Divergence Theorem and Mean Curvature as applied to Tension in a Membrane

Question

I'm reading a paper on tension in a membrane and am currently stuck at this part.

The paper so far reads: We consider a portion $S^M$ of a membrane $\Omega^M$, where $\hat k$ denotes the unit vector pointing out of $S^M$ which is normal to $\partial S^M$ and $\hat n$ denoting the unit normal to $\Omega^M$.

Now the paper states that, using the divergence theorem,

$$\int_{\partial S^M} \hat k\,ds=\int_{S^M} (-\nabla\cdot \hat n)\,\hat n\,dS$$

The proof in the paper uses the idea of local parameters, which I am unable to understand (despite reading the outstanding answer here.

Is there:

1. Any proof that does not require applying the idea of local parameters, or
2. Any text that I should read first to better understand the abstract algebra behind the idea of local parameters?

As a side note, the paper also states that $-\nabla\cdot \hat n$ is equal to the mean curvature of the membrane. The wikipedia link suggests that it in fact is equal to twice the mean curvature. Which is correct?

Answer 1

I don't know the answer to your main question. I do know the answer to your side note. Depending on the community/journal where the paper was published, the definition of mean curvature may be $H = k_1 + k_2$ or $2H = k_1 + k_2$, where $k_1$ and $k_2$ are the two principal curvatures in 3D. The latter is technically the correct one, the former is used for convenience (often in the CFD world).

Source: my own experience with these differing definitions.

Answer 2

I accidentally saw your question when I was reading "Introduction to tensor analysis and the calculus of moving surfaces" by Pavel Grinfeld. I believe the section 14.6 in this book provides the proof of you question. Pretty much, the proof is an application of Gauss's theorem: the right hand side of your equation, i.e., $(−\nabla \cdot \hat n) \hat n = \nabla \hat s = \nabla \nabla R$ where $R$ is the position vector. That is to say, $-H \hat n$ is simply the second derivative of the position vector (the first derivatives of the position vector defines the base of tangential vectors of the surface). If you apply Gauss' theorem, the integration of $\nabla \nabla R$ over the surface is equals to the integration of $k \nabla R$ over the boundary of the surface, where $k$ is the projection of the normal of the boundary curve over the surface tangential vectors. In other words, $k \nabla R$ combined will be $\hat k$. Therefore, $\int _{\partial S^M} \hat k \ \mathrm d s = \int _{S^M}(−\nabla ⋅ \hat n) \hat n \ \mathrm dS$ is true.