# A Fluids -- Surface Energy

1. Oct 30, 2017

### joshmccraney

I'm studying thin fluid films, and the text writes free surface energy of a film (puddle) over domain $(0,X)$ can be expressed as $$E=\int_0^X \left[\frac{h_x^2}{2}+\omega(h)+G\frac{h^2}{2} \right]\, dx$$ where $X$ is a length that the thin film (puddle) rests on, $h$ is the height of the film (puddle), $G$ is a gravity term (0 for no gravity and 1 for gravity), and $\omega(h)$ is energy density due to van der waals forces. Firstly, can someone explain the derivation of this integral to me?

However, that's not my main question. This integral is subject to the constraint $$A = \int_0^X h\, dx$$ where $A$ is area. To minimize the energy subject to the constraint yields the funcitonal $$F = \int_0^X \left[\frac{h_x^2}{2}+\omega(h)+G\frac{h^2}{2} -p(\bar{h})h\right]\, dx$$ where $p$ is a Lagrange multiplier. Here's where I'm stuck: the author proceeds by saying "neglecting the energy in the narrow transition regions at the extremes of the plateau region of thickness $h_0$ and length $L$ (the puddle length, not confused with domain $(0,X)$), the energy per unit length can be calculated as $$g(h_0) = \omega(h_0)+G\frac{h_0^2}{2} -p(\bar{h})h_0.$$
Why is this the energy per unit length when we said $E$ was the free energy?

2. Nov 4, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.