# 拉普拉斯-龍格-冷次向量

## 概論

${\displaystyle \mathbf {r} \cdot \mathbf {L} =0}$

LRL向量${\displaystyle \mathbf {A} }$ ，也肯定地包含於粒子的運動平面。可是，只有當連心力遵守平方反比定律時，${\displaystyle \mathbf {A} }$ 才是常數向量[1]。對於別種連心力，${\displaystyle \mathbf {A} }$ 不是常數向量，其大小與方向都會改變。假若連心力近似地遵守平方反比定律，則${\displaystyle \mathbf {A} }$ 的大小近似常數，而方向會緩慢地轉動。對於所有的連心力，可以定義一個廣義LRL向量，但是，這廣義向量通常並沒有解析解，假若有，也會是一個非常複雜的函數[8][9]

## 數學定義

${\displaystyle \mathbf {F} (r)=-{\frac {k}{r^{2}}}\mathbf {\hat {r}} }$

${\displaystyle \mathbf {A} =\mathbf {p} \times \mathbf {L} -mk\mathbf {\hat {r}} }$

${\displaystyle {\frac {\mathrm {d} E}{\mathrm {d} t}}={\frac {p}{m}}{\dot {p}}+{\frac {k}{r^{2}}}{\dot {r}}=0}$

${\displaystyle \mathbf {e} ={\frac {\mathbf {A} }{mk}}={\frac {1}{mk}}(\mathbf {p} \times \mathbf {L} )-\mathbf {\hat {r}} }$

## 开普勒軌道導引

${\displaystyle \mathbf {A} \cdot \mathbf {r} =Ar\cos \theta =\mathbf {r} \cdot \left(\mathbf {p} \times \mathbf {L} \right)-mkr}$

${\displaystyle \mathbf {r} \cdot \left(\mathbf {p} \times \mathbf {L} \right)=\mathbf {L} \cdot \left(\mathbf {r} \times \mathbf {p} \right)=\mathbf {L} \cdot \mathbf {L} =L^{2}}$

${\displaystyle Ar\cos \theta =L^{2}-mkr}$

${\displaystyle {\frac {1}{r}}={\frac {mk}{L^{2}}}\left(1+{\frac {A}{mk}}\cos \theta \right)}$

${\displaystyle e={\frac {A}{mk}}={\frac {\left|\mathbf {A} \right|}{mk}}}$

{\displaystyle {\begin{aligned}\mathbf {A} \cdot \mathbf {A} &=(\mathbf {p} \times \mathbf {L} -mk\mathbf {\hat {r}} )\cdot (\mathbf {p} \times \mathbf {L} -mk\mathbf {\hat {r}} )\\&=p^{2}L^{2}+m^{2}k^{2}-2mk{\hat {\mathbf {r} }}\cdot (\mathbf {p} \times \mathbf {L} )\\&=\left(2mE+{\frac {2mk}{r}}\right)L^{2}+m^{2}k^{2}-{\frac {2mk}{r}}L^{2}\\\end{aligned}}}

${\displaystyle A^{2}=m^{2}k^{2}+2mEL^{2}}$

${\displaystyle e^{2}=1+{\frac {2L^{2}}{mk^{2}}}E}$

## 圓形的速端曲線

${\displaystyle L^{2}\mathbf {p} =\mathbf {L} \times \mathbf {A} -mk{\hat {\mathbf {r} }}\times \mathbf {L} }$

${\displaystyle p_{x}^{2}+\left(p_{y}-A/L\right)^{2}=\left(mk/L\right)^{2}}$

## 在微擾勢下的系統演化

${\displaystyle h(r)=-\ {\frac {h}{r^{n}}}}$

{\displaystyle {\begin{aligned}{\bar {\Omega }}={\frac {\partial }{\partial L}}\langle h(r)\rangle &={\frac {\partial }{\partial L}}\left\{{\frac {1}{T}}\int _{0}^{T}h(r)\,\mathrm {d} t\right\}\\&={\frac {\partial }{\partial L}}\left\{{\frac {m}{TL}}\int _{0}^{2\pi }r^{2}h(r)\,\mathrm {d} \theta \right\}\\\end{aligned}}}

${\displaystyle h(r)={\frac {kL^{2}}{m^{2}c^{2}}}\left({\frac {1}{r^{3}}}\right)}$

${\displaystyle {\frac {1}{r}}={\frac {mk}{L^{2}}}\left(1+{\frac {A}{mk}}\cos \theta \right)}$

${\displaystyle {\bar {\Omega }}={\frac {6\pi k^{2}}{TL^{2}c^{2}}}}$

## 帕松括號

${\displaystyle \{L_{i},L_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}L_{s}}$

${\displaystyle \mathbf {D} ={\frac {\mathbf {A} }{\sqrt {2m\left|E\right|}}}}$

${\displaystyle \{D_{i},L_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}D_{s}}$

${\displaystyle \{D_{i},D_{j}\}=-\sum _{s=1}^{3}\epsilon _{ijs}L_{s}}$

${\displaystyle \{D_{i},D_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}L_{s}}$

${\displaystyle \{L_{i},L_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}L_{s}}$
${\displaystyle \{D_{i},L_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}D_{s}}$
${\displaystyle \{D_{i},D_{j}\}=\sum _{s=1}^{3}\epsilon _{ijs}L_{s}}$

${\displaystyle C_{1}=\mathbf {D} \cdot \mathbf {D} +\mathbf {L} \cdot \mathbf {L} ={\frac {mk^{2}}{2\left|E\right|}}}$
${\displaystyle C_{2}=\mathbf {D} \cdot \mathbf {L} =0}$

${\displaystyle \{C_{1},D_{i}\}=\{C_{2},D_{i}\}=0}$

${\displaystyle \{C_{1},L_{i}\}=\{C_{2},L_{i}\}=0}$

## 氫原子量子力學

LRL向量${\displaystyle \mathbf {A} }$ 的量子算符有一個奧妙之處，那就是動量算符與角動量算符並不對易。動量與角動量的叉積必須仔細地加以定義[26]。LRL向量的直角座標分量典型地定義為

${\displaystyle A_{k}\equiv -m_{e}\alpha {\hat {r}}_{k}+{\frac {1}{2}}\sum _{i=1}^{3}\sum _{j=1}^{3}\epsilon _{ijk}\left(p_{i}l_{j}+l_{j}p_{i}\right)}$

${\displaystyle \mathbf {A} =-m_{e}\alpha {\hat {r}}+{\frac {1}{2}}(\mathbf {p} \times \mathbf {L} -\mathbf {L} \times \mathbf {p} )}$

${\displaystyle H={\frac {\mathbf {p} ^{2}}{2m_{e}}}-{\frac {\alpha }{r}}}$

${\displaystyle \mathbf {A} }$ 向量成正比的${\displaystyle \mathbf {D} }$ 向量則是

${\displaystyle \mathbf {D} ={\frac {\mathbf {A} }{\sqrt {-2m_{e}H}}}}$

${\displaystyle \{L_{i},\,L_{j}\}=i\hbar \epsilon _{ijk}L_{k}}$
${\displaystyle \{L_{i},\,D_{j}\}=i\hbar \epsilon _{ijk}D_{k}}$
${\displaystyle \{D_{i},\,D_{j}\}=i\hbar \epsilon _{ijk}L_{k}}$
${\displaystyle \{H,\,D_{i}\}=0}$

${\displaystyle J_{0}\equiv D_{3}}$
${\displaystyle J_{\pm 1}\equiv \mp {\frac {1}{\sqrt {2}}}\left(D_{1}\pm iD_{2}\right)}$

${\displaystyle C_{1}\equiv \mathbf {D} ^{2}+\mathbf {L} ^{2}={\frac {m_{e}\alpha ^{2}}{-2H}}-\hbar ^{2}}$

${\displaystyle \{J_{+1},J_{-1}\}=i\{D_{1},\,D_{2}\}=-\hbar L_{3}}$

${\displaystyle J_{0}|l,\,m\rangle =i{\sqrt {l^{2}-m^{2}}}\ {\mathfrak {C}}_{l}|l-1,\,m\rangle -i{\sqrt {(l+1)^{2}-m^{2}}}\ {\mathfrak {C}}_{l+1}|l+1,\,m\rangle }$
${\displaystyle J_{+1}|l,\,m\rangle =-i{\sqrt {(l-m)(l-m-1)/2}}\ {\mathfrak {C}}_{l}|l-1,\,m+1\rangle -i{\sqrt {(l+m+1)(l+m+2)/2}}\ {\mathfrak {C}}_{l+1}|l+1,\,m+1\rangle }$
${\displaystyle J_{-1}|l,\,m\rangle =-i{\sqrt {(l+m)(l+m-1)/2}}\ {\mathfrak {C}}_{l}|l-1,\,m-1\rangle -i{\sqrt {(l-m+1)(l-m+2)/2}}\ {\mathfrak {C}}_{l+1}|l+1,\,m-1\rangle }$

{\displaystyle {\begin{aligned}\{J_{+1},\,J_{-1}\}|l,\,m\rangle &=-m[(2l-1){\mathfrak {C}}_{l}^{2}-(2l+3){\mathfrak {C}}_{l+1}^{2}]|l,\,m\rangle \\&=-\hbar L_{3}|l,\,m\rangle =-m\hbar ^{2}\\\end{aligned}}}

${\displaystyle (2l-1){\mathfrak {C}}_{l}^{2}-(2l+3){\mathfrak {C}}_{l+1}^{2}=\hbar ^{2}}$

${\displaystyle {\mathfrak {C}}_{l}={\sqrt {\frac {n^{2}-l^{2}}{4l^{2}-1}}}\ \hbar }$

{\displaystyle {\begin{aligned}D^{2}|n,\,l,\,m\rangle &=[J_{+1}J_{-1}+J_{-1}J_{+1}+J_{0}^{2}]|n,\,l,\,m\rangle \\&=(n^{2}-l^{2}-l-1)\hbar ^{2}|n,\,l,\,m\rangle \\\end{aligned}}}

${\displaystyle C_{1}|n,\,l,\,m\rangle =(D^{2}+L^{2})|n,\,l,\,m\rangle =(n^{2}-1)\hbar ^{2}|n,\,l,\,m\rangle }$

${\displaystyle E_{n}=-{\frac {m_{e}\alpha ^{2}}{2\hbar ^{2}n^{2}}}=-{\frac {m_{e}e^{4}}{2n^{2}(4\pi \epsilon _{0})^{2}\hbar ^{2}}}}$

## 保守性與對稱性

${\displaystyle \left|\mathbf {e} \right|^{2}=e_{1}^{2}+e_{2}^{2}+e_{3}^{2}+e_{4}^{2}}$

1935年，弗拉基米尔·福克表明，在量子力學裏，束縛的开普勒問題等價於一個粒子自由地移動於四維空間的三維單位球[4]。更具體地，佛克表明，在开普勒問題的動量空間，薛丁格波函數球諧函數球極平面投影。圓球的旋轉與重複射影造成了橢圓軌域的連續映射，同時維持能量不變；這對應於主量子數${\displaystyle n}$ 相同的軌域的混合。隨後，華倫泰·巴格曼注意到，跟LRL向量成比例的向量${\displaystyle \mathbf {D} }$ 與角動量${\displaystyle \mathbf {L} }$ 帕松括號形成SO(4)的李代數[5]。簡單地說，${\displaystyle \mathbf {D} }$ ${\displaystyle \mathbf {L} }$ 的六個物理量對應於在四維空間裏的六個保守的角動量分量，相伴於在四維空間裏的六個合法的簡單旋轉（從四個軸中，選兩個軸為旋轉軸。一共有六種可能）。這結論並不意示宇宙是一個三維球面；而只是說，這個特別的物理問題（开普勒問題），在數學上，等價於移動於三維球面的一個自由粒子。

${\displaystyle ds^{2}=e_{1}^{2}+e_{2}^{2}+e_{3}^{2}-e_{4}^{2}}$

## 旋轉對稱性在四維空間

${\displaystyle {\boldsymbol {\eta }}=\displaystyle {\frac {p^{2}-p_{0}^{2}}{p^{2}+p_{0}^{2}}}\mathbf {\hat {w}} +{\frac {2p_{0}}{p^{2}+p_{0}^{2}}}\mathbf {p} }$

${\displaystyle {\boldsymbol {\eta }}={\hat {\boldsymbol {\eta }}}}$

${\displaystyle \mathbf {p} }$ ${\displaystyle {\hat {\boldsymbol {\eta }}}}$ 映射有一個獨特唯一的逆反；例如，動量${\displaystyle \mathbf {p} }$ 的x-軸分量是

${\displaystyle p_{x}=p_{0}{\frac {\eta _{x}}{1-\eta _{w}}}}$

${\displaystyle p_{y}}$ ${\displaystyle p_{z}}$ 也有類似的公式。換句話說，三維動量向量${\displaystyle \mathbf {p} }$ 是四維單位向量${\displaystyle {\hat {\boldsymbol {\eta }}}}$ 球極平面投影，其比例因子為${\displaystyle p_{0}}$

${\displaystyle p_{x}=0}$
${\displaystyle p_{y}=p=(A+mk)/L}$
${\displaystyle \eta _{y}=\cos(\alpha )={\frac {2p_{0}p_{y}}{p_{y}^{2}+p_{0}^{2}}}}$

{\displaystyle {\begin{aligned}\sin(\alpha )&={\frac {p_{y}^{2}-p_{0}^{2}}{p_{y}^{2}+p_{0}^{2}}}\\&={\frac {(A+mk)^{2}-2m|E|L^{2}}{(A+mk)^{2}+2m|E|L^{2}}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}\sin(\alpha )&={\frac {(A+mk)^{2}+2mEL^{2}}{(A+mk)^{2}-2mEL^{2}}}\\&={\frac {A}{mk}}=e\\\end{aligned}}}

${\displaystyle \eta _{w}=\mathrm {cn} \,\chi \ \mathrm {cn} \,\psi }$
${\displaystyle \eta _{x}=\mathrm {sn} \,\chi \ \mathrm {dn} \,\psi \ \cos \phi }$
${\displaystyle \eta _{y}=\mathrm {sn} \,\chi \ \mathrm {dn} \,\psi \ \sin \phi }$
${\displaystyle \eta _{z}=\mathrm {dn} \,\chi \ \mathrm {sn} \,\psi }$

## 开普勒問題LRL向量恆定的證明

### 直接證明

${\displaystyle {\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}=f(\mathbf {r} ){\hat {\mathbf {r} }}}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\mathbf {L} =0}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {p} \times \mathbf {L} \right)={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}\times \mathbf {L} =f(\mathbf {r} )\mathbf {\hat {r}} \times \left(\mathbf {r} \times m{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\right)=f(\mathbf {r} ){\frac {m}{r}}\left[\mathbf {r} \left(\mathbf {r} \cdot {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\right)-r^{2}{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\right]}$

${\displaystyle \mathbf {r} \cdot {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}={\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \cdot \mathbf {r} \right)={\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} t}}\left(r^{2}\right)=r{\frac {\mathrm {d} r}{\mathrm {d} t}}}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {p} \times \mathbf {L} \right)=-mf(\mathbf {r} )r^{2}\left[{\frac {1}{r}}{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}-{\frac {\mathbf {r} }{r^{2}}}{\frac {\mathrm {d} r}{\mathrm {d} t}}\right]=-mf(\mathbf {r} )r^{2}{\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\mathbf {r} }{r}}\right)}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {p} \times \mathbf {L} \right)=mk{\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\mathbf {r} }{r}}\right)={\frac {\mathrm {d} }{\mathrm {d} t}}\left(mk\mathbf {\hat {r}} \right)}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\mathbf {A} ={\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {p} \times \mathbf {L} \right)-{\frac {\mathrm {d} }{\mathrm {d} t}}\left(mk\mathbf {\hat {r}} \right)=0}$

### 哈密頓-雅可比方程式

${\displaystyle \xi =r+x}$
${\displaystyle \eta =r-x}$

${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$

${\displaystyle x={\frac {1}{2}}\left(\xi -\eta \right)}$
${\displaystyle y={\sqrt {\xi \eta }}}$

{\displaystyle {\begin{aligned}H&={\frac {1}{2}}m{\dot {x}}^{2}+{\frac {1}{2}}m{\dot {y}}^{2}-{\frac {k}{r}}\\&={\frac {2\xi p_{\xi }^{2}}{m(\xi +\eta )}}+{\frac {2\eta p_{\eta }^{2}}{m(\xi +\eta )}}-{\frac {2k}{\xi +\eta }}\\\end{aligned}}}

${\displaystyle 2\xi p_{\xi }^{2}-mk-mE\xi =-2\eta p_{\eta }^{2}+mk+mE\eta }$

${\displaystyle 2\xi p_{\xi }^{2}-mk-mE\xi =-\Gamma }$
${\displaystyle 2\eta p_{\eta }^{2}-mk-mE\eta =\Gamma }$

{\displaystyle {\begin{aligned}A_{x}&=p_{y}(xp_{y}-yp_{x})-mk{\frac {x}{r}}\\&=xp_{y}^{2}-yp_{x}p_{y}-mk+m\eta {\frac {k}{r}}\\\end{aligned}}}

${\displaystyle A_{x}=xp_{y}^{2}-yp_{x}p_{y}+{\frac {1}{2}}m^{2}v^{2}\eta -mk-mE\eta }$

${\displaystyle xp_{y}^{2}-yp_{x}p_{y}+{\frac {1}{2}}m^{2}v^{2}\eta ={\frac {m^{2}}{8}}{\dot {\eta }}^{2}{\frac {(\eta +\xi )^{2}}{\eta }}=2\eta p_{\eta }^{2}}$

${\displaystyle A_{x}=\Gamma }$

### 諾特定理

LRL向量的保守性與前面所提的旋轉對稱性，兩者之間的關係，可以用諾特定理來做連結分析。諾特定理也可以用來辨明LRL向量是運動常數。諾特定理表明[36]：在一個物理系統裏，對於廣義坐標${\displaystyle q_{i}}$ 的微小變分${\displaystyle \delta q_{i}=\epsilon g_{i}(\mathbf {q} ,\ \mathbf {\dot {q}} ,\ t)}$ ，假若，取至微小參數${\displaystyle \epsilon }$ 的一階，拉格朗日量${\displaystyle {\mathcal {L}}}$ 變分${\displaystyle \delta {\mathcal {L}}}$

${\displaystyle \delta {\mathcal {L}}=\epsilon {\frac {\mathrm {d} }{\mathrm {d} t}}G(\mathbf {q} ,\ t)}$

${\displaystyle \Gamma =-G+\sum _{i}g_{i}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}\right)}$

${\displaystyle \delta x_{i}={\frac {\epsilon }{2}}\left[2p_{i}x_{s}-x_{i}p_{s}-(\mathbf {r} \cdot \mathbf {p} )\delta _{is}\right]}$

${\displaystyle {\mathcal {L}}=\sum _{i}\left({\frac {1}{2}}m{\dot {x}}_{i}{\dot {x}}_{i}\right)+{\frac {k}{r}}}$

${\displaystyle m{\ddot {x}}_{i}+k{\frac {x_{i}}{r^{3}}}=0}$

{\displaystyle {\begin{aligned}\delta {\dot {x}}_{i}&={\frac {\epsilon }{2}}\left[2{\dot {p}}_{i}x_{s}-x_{i}{\dot {p}}_{s}+p_{i}{\dot {x}}_{s}-{\frac {p^{2}}{m}}\delta _{is}-(\mathbf {r} \cdot {\dot {\mathbf {p} }})\delta _{is}\right]\\&={\frac {\epsilon }{2}}\left[-{\frac {k}{r^{3}}}x_{i}x_{s}+p_{i}{\dot {x}}_{s}-{\frac {p^{2}}{m}}\delta _{is}+{\frac {k}{r}}\delta _{is}\right]\\\end{aligned}}}

{\displaystyle {\begin{aligned}\delta {\mathcal {L}}&=\sum _{i}\left({\frac {\partial {\mathcal {L}}}{\partial x_{i}}}\delta x_{i}+{\frac {\partial {\mathcal {L}}}{\partial {\dot {x}}_{i}}}\delta {\dot {x}}_{i}\right)\\&=\sum _{i}\left(-{\frac {kx_{i}}{r^{3}}}\delta x_{i}+m{\dot {x}}_{i}\delta {\dot {x}}_{i}\right)\\\end{aligned}}}

${\displaystyle \delta {\mathcal {L}}=\epsilon mk{\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {x_{s}}{r}}\right)}$

${\displaystyle \Gamma =p^{2}x_{s}-p_{s}\left(\mathbf {r} \cdot \mathbf {p} \right)-{\frac {mkx_{s}}{r}}=\left[\mathbf {p} \times \mathbf {L} -mk{\hat {\mathbf {r} }}\right]_{s}}$

### 李變換

${\displaystyle t\rightarrow \lambda ^{3}t,\ \mathbf {r} \rightarrow \lambda ^{2}\mathbf {r} ,\ \mathbf {p} \rightarrow {\frac {1}{\lambda }}\mathbf {p} }$

${\displaystyle L\rightarrow \lambda L,\ E\rightarrow {\frac {1}{\lambda ^{2}}}E}$

${\displaystyle A^{2}=m^{2}k^{2}e^{2}=m^{2}k^{2}+2mEL^{2}}$

## 推廣至別種位勢和相對論

LRL向量可以推廣至其他狀況；可以用來辨認在其他狀況下的保守值。

${\displaystyle {\mathcal {A}}=\mathbf {A} +{\frac {mq}{2}}\left[\left(\mathbf {r} \times \mathbf {E} \right)\times \mathbf {r} \right]}$

${\displaystyle {\mathcal {A}}=\left({\frac {\partial \xi }{\partial u}}\right)\left(\mathbf {p} \times \mathbf {L} \right)+\left[\xi -u\left({\frac {\partial \xi }{\partial u}}\right)\right]L^{2}\mathbf {\hat {r}} }$

${\displaystyle \theta =L\int ^{u}{\frac {du}{\sqrt {m^{2}c^{2}\left(\gamma ^{2}-1\right)-L^{2}u^{2}}}}}$

${\displaystyle {\mathcal {B}}=\mathbf {L} \times {\mathcal {A}}}$

${\displaystyle {\mathcal {W}}=\alpha {\mathcal {A}}\otimes {\mathcal {A}}+\beta \,{\mathcal {B}}\otimes {\mathcal {B}}}$

${\displaystyle {\mathcal {W}}={\frac {1}{2m}}\mathbf {p} \otimes \mathbf {p} +{\frac {k}{2}}\,\mathbf {r} \otimes \mathbf {r} }$

${\displaystyle {\mathcal {A}}={\frac {1}{\sqrt {mr^{2}\omega _{0}A-mr^{2}E+L^{2}}}}\left\{\left(\mathbf {p} \times \mathbf {L} \right)+\left(mr\omega _{0}A-mrE\right)\mathbf {\hat {r}} \right\}}$

## 別種比例與表述

${\displaystyle \mathbf {e} ={\frac {1}{mk}}\left(\mathbf {p} \times \mathbf {L} \right)-\mathbf {\hat {r}} ={\frac {m}{k}}\left(\mathbf {v} \times \mathbf {L} \right)-\mathbf {\hat {r}} }$

${\displaystyle \mathbf {M} =\mathbf {v} \times \mathbf {L} -k\mathbf {\hat {r}} }$

${\displaystyle \mathbf {D} ={\frac {\mathbf {A} }{P_{0}}}={\frac {1}{\sqrt {2m\left|E\right|}}}\left\{\mathbf {p} \times \mathbf {L} -mk\mathbf {\hat {r}} \right\}}$

${\displaystyle \mathbf {D} }$ 與角動量${\displaystyle \mathbf {L} }$ 的單位相同。在非常稀有的狀況，LRL向量的正負號會改變。這些，都不會影響它是運動常數的事實。

${\displaystyle \mathbf {B} =\mathbf {p} -\left({\frac {mk}{L^{2}r}}\right)\ \left(\mathbf {L} \times \mathbf {r} \right)}$

${\displaystyle {\mathcal {W}}=\alpha \mathbf {A} \otimes \mathbf {A} +\beta \,\mathbf {B} \otimes \mathbf {B} }$

${\displaystyle {\mathcal {W}}_{ij}=\alpha A_{i}A_{j}+\beta B_{i}B_{j}}$

${\displaystyle \mathbf {L} \cdot {\mathcal {W}}=\alpha \left(\mathbf {L} \cdot \mathbf {A} \right)\mathbf {A} +\beta \left(\mathbf {L} \cdot \mathbf {B} \right)\mathbf {B} =0}$

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