Deriving the pressure difference across a gas-liquid interface

Question

The question I'm interested in is deriving the pressure difference across a gas-liquid interface $z = h(x,y)$ is $$\Delta p = \frac{\sigma h_{xx}}{(1+h_{x}^2)^{3/2}} + \frac{\sigma h_{yy}}{(1+h_{y}^2)^{3/2}} $$

I know that: $p_i + p_o = \Delta p = \sigma (\frac{1}{R_1} + \frac{1}{R_2})$, and I think I can use this as a starting point in the derivation of the above equation.

Any help would be greatly appreciated, thanks.

Answer 1

The pressure difference balances the capillary force which is surface tension $\sigma$ times twice the mean curvature

$$2H = \frac{1}{R_1} + \frac{1}{R_2} = \nabla \cdot \mathbf{n}$$

Here the unit normal vector $\mathbf{n}$ is given by

$$\mathbf{n} = \frac{\nabla(z - h(x,y))}{\|\nabla(z - h(x,y)) \|} = \frac{- \frac{\partial h}{\partial x}\mathbf{e}_x - \frac{\partial h}{\partial y} \mathbf{e}_y + \mathbf{e}_z}{\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^2+ \left(\frac{\partial h}{\partial y}\right)^2}}$$

Take the divergence of $\mathbf{n}$ to finish.