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In determining the shape of a meniscus, we have to minimize the energy per unit length along the direction perpendicular to the cross-section of the meniscus: $$\frac{E}{L}=\int^L_0 dx [\gamma \sqrt{1+(\partial_x h)^2}+\frac{1}{2}\Delta\rho g h^2]$$ where $h(x)$ is the height of the meniscus at $x$, $\gamma$ is the surface tension factor and $\Delta \rho$ is the difference between the densities of the fluid and the vapor above it.

I understand where the first term comes from -- it's the contribution of energy due to surface tension. But I don't understand how the second term is obtained. It looks like a contribution of gravitational potential energy, but I don't know how this precise form is arrived at. In particular, I don't understand why $h$ is squared.

Thank you.

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Elastic potential energy – David H Oct 18 '13 at 7:55
@DavidH: The multiplyer $g$ makes it very unlikely. – user23660 Oct 18 '13 at 8:02
@user23660 That was intended more as a cue to look for a restoring force. Take your solution below for example. It's not that difficult to learn how to "reverse engineer" derivations like these just by knowing what you need to end up with. It's a quick and dirty way to arrive at what the forces must be, even if you haven't the slightest clue why. – David H Oct 18 '13 at 8:39
up vote 5 down vote accepted

Indeed, the second term is the potential gravitational energy. Gravitational energy for a continuous body could be calculated as $$ E_g = \int_\text{body} \rho g\, z \,d^3x, $$ where we assume that gravitational acceleration $g$ has only $z$-component. The 'body' you have in your problem would stretch to length $L$ along the $y$ axis and lays between the surfaces $z=0$ and $z= h(x)$. So the volumetric integral reduces to: $$ E_g = \int_0^L dy \int_{x_1}^{x_2}\left(\int_0^{h(x)} \rho \,g\, z \,dz\,\right)dx=L \int_{x_1}^{x_2} \frac12 \rho\, g\,h^2 dx. $$ In order to receive your expression we need to substitute $\rho$ with $\Delta \rho$ to account for the fact that fluid displaces gas, which also has potential energy.

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Thank you very much! This is very helpful! :) – Alex Oct 18 '13 at 14:44

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