I was reading this thesis about surface tension and its role in floating bodies. I couldn't quite understand at page 10 how the author applied the 2D divergence theorem outside the region bounded by the contour. Shouldn't you use the divergence theorem for a region inside the boundary? The line integral is evaluated around a boundary C \pi, but the double integral is evaluated outside the region bounded by C\pi

The line integral is evaluated around a boundary C \pi, but the double integral is evaluated outside the region bounded by C\pi

Here is my thinking. The divergence theorem in 2D is:

$\iint_S \nabla \cdot \textbf{F}\hspace{0.2cm} dA = \int_{\partial S} \textbf{F}\cdot d\textbf{n}$

Therefore, if $C_\pi$ the boundary of our region ($\partial S$ in the Divergence theorem), then I would expect the integral:

$\int_{C_\pi} \textbf{n}\cdot d\textbf{n}$

to be equal to:

$\iint_{W_\pi} \nabla \cdot \textbf{n} \hspace{0.2cm} dA$

where $W_\pi$ is the region bounded by $C_\pi$. However, the author sets this equal to:

$\iint_{\mathbb{R}^2 \setminus W_\pi} \nabla \cdot \textbf{n} \hspace{0.2cm} dA$

Then that implies that: $\iint_{\mathbb{R}^2 \setminus W_\pi} \nabla \cdot \textbf{n} \hspace{0.2cm} dA=\iint_{W_\pi} \nabla \cdot \textbf{n} \hspace{0.2cm} dA$

which can't possibly be right. Is there something I am missing?

  • $\begingroup$ You could consider a bigger surface $W_{big}$ and then remove your surface from it $W_{big}\setminus W_{\pi}$. Provided the integrand (the stuff in the surface integral) vanishes at infinity, there is no pentalty in extending the big surface to fill the whole space $W_{big} \setminus W_{\pi}\to\mathbb{R}^2 \setminus W_\pi$. You could apply your divergence theorem on this bigger surface, but since the integral is zero at infinity you would only get non-zero edge integral from $\mathcal{C}\pi$ $\endgroup$ – Cryo Jul 19 '19 at 13:26
  • $\begingroup$ @Cryo So I can extend the surface as much as I want since the integrand converges to zero? But then wouldn't the surface integral evaluated inside the boundary be equal to that evaluated outside? $\endgroup$ – Kouta Dagnino Jul 19 '19 at 13:33
  • $\begingroup$ I didnt say that integrand is zero immediately outside the boundary. I only said that integrand needs to go to zero sufficiently fast infinitely far away from the boundary. The value of the integrand immediatelly outside the boundary can be non-trivial. So surface integral may be different. I am not sure if I am answering your (new) question correctly. Could you put it in mathematical form? $\endgroup$ – Cryo Jul 19 '19 at 13:40
  • $\begingroup$ sure I will edit the question and add some more stuff so that it is clearer. Thank you. $\endgroup$ – Kouta Dagnino Jul 19 '19 at 13:43

IMHO your notation is misleading.

What do you mean by $\int_{\partial S} d\mathbf{n}\cdot\mathbf{F}$? I guess this is an integral along the edge of your surface, and then $\mathbf{n}$ is the normal to the surface, point outwards, but then I would say:

$\oint_{\partial S} dl \,\mathbf{\hat{n}}\cdot \mathbf{F}$, i.e. you need to have $dl$ as the small step along your edge. The normalized normal to the edge is here $\mathbf{\hat{n}}$. Moving on to the next integral

$\int_{C_\pi} \mathbf{n}.d\mathbf{n}$

Again, what do you mean? Is the $\mathbf{n}$ the edge normal in both cases? In this case you are basically interested in:

$\oint_{C_\pi} dl\, \mathbf{\hat{n}}.\mathbf{\hat{n}}=\oint_{C_\pi} dl=circumference$

Since $\mathbf{\hat{n}}$ is the normalized normal. By the way, the next integral, $\iint_{W_\pi} dA \,\boldsymbol{\nabla}\cdot\mathbf{n}$ makes no sence at all, in this case. The normal to the edge is only defined on the edge. How do you evaluate it in the bulk?

Perhaps you meant something like:

$\oint_{C_\pi} dl\,\mathbf{\hat{n}}\cdot\mathbf{F}$, for some generic field $\mathbf{F}$

In this case $\oint_{C_\pi} dl\,\mathbf{\hat{n}}\cdot\mathbf{F}=\int_{W_\pi} d^2 r \boldsymbol{\nabla}.\mathbf{F}$

But also, if you invert the boundary normal ($\mathbf{\hat{n}}\to -\mathbf{\hat{n}}$) and assume that $\mathbf{F}$ vanishes at infinity then the edge integral may be taken as the edge integral of $\mathbb{R}^2$ with $W_\pi$ removed:

$\oint_{C_\pi} dl\,\mathbf{\hat{n}}\cdot\mathbf{F}=- \oint_{\partial \left(\mathbb{R}^2\setminus W_\pi\right)} dl\,\mathbf{\hat{n}}\cdot\mathbf{F}=-\int_{\mathbb{R}^2\setminus W_\pi} d^2 r \boldsymbol{\nabla}.\mathbf{F}$

This suggestes that $\int_{\mathbb{R}^2} \boldsymbol{\nabla}.\mathbf{F} = \int_{\mathbb{R}^2\setminus W_\pi} d^2 r \boldsymbol{\nabla}.\mathbf{F} + \int_{W_\pi} d^2 r \boldsymbol{\nabla}.\mathbf{F} = 0$

Which makes sense since (with slight abuse of notation)

$\int_{\mathbb{R}^2} \boldsymbol{\nabla}.\mathbf{F} = \oint_{\partial \mathbb{R}^2} dl\,\mathbf{\hat{n}}.\mathbf{F} $

and we agreed that $\mathbf{F}$ vanishes at infinity.

  • $\begingroup$ Maybe I should write to the author because from a physical perspective it doesn't really make sense XD It would imply that the archimedean force is equivalent to the surface tension force for any object of any size... $\endgroup$ – Kouta Dagnino Jul 20 '19 at 8:50
  • $\begingroup$ Anyways, I have finally solved the problem (after a couple months : |). Turns out they are the same. $\endgroup$ – Kouta Dagnino Dec 21 '19 at 10:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.