# 伯特蘭定理

${\displaystyle V(r)={\frac {-k}{r}}}$
${\displaystyle V(r)={\frac {1}{2}}kr^{2}}$

1687年，物理学家艾薩克·牛頓在著作《自然哲學的數學原理》裏提出了萬有引力定律，解釋了行星繞著太陽的公轉为何遵守克卜勒定律。此后許多科學家開始研究，當行星的運動稍許偏離了這軌道時，可能會發生的狀況。其中一個問題為軌道是否仍舊閉合。但經過多年的探討亦無法給出合理的解答。直到1873年，法國數學家約瑟·伯特蘭發表伯特蘭定理，才正確解析此問題。该定理對於經典天體力學研究非常重要，伯特蘭定理給予實驗者一個精確的方法，來測試萬有引力的平方反比性質。

## 前論

${\displaystyle {\mathcal {L}}={\frac {1}{2}}m{\dot {r}}^{2}+{\frac {1}{2}}mr^{2}{\dot {\theta }}^{2}-V(r)}$

${\displaystyle m{\ddot {r}}-mr{\dot {\theta }}^{2}+{\frac {dV}{dr}}=0}$
${\displaystyle {\frac {d}{dt}}(mr^{2}{\dot {\theta }})=0}$

${\displaystyle \ell ={\frac {\partial {\mathcal {L}}}{\partial {\dot {\theta }}}}=mr^{2}{\dot {\theta }}}$

${\displaystyle m{\ddot {r}}-{\frac {\ell ^{2}}{mr^{3}}}+{\frac {dV}{dr}}=0}$

${\displaystyle {\frac {d}{dt}}={\frac {\ell }{mr^{2}}}{\frac {d}{d\theta }}}$

${\displaystyle {\frac {\ell }{r^{2}}}{\frac {d}{d\theta }}\left({\frac {\ell }{mr^{2}}}{\frac {dr}{d\theta }}\right)-{\frac {\ell ^{2}}{mr^{3}}}=-{\frac {dV}{dr}}}$

${\displaystyle {\frac {d^{2}u}{d\theta ^{2}}}+u=-{\frac {m}{\ell ^{2}}}{\frac {d}{du}}V(1/u)}$

## 導引

${\displaystyle J(u)=-{\frac {m}{\ell ^{2}}}{\frac {d}{du}}V(1/u)=-{\frac {m}{\ell ^{2}u^{2}}}f(1/u)}$ (1)

${\displaystyle {\frac {d^{2}u}{d\theta ^{2}}}+u=J(u)}$

${\displaystyle u_{0}=J(u_{0})}$

${\displaystyle J(u)\approx u_{0}+\eta J^{\prime }(u_{0})+{\frac {1}{2}}\eta ^{2}J^{\prime \prime }(u_{0})+{\frac {1}{6}}\eta ^{3}J^{\prime \prime \prime }(u_{0})+\ldots }$

${\displaystyle {\frac {d^{2}\eta }{d\theta ^{2}}}+\eta =\eta J^{\prime }(u_{0})+{\frac {1}{2}}\eta ^{2}J^{\prime \prime }(u_{0})+{\frac {1}{6}}\eta ^{3}J^{\prime \prime \prime }(u_{0})\ldots }$

${\displaystyle {\frac {d^{2}\eta }{d\theta ^{2}}}+\beta ^{2}\eta ={\frac {1}{2}}\eta ^{2}J^{\prime \prime }(u_{0})+{\frac {1}{6}}\eta ^{3}J^{\prime \prime \prime }(u_{0})\ldots }$ (2)

${\displaystyle {\frac {d^{2}\eta }{d\theta ^{2}}}+\beta ^{2}\eta =0}$

${\displaystyle \beta ^{2}}$ 必須是個非負數；否則，軌道的半徑會呈指數方式遞增。一階微擾解答為

${\displaystyle \eta (\theta )=h_{1}\cos(\beta \theta )}$

{\displaystyle {\begin{aligned}J^{\prime }(u_{0})&=-{\frac {m}{\ell ^{2}}}\left(-\left.{\frac {2f(1/u)}{u^{3}}}\right|_{u_{0}}+\left.{\frac {1}{u^{2}}}{\frac {df(1/u)}{du}}\right|_{u_{0}}\right)\\&=-\left.{\frac {2J(u)}{u}}\right|_{u_{o}}+{\frac {J(u)}{f(1/u)}}\left.{\frac {df}{du}}\right|_{u_{0}}=-2+{\frac {u_{0}}{f(1/u_{0})}}\left.{\frac {df}{du}}\right|_{u_{0}}=1-\beta ^{2}\\\end{aligned}}}

${\displaystyle {\frac {df}{du}}=-(\beta ^{2}-3){\frac {f(1/u)}{u}}}$

${\displaystyle {\frac {df}{dr}}=\left(\beta ^{2}-3\right){\frac {f}{r}}}$

${\displaystyle f(r)=-{\frac {k}{r^{3-\beta ^{2}}}}}$

${\displaystyle J(u)={\frac {mk}{\ell ^{2}}}u^{1-\beta ^{2}}}$ (3)

${\displaystyle \eta (\theta )=h_{0}+h_{1}\cos(\beta \theta )+h_{2}\cos(2\beta \theta )+h_{3}\cos(3\beta \theta )+\ldots }$

${\displaystyle h_{0}=h_{1}^{2}{\frac {J^{\prime \prime }(u_{0})}{4\beta ^{2}}}}$ (4)
${\displaystyle 0=(2h_{1}h_{0}+h_{1}h_{2}){\frac {J^{\prime \prime }(u_{0})}{2}}+h_{1}^{3}{\frac {J^{\prime \prime \prime }(u_{0})}{8}}}$ (5)
${\displaystyle h_{2}=-h_{1}^{2}{\frac {J^{\prime \prime }(u_{0})}{12\beta ^{2}}}}$ (6)

${\displaystyle J(u)}$ ${\displaystyle u_{0}}$ 對於${\displaystyle u}$ 的微分：

${\displaystyle J^{\prime }(u_{0})=(1-\beta ^{2}){\frac {mk}{\ell ^{2}}}u_{0}^{-\beta ^{2}}=1-\beta ^{2}}$
${\displaystyle J^{\prime \prime }(u_{0})=(1-\beta ^{2})(-\beta ^{2}){\frac {mk}{\ell ^{2}}}u_{0}^{-\beta ^{2}-1}=-{\frac {\beta ^{2}(1-\beta ^{2})}{u_{0}}}}$ (7)
${\displaystyle J^{\prime \prime \prime }(u_{0})=(1-\beta ^{2})(-\beta ^{2})(-\beta ^{2}-1){\frac {mk}{\ell ^{2}}}u_{0}^{-\beta ^{2}-2}={\frac {\beta ^{2}(1-\beta ^{2})(1+\beta ^{2})}{u_{0}^{2}}}}$ (8)

${\displaystyle h_{0}=-{\frac {(1-\beta ^{2})h_{1}^{2}}{4u_{0}}}}$ (9)
${\displaystyle h_{2}={\frac {(1-\beta ^{2})h_{1}^{2}}{12u_{0}}}}$ (10)

${\displaystyle \beta ^{2}(1-\beta ^{2})(4-\beta ^{2})=0}$

## 平方反比力（克卜勒問題）

${\displaystyle V(\mathbf {r} )={\frac {-k}{r}}=-ku}$

${\displaystyle {\frac {d^{2}u}{d\theta ^{2}}}+u=-{\frac {m}{\ell ^{2}}}{\frac {d}{du}}V(1/u)={\frac {km}{\ell ^{2}}}}$

${\displaystyle u={\frac {km}{\ell ^{2}}}\left[1+e\cos \left(\theta -\theta _{0}\right)\right]}$

${\displaystyle e={\sqrt {1+{\frac {2E\ell ^{2}}{k^{2}m}}}}}$

## 徑向諧振子

${\displaystyle V(\mathbf {r} )={\frac {1}{2}}kr^{2}={\frac {1}{2}}k(x^{2}+y^{2}+z^{2})}$

${\displaystyle {\mathcal {L}}={\frac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2}+{\dot {z}}^{2})+{\frac {1}{2}}k(x^{2}+y^{2}+z^{2})}$

${\displaystyle {\frac {d^{2}x}{dt^{2}}}+\omega _{0}^{2}x=0}$
${\displaystyle {\frac {d^{2}y}{dt^{2}}}+\omega _{0}^{2}y=0}$
${\displaystyle {\frac {d^{2}z}{dt^{2}}}+\omega _{0}^{2}z=0}$

${\displaystyle x=A_{x}\cos \left(\omega _{0}t+\phi _{x}\right)}$
${\displaystyle y=A_{y}\cos \left(\omega _{0}t+\phi _{y}\right)}$
${\displaystyle z=A_{z}\cos \left(\omega _{0}t+\phi _{z}\right)}$

## 牛頓旋轉軌道定理

${\displaystyle \Delta F(r)={\frac {L_{1}^{2}}{mr^{3}}}\left(1-\alpha ^{2}\right)}$

## 參考文獻

1. ^ Bertrand, J. Théorème relatif au mouvement d'un point attiré vers un centre fixe. C. R. Acad. Sci. 1873, 77: 849–853.
Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 89–92. ISBN 0201657023 （英语）.
Grandati, Yves; Bérard, Alain, Inverse problem and Bertrand's theorem, American Journal of Physics, August 2008, 76 (8): pp. 782–787
Tikochinsky, Yoel, A simplified proof of Bertrand's theorem, American Journal of Physics, December 1988, 56 (12): pp. 1063–1157
Zarmi, Yair, The Bertrand theorem revisited, American Journal of Physics, April 2002, 70 (4): pp. 446–449