# A Young-Laplace Equation

1. Jul 25, 2016

### joshmccraney

Hi PF!

Here I'm trying to derive the Young-Laplace equation, which states $$\Delta P = \gamma \left(\frac{1}{R_1} + \frac{1}{R_2} \right)$$ where $\Delta P$ is change in pressure, $\gamma$ is a proportionality constant, and $R$ is a radius of curvature, both of which are orthogonal to the other and work so as to parameterize a fluid surface. The derivation follows:

Given a surface of fluid we know (how?) that the change in surface energy $d E_s$ equals the change of work done $dW$. Notice $dW = \Delta P dV = \Delta P x y dz$ (cartesian coordinates). Now $d E_s = \gamma dA$. Then $A = xy$ and $A' = (x+dx)(y+dy) = xy +ydx + x dy$ ignoring higher order infinitesimals. Then $dA = ydx + x dy$. Suppose the surface $dA$ is parameterized by two radii of curvature $R_1$ and $R_2$. Then, by similar triangles, we have $$\frac{R_1+dz}{R_1} = \frac{x+dx}{x} \implies \\ dx = \frac{x dz}{R_1}.$$ Substitute this into the expression for $dA$ to arrive at $$dA = \frac{yxdz}{R_1} + x dy.$$ Identical logic applied to $dy$ implies the final relation for area $$dA = xy dz \left( \frac{1}{R_1}+\frac{1}{R_2} \right) \implies \\ d E_s = xy dz \gamma \left( \frac{1}{R_1}+\frac{1}{R_2} \right).$$ Since $dW = d E_s$ we then have $$\Delta P = \gamma \left( \frac{1}{R_1}+\frac{1}{R_2} \right)$$ where the $xy dz$ term cancels.

My question is, is the following relation correct: $d E_s = \gamma d A$ and $d W = \Delta P dV = \Delta P dA dz$ which implies $\gamma = \Delta P dz$, which, when compared with the above result, yields $1/dz = 1/R_1 + 1/R_2$. Is this still correct?

2. Jul 25, 2016