Surface area and its variation

A function that gives surface area is one of the basic funcationals and it is expressed as:

(1)
\begin{align} a=\int_a{da}, \end{align}

where $a$ is the area, $\int_a$ represents an integration over the surface domain and $da$ is an area element.
While this expression is easy to understand, we cannot perform anything with it unless we know the detailed expression of $da$.
Hence, we want to rewrite it as

(2)
\begin{align} a=\int_a{f(\theta^1,\theta^2)d\theta^1d\theta^2} \end{align}

and identify the concrete expression of $f(\theta^1,\theta^2)$.
Consequently, we obtain

(3)
\begin{align} f=\sqrt{g_{11}g_{22}-g_{12}^2}, \end{align}

where $g_{ij}$ is a metric.
This has been historically expressed as

(4)
\begin{align} f=\sqrt{EF-G^2}. \end{align}

If we suppose that $g_{ij}$ represents a matrix as well as representing its components, i.e.,

(5)
\begin{align} g_{ij}=\left[\begin{array}{cc}g_{11}g_{12}\\g_{21}g_{22}\end{array}\right], \end{align}

we can write

(6)
\begin{align} f=\sqrt{\mathrm{det}g_{ij}}. \end{align}

The variation of the surface area can be written as

(7)
\begin{align} \delta a=\int_{a}{\delta f d\theta^{1}d\theta^{2}}. \end{align}

Thus, we calculate the variation of $f$.

(8)
\begin{align} \delta f =\delta\sqrt{\mathrm{det}g_{ij}}=\frac{1}{2}\frac{\delta\mathrm{det}g_{ij}}{\sqrt{\mathrm{det}g_{ij}}} \end{align}
(9)
\begin{align} \because \delta y^{p}=py^{p-1}\delta y. \end{align}

Thus, we calculate the variation of $\mathrm{det}g_{ij}$.
This calculation is extremely complicated but it is like as follows:

(10)
\begin{align} \delta\mathrm{det}g_{ij}=\mathrm{det}g^{ij}\frac{\delta\mathrm{det}g_{ij}}{\mathrm{det}g^{ij}}=\mathrm{det}g^{ij}\mathrm{det}\left(\hat{g}^{ik}\delta g_{kj}\right). \end{align}

Here, $\hat{\cdot}$ on $\hat{g}^{jk}$ means that $g^{jk}$ does not subject variation.
The calculation continues as follows:

(11)
\begin{align} =\mathrm{det}g_{ij}\left(\mathrm{det}\left(\delta^i_{\cdot j}+\hat{g}^{ik}\delta g_{kj}\right)-1\right) \end{align}

and

(12)
\begin{align} =\mathrm{det}g_{ij}g^{ij}\delta g_{ji}. \end{align}

As the result, we obtain

(13)
\begin{align} \delta a=\frac{1}{2}\int{g^{ij}\delta g_{ji}\sqrt{\mathrm{det}g_{ij}}d\theta^1d\theta^2} \end{align}
(14)
\begin{align} \therefore\delta a=\frac{1}{2}\int{g^{ij}\delta g_{ji}\mathrm{da}}. \end{align}

This is dimension-free expression so it can be easily generalized to an arbitrary dimension.

(15)
\begin{align} \delta\sqrt{\det{g_{ij}}}=\frac{1}{2}g^{\alpha\beta}\delta g_{\alpha\beta}\sqrt{\det{g_{ij}}} \end{align}

is a relation that is frequently used in relativity.
Here, because we have

(16)
\begin{align} \frac{1}{2}\delta g_{ij}=\frac{1}{2}\left(\delta \boldsymbol{g}_i \cdot \boldsymbol{g}_j+\boldsymbol{g}_i\cdot\delta \boldsymbol{g}_j\right), \end{align}

we obtain

(17)
\begin{align} \delta a=\frac{1}{2}\left(\int{ g^{ij}\delta\boldsymbol{g}_i\cdot \boldsymbol{g}_j\mathrm{da}}+\int{ g^{ij}\delta\boldsymbol{g}_j\cdot \boldsymbol{g}_i\mathrm{da}}\right)=\frac{1}{2}\left(\int{\delta \boldsymbol{g}_i\cdot \boldsymbol{g}^i\mathrm{da}}+\int{\delta \boldsymbol{g}_j\cdot \boldsymbol{g}^j\mathrm{da}}\right) \end{align}
(18)
\begin{align} =\int{\delta \boldsymbol{g}_i\cdot \boldsymbol{g}^i\mathrm{da}} \end{align}
(19)
\begin{align} =\int{\frac{\partial\delta\boldsymbol{u}}{\partial \theta^i}\cdot \boldsymbol{g}^i\mathrm{da}} \end{align}
(20)
\begin{align} =\int{\mathrm{div}\delta\boldsymbol{u}\mathrm{da}}. \end{align}
page revision: 21, last edited: 13 May 2015 06:20