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About Young-laplacian equation.Hi guys,
In physics, the Young–Laplace equation is a nonlinear partial differential equation that describes the capillary pressure difference sustained across the interface between two static fluids, such as water and air, due to the phenomenon of surface tension. But how can I deduce this equation?http://en.wikipedia.org/wiki/Young%E...place_equation |

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Thanks alot, after draw the shell element, I know , but how can I turn it to ? Now I only know Could it be said that ? |

Hi, follow the variation of the normal vector along the element. Imagine that, at the center of the shell element, the normal is point upwards but, if you move along one side in the, let's say, the x direction, the normal at the edge of the element will be crooked because of the curvature of the element.
At the edge the normal will not be pointing exactly upwards but somewhat along the x direction. How much is "somewhat"? => . The variation in the x coordinate is dx/2 (half length of the element along the x coordinate). Therefore the variation of the x component of the normal vector as we move along the x coordinate is: and re-arranging: Thus expresses the variation of the normal components due to curvature of the fluids interface. It's just some geometry and some algebraic manipulations. Cheers, Michujo. |

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I just came back. Thats beautiful!! I need time to see it deeply, Thanks very much Michujo!!! |

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No worries. I drew it so it's much easier to see instead of talking about vectors and differentials...
Cheers, Michujo. |

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I just think about it deeply, Thats more clear now. Thats very kool. But there is only one thing left now: I dont understand why the increment in x-component of the normal vector equals , could you explain it for a little bit? Thanks for your patience!!Or you can recommend me some papers about this. In this image, q's magnitude equals ? |

Hi, yes, in the image the magnitude of q is .
It's all geometry: 1) The interface element spans and along the x1 and x2 coordinates. 2) Also, you know that the modulus of the normal vector is 1. 3) Finally, as the normal vector is always perpendicular to the interface, by geometry, as the element turns and angle of so will the normal vector. Therefore the components of the normal vector will be and . For that you have moved a distance dx1/2 from the center of the element. 4) tends to as tends to zero. I cannot think of any reference now but I'm sure there must be tons of information out there somewhere. Cheers, Michujo. |

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Cheers! |

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