直交曲線座標とデカルト座標の微分変数の交換

直交曲線座標とデカルト座標の微分変数の公式

ここでは直交曲線座標 \(\mathbf r = (q_1, q_2, q_3)\) とデカルト座標 \(\mathbf r = (x, y, z)\) の間に以下の関係式が成立することを証明します。
\[\begin{align}
\frac{\partial q_i}{\partial x} = \frac{1}{h_i^2} \frac{\partial x}{\partial q_i}
\end{align}\]

式としては単純なのですが、意外と証明は面倒くさいです。

直交曲線座標系における偏微分

直交曲線座標 \(\mathbf r = (q_1, q_2, q_3)\) でデカルト座標 \(\mathbf r = (x, y, z)\) を偏微分します。
\[\begin{align}
\frac{\partial \mathbf r}{\partial q_1} &=
\frac{\partial x}{\partial q_1} \mathbf e_x + \frac{\partial y}{\partial q_1} \mathbf e_y + \frac{\partial z}{\partial q_1}\mathbf e_z\\
\frac{\partial \mathbf r}{\partial q_2} &=
\frac{\partial x}{\partial q_2} \mathbf e_x + \frac{\partial y}{\partial q_2} \mathbf e_y + \frac{\partial z}{\partial q_2}\mathbf e_z\\
\frac{\partial \mathbf r}{\partial q_3} &=
\frac{\partial x}{\partial q_3} \mathbf e_x + \frac{\partial y}{\partial q_3} \mathbf e_y + \frac{\partial z}{\partial q_3}\mathbf e_z\\
\end{align}\]

直交基底なのでこれらの 3 式は直交していて、それぞれ内積をとると 0 です。
\[\begin{align}
\left(
\frac{\partial x}{\partial q_1},
\frac{\partial y}{\partial q_1},
\frac{\partial z}{\partial q_1}
\right) \cdot
\left(
\frac{\partial x}{\partial q_2},
\frac{\partial y}{\partial q_2},
\frac{\partial z}{\partial q_2}
\right) &= 0 \label{eq:curvilinear-orthognal-1}\\
\left(
\frac{\partial x}{\partial q_1},
\frac{\partial y}{\partial q_1},
\frac{\partial z}{\partial q_1}
\right) \cdot
\left(
\frac{\partial x}{\partial q_3},
\frac{\partial y}{\partial q_3},
\frac{\partial z}{\partial q_3}
\right) &=0 \label{eq:curvilinear-orthognal-2}\\
\end{align}\]

また、 \((q_1, q_2, q_3)\) は独立関数なので、互いに微分しても 0 です。
\[\begin{align}
\frac{\partial q_1}{\partial q_2} &=
\frac{\partial x}{\partial q_2}\frac{\partial q_1}{\partial x} +
\frac{\partial y}{\partial q_2}\frac{\partial q_1}{\partial y} +
\frac{\partial z}{\partial q_2}\frac{\partial q_1}{\partial z} = 0 \\
\frac{\partial q_1}{\partial q_3} &=
\frac{\partial x}{\partial q_3}\frac{\partial q_1}{\partial x} +
\frac{\partial y}{\partial q_3}\frac{\partial q_1}{\partial y} +
\frac{\partial z}{\partial q_3}\frac{\partial q_1}{\partial z} = 0 \\
\end{align}\]

この 2 つの式は内積形式に書き直せます。
\[\begin{align}
\left(
\frac{\partial q_1}{\partial x},
\frac{\partial q_1}{\partial y},
\frac{\partial q_1}{\partial z}
\right) \cdot
\left(
\frac{\partial x}{\partial q_2},
\frac{\partial y}{\partial q_2},
\frac{\partial z}{\partial q_2}
\right) &= 0 \label{eq:curvilinear-independent-1}\\
\left(
\frac{\partial q_1}{\partial x},
\frac{\partial q_1}{\partial y},
\frac{\partial q_1}{\partial z}
\right) \cdot
\left(
\frac{\partial x}{\partial q_3},
\frac{\partial y}{\partial q_3},
\frac{\partial z}{\partial q_3}
\right) &=0 \label{eq:curvilinear-independent-2}\\
\end{align}\]

式\eqref{eq:curvilinear-orthognal-1}、\eqref{eq:curvilinear-orthognal-2}、\eqref{eq:curvilinear-independent-1}、\eqref{eq:curvilinear-independent-2} を見ると、\(\left(\displaystyle \frac{\partial x}{\partial q_1}, \displaystyle \frac{\partial y}{\partial q_1}, \displaystyle \frac{\partial z}{\partial q_1} \right)\) と \(\left(\displaystyle \frac{\partial q_1}{\partial x}, \displaystyle \frac{\partial q_1}{\partial y}, \displaystyle \frac{\partial q_1}{\partial z} \right)\) が同じ 2 つのベクトルに垂直なので平行であることがわかります。

そこで、この間の比 \(k_i\) を以下の様に定義します。
\[
\frac{\displaystyle \frac{\partial x}{\partial q_i}}{\displaystyle \frac{\partial q_i}{\partial x}} =
\frac{\displaystyle \frac{\partial y}{\partial q_i}}{\displaystyle \frac{\partial q_i}{\partial y}} =
\frac{\displaystyle \frac{\partial z}{\partial q_i}}{\displaystyle \frac{\partial q_i}{\partial z}} =
k_i
\]

\(q_i\) を自分自身で微分してデカルト座標に変数変換すると、
\[\begin{align}
1 = \frac{\partial q_1}{\partial q_1}
&= \frac{\partial q_1}{\partial x}\frac{\partial x}{\partial q_1} +
\frac{\partial q_1}{\partial y}\frac{\partial y}{\partial q_1} +
\frac{\partial q_1}{\partial z}\frac{\partial z}{\partial q_1} \nonumber\\
&= \frac{1}{k_1}\left(\frac{\partial x}{\partial q_1}\right)^2+
\frac{1}{k_1}\left(\frac{\partial y}{\partial q_1}\right)^2+
\frac{1}{k_1}\left(\frac{\partial z}{\partial q_1}\right)^2 \nonumber\\
&= \frac{h_1^2}{k_1} \\
\end{align}\]

よって以下の等式が成立します。
\[\begin{align}
k_i = h_i^2
\end{align}\]

ゆえに
\[\begin{align}
\frac{\partial q_i}{\partial x} = \frac{1}{h_i^2} \frac{\partial x}{\partial q_i}
\end{align}\]

補足:同じ2つのベクトルに垂直なら平行であることの証明

式\eqref{eq:curvilinear-orthognal-1}、\eqref{eq:curvilinear-orthognal-2}、\eqref{eq:curvilinear-independent-1}、\eqref{eq:curvilinear-independent-2} から \(\left(\displaystyle \frac{\partial q_1}{\partial x}, \displaystyle \frac{\partial q_1}{\partial y}, \displaystyle \frac{\partial q_1}{\partial z} \right)\) と \(\left(\displaystyle \frac{\partial x}{\partial q_1}, \displaystyle \frac{\partial y}{\partial q_1}, \displaystyle \frac{\partial z}{\partial q_1} \right)\) が平行であることを証明します。

式\eqref{eq:curvilinear-orthognal-1}、\eqref{eq:curvilinear-orthognal-2} を再掲します。
\[\begin{align}
\left(
\frac{\partial x}{\partial q_1},
\frac{\partial y}{\partial q_1},
\frac{\partial z}{\partial q_1}
\right) \cdot
\left(
\frac{\partial x}{\partial q_2},
\frac{\partial y}{\partial q_2},
\frac{\partial z}{\partial q_2}
\right) &= 0 \\
\left(
\frac{\partial x}{\partial q_1},
\frac{\partial y}{\partial q_1},
\frac{\partial z}{\partial q_1}
\right) \cdot
\left(
\frac{\partial x}{\partial q_3},
\frac{\partial y}{\partial q_3},
\frac{\partial z}{\partial q_3}
\right) &= 0 \\
\end{align}\]
この 2 式から \(\displaystyle \frac{\partial x}{\partial q_1}\) を消すと
\[\begin{align}
\frac{\partial q_1}{\partial y}
\left(
\frac{\partial x}{\partial q_3}
\frac{\partial y}{\partial q_2}
– \frac{\partial x}{\partial q_2}
\frac{\partial y}{\partial q_3}
\right)
+
\frac{\partial q_1}{\partial z}
\left(
\frac{\partial x}{\partial q_3}
\frac{\partial z}{\partial q_2}
– \frac{\partial x}{\partial q_2}
\frac{\partial z}{\partial q_3}
\right) = 0 \label{eq:parallel-intermediate-1}\\
\end{align}\]

次に \eqref{eq:curvilinear-independent-1}、\eqref{eq:curvilinear-independent-2} を再掲します。
\[\begin{align}
\frac{\partial q_1}{\partial q_2} &=
\frac{\partial x}{\partial q_2}\frac{\partial q_1}{\partial x} +
\frac{\partial y}{\partial q_2}\frac{\partial q_1}{\partial y} +
\frac{\partial z}{\partial q_2}\frac{\partial q_1}{\partial z} = 0 \\
\frac{\partial q_1}{\partial q_3} &=
\frac{\partial x}{\partial q_3}\frac{\partial q_1}{\partial x} +
\frac{\partial y}{\partial q_3}\frac{\partial q_1}{\partial y} +
\frac{\partial z}{\partial q_3}\frac{\partial q_1}{\partial z} = 0 \\
\end{align}\]

この 2 式から \(\displaystyle \frac{\partial q_1}{\partial x}\) を消すと
\[\begin{align}
\frac{\partial y}{\partial q_1}
\left(
\frac{\partial x}{\partial q_3}
\frac{\partial y}{\partial q_2}
– \frac{\partial x}{\partial q_2}
\frac{\partial y}{\partial q_3}
\right)
+
\frac{\partial z}{\partial q_1}
\left(
\frac{\partial x}{\partial q_3}
\frac{\partial z}{\partial q_2}
– \frac{\partial x}{\partial q_2}
\frac{\partial z}{\partial q_3}
\right) = 0 \label{eq:parallel-intermediate-2}\\
\end{align}\]

式 \eqref{eq:parallel-intermediate-1} と \eqref{eq:parallel-intermediate-2} から
\[\begin{align}
\frac{
\displaystyle \frac{\partial y}{\partial q_1}
}{
\displaystyle \frac{\partial z}{\partial q_1}
}
=
\frac{
\displaystyle \frac{\partial q_1}{\partial y}
}{
\displaystyle \frac{\partial q_1}{\partial z}
}
=
\frac{
\displaystyle
\frac{\partial z}{\partial q_3}
\frac{\partial x}{\partial q_2}
– \frac{\partial z}{\partial q_2}
\frac{\partial x}{\partial q_3}
}{
\displaystyle
\frac{\partial x}{\partial q_3}
\frac{\partial y}{\partial q_2}
– \frac{\partial x}{\partial q_2}
\frac{\partial y}{\partial q_3}
}
\end{align}\]
ゆえに
\[\begin{align}
\frac{\partial y}{\partial q_1}: \frac{\partial q_1}{\partial z}
=
\frac{\partial q_1}{\partial y} : \frac{\partial z}{\partial q_1}
\end{align}\]
他の座標軸についても同様の関係が成立するので、結局
\[\begin{align}
\frac{\partial x}{\partial q_1} : \frac{\partial y}{\partial q_1} : \frac{\partial z}{\partial q_1}
=
\frac{\partial q_1}{\partial x} : \frac{\partial q_1}{\partial y} : \frac{\partial q_1}{\partial z}
\end{align}\]
よって、\(\left(\displaystyle \frac{\partial q_1}{\partial x}, \displaystyle \frac{\partial q_1}{\partial y}, \displaystyle \frac{\partial q_1}{\partial z} \right)\) と \(\left(\displaystyle \frac{\partial x}{\partial q_1}, \displaystyle \frac{\partial y}{\partial q_1}, \displaystyle \frac{\partial z}{\partial q_1} \right)\) は平行になります。

直交曲線座標系における勾配 (grad)
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