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Surface tension and excess pressure

  1. Oct 27, 2011 #1
    Let σ be the surface tension ......
    I got to know that the excess pressure for a liquid-gas interface with radii of curvature (see http://en.wikipedia.org/wiki/Surface_tension#Surface_curvature_and_pressure .....the part on Surface curvature and pressure and Young-Laplace equation).

    is given by ΔP= σ(1/R1 + 1/R2)

    I have 3 cases as shown in my attached figure... in there cases, how do we know which radii/lengths to plug into the above formula?

    how do we/on what basis do we make the selection..?

    I'm guessing that we just look at the liquid-gas interface and see what radii are on the opposite sides....but that doesn't work for the rectangular plate....

    also, especially in regard to the liquid droplet, if we take the inner radius to be one radius, on the other side of the boundary of the droplet, we have an infinite radius!!

    Please help!!

    Attached Files:

  2. jcsd
  3. Oct 27, 2011 #2


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    As the Wikipedia article states, the quantity in parentheses is known as the "mean curvature" of the surface. 1/R1 and 1/R2 are known as the principal curvatures. For a smooth surface if you look at all curves on the surface that pass through the central point, their curvature will vary. There will be two directions in which the curvature takes an extreme value, and by definition these are the principal curvatures. The two principal directions will be perpendicular to each other.

    In your examples,
    1) You've correctly identified r and r', only remember that r' is negative.
    2) For the rectangular plate the curvature will be different at different points around the plate. Along one of the straight sides, you've identified one principal curvature r' (it's negative). In the other principal direction parallel to the edge the curvature is zero (r is infinite, that's fine)
    3) For the bubble, both curvatures have the same value, 1/R1.
  4. Oct 28, 2011 #3

    Andy Resnick

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    I'm a little confused by the diagrams- if the fluid is not pinned to the edges of the plate, the mean curvature is simple to understand- one principal radius is r', and the other is r, and note that r varies with height. This means r' has to vary with height as well: IIRC, if there is no gravity the profile is a catenary, and if gravity is present you get an 'amphora' shape, and the boundary condition is given by the contact angle.

    Now, if the fluid is pinned to the plate edges, things change considerably: for the circle plate there's no conceptual issue (except the contact angle can vary), but the rectangular plate *is* much more complicated because of the corners. The radius of curvature there is infinite, so the fluid can't stay pinned there. I'm unaware of a solution to this problem, although the fluid shape is a minimal surface, so there may be a solution out there.
  5. Oct 28, 2011 #4

    I guess the bolded parts are important to remember....

    I don't understand the bold parts.....why is the radius of curvature in one direction zero? How does the curvature vary around the rectangle?

    Also, you said that the curvatures are supposed to be taken perpendicular to one-another...this does not happen here...neither for the circular disc case for that matter...could you elaborate more on this example,please?
  6. Oct 28, 2011 #5
    How are the radii of curvature at the corners infinite? For that matter, I don't think I understand what happens along the edges either..what about if it were an infinite rectangular plate?

    The liquid isn't pinned, I guess....we're just sorta balancing the plate on a film of liquid...much like in a contact lens-eye interface...
  7. Oct 28, 2011 #6

    Andy Resnick

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    No, the curvature is infinite- the radius is zero (k = 1/r) because the surface is discontinuous at an edge or corner.

    If the liquid isn't pinned, then the shape of the plate doesn't matter.
  8. Oct 29, 2011 #7
    I think I'm having a basic problem in how to decide which radii I'm talking about...please could you also look at post #4....maybe if you answered those questions first, it would be easier for me....
  9. Oct 29, 2011 #8
    If you look the derivation it would be easy.
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