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A Young-Laplace Equation

  1. Jul 25, 2016 #1

    joshmccraney

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    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. jcsd
  3. Jul 25, 2016 #2

    Andy Resnick

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