))^{2}+(t\zeta-\tfrac{1}{60}\sqrt{30+6\sqrt{5}}(5+\sqrt{5}))^{2}} \\
=&a(1-\sqrt{3}(\sqrt{10+2\sqrt{5}}(\dfrac{3\sqrt{5}-5}{60}\xi+\dfrac{5+\sqrt{5}}{60}\zeta)+\dfrac{1}{3}\eta)t
+(\dfrac{2\sqrt{5}-5}{30}\xi^{2}-\dfrac{3\sqrt{5}-5}{60}\xi\eta-\dfrac{\sqrt{5}}{15}\xi\zeta-\dfrac{1}{6}\eta^{2}-\dfrac{\sqrt{5}+5}{60}\eta\zeta-\dfrac{2\sqrt{5}+5}{30}\zeta^{2}+\dfrac{1}{2})t^{2}+\cdots ),
\end{align*}
\begin{align*}
\tilde{L}_{12}=&\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}}\\
=&\ln\dfrac
{2+\tfrac{\sqrt{5}-1}{\sqrt{3}}-(\tfrac{\sqrt{5}+5}{60}\sqrt{30+6\sqrt{5}}(\xi+2\zeta)+\tfrac{1}{\sqrt{3}}\eta)t
+(\tfrac{4\sqrt{5}-15}{30}\xi^{2}+\tfrac{1}{6}\eta^{2}-\frac{1}{15}(2\sqrt{5}+5)(\xi\zeta+\zeta^{2})
+\frac{\sqrt{10+2\sqrt{5}}}{60}((-3\sqrt{5}+5)\xi\eta-(\sqrt{5}+5)\eta\zeta)+1)t^{2}+\cdots}
{2-\tfrac{\sqrt{5}-1}{\sqrt{3}}-(\tfrac{\sqrt{5}+5}{60}\sqrt{30+6\sqrt{5}}(\xi+2\zeta)+\tfrac{1}{\sqrt{3}}\eta)t
+(\tfrac{4\sqrt{5}-15}{30}\xi^{2}+\tfrac{1}{6}\eta^{2}-\frac{1}{15}(2\sqrt{5}+5)(\xi\zeta+\zeta^{2})
+\frac{\sqrt{10+2\sqrt{5}}}{60}((-3\sqrt{5}+5)\xi\eta-(\sqrt{5}+5)\eta\zeta)+1)t^{2}+\cdots} \\
=&\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}
+(\frac{3\sqrt{5}-5}{20}\sqrt{10+2\sqrt{5}}(\xi+2\zeta)+(\sqrt{5}-2)\eta)t \\
&+\sqrt{3}(\frac{11\sqrt{5}-23}{12}\xi^{2}+\frac{2\sqrt{5}-3}{3}\xi\zeta
+\frac{17\sqrt{5}-37}{12}\eta^{2}+\frac{2\sqrt{5}-3}{3}\zeta^{2}+(2-\sqrt{5})
+(\sqrt{10+2\sqrt{5}}(\frac{\sqrt{5}-2}{6}\xi\eta+\frac{9\sqrt{5}-19}{12}\eta\zeta))t^{2}\cdots ,
\end{align*}
\begin{align*}
f_{L}(x,y,z,a)=&\frac{h-z}{2}(-\dfrac{1+\sqrt{5}}{4}x+\dfrac{(1-\sqrt{5})\sqrt{10+2\sqrt{5}}}{8}y+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}a)\tilde{L}_{12} \\
=&a^{2}(\dfrac{\sqrt{3}}{120}(10+2\sqrt{5})^{3/2}-\zeta)(-\dfrac{1+\sqrt{5}}{4}\xi+\dfrac{(1-\sqrt{5})\sqrt{10+2\sqrt{5}}}{8}\eta+\dfrac{\sqrt{15}\sqrt{10+2\sqrt{5}}}{30}) \\
=&a^{2}[\frac{3\sqrt{5}+5}{30}\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)} \\
&+(\frac{\sqrt{3}}{120}(\sqrt{10+2\sqrt{5}}(-(3\sqrt{5}+5)\xi+8\zeta)+4(5-\sqrt{5})\eta)\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}+\frac{\sqrt{10+2\sqrt{5}}}{30}(\xi+2\zeta)+\frac{-\sqrt{5}+5}{30}\eta)t \\
&+(\sqrt{3}\{\frac{-\sqrt{5}+1}{18}\xi^{2}+\frac{\sqrt{5}-15}{90}\xi\zeta+\frac{\sqrt{5}-5}{90}\eta^{2}+\frac{7\sqrt{5}-15}{90}\zeta^{2}+\frac{\sqrt{5}-5}{30}
+\frac{\sqrt{10+2\sqrt{5}}}{90}((\sqrt{5}-5)\xi\eta+(\sqrt{5}-15)\eta\zeta)\} \\
&+(\frac{\sqrt{5}+1}{4}\xi\zeta+\frac{\sqrt{5}-1}{8}\sqrt{10+2\sqrt{5}}\eta\zeta)\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)})t^{2}+\cdots]
\end{align*}
\[u_{L}(x,y,z,a)=\sum_{j=0}^{4}f_{L}(x_{j}, y_{j}, z_{j}, a),\
\begin{pmatrix} x_{j} \\ y_{j} \\ z_{j} \end{pmatrix}=(Z^{j})^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\]
より、
\begin{align*}
u_{L}(x,y,z,a)=&a^{2}[\frac{3\sqrt{5}+5}{6}\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}
+(\frac{\sqrt{10+2\sqrt{5}}}{3}-\frac{\sqrt{150+30\sqrt{5}}}{6}\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)})\zeta t
\\
&+(-\frac{\sqrt{15}}{9}(\xi^{2}+\eta^{2})+\frac{7\sqrt{15}-15\sqrt{3}}{18}\zeta^{2}+\frac{\sqrt{15}-5\sqrt{3}}{6})t^{2}+\cdots]
\end{align*}
\[\sum_{i=0}^{11}u_{L}(x_{i}, y_{i}, z_{i}, a)
=a^{2}[(3\sqrt{5}+5)\ln\frac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}
+\frac{4}{3}(2\sqrt{15}-10\sqrt{3})t^{2}+\cdots]\]
である。次に、\(u_{A}(x, y, z, a)\)については、
\begin{align*}
S_{1}=&\arctan \dfrac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}(h-z)r_{1}}{-[2x^{2}-4bx+(4\sqrt{5}-10)y^{2}+2b^{2}+4(\sqrt{5}-2)(h-z)^{2}]}\\
=&\arctan \dfrac{\sqrt{3}(1+\tfrac{\sqrt{5}-3}{4}\sqrt{30+6\sqrt{5}}\zeta t)
\sqrt{
1-\tfrac{1}{60}\sqrt{30+6\sqrt{5}}(10-2\sqrt{5})\xi+(5+\sqrt{5}){60}\zeta)t
+\tfrac{1}{30}((2\sqrt{5}-10)\xi^{2}-(2\sqrt{5}+10)\xi\zeta-(2\sqrt{5}+5)\zeta^{2}+15)t^{2}+\cdots
}}
{-[1+\tfrac{\sqrt{5}-5}{10}\sqrt{30+6\sqrt{5}}\xi t+(\tfrac{3}{2}\xi^{2}+\tfrac{1}{2}(6\sqrt{5}-15)\eta^{2}+(3\sqrt{5}-6)\zeta^{2})t^{2}]} \\
=&\arctan [-\sqrt{3}
+\sqrt{10+2\sqrt{5}}(\frac{\sqrt{5}-5}{5}\xi+\frac{-8\sqrt{5}+20}{5}\zeta)t
+\sqrt{3}(\frac{22\sqrt{5}-65}{30}\xi^{2}+\frac{-116\sqrt{5}+260}{15}\xi\zeta+\frac{6\sqrt{5}-15}{2}\eta^{2}+\frac{428\sqrt{5}-925}{30}\zeta^{2}-\frac{1}{2})t^{2}+\cdots] \\
=&\frac{2\pi}{3}
+\sqrt{10+2\sqrt{5}}(\frac{\sqrt{5}-5}{20}\xi+\frac{-2\sqrt{5}+5}{5}\zeta)t
+\sqrt{3}(\frac{2\sqrt{5}-1}{24}\xi^{2}+\frac{-\sqrt{5}+1}{3}\xi\zeta+\frac{6\sqrt{5}-15}{8}\eta^{2}+\frac{-20\sqrt{5}+55}{24}\zeta^{2}-\frac{1}{8})t^{2}+\cdots
\end{align*}
\begin{align*}
g(x,y,z,a)=\frac{(h-z)^{2}}{2}S_{1}=&a^{2}[\frac{\pi}{45}(2\sqrt{5}+5)
+\sqrt{10+2\sqrt{5}}(-\frac{\sqrt{5}+3}{120}\xi+\frac{1}{30}\zeta-\frac{\sqrt{15}+\sqrt{5}}{60}\pi\zeta)t \\
&+(\sqrt{3}\{\frac{8\sqrt{5}+15}{720}\xi^{2}+\frac{1}{9}\xi\zeta-\frac{1}{16}\eta^{2}+\frac{58\sqrt{5}-165}{720}\zeta^{2}
-\frac{2\sqrt{5}+5}{240}\}+\frac{\pi}{3}\zeta^{2})t^{2}+\cdots]
\end{align*}
\begin{align*}
&\dfrac{(h-z)^{2}}{2}(S_{1}+S_{2}+S_{3}+S_{4}+S_{5})=\sum_{j=0}^{4}g(x_{j}, y_{j}, z_{j}, a) \\
=&a^{2}[\frac{\pi}{9}(2\sqrt{5}+5)
+\sqrt{10+2\sqrt{5}}(\frac{1}{6}\zeta-\frac{\sqrt{15}+\sqrt{5}}{12}\pi\zeta)t
+(\sqrt{3}\{\frac{4\sqrt{5}-15}{144}(\xi^{2}+\eta^{2})+\frac{58\sqrt{5}-165}{144}\zeta^{2}
-\frac{2\sqrt{5}+5}{48}\}+\frac{5\pi}{3}\zeta^{2})t^{2}+\cdots]
\end{align*}
より、
\[
u_{A}(x,y,z,a)=-a^{2}[\frac{\pi}{90}(2\sqrt{5}+5)
+\sqrt{10+2\sqrt{5}}(\frac{1}{6}\zeta-\frac{\sqrt{15}+\sqrt{5}}{12}\pi\zeta)t
+(\sqrt{3}\{\frac{4\sqrt{5}-15}{144}(\xi^{2}+\eta^{2})+\frac{58\sqrt{5}-165}{144}\zeta^{2}
-\frac{2\sqrt{5}+5}{48}\}+\frac{\pi}{6}\zeta^{2})t^{2}+\cdots]
\]
\[\sum_{i=0}^{11}u_{A}(x_{i}, y_{i}, z_{i}, a)
=-a^{2}[\frac{\pi}{15}(4\sqrt{5}+10)
+(\frac{4}{3}(2\sqrt{15}-10\sqrt{3})+\frac{2\pi}{3})t^{2}+\cdots]\]
である。\(u_{L}(x, y, z, a)\)と\(u_{A}(x, y, z, a)\)の結果をまとめると、
\[
U(x,y,z)=G\rho [(-(5+3\sqrt{5})\ln{\dfrac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}}+\dfrac{10+4\sqrt{5}}{15}\pi)a^{2} + \dfrac{2\pi}{3}r^{2}+\cdots]
\]
となる。さらに高次まで\(\rm{Taylor}\)展開を行うと以下のようになる。
\begin{align*}
U(x,y,z)=&G\rho \{(-(5+3\sqrt{5})\ln{\dfrac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}}
+\frac{(10+4\sqrt{5})\pi}{15})a^{2}+\frac{2\pi}{3}r^{2} \\
&+\dfrac{1}{a^{4}}\cdot\dfrac{219\sqrt{15}-473\sqrt{3}}{432}
[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}]\\
&+\dfrac{1}{a^{8}}\cdot\dfrac{2467569\sqrt{15}-5515985\sqrt{3}}{777600}
[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}] \\
&+\dfrac{1}{a^{10}}\cdot\dfrac{413507010\sqrt{15}-924768227\sqrt{3}}{28512000}
[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}]+\cdots\}
\end{align*}
内部ポテンシャルについても、外部ポテンシャルと同様に正二十面体の対称性をもつ球面調和関数を用いて表すと以下のようになる。
\begin{align*}
U(x,y,z)=&G\rho a^{2}[(-(5+3\sqrt{5})\ln{\dfrac{2\sqrt{3}+(\sqrt{5}-1)}{2\sqrt{3}-(\sqrt{5}-1)}}
+\frac{(10+4\sqrt{5})\pi}{15})
+\frac{2\pi}{3}\Bigl(\dfrac{r}{a}\Bigr)^{2} \\
&-\Bigl(\dfrac{r}{a}\Bigr)^{6}\cdot\dfrac{(438\sqrt{15}-946\sqrt{3})}{27}\sqrt{\dfrac{\pi}{143}}Ic_{6}(\theta,\varphi+\pi)
+\Bigl(\dfrac{r}{a}\Bigr)^{10}\cdot\dfrac{79599\sqrt{15}-177935\sqrt{3}}{1215}\sqrt{\dfrac{\pi}{1729}}Ic_{10}(\theta,\varphi+\pi) \\
&-\Bigl(\dfrac{r}{a}\Bigr)^{12}\cdot\dfrac{12578160\sqrt{15}-28129832\sqrt{3}}{66825}\sqrt{\dfrac{5\pi}{119}}Ic_{12}(\theta,\varphi+\pi)+\cdots ]
\end{align*}