Geodesic equation
\[
\frac{d^2 x^\lambda}{dt^2}
+ \Gamma^\lambda_{\mu \nu}
\frac{dx^\mu}{dt} \frac{dx^\nu}{dt} = 0
\]
半径 \(r\) の球面を極座標 \((\theta, \phi)\) で表す:
\[\begin{split}\displaystyle \left[\begin{matrix}\sin{\left(\mathbf{\theta} \right)} \cos{\left(\mathbf{\phi} \right)} \mathbf{r}\\\sin{\left(\mathbf{\theta} \right)} \sin{\left(\mathbf{\phi} \right)} \mathbf{r}\\\cos{\left(\mathbf{\theta} \right)} \mathbf{r}\end{matrix}\right]\end{split}\]
\(\theta\) と \(\phi\) を媒介変数 \(t\) の関数とみなす:
\[\displaystyle \theta{\left(t \right)}\]
\[\displaystyle \phi{\left(t \right)}\]
\[\displaystyle \frac{d}{d t} \theta{\left(t \right)}\]
\[\displaystyle \frac{d}{d t} \phi{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \phi{\left(t \right)}\]
測地線の公式に代入する:
\[
\frac{d^2 x^\lambda}{dt^2}
+ \Gamma^\lambda_{\mu \nu}
\frac{dx^\mu}{dt} \frac{dx^\nu}{dt} = 0
\]
\[\begin{split}\displaystyle \left[\begin{matrix}\left[\begin{matrix}0 & 0\\0 & - \sin{\left(\theta \right)} \cos{\left(\theta \right)}\end{matrix}\right] & \left[\begin{matrix}0 & \frac{\cos{\left(\theta \right)}}{\sin{\left(\theta \right)}}\\\frac{\cos{\left(\theta \right)}}{\sin{\left(\theta \right)}} & 0\end{matrix}\right]\end{matrix}\right]\end{split}\]
\[\displaystyle - \sin{\left(\theta \right)} \cos{\left(\theta \right)} \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} + \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \phi{\left(t \right)} + \frac{2 \cos{\left(\theta \right)} \frac{d}{d t} \phi{\left(t \right)} \frac{d}{d t} \theta{\left(t \right)}}{\sin{\left(\theta \right)}}\]
\[\begin{split}
\begin{align}
\theta'' - \sin(\theta) \cos(\theta) (\phi')^2 & = 0 \\
\phi'' + 2 \cot(\theta) \phi' \theta' & = 0
\end{align}
\end{split}\]
測地線は一次元なので、\(\theta\) を \(\phi = \phi(t)\) の関数とみなし、媒介変数 \(t\) で微分する。
\[\displaystyle \theta{\left(\phi{\left(t \right)} \right)}\]
\[\displaystyle \theta{\left(\phi{\left(t \right)} \right)}\]
\[\displaystyle \phi{\left(t \right)}\]
\[\displaystyle \frac{d}{d t} \phi{\left(t \right)}\]
\[\displaystyle \frac{d}{d t} \phi{\left(t \right)} \frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}\]
\[\displaystyle \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} \frac{d^{2}}{d \phi{\left(t \right)}^{2}} \theta{\left(\phi{\left(t \right)} \right)} + \frac{d^{2}}{d t^{2}} \phi{\left(t \right)} \frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}\]
\[\displaystyle - \sin{\left(\theta \right)} \cos{\left(\theta \right)} \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} \frac{d^{2}}{d \phi{\left(t \right)}^{2}} \theta{\left(\phi{\left(t \right)} \right)} + \frac{d^{2}}{d t^{2}} \phi{\left(t \right)} \frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \phi{\left(t \right)} + \frac{2 \cos{\left(\theta \right)} \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} \frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}}{\sin{\left(\theta \right)}}\]
次のように簡略化する:
\[\displaystyle \frac{d^{2}}{d t^{2}} \phi{\left(t \right)}\]
二つ目の式 geodesic[1]
を \(\phi''\) (phi.diff().diff()
) について解く:
\[\displaystyle - \frac{2 \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2} \frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}}{\tan{\left(\theta \right)}}\]
一つ目の式 geodesic[0]
に代入する:
\[\displaystyle \left(- \frac{\sin{\left(2 \theta \right)}}{2} + \frac{d^{2}}{d \phi{\left(t \right)}^{2}} \theta{\left(\phi{\left(t \right)} \right)} - \frac{2 \left(\frac{d}{d \phi{\left(t \right)}} \theta{\left(\phi{\left(t \right)} \right)}\right)^{2}}{\tan{\left(\theta \right)}}\right) \left(\frac{d}{d t} \phi{\left(t \right)}\right)^{2}\]
次の簡略化された測地線の方程式が得られる:
\[
\theta'' - 2 \cot(\theta) (\theta')^2 - \cos(\theta)\sin(\theta) = 0
\]
さらに、\(\omega\) を次のように定義する:
\[
\omega(t) = \cot(\theta(t))
\]
\[\displaystyle \frac{d}{d t} \theta{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}\]
\[\displaystyle \cot{\left(\theta{\left(t \right)} \right)}\]
\[\displaystyle - \frac{\frac{d}{d t} \theta{\left(t \right)}}{\sin^{2}{\left(\theta{\left(t \right)} \right)}}\]
\[\displaystyle \frac{2 \cot{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} - \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}}{\sin^{2}{\left(\theta{\left(t \right)} \right)}}\]
\[\displaystyle \omega{\left(t \right)}\]
\[\displaystyle \frac{d}{d t} \omega{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \omega{\left(t \right)}\]
\(\omega\) に式を代入してしまうと、\(\omega\) 自身やその微分について、評価を保留して記号として参照できないため、次のように間接的な手法で値を求めていく:
定義式の左辺 \(\cot(\theta(t))\) を微分すると \(\omega'\) : omega.diff()
が得られる:
\[\displaystyle - \frac{\frac{d}{d t} \theta{\left(t \right)}}{\sin^{2}{\left(\theta{\left(t \right)} \right)}}\]
上式を solve()
を使って \(\theta'\) : theta.diff()
について解き、\(\omega'\) で表す:
\[\displaystyle - \sin^{2}{\left(\theta{\left(t \right)} \right)} \frac{d}{d t} \omega{\left(t \right)}\]
\(\theta' = - \omega' \sin^2(\theta)\) が得られる。
同じように \(\cot(\theta(t))\) を二回微分すると \(\omega''\) : omega.diff().diff()
が得られる:
\[\displaystyle \frac{2 \cot{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} - \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}}{\sin^{2}{\left(\theta{\left(t \right)} \right)}}\]
上式を \(\theta''\) : theta.diff().diff()
について解き、\(\omega''\), \(\omega'\) で表す:
\[\displaystyle \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}\]
\[\displaystyle - \sin^{2}{\left(\theta{\left(t \right)} \right)} \frac{d^{2}}{d t^{2}} \omega{\left(t \right)} + \frac{2 \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{\tan{\left(\theta{\left(t \right)} \right)}}\]
\[\displaystyle \left(2 \sin{\left(\theta{\left(t \right)} \right)} \cos{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \omega{\left(t \right)}\right)^{2} - \frac{d^{2}}{d t^{2}} \omega{\left(t \right)}\right) \sin^{2}{\left(\theta{\left(t \right)} \right)}\]
上で得られた \(\theta''\) : theta2
, \(\theta'\) : theta1
を測地線の方程式に代入する:
\[
\theta'' - 2 \cot(\theta) (\theta')^2 - \cos(\theta)\sin(\theta) = 0
\]
\[\displaystyle \left(2 \sin{\left(\theta{\left(t \right)} \right)} \cos{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \omega{\left(t \right)}\right)^{2} - \frac{d^{2}}{d t^{2}} \omega{\left(t \right)}\right) \sin^{2}{\left(\theta{\left(t \right)} \right)} - 2 \sin^{4}{\left(\theta{\left(t \right)} \right)} \cot{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \omega{\left(t \right)}\right)^{2} - \sin{\left(\theta{\left(t \right)} \right)} \cos{\left(\theta{\left(t \right)} \right)}\]
式の簡略化に先立って、三角関数の倍角公式の適用を避けるため \(\sin^2(\theta)\) で割っておく:
\[\displaystyle - 2 \sin^{2}{\left(\theta{\left(t \right)} \right)} \cot{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \omega{\left(t \right)}\right)^{2} + 2 \sin{\left(\theta{\left(t \right)} \right)} \cos{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \omega{\left(t \right)}\right)^{2} - \frac{d^{2}}{d t^{2}} \omega{\left(t \right)} - \frac{\cos{\left(\theta{\left(t \right)} \right)}}{\sin{\left(\theta{\left(t \right)} \right)}}\]
\[\displaystyle - \frac{d^{2}}{d t^{2}} \omega{\left(t \right)} - \frac{1}{\tan{\left(\theta{\left(t \right)} \right)}}\]
\(\omega(t) = \cot(\theta(t))\) と置いていたので、測地線の方程式は次の微分方程式と等価になる:
\[
\omega''(t) + \omega(t) = 0
\]
微分方程式を解く:
\[\displaystyle \omega{\left(t \right)}\]
\[\displaystyle \frac{d^{2}}{d t^{2}} \omega{\left(t \right)}\]
\[\displaystyle \omega{\left(t \right)} = C_{1} \sin{\left(t \right)} + C_{2} \cos{\left(t \right)}\]
球面上の大円を表す方程式となる。