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本文鏈接:拉普拉斯—龍格—楞次矢量(排版更漂亮,敘述更清晰)

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預備知識 開普勒問題

在開普勒問題中, 我們定義拉普拉斯—龍格—楞次矢量(Laplace-Runge-Lenz Vector) (通常簡稱為 LRL 矢量)為


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ewcommand{arccosRound}[2][{}]{arccos^{#1}left(#2
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ewcommand{arctanRound}[2][{}]{arctan^{#1}left(#2
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ewcommand{order}[1]{mathcal{O}left(#1
ight)}  egin{equation} vec A = vec p cross vec L - mk uvec r end{equation}

 
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ight.} 
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ewcommand{sinhRound}[2][{}]{sinh^{#1}left(#2
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ewcommand{coshRound}[2][{}]{cosh^{#1}left(#2
ight)} egin{equation} mathbf A = mathbf p cross mathbf L - mk uvec r  end{equation}


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ight.} 
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ewcommand{dv}[2][{}]{frac{mathrm{d}^{#1}}{mathrm{d}{#2}^{#1}}}  dot{mathbf A} = dot{mathbf p}cross mathbf L  - mkdot{uvec r}


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ight.} 
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enewcommand{Re}{mathrm{Re}} 
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ewcommand{dvTwo}[3][{}]{frac{mathrm{d}^{#1}{#2}}{mathrm{d}{#3}^{#1}}}  egin{equation} dot{mathbf p} = mathbf F = - frac{k}{r^2}uvec r end{equation}

其中 vec p 為質點動量, vec L 為軌道角動量, k = GMmvec r 為質點位矢 vec r 的單位矢量. 在開普勒問題中, 可以證明 vec A 是一個守恆量.

守恆證明

   我們下面證明 dot{vec A} = 0 . 對式 1 求時間導數, 考慮到中心力場中質點角動量 LL 守恆, 有

egin{equation} dot{vec A} = dot{vec p}	imesvec L  - mkdot{vec r} end{equation}

其中由牛頓第二定律和萬有引力定律, 有

˙  dot{vec p} = vec F = - frac{k} {r^2} vec r

又由「極坐標中單位矢量的偏導」 得

最後由式 5, 極坐標系中的角動量等於

egin{equation} vec L = mr^2dot 	heta vec z end{equation}

將式 3至式 5代入式 2得

egin{equation} dot{vec A} = -frac{k}{r^2}vec r 	imes (mr^2dot	hetavec z) - mkdot	hetavec	heta =-mkdot	heta (vec r	imes vec z + vec 	heta) = vec 0 end{equation}

最後一個等號成立是因為 vec r	imesvec z = -vec	heta , 可以類比直角坐標系中的 vec x	imesvec z = -vec y . 證畢.

-----------------------------------------------------------------------------------------------更詳細內容請點擊:小時物理百科。

本文鏈接:拉普拉斯—龍格—楞次矢量(排版更漂亮,敘述更清晰)


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