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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 $\delta S^M$ (defined as the boundary of the $S^M$ in this paper) and $\hat n$ denoting the unit normal to $\Omega^M$.

Now the paper states that, using the divergence theorem,

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

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?

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Could you provide a link to the paper? –  user26872 Mar 11 '13 at 17:02
    
@oen - let me see whether I can find one available on the internet. –  Vincent Tjeng Mar 12 '13 at 0:50
1  
@oen - the paper is here, but unfortunately it's behind a paywall, and I'm really unable to find a free version. sciencedirect.com/science/article/pii/0362546X86900039. I hope you have access from your side, though. –  Vincent Tjeng Mar 13 '13 at 15:28

1 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.

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CFD=computational fluid dynamics? –  Vincent Tjeng Mar 11 '13 at 7:34
    
@Vincent Yes, that's right. –  vergere6 Mar 12 '13 at 2:14

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