Mean curvature

In this section, we prove that the divergence of a unit tensor of a surface is the double of the mean curvature vector.
When this is proved, consequently, we can confirm that a minimal surface is a surface which its mean curvature is zero everywhere.

(1)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)=\boldsymbol{x}(\theta^1,\theta^2)+\boldsymbol{I}\cdot d\boldsymbol{x}+\frac{1}{2}d\boldsymbol{x}\cdot\boldsymbol{H}\cdot d\boldsymbol{x}+\cdots. \end{align}

Here, $d\boldsymbol{x},\boldsymbol{I},\boldsymbol{H}$ are respectively defined by

(2)
\begin{align} d\boldsymbol{x}\equiv d\theta^i\boldsymbol{g}_i \end{align}
(3)
\begin{align} \boldsymbol{I}\equiv \boldsymbol{x}\otimes\nabla=\frac{\partial \boldsymbol{x}}{\partial \theta^i}\otimes\boldsymbol{g}^i=\boldsymbol{g}_i\otimes\boldsymbol{g}^i=\boldsymbol{g}^i\otimes\boldsymbol{g}_i \end{align}
(4)
\begin{align} \boldsymbol{H}\equiv \nabla\otimes\left(\boldsymbol{x}\otimes\nabla\right)=\boldsymbol{g}^\alpha\otimes\frac{\partial}{\partial \theta^\alpha}\left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right). \end{align}

The derivation is written in Taylor expansion of a surface.

Now, an arbitrary tangent vector on a surface can be represented by $\boldsymbol{I}\cdot d\boldsymbol{x}=d\boldsymbol{x}$; hence when we omit the higher terms, we obtain

(5)
\begin{align} \boldsymbol{x}(\theta^1+d\theta^1,\theta^2+d\theta^2)\simeq\boldsymbol{x}(\theta^1,\theta^2)+d\boldsymbol{x}. \end{align}

This represents a tangent plane.
We define

(6)
\begin{align} \boldsymbol{N}\equiv\boldsymbol{g}_1\cdot\boldsymbol{H}\cdot\boldsymbol{g}^1+\boldsymbol{g}_2\cdot\boldsymbol{H}\cdot\boldsymbol{g}^2=\boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i. \end{align}

Let us rewrite the right-hand side as

(7)
\begin{align} \mathrm{tr}\boldsymbol{H}\equiv\boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i. \end{align}

Such an operation is called contraction. Particularly, the above operation is called a trace.
Interestingly, the following relation holds:

(8)
\begin{align} \mathrm{div}\boldsymbol{I}=\boldsymbol{N}. \end{align}

This can be verified by comparing

(9)
\begin{align} \mathrm{div}\boldsymbol{I}\equiv\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i \end{align}

and

(10)
\begin{align} \boldsymbol{g}_i\cdot\boldsymbol{H}\cdot\boldsymbol{g}^i=\boldsymbol{g}_i\cdot\boldsymbol{g}^\alpha\otimes\frac{\partial\boldsymbol{I}}{\partial \theta^\alpha}\cdot\boldsymbol{g}^i=\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i. \end{align}

In summary, the following relations holds:

(11)
\begin{align} \boldsymbol{N}=\mathrm{div}\boldsymbol{I}=\mathrm{tr}\boldsymbol{H}. \end{align}

It is very interesting that $\boldsymbol{N}$ represents a normal (orthogonal to the tangent plane) vector and its norm is the double of a quantity called mean curvature.
We first prove that it represents a normal vector.
The calculation is like as follows:

(12)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{I}}{\partial \theta^i}\cdot\boldsymbol{g}^i=\frac{\partial \left(\boldsymbol{g}_i\otimes\boldsymbol{g}^i\right)}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha=\frac{\partial\left( \boldsymbol{g}^i\otimes\boldsymbol{g}_i\right)}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha=\frac{\partial\boldsymbol{g}^i}{\partial \theta^\alpha}\otimes\boldsymbol{g}_i\cdot\boldsymbol{g}^\alpha+\boldsymbol{g}^i\otimes\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha \end{align}

Hence,

(13)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}+\boldsymbol{g}^i\otimes\frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot\boldsymbol{g}^\alpha \end{align}

ここで、少し複雑な法則を用います。共変基底、反変基底の関係$\boldsymbol{g}_i\cdot\boldsymbol{g}^j=\delta_{i}^{\cdot j}$を思い出しましょう。いずれの場合も0か1ですから、任意の共変基底と反変基底、任意の座標パラメータ$\theta^\alpha$について

(14)
\begin{align} \frac{\partial\left( \boldsymbol{g}_i\cdot\boldsymbol{g}^j\right)}{\partial \theta^\alpha}=0 \end{align}

が成り立ちます。従って、

(15)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j+\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha}=0 \end{align}

さらに、

(16)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^\alpha}\cdot \boldsymbol{g}^j=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^\alpha} \end{align}

(17)
\begin{align} \frac{\partial \boldsymbol{g}_i}{\partial \theta^j}\cdot \boldsymbol{g}^j=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{g}^j}{\partial \theta^j} \end{align}

が成り立ちます。代入して

(18)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}-\boldsymbol{g}^j\otimes\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j \end{align}

さて、第二項ですが、$\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j$がスカラーですから、

(19)
\begin{align} \left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j\right)\boldsymbol{g}^j \end{align}

と書けます。従って、

(20)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}-\left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{g}_j\right)\boldsymbol{g}^j \end{align}

となります。

(21)
\begin{align} \boldsymbol{n} \end{align}

を定義します。大きさが1で、接平面と直交していますから、共変基底、反変基底の区別を付ける必要がありません。
この単位法線ベクトルを用いると空間の単位テンソル

(22)
\begin{align} \boldsymbol{D}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2+\boldsymbol{n}\otimes\boldsymbol{n} \end{align}

を記述できます。

(23)
\begin{align} \boldsymbol{I}=\boldsymbol{g}_1\otimes\boldsymbol{g}^1+\boldsymbol{g}_2\otimes\boldsymbol{g}^2 \end{align}

と比べてみると、両者の関係がよくわかります。つまり、

(24)
\begin{align} \boldsymbol{D}-\boldsymbol{I}=\boldsymbol{n}\otimes\boldsymbol{n} \end{align}

です。すると

(25)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}(\boldsymbol{D}-\boldsymbol{I}) \end{align}

と書けます。他に

(26)
\begin{align} \boldsymbol{N}=\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}(\boldsymbol{n}\otimes\boldsymbol{n}) \end{align}

(27)
\begin{align} \boldsymbol{N}=\left(\frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot\boldsymbol{n}\right)\boldsymbol{n} \end{align}

とかけますが、いずれも同じことを表しています。つまり

(28)
\begin{align} \frac{\partial \boldsymbol{g}^1}{\partial \theta^1}+\frac{\partial \boldsymbol{g}^2}{\partial \theta^2} \end{align}

から接平面成分が削除され、法線成分のみ取り出したものが$\boldsymbol{N}$です。

さてこの法線成分について

(29)
\begin{align} \frac{\partial \boldsymbol{g}^i}{\partial \theta^i}\cdot \boldsymbol{n}=-\boldsymbol{g}^i\frac{\partial\boldsymbol{n}}{\partial \theta^i}=-g^{ij}\boldsymbol{g}_j\frac{\partial\boldsymbol{n}}{\partial \theta^i} \end{align}

について計算を続けます。

(30)
\begin{align} \mathrm{II}\equiv\frac{1}{2}h_{ij}d\theta^i d\theta^j \end{align}

において

(31)
\begin{align} h_{ij}=\frac{\partial \boldsymbol{g}_j}{\partial \theta^i}\cdot\boldsymbol{n}=-\boldsymbol{g}_i\cdot\frac{\partial \boldsymbol{n}}{\partial \theta^j} \end{align}

だったことを思い出すと、

(32)
$$=g^{ij}h_{ij}$$

と書けます。

(33)
\begin{align} \frac{\partial g^{ij}\boldsymbol{g}_i}{\partial \theta^j}\cdot\boldsymbol{n}=\left(\frac{\partial g^{ij}}{\partial \theta^j}\boldsymbol{g}_i\right)\cdot\boldsymbol{n}+h_{ij}g^{ij} \end{align}

として、第一項が零であることを用いても同じ結果を得ます。
さて、曲面上の曲線と主曲率で平均曲率は

(34)
\begin{align} H=\frac{1}{2}(\lambda_1+\lambda_2)=\frac{LG-2MF+EN}{2(EG-F^2)} \end{align}

として与えられましたが、同時に

(35)
\begin{align} H=\frac{g^{ij}h_{ij}}{2} \end{align}

とも書けました。

page revision: 30, last edited: 03 Jun 2015 05:40