正四面体惑星の重力ポテンシャルの\(\rm{Taylor}\)展開

　正四面体惑星の重力ポテンシャルについて、 \[u_{1}(x, y, z, a)=u_{L}(x, y, z, a)+u_{A}(x, y, z, a)\] \[u_{L}(x, y, z, a) = \frac{a - (x - y + z)}{6\sqrt{2}} [(x - y - 2 z + 2 a)\tilde{L}_{12}+(x + 2 y + z + 2 a)\tilde{L}_{23}+(-2 x - y + z + 2 a)\tilde{L}_{31}]\] \[u_{A}(x, y, z, a) =-\frac{(a - (x - y + z))^2}{6}[S_{1}+S_{2}+S_{3}-\pi\mathrm{sgn}(a - (x - y + z))]\] とする。

外部ポテンシャルについての\(\rm{Taylor}\)展開

\(r = \sqrt{x^{2} + y^{2} + z^{2}} \gg a\)のとき、\(\xi =x/r,\ \)\(\eta =y/r,\ \)\(\zeta =z/r,\ \)\(\alpha = a/r\)として、\(\alpha\)について\(\rm{Taylor}\)展開をすると、まず、\(u_{L}(x, y, z, a)\)については、 \[r_{1}=r\sqrt{(\xi -\alpha)^{2}+(\eta -\alpha)^{2}+(\zeta -\alpha) ^{2}}=r\sqrt{1-2\alpha(\xi+\eta+\zeta)+3\alpha^{2}} =r(1-(\xi+\eta+\zeta)\alpha+(\frac{3}{2}-\frac{1}{2}(\xi+\eta+\zeta)^{2})\alpha^{2}+\cdots ), \] \[r_{2}=r\sqrt{(\xi +\alpha)^{2}+(\eta +\alpha)^{2}+(\zeta -\alpha)^{2}}=r\sqrt{1+2\alpha(\xi+\eta-\zeta)+3\alpha^{2}} =r(1+(\xi+\eta-\zeta)\alpha+(\frac{3}{2}-\frac{1}{2}(\xi+\eta-\zeta)^{2})\alpha^{2}+\cdots ), \] \[r_{3}=r\sqrt{(\xi -\alpha)^{2}+(\eta +\alpha)^{2}+(\zeta +\alpha)^{2}}=r\sqrt{1+2\alpha(\eta+\zeta-\xi)+3\alpha^{2}} =r(1+(\eta+\zeta-\xi)\alpha+(\frac{3}{2}-\frac{1}{2}(\eta+\zeta-\xi)^{2})\alpha^{2}+\cdots ), \] \begin{align*} \tilde{L}_{12}&=\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}} =\ln\dfrac{2+(2\sqrt{2}-2\zeta)\alpha+(3-((\xi+\eta)^{2}+\zeta^{2}))\alpha^{2}+\cdots} {2-(2\sqrt{2}+2\zeta)\alpha+(3-((\xi+\eta)^{2}+\zeta^{2}))\alpha^{2}+\cdots } \\ &=2\sqrt{2}\alpha+\sqrt{2}\zeta\alpha^{2}+\sqrt{2}((\xi+\eta)^{2}+3\zeta^{2}-\frac{5}{3})\alpha^{3}+\cdots , \end{align*} \begin{align*} \tilde{L}_{23}&=\ln{\frac{r_{2}+r_{3}+r_{23}}{r_{2}+r_{3}-r_{23}}} =\ln\dfrac{2+(2\sqrt{2}+2\eta)\alpha+(3-((\zeta-\xi)^{2}+\eta^{2}))\alpha^{2}+\cdots} {2-(2\sqrt{2}-2\eta)\alpha+(3-((\zeta-\xi)^{2}+\eta^{2}))\alpha^{2}+\cdots } \\ &=2\sqrt{2}\alpha-\sqrt{2}\eta\alpha^{2}+\sqrt{2}((\zeta-\xi)^{2}+3\eta^{2}-\frac{5}{3})\alpha^{3}+\cdots , \end{align*} \begin{align*} \tilde{L}_{31}&=\ln{\frac{r_{3}+r_{1}+r_{31}}{r_{3}+r_{1}-r_{31}}} =\ln\dfrac{2+(2\sqrt{2}-2\xi)\alpha+(3-((\eta+\zeta)^{2}+\xi^{2}))\alpha^{2}+\cdots }{2-(2\sqrt{2}+2\xi)\alpha+(3-((\eta+\zeta)^{2}+\xi^{2}))\alpha^{2}+\cdots } \\ &=2\sqrt{2}\alpha+\sqrt{2}\xi\alpha^{2}+\sqrt{2}((\eta+\zeta)^{2}+3\xi^{2}-\frac{5}{3})\alpha^{3}+\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =&\frac{r^{2}}{6\sqrt{2}}(\alpha-(\xi -\eta +\zeta ))\times \\ &[(\xi-\eta-2\zeta+2\alpha)(2\sqrt{2}\alpha+\sqrt{2}\zeta\alpha^{2}+\sqrt{2}((\xi+\eta)^{2}+3\zeta^{2}-\frac{5}{3})\alpha^{3}+\cdots)\\ &+(\xi+2\eta+\zeta+2\alpha)(2\sqrt{2}\alpha-\sqrt{2}\eta\alpha^{2}+\sqrt{2}((\zeta-\xi)^{2}+3\eta^{2}-\frac{5}{3})\alpha^{3}+\cdots)\\ &+(-2\xi-\eta+\zeta+2\alpha)(2\sqrt{2}\alpha+\sqrt{2}\xi\alpha^{2}+\sqrt{2}((\eta+\zeta)^{2}+3\xi^{2}-\frac{5}{3})\alpha^{3}+\cdots)]\\ =&r^{2}[(\frac{2}{3}(\xi^{3}-\eta^{3}+\zeta^{3})+2(\xi\eta\zeta-\xi+\eta-\zeta))\alpha^{2}\\ &+(\frac{2}{3}(\xi^{3}-\eta^{3}+\zeta^{3})(\xi-\eta+\zeta)+2\xi\eta\zeta(\xi-\eta+\zeta)-\frac{2}{3}(\xi\eta+\eta\zeta-\zeta\xi)+\frac{2}{3})\alpha^{3}+\cdots] \end{align*} \begin{align*} &u_{L}(x, y, z, a) + u_{L}(-x, -y, z, a) + u_{L}(-x, y, -z, a) + u_{L}(x, -y, -z, a) \\ =&r^{2}[8\xi\eta\zeta\alpha^{2}+\frac{8}{3}(\xi^{4}+\eta^{4}+\zeta^{4}+1)\alpha^{3}+\cdots] \end{align*} 　である。次に、\(u_{A}(x, y, z, a)\)については、三角形である各面の頂点の位置ベクトルを\(\mathbf{v}_{1},\ \)\(\mathbf{v}_{2},\ \)\(\mathbf{v}_{3},\ \)として、\(\mathbf{r}=(x,y,z),\ \)\(\mathbf{r}_{1}=\mathbf{r}-\mathbf{v}_{1},\ \)\(\mathbf{r}_{2}=\mathbf{r}-\mathbf{v}_{2},\ \)\(\mathbf{r}_{3}=\mathbf{r}-\mathbf{v}_{3}\ \)、\(\mathbf{n}(|\mathbf{n}|=1)\)を面の法線ベクトルとすると、 \[ u_{A}(x, y, z, a)=-(\mathbf{r}_{1}\cdot\mathbf{n})^{2}\arctan\frac{ [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}} {|\mathbf{r}_{1}||\mathbf{r}_{2}||\mathbf{r}_{3}|+(\mathbf{r}_{1}\cdot\mathbf{r}_{2})|\mathbf{r}_{3}|+(\mathbf{r}_{2}\cdot\mathbf{r}_{3})|\mathbf{r}_{1}|+(\mathbf{r}_{3}\cdot\mathbf{r}_{1})|\mathbf{r}_{2}|} \] と表すこともできて、\(\mathbf{v}_{1}=(a,a,a),\ \)\(\mathbf{v}_{2}=(-a,-a,a),\ \)\(\mathbf{v}_{3}=(a,-a,-a)\ \)とすると、 \[ \mathbf{n}=(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}),\ \mathbf{r}_{1}\cdot\mathbf{n}=\frac{1}{\sqrt{3}}(x-y+z-a)=\frac{1}{\sqrt{3}}r(\xi-\eta+\zeta-\alpha) \] \[ (\mathbf{r}_{1}\cdot\mathbf{r}_{2})=|\mathbf{r}|^{2}-((\mathbf{v}_{1}+\mathbf{v}_{2})\cdot\mathbf{r})+(\mathbf{v}_{1}\cdot\mathbf{v}_{2})=r^{2}-2az-a^{2}=r^{2}(1-\alpha\zeta-\alpha^{2}), \] \[ (\mathbf{r}_{2}\cdot\mathbf{r}_{3})=|\mathbf{r}|^{2}-((\mathbf{v}_{2}+\mathbf{v}_{3})\cdot\mathbf{r})+(\mathbf{v}_{2}\cdot\mathbf{v}_{3})=r^{2}+2ay-a^{2}=r^{2}(1+\alpha\eta-\alpha^{2}), \] \[ (\mathbf{r}_{3}\cdot\mathbf{r}_{1})=|\mathbf{r}|^{2}-((\mathbf{v}_{3}+\mathbf{v}_{1})\cdot\mathbf{r})+(\mathbf{v}_{3}\cdot\mathbf{v}_{1})=r^{2}-2ax-a^{2}=r^{2}(1-\alpha\xi-\alpha^{2}), \] \begin{align*} |\mathbf{r}_{1}|&=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{1}\cdot\mathbf{r}+|\mathbf{v}_{1}|^{2})} =\sqrt{r^{2}-2a(x+y+z)+3a^{2}}=r\sqrt{1-2\alpha(\xi+\eta+\zeta)+3\alpha^{2}}\\ &=r(1-(\xi+\eta+\zeta)\alpha+(\frac{3}{2}-\frac{1}{2}(\xi+\eta+\zeta)^{2})\alpha^{2}+\cdots), \end{align*} \begin{align*} |\mathbf{r}_{2}|&=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{2}\cdot\mathbf{r}+|\mathbf{v}_{2}|^{2})} =\sqrt{r^{2}+2a(x+y-z)+3a^{2}}=r\sqrt{1+2\alpha(\xi+\eta-\zeta)+3\alpha^{2}}\\ &=r(1-(\xi+\eta-\zeta)\alpha+(\frac{3}{2}-\frac{1}{2}(\xi+\eta-\zeta)^{2})\alpha^{2}+\cdots), \end{align*} \begin{align*} |\mathbf{r}_{3}|&=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{3}\cdot\mathbf{r}+|\mathbf{v}_{3}|^{2})} =\sqrt{r^{2}+2a(-x+y+z)+3a^{2}}=r\sqrt{1+2\alpha(-\xi+\eta+\zeta)+3\alpha^{2}}\\ &=r(1-(-\xi+\eta+\zeta)\alpha+(\frac{3}{2}-\frac{1}{2}(\eta+\zeta-\xi)^{2})\alpha^{2}+\cdots), \end{align*} \begin{align*} [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}=&[\mathbf{v}_{1}\times\mathbf{v}_{2}]\cdot\mathbf{v}_{3} -(\mathbf{v}_{1}\times\mathbf{v}_{2}+\mathbf{v}_{2}\times\mathbf{v}_{3}+\mathbf{v}_{3}\times\mathbf{v}_{1})\cdot\mathbf{r}\\ =&4a^{3}-4(x-y+z)a^{2}=4r^{3}(\alpha^{3}-(\xi-\eta+\zeta)\alpha^{2}) \end{align*} より、 \[ \arctan\frac{ [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}} {|\mathbf{r}_{1}||\mathbf{r}_{2}||\mathbf{r}_{3}|+(\mathbf{r}_{1}\cdot\mathbf{r}_{2})|\mathbf{r}_{3}|+(\mathbf{r}_{2}\cdot\mathbf{r}_{3})|\mathbf{r}_{1}|+(\mathbf{r}_{3}\cdot\mathbf{r}_{1})|\mathbf{r}_{2}|} \\ =-(-\xi+\eta-\zeta)\alpha^{2}-((-\xi+\eta-\zeta)^{2}-1)\alpha^{3}+\cdots\] \[ u_{A}(x, y, z, a)=-\frac{1}{3}(-\xi+\eta-\zeta)^{3}\alpha^{2} +(\frac{1}{3}(-\xi+\eta-\zeta)^{4}-(-\xi+\eta-\zeta)^{2})\alpha^{3}+\cdots \] \begin{align*} &u_{A}(x, y, z, a) + u_{A}(-x, -y, z, a) + u_{A}(-x, y, -z, a) + u_{A}(x, -y, -z, a)\\ &=r^{2}[-8\xi\eta\zeta\alpha^{2} +(-\frac{8}{3}(\xi^{4}+\eta^{4}+\zeta^{4})+4(\xi^{2}+\eta^{2}+\zeta^{2})^{2}-4)\alpha^{3}+\cdots] \end{align*} 　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、 \[ U(x,y,z)=-G\rho r^{2}[\frac{8}{3}\alpha^{3}+\cdots]=-G\rho [\frac{8a^{3}}{3}r^{-1}+\cdots] \] 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\{\dfrac{8a^{3}}{3r} + \dfrac{8a^{6}}{3r^{7}}xyz -\dfrac{a^{7}}{r^{9}}\cdot\dfrac{4}{15}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &+\dfrac{a^{9}}{r^{13}}\cdot\dfrac{16}{63}[(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{a^{10}}{r^{15}}\cdot\dfrac{52}{5}xyz[(x^{4}+y^{4}+z^{4})-\frac{5}{3}(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &+\dfrac{a^{11}}{r^{17}}\cdot\dfrac{1}{5}[(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})] \\ &+\dfrac{a^{12}}{r^{19}}\cdot 16xyz[(x^{6}+y^{6}+z^{6})-\frac{7}{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})+\frac{70}{3}x^{2}y^{2}z^{2}] \\ &-\dfrac{a^{13}}{r^{21}}\cdot\dfrac{8}{11}[(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})]\\ &+\dfrac{a^{14}}{r^{23}}\cdot\dfrac{255}{7}xyz[(x^{8}+y^{8}+z^{8})-6(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+\dfrac{63}{5}(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})] \\ &+\dfrac{a^{15}}{r^{25}}\cdot\dfrac{829}{2730}[(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{112690}{829}(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{507031}{829}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{1907205}{829}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{4450145}{829}(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{22250725}{829}x^{4}y^{4}z^{4}]+\cdots\} \end{align*} 　多重極展開と同じ結果となる。

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\(\rm{Taylor}\)展開をすると、まず、\(u_{L}(x, y, z, a)\)については、 \[r_{1}=a\sqrt{(1-t\xi)^{2}+(1-t\eta)^{2}+(1-t\zeta)^{2}}=a\sqrt{3-2t(\xi+\eta+\zeta) +t^{2}} =\sqrt{3}a(1-\frac{1}{3}(\xi+\eta+\zeta)t+\frac{1}{9}(1-\xi\eta-\eta\zeta-\zeta\xi)t^{2}+\cdots ),\] \[r_{2}=a\sqrt{(1+t\xi)^{2}+(1+t\eta)^{2}+(1-t\zeta)^{2}}=a\sqrt{3+2t(\xi+\eta-\zeta) +t^{2}} =\sqrt{3}a(1+\frac{1}{3}(\xi+\eta-\zeta)t+\frac{1}{9}(1-\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots ),\] \[r_{3}=a\sqrt{(1-t\xi)^{2}+(1+t\eta)^{2}+(1+t\zeta)^{2}}=a\sqrt{3+2t(-\xi+\eta+\zeta)+t^{2}} =\sqrt{3}a(1+\frac{1}{3}(-\xi+\eta+\zeta)t+\frac{1}{9}(1+\xi\eta-\eta\zeta+\zeta\xi)t^{2}+\cdots ),\] \begin{align*} \tilde{L}_{12}&=\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}} =\ln\dfrac{2\sqrt{3}+2\sqrt{2}-(2\zeta /\sqrt{3})t+((2-2\xi\eta)/3\sqrt{3})t^{2}+\cdots} {2\sqrt{3}-2\sqrt{2}-(2\zeta /\sqrt{3})t+((2-2\xi\eta)/3\sqrt{3})t^{2}+\cdots } \\ &=\ln(5+2\sqrt{6})+\frac{2\sqrt{6}}{3}\zeta t +(\frac{2\sqrt{6}}{9}\xi\eta +\frac{2\sqrt{6}}{3}\zeta ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots , \end{align*} \begin{align*} \tilde{L}_{23}&=\ln{\frac{r_{2}+r_{3}+r_{23}}{r_{2}+r_{3}-r_{23}}} =\ln\dfrac{2\sqrt{3}+2\sqrt{2}+(2\eta /\sqrt{3})t+((2+2\zeta\xi)/3\sqrt{3})t^{2}+\cdots} {2\sqrt{3}-2\sqrt{2}+(2\eta /\sqrt{3})t+((2+2\zeta\xi)/3\sqrt{3})t^{2}+\cdots } \\ &=\ln(5+2\sqrt{6})-\frac{2\sqrt{6}}{3}\eta t +(-\frac{2\sqrt{6}}{9}\zeta\xi +\frac{2\sqrt{6}}{3}\eta ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots , \end{align*} \begin{align*} \tilde{L}_{31}&=\ln{\frac{r_{3}+r_{1}+r_{31}}{r_{3}+r_{1}-r_{31}}} =\ln\dfrac{2\sqrt{3}+2\sqrt{2}-(2\xi /\sqrt{3})t+((2-2\eta\zeta)/3\sqrt{3})t^{2}+\cdots} {2\sqrt{3}-2\sqrt{2}-(2\xi /\sqrt{3})t+((2-2\eta\zeta)/3\sqrt{3})t^{2}+\cdots } \\ &=\ln(5+2\sqrt{6})+\frac{2\sqrt{6}}{3}\xi t +(\frac{2\sqrt{6}}{9}\eta\zeta +\frac{2\sqrt{6}}{3}\xi ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =& \frac{a^{2}}{6\sqrt{2}}(1-(\xi -\eta +\zeta )t)\times \\ &\{[((\xi -\eta-2\zeta)t +2)(\ln(5+2\sqrt{6})+\frac{2\sqrt{6}}{3}\zeta t +(\frac{2\sqrt{6}}{9}\xi\eta +\frac{2\sqrt{6}}{3}\zeta ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots)]\\ &+[(( \xi+2\eta +\zeta)t +2)(\ln(5+2\sqrt{6})-\frac{2\sqrt{6}}{3}\eta t +(-\frac{2\sqrt{6}}{9}\zeta\xi +\frac{2\sqrt{6}}{3}\eta ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots)]\\ &+[((-2\xi-\eta +\zeta)t +2)(\ln(5+2\sqrt{6})+\frac{2\sqrt{6}}{3}\xi t +(\frac{2\sqrt{6}}{9}\eta\zeta +\frac{2\sqrt{6}}{3}\xi ^{2}-\frac{2\sqrt{6}}{9})t^{2}+\cdots)]\}\\ =& a^{2}[\frac{1}{\sqrt{2}}\ln(5+2\sqrt{6}) +(-\frac{1}{\sqrt{2}}\ln(5+2\sqrt{6}+\frac{2\sqrt{3}}{9})(\xi-\eta+\zeta))t +(\frac{8\sqrt{3}}{27}(\xi\eta+\eta\zeta-\zeta\xi)-\frac{4\sqrt{3}}{9})t^{2}+\cdots] \end{align*} \begin{align*} &u_{L}(x, y, z, a)+u_{L}(x, -y, -z, a)+u_{L}(-x, y, -z, a)+u_{L}(-x, -y, z, a)\\ =&a^{2}[2\sqrt{2}\ln(5+2\sqrt{6})-\frac{16\sqrt{3}}{9}]t^{2}+\cdots =2\sqrt{2}\ln(5+2\sqrt{6})a^{2}-\frac{16\sqrt{3}}{9}r^{2}+\cdots \end{align*} 　である。次に、\(u_{A}(x, y, z, a)\)については、 \begin{align*} S_{1}&=\arctan{\frac{[a - (x - y + z)]r_{1}}{-[(a - x + y - z) (z + x + a) + 3 z x - 3 a y]}} =\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3-2t(\xi+\eta+\zeta) +t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 + \zeta t + \xi t) + 3\zeta\xi t^{2} - 3\eta t]}}\\ &=\arctan[-\sqrt{3}+\frac{4\sqrt{3}}{3}(\xi-2\eta+\zeta)t +(-\frac{13\sqrt{3}}{18}-\frac{11\sqrt{3}}{3}\eta^{2}+\frac{34\sqrt{3}}{9}(\xi\eta+\eta\zeta)+\frac{4\sqrt{3}}{9}\zeta\xi)t^{2}+\cdots]\\ &=\frac{2\pi}{3}+\frac{\sqrt{3}}{3}(\xi-2\eta+\zeta)t +(-\frac{\sqrt{3}}{36}+\frac{\sqrt{3}}{12}\eta^{2}+\frac{7\sqrt{3}}{18}(-\xi\eta-\eta\zeta+2\zeta\xi))t^{2}+\cdots, \end{align*} \begin{align*} S_{2}&=\arctan{\frac{[a - (x - y + z)]r_{2}}{-[(a - x + y - z) (-y + z + a) - 3 y z + 3 a x]}} =\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3+2t(\xi+\eta-\zeta) +t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 - \eta t + \zeta t) - 3\eta\zeta t^{2} + 3\xi t]}}\\ &=\arctan[-\sqrt{3}+\frac{4\sqrt{3}}{3}(2\xi-\eta+\zeta)t +(-\frac{13\sqrt{3}}{18}-\frac{11\sqrt{3}}{3}\xi^{2}+\frac{34\sqrt{3}}{9}(-\zeta\xi+\xi\eta)+\frac{4\sqrt{3}}{9}\eta\zeta)t^{2}+\cdots]\\ &=\frac{2\pi}{3}+\frac{\sqrt{3}}{3}(2\xi-\eta+\zeta)t +(-\frac{\sqrt{3}}{36}+\frac{\sqrt{3}}{12}\xi^{2}+\frac{7\sqrt{3}}{18}(-\xi\eta-2\eta\zeta+\zeta\xi))t^{2}+\cdots, \end{align*} \begin{align*} S_{3}&=\arctan{\frac{[a - (x - y + z)]r_{3}}{-[(a - x + y - z) (x - y + a) - 3 x y + 3 a z]}} =\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3+2t(-\xi+\eta+\zeta)+t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 + \xi t - \eta t) - 3\xi\eta t^{2}+ 3\zeta t]}}\\ &=\arctan[-\sqrt{3}+\frac{4\sqrt{3}}{3}(\xi-\eta+2\zeta)t +(-\frac{13\sqrt{3}}{18}-\frac{11\sqrt{3}}{3}\zeta^{2}+\frac{34\sqrt{3}}{9}(\eta\zeta-\zeta\xi)-\frac{4\sqrt{3}}{9}\xi\eta)t^{2}+\cdots]\\ &=\frac{2\pi}{3}+\frac{\sqrt{3}}{3}(\xi-\eta+2\zeta)t +(-\frac{\sqrt{3}}{36}+\frac{\sqrt{3}}{12}\zeta^{2}+\frac{7\sqrt{3}}{18}(-2\xi\eta-\eta\zeta+\zeta\xi))t^{2}+\cdots, \end{align*} \[S_{1}+S_{2}+S_{3}-\pi\mathrm{sgn}(a-(x+y+z)) =\pi+\frac{4\sqrt{3}}{3}(\xi-\eta+\zeta)t+\frac{14\sqrt{3}}{9}(-\xi\eta-\eta\zeta+\zeta\xi)t^{2}+\cdots\] より、 \begin{align*} u_{A}(x, y, z, a)=&-\frac{a^{2}}{6}(1-(\xi-\eta+\zeta)t)^{2}[\pi+\frac{4\sqrt{3}}{3}(\xi-\eta+\zeta)t+\frac{14\sqrt{3}}{9}(-\xi\eta-\eta\zeta+\zeta\xi)t^{2}+\cdots]\\ =&a^{2}[-\frac{\pi}{6}(1-(\xi-\eta+\zeta)t)^{2}-\frac{2\sqrt{3}}{9}(\xi-\eta+\zeta)t+(\frac{4\sqrt{3}}{9}+\frac{17\sqrt{3}}{27}(-\xi\eta-\eta\zeta+\zeta\xi))t^{2}+\cdots], \end{align*} \begin{align*} &u_{A}(x, y, z, a)+u_{A}(x, -y, -z, a)+u_{A}(-x, y, -z, a)+u_{A}(-x, -y, z, a)\\ =&a^{2}[-\frac{2\pi}{3}(1+t^{2})+\frac{16\sqrt{3}}{9}t^{2}+\cdots] =-\frac{2\pi}{3}(a^{2}+r^{2})+\frac{16\sqrt{3}}{9}r^{2}+\cdots \end{align*} 　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、 \[ U(x,y,z)=G\rho[(-2\sqrt{2}\ln(5+2\sqrt{6})+\frac{2\pi}{3})a^{2}+\frac{2\pi}{3}r^{2}+\cdots] \] 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\{(-2\sqrt{2}\ln(5+2\sqrt{6})+\frac{2\pi}{3})a^{2}+\frac{2\pi}{3}r^{2} + \dfrac{8\sqrt{3}}{3a}xyz -\dfrac{1}{a^{2}}\cdot\dfrac{4\sqrt{3}}{27}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &-\dfrac{1}{a^{4}}\cdot\dfrac{16\sqrt{3}}{1215}[(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{1}{a^{5}}\cdot\dfrac{404\sqrt{3}}{1215}xyz[(x^{4}+y^{4}+z^{4})-\frac{5}{3}(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &-\dfrac{1}{a^{6}}\cdot\dfrac{37\sqrt{3}}{5103}[(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})] \\ &-\dfrac{1}{a^{7}}\cdot\dfrac{592\sqrt{3}}{5103}xyz[(x^{6}+y^{6}+z^{6})-\frac{7}{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})+\frac{70}{3}x^{2}y^{2}z^{2}] \\ &-\dfrac{1}{a^{8}}\cdot\dfrac{56\sqrt{3}}{19683}[(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})]\\ &-\dfrac{1}{a^{9}}\cdot\dfrac{2675\sqrt{3}}{19683}xyz[(x^{8}+y^{8}+z^{8})-6(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+\dfrac{63}{5}(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})] \\ &-\dfrac{1}{a^{10}}\cdot\dfrac{3013\sqrt{3}}{1771470}[(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{474750}{3013}(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{633297}{3013}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{1625805}{3013}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{3793545}{3013}(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{18967725}{3013}x^{4}y^{4}z^{4}]+\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四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(r,\theta,\varphi)=&G\rho a^{2}[(-2\sqrt{2}\ln(5+2\sqrt{6})+\frac{2\pi}{3}) +\frac{2\pi}{3}\Bigl(\dfrac{r}{a}\Bigr)^{2} +\Bigl(\dfrac{r}{a}\Bigr)^{3}\cdot\dfrac{16}{3}\sqrt{\dfrac{315}{\pi}}Te_{3}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{4}\cdot\dfrac{16}{27}\sqrt{\dfrac{\pi}{7}}Te_{4}(\theta,\varphi)\\ &-\Bigl(\dfrac{r}{a}\Bigr)^{6}\cdot\dfrac{128}{1215}\sqrt{\dfrac{3\pi}{26}}Te_{6,o}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{7}\cdot\dfrac{1616}{3645}\sqrt{\dfrac{\pi}{455}}Te_{7}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{8}\cdot\dfrac{592}{5103}\sqrt{\dfrac{\pi}{187}}Te_{8}(\theta,\varphi)\\ &-\Bigl(\dfrac{r}{a}\Bigr)^{9}\cdot\dfrac{592}{1701}\sqrt{\dfrac{\pi}{1045}}Te_{9,t}(\theta,\varphi) +\Bigl(\dfrac{r}{a}\Bigr)^{10}\cdot\dfrac{28}{6561}\sqrt{\dfrac{3\pi}{910}}Te_{10,o}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{11}\cdot\dfrac{16}{535}\sqrt{\dfrac{3\pi}{55913}}Te_{11}(\theta,\varphi)\\ &-\Bigl(\dfrac{r}{a}\Bigr)^{12}\cdot( \dfrac{5202576}{8170905375}\sqrt{\dfrac{123\pi}{11}}Te_{12,o8}(\theta,\varphi) +\dfrac{28602896}{302626125}\sqrt{\dfrac{82\pi}{676039}}Te_{12,o12}(\theta,\varphi))+\cdots] \end{align*}
「正四面体惑星の重力ポテンシャルの多重極展開」へ戻る 目次へ戻る 「正八面体惑星の重力ポテンシャルの多重極展開」へ進む