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Surface tension question.

  1. Feb 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Two bubbles, one of radius R and the other of radius 3R come into contact with each other. What is the distance between the centres of the two bubbles? Ignore the weight of the bubbles.

    2. Relevant equations

    p(inside bubble)= p(atm) + 4T/R where T is the surface tension and R the radius of the bubble

    3. The attempt at a solution

    The radius of the common surface of the bubbles comes out to be 3R/2. I've taken the radius of the circle that is common to both bubbles as r. Now I've balanced the forces first on this common surface which gave me:

    (2r)/(3R) = sin theta where theta is the semi vertical angle of the cone subtended by this common surface onto the centre of the imaginary sphere of which the common surface is a part of.

    Then I balanced forces on the surfaces of the two bubbles which gave me:

    r/R = sin alpha where alpha is the semi-vertical angle of the cone subtended by the common surface onto the centre of the smaller bubble.

    r/(3R) = sin beta where beta is the semi-vertical angle of the cone subtended by the common surface onto the centre of the larger bubble.

    I don't know what to do next. I have a feeling that the question is missing some information.
  2. jcsd
  3. Feb 11, 2014 #2


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    Seems to me you're misreading the question. I read it as a simple geometry question, nothing to do with surface tension. It doesn't say the bubbles deform or merge. Did you draw a diagram?
    What do you mean by the 'common surface' of two touching spheres? How do you get 3R/2?
  4. Feb 13, 2014 #3
    The bubbles do deform. The question is in the Surface Tension section in my text book. I got 3R/2 by calculating pressure difference across the common surface and equating it with 4T/R'. There was a question before this one where we had to do this.
    Last edited: Feb 13, 2014
  5. Feb 13, 2014 #4


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    OK, I see. To get the 3R/2 you have to assume that the difference in pressure stays constant, yes?
    There would certainly be enough information if we knew the external pressure, but it seems to me the problem is very messy unless we can make some simplifying approximation. If we assume external pressure is relatively large then we can ignore any change in bubble volume. That in turn means the pressures do not change. But it does imply the radii of the bubbles where they are not in contact must increase.
    So we have two unknowns, the two new radii, and two equations representing that the volumes do not change. The algebra looks horrendous, though.
    Does it help to note that the angle between the three surfaces is 2π/3?

    Edit: Some of what I wrote above cannot be right. If the volumes stay the same then the pressures stay the same, so the external radii stay the same, so the volumes reduce!
    Seems to me now that the radii must increase, so the pressures reduce, so the volumes increase.
    Last edited: Feb 13, 2014
  6. Feb 14, 2014 #5


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    More thoughts...

    There are four unknowns: the three radii and the separation of the centres.
    Consideration of pressure differences gives a very simple equation relating the three curvatures.
    Using the 120 degree angles at which the membranes meet gives two fairly simple equations relating the curvatures and the separation (but I'm not sure these three equations are independent).
    Let the ambient pressure be Pamb. If the initial pressure in a bubble is Pamb + P, where P << Pamb, and the reduction in pressure after contact is ΔP, then the fractional reduction in pressure is approximately ΔP/Pamb, a very small quantity. So the fractional increase in volume is correspondingly small. But the change in external radius is related to ΔP/P, a much larger value. Hence we can take the volumes as constant, but not the external radii.
    This gives us two more (but much messier) equations - an embarrassment since we now have more equations than unknowns.
  7. Feb 14, 2014 #6
    Is there a fairly simple equation for the volume of the two bubbles once they have touched? There is a relationship for the three radii of a double bubble 1/Rcommon = 1/Rsmall - 1/Rlarge . Which can give another equation that relates the change in radius of the smaller bubble to the larger one. [ If the small bubble radius increases to aR then the larger bubble increases to 3aR/(3-2a) for the given 3R/2 radius of the common surface ]

    I haven't managed to solve it myself. It seems like it ought to be easier than the pages of messy algebra that I have currently.
  8. Feb 14, 2014 #7


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    In my post #5 I avoided assuming the 3R/2 radius. That relies on the assumption that the pressure difference remains constant, and I don't see a good argument for that. Yes, transparent says that was the answer to an earlier question, but since I've not seen that question I can't be sure the conditions match.
    However, if you're happy to assume 3R/2, then you can get your third and fourth equations from the 120 degree angles, as I mentioned in post #5. Worth a try.
  9. Feb 15, 2014 #8
    That earlier question was:
    Two bubbles of radius R1 and R2 (R1>R2) come in contact with each other and deform. What is the Radius of curvature of the common surface?
    The answer was (R1R2)/(R1-R2)
    We do ignore the change in volume.

    There was another question which said:
    Two bubbles, each of radius R, come into contact with each other. What is the distance between their centres?

    I managed to solve that one (I have also attached a (poorly made) diagram):
    I assumed an element of length dx on the common boundary. The common surface applies a force 2Tdx on the element and so do the other two surfaces. Since it is a symmetric arrangement, The direction of the forces of the other two surfaces will be symmetric about the direction of the force applied by the common surface. I resolved the vectors and made the net force equal to zero. This gave me the angle between the vector as 120 degrees. On solving, the distance came out to be R.

    But in this question, it is not a symmetric arrangement and hence the angle will not be 120 degrees. Hence the trouble.

    Attached Files:

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    Last edited: Feb 15, 2014
  10. Feb 15, 2014 #9


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    But you see the difficulty: if the volume does not change then the external radii must increase appreciably. So the pressure differences (above ambient) change appreciably. This means we can no longer be sure what the pressure difference is between the bubbles, so there is no sound basis for the result (R1R2)/(R1-R2). (If the new external radii are R1' and R2', then certainly the internal radius is (R1'R2')/(R1'-R2').)
    Maybe (R1R2)/(R1-R2) is correct, but there needs to be some theoretical justification for saying that the difference in pressure between the bubbles hardly changes.
    It's always 120 degrees. Consider the force diagram in a short section of where the two bubbles meet. There are three membranes meeting in a line, and the same tension in each membrane.
  11. Feb 15, 2014 #10
    I tried the simpler problem of two bubbles of the same radius which makes the common surface a flat disc. My algebra got messy on that one too, so I did some approximating and ended with each bubble's radius increasing by approx 6%. Which means I'd be offering 1.06R as the distance between the centres of the two now slightly enlarged bubbles. As an engineer I'd be ok with the approximation but both the mathematician and physicist within are uncomfortable with accepting 1.06R for the symmetric double bubble. It is an intriguing problem and I keep thinking that I must be missing some simplification somewhere.

    I do have an upper and lower bound for the distance between the R and 3R bubbles. My upper bound is the trivial 4R where the bubbles have just touched and lower bound of √7 R by assuming the radii of the bubbles don't change. I know the lower bound is too low as the pressure within both bubbles after they join must go down.
    Last edited: Feb 15, 2014
  12. Feb 15, 2014 #11
    I suppose you're right. Three forces acting on a point will give zero iff the angles between them are 120 degrees. Thanks.

    The answer is in fact √7 R. Could you show me how you got that?
  13. Feb 15, 2014 #12
    Cosine rule.
    √7 R may be the answer in the book but I'm confident that particular answer is incorrect.
  14. Feb 15, 2014 #13


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    Well, inexact.
  15. Feb 16, 2014 #14
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