立方体惑星の重力ポテンシャルの多重極展開
立方体惑星の重力ポテンシャル
立方体の\(8\)つの頂点を\((\pm a,\pm a,\pm a)\)とすると、立方体惑星の重力ポテンシャルは以下の形で与えられる。 \begin{align*} U(x,y,z)=&\Phi(x-a,y-a,z-a)-\Phi(x+a,y-a,z-a)-\Phi(x-a,y+a,z-a)+\Phi(x+a,y+a,z-a) \\ &-\Phi(x-a,y-a,z+a)+\Phi(x+a,y-a,z+a)+\Phi(x-a,y+a,z+a)-\Phi(x+a,y+a,z+a) \end{align*} \begin{align*} \Phi(x,y,z) =&-\dfrac{x^{2}}{2}\arctan\frac{yz}{x\sqrt{x^{2}+y^{2}+z^{2}}} -\dfrac{y^{2}}{2}\arctan\frac{zx}{y\sqrt{x^{2}+y^{2}+z^{2}}} -\dfrac{z^{2}}{2}\arctan\frac{xy}{z\sqrt{x^{2}+y^{2}+z^{2}}}\\ &+xy\ln[\sqrt{x^{2}+y^{2}+z^{2}}-z] +yz\ln[\sqrt{x^{2}+y^{2}+z^{2}}-x] +zx\ln[\sqrt{x^{2}+y^{2}+z^{2}}-y] \end{align*}
重力ポテンシャルの多重極展開 立方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) \\ =&\int_{-a}^{a}\Big[\int_{-a}^{a}\Big[\int_{-a}^{a}(rr')^{l}P_{l}\Bigl(\dfrac{\vec{r}\cdot\vec{r}'}{rr'}\Bigr)dx'\Big]dy'\Big]dz' \end{align*} 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、 \[-\dfrac{G\rho}{r}\iiint_{V}P_{0}(cos \theta ')dx'dy'dz'=-\dfrac{8G\rho a^{3}}{r}\] \[-\dfrac{G\rho}{r^{5}}\iiint_{V}(rr')^{2}P_{2}(cos \theta ')dx'dy'dz'= 0\] \[-\dfrac{G\rho}{r^{9}}\iiint_{V}(rr')^{4}P_{4}(cos \theta ')dx'dy'dz' =\dfrac{G\rho a^{7}}{r^{9}}\cdot\dfrac{28}{15}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})]\] \[-\dfrac{G\rho}{r^{13}}\iiint_{V}(rr')^{6}P_{6}(cos \theta ')dx'dy'dz' =-\dfrac{G\rho a^{9}}{r^{13}}\cdot\dfrac{16}{21}[(x^{6}+y^{6}+z^{6})-\frac{15}{2}(x^{4}y^{2}+x^{2}y^{4}+y^{4}z^{2}+y^{2}z^{4}+z^{4}x^{2}+z^{2}x^{4})+90x^{2}y^{2}z^{2}]\] \[-\dfrac{G\rho}{r^{17}}\iiint_{V}(rr')^{8}P_{8}(cos \theta ')dx'dy'dz' =-\dfrac{G\rho a^{11}}{r^{17}}\cdot\dfrac{11}{5}[(x^{8}+y^{8}+z^{8})-14(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+35(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})]\] \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{104}{33}[(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{1999}{910}[(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{772090}{1999}(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{1700083}{1999}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})-\frac{1664025}{1999}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &+\frac{3882725}{1999}(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{19413625}{1999}x^{4}y^{4}z^{4}] \end{align*} 立方体惑星の総質量は\(M=8\rho a^{3}\)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{14}{15}\sqrt{\dfrac{\pi}{21}}Te_{4}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{6}\cdot\dfrac{16}{21}\sqrt{\dfrac{\pi}{26}}Te_{6,o}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{8}\cdot\dfrac{22}{5}\sqrt{\dfrac{\pi}{561}}Te_{8}(\theta,\varphi)\\ &+\big(\dfrac{a}{r}\big)^{10}\cdot\dfrac{26}{99}\sqrt{\dfrac{\pi}{910}}Te_{10,o}(\theta,\varphi) +\big(\dfrac{a}{r}\big)^{12}\cdot (\dfrac{133122}{1492400}\sqrt{\dfrac{41\pi}{11}}Te_{12,o8}(\theta,\varphi) - \dfrac{7429}{16400}\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、 \[-\dfrac{14}{15}\sqrt{\dfrac{\pi}{21}}=-0.36099573\cdots, \ \dfrac{16}{21}\sqrt{\dfrac{\pi}{26}}=0.26484327\cdots, \ \dfrac{22}{5}\sqrt{\dfrac{\pi}{561}}=0.32926546\cdots,\] \[\dfrac{26}{99}\sqrt{\dfrac{\pi}{910}}=0.01543094\cdots, \ \dfrac{133122}{1492400}\sqrt{\dfrac{41\pi}{11}}=0.30523555\cdots, \ -\dfrac{7429}{16400}\sqrt{\dfrac{246\pi}{676039}}=0.01531592\cdots, \] 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\(\rm{Taylor}\)展開をすると、 \begin{align*} &\sqrt{(x-a)^{2}+(y-a)^{2}+(z-a)^{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}{\sqrt{3}}(\xi+\eta+\zeta)t+\frac{1}{3}(1-\xi\eta-\eta\zeta-\zeta\xi)t^{2}+\cdots ), \end{align*} \begin{align*} &\ln[\sqrt{(x-a)^{2}+(y-a)^{2}+(z-a)^{2}}-(z-a)]\\ =&\ln(a)+\ln[\sqrt{3}(1-\frac{1}{\sqrt{3}}(\xi+\eta+\zeta)t+\frac{1}{3}(1-\xi\eta-\eta\zeta-\zeta\xi)t^{2}+\cdots ) -(\zeta t -1)] \\ =&\ln(a)+\ln(\sqrt{3}+1) +(-\frac{3-\sqrt{3}}{6}(\xi+\eta)-\frac{1}{\sqrt{3}}\zeta)t +(-\frac{9-4\sqrt{3}}{36}(\xi+\eta)^{2}-\frac{\sqrt{3}}{9}(\xi+\eta)\zeta-\frac{9-\sqrt{3}}{36}\zeta ^{2}+\frac{3-\sqrt{3}}{12})t^{2}+\cdots , \end{align*} \begin{align*} &\sqrt{(x-a)^{2}+(y-a)^{2}+(z+a)^{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}{\sqrt{3}}(\xi+\eta-\zeta)t+\frac{1}{3}(1-\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots ), \end{align*} \begin{align*} &\ln[\sqrt{(x-a)^{2}+(y-a)^{2}+(z+a)^{2}}-(z+a)]\\ =&\ln(a)+\ln[\sqrt{3}(1-\frac{1}{\sqrt{3}}(\xi+\eta-\zeta)t+\frac{1}{3}(1-\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots ) -(\zeta t +1)] \\ =&\ln(a)+\ln(\sqrt{3}-1) +(-\frac{3+\sqrt{3}}{6}(\xi+\eta)-\frac{1}{\sqrt{3}}\zeta)t +(-\frac{9+4\sqrt{3}}{36}(\xi+\eta)^{2}-\frac{\sqrt{3}}{9}(\xi+\eta)\zeta-\frac{9+\sqrt{3}}{36}\zeta ^{2}+\frac{3+\sqrt{3}}{12})t^{2}+\cdots , \end{align*} \begin{align*} &\arctan\frac{(y+a)(z+a)}{(x+a)\sqrt{(x+a)^{2}+(y+a)^{2}+(z+a)^{2}}} =\arctan\frac{(t\eta+1)(t\zeta+1)}{(t\xi+1)\sqrt{3+2t(\xi+\eta+\zeta) +t^{2}}} \\ =&\arctan [\frac{1}{\sqrt{3}}+(-\frac{4\sqrt{3}}{9}\xi+\frac{2\sqrt{3}}{9}\eta+\frac{2\sqrt{3}}{9}\zeta)t +(\frac{\sqrt{3}}{2}\xi^{2}+\frac{2\sqrt{3}}{9}(-\xi\eta+\eta\zeta-\zeta\xi)-\frac{\sqrt{3}}{18}(\eta^{2}+\zeta^{2}+1))t^{2} +\cdots ]\\ =&\frac{\pi}{6}+(-\frac{\sqrt{3}}{3}\xi+\frac{\sqrt{3}}{6}\eta+\frac{\sqrt{3}}{6}\zeta)t +(\frac{19\sqrt{3}}{72}\xi^{2}+\frac{\sqrt{3}}{18}(-\xi\eta+2\eta\zeta-\zeta\xi)-\frac{5\sqrt{3}}{72}(\eta^{2}+\zeta^{2})-\frac{\sqrt{3}}{24})t^{2} +\cdots , \end{align*} \begin{align*} &\arctan\frac{(y+a)(z+a)}{(x-a)\sqrt{(x-a)^{2}+(y+a)^{2}+(z+a)^{2}}} =\arctan\frac{(t\eta+1)(t\zeta+1)}{(t\xi-1)\sqrt{3+2t(-\xi+\eta+\zeta) +t^{2}}} \\ =&\arctan [-\frac{1}{\sqrt{3}}+(-\frac{4\sqrt{3}}{9}\xi-\frac{2\sqrt{3}}{9}\eta-\frac{2\sqrt{3}}{9}\zeta)t +(-\frac{\sqrt{3}}{2}\xi^{2}-\frac{2\sqrt{3}}{9}(\xi\eta+\eta\zeta+\zeta\xi)+\frac{\sqrt{3}}{18}(\eta^{2}+\zeta^{2}+1))t^{2} +\cdots ]\\ =&-\frac{\pi}{6}+(-\frac{\sqrt{3}}{3}\xi-\frac{\sqrt{3}}{6}\eta-\frac{\sqrt{3}}{6}\zeta)t +(-\frac{19\sqrt{3}}{72}\xi^{2}-\frac{\sqrt{3}}{18}(\xi\eta+2\eta\zeta+\zeta\xi)+\frac{5\sqrt{3}}{72}(\eta^{2}+\zeta^{2})+\frac{\sqrt{3}}{24})t^{2} +\cdots , \end{align*} 等より、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\{(-12\ln(2+\sqrt{3})+2\pi)a^{2}+\frac{2\pi}{3}r^{2} \\ &+\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{\sqrt{3}}{486}[(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^{6}}\cdot\dfrac{17\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^{8}}\cdot\dfrac{13\sqrt{3}}{314928}[(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^{10}}\cdot\dfrac{4307}{19486170}[(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{1052370}{4307}(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{1951719}{4307}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{81675}{4307}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{190575}{4307}(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{952875}{4307}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}[(-12\ln(2+\sqrt{3})+2\pi) +\frac{2\pi}{3}\Bigl(\dfrac{r}{a}\Bigr)^{2} +\Bigl(\dfrac{r}{a}\Bigr)^{4}\cdot\dfrac{16\sqrt{3}}{27}\sqrt{\dfrac{\pi}{21}}Te_{4}(\theta,\varphi)\\ &+\Bigl(\dfrac{r}{a}\Bigr)^{6}\cdot\dfrac{4\sqrt{3}}{243}\sqrt{\dfrac{\pi}{26}}Te_{6,o}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{8}\cdot\dfrac{272\sqrt{3}}{5103}\sqrt{\dfrac{\pi}{561}}Te_{8}(\theta,\varphi) +\Bigl(\dfrac{r}{a}\Bigr)^{10}\cdot\dfrac{26\sqrt{3}}{944784}\sqrt{\dfrac{\pi}{910}}Te_{10,o}(\theta,\varphi)\\ &+\Bigl(\dfrac{r}{a}\Bigr)^{12}\cdot( \dfrac{176}{60525225}\sqrt{\dfrac{41\pi}{11}}Te_{12,o8}(\theta,\varphi) +\dfrac{13917872}{1997332425}\sqrt{\dfrac{246\pi}{676039}}Te_{12,o12}(\theta,\varphi)+\cdots] \end{align*}
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