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I'm currently reading a paper about incompressible Euler's equation, and I don't understand how the surface element expand. So here comes the question.

Let $\Omega$ be a Riemannian manifold with metric $g_{ab}$ and $N^{a}$ be the unit normal to $\partial\Omega$. We know the volume element is $$ dV_{g}=(det(g))^{\frac{1}{2}}\,dV $$ where $dV$ is the volume element under Eucidean meansure. Now we define the induced measure $\gamma_{ab}=g_{ab}-N_{a}N_{b}$ on the tangent space to the boundary $\partial\Omega$.Then the author claims that $$ dS_{\gamma}=(det(g))^{\frac{1}{2}}(\sum_{n}N_{n}^{2})^{-\frac{1}{2}}\,dS $$ I have no idea how the formula is derived. Since, intuitively, the measure change under different metric should be the square of the determinant of the metric and it seems not the case for the induced surface measure.

CC_Azusa
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1 Answers1

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Let $(u,v)$ be an orthonormal basis (with respect to $g$) for the tangent space at some point of $\partial M$.

Then $$dS_{\gamma} = |u\times v|\,dS = (u\times v)\cdot \tilde{N}\,dS$$ where $\tilde{N}$ is the normal, under the Euclidean metric, to the surface. $N$ and $\tilde{N}$ are parallel; in particular $\tilde{N} = N/\|N\|$. Therefore $$dS_{\gamma} = \frac{(u\times v)\cdot N}{\|N\|}\,dS = \frac{dV_g}{\|N\|dV}\,dS = \frac{\sqrt{\det g}}{{\|N\|}}\,dS.$$

user7530
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  • This is a very good intuition. However, it only solves the $\mathbb{R}^{3}$ case. I was wondering if there is a general understanding for the $n$ dimensional case. – CC_Azusa Mar 26 '14 at 16:26
  • So based on your example, if we choose local coordinate in which the boundary is flat,i.e., the normal $N=(0,0,\cdots,N_{n})$, I wonder is we can come up with analogous solution. – CC_Azusa Mar 26 '14 at 16:37
  • In $n$ dimensions I believe the situation is analogous, with the Euclidean volume of the parallelapiped $\det \left[u_1\ u_2\ \ldots\ u_n\ \tilde{N}\right]$ replacing the triple product. – user7530 Mar 26 '14 at 18:09