Laplace pressure is given by $$\Delta p=\gamma \left(\frac{1}{R}+\frac{1}{R'}\right)$$ where $R$ and $R'$ are the radii of the curvature of the surface.

Using the following diagramcapillary bridge

the book I'm trying to learn from says the Laplace pressure of the drop is given by $$\Delta p=\gamma \left(\frac{1}{R}-\frac{\cos(\theta_E)}{H/2}\right)\approx \frac{-2\gamma \cos(\theta_E)}{H}$$ with an angle $\theta_E<\pi/2$. How does the expression that replaces $1/R'$ become negative?


Hmmm, the question seems to have changed while I was typing this answer!

Just to be clear: the liquid bridge is rotationally symmetric about the vertical line of the two arrows that are pointing towards one another. This is why $R$ is one of the principal radii of curvature.

The value of $R'$ is just trigonometry. Given the contact angle $\theta_E$ on the diagram, it's fairly easy to see that the radius of the circular interface shown on the diagram $R'$ satisfies $\cos\theta_E=\frac{1}{2}H/R'$.

The radius of curvature carries a sign: $R$ is with respect to a centre within the liquid, $R'$ is curving the opposite way, with a centre of curvature out in the gas phase. So, it gives a negative contribution to the Laplace pressure in this case.

[EDIT following OP comment]

Calculating the force is a bit subtle. It isn't just the Laplace pressure times the area of contact between liquid and plate. This chapter by Kralchevsky (from his publication list, but also available in the book by Kralchevsky and K. Nagayama, Particles at Fluid Interfaces and Membranes (Elsevier, 2001)) describes how the force is evaluated at the mid-plane between the plates. There is a Laplace pressure contribution, which acts on the area $\pi R^2$ in your diagram, plus a liquid-vapour surface tension contribution, which acts on the circumference $2\pi R$. They go on to say that the force could be evaluated by taking any cross-section parallel to the plates, giving the same answer, but the mid-plane means that you don't need to worry about the angles at which the forces act. You could do it at the surface of the plate (with a larger cross-sectional area, as you say) but the surface tension force (at an angle) still contributes, and the answer turns out to be the same.

Have a look at section 11.2 in that chapter.

  • $\begingroup$ Aha, yes, I had managed to wrap my head around the geometry of the situation not long after I had posted (and after scribbling all over a side of A4)! Does this mean that for a bridge that's bulging outwards both $R$ and $R'$ will be positive? $\endgroup$ – Nemon27 Nov 22 '18 at 20:17
  • 1
    $\begingroup$ Yes, that should be correct. So, just like a spherical droplet, where $R=R'$, both give positive contributions to the Laplace pressure. In principle this is possible if the contact angle exceeds $\pi/2$ (the fluid is in the "partial drying" regime, i.e. it dislikes the surface, as opposed to the "partial wetting" regime). I see a picture of this near the section "Convex capillary bridge" on the Wikipedia page. I don't know how stable that arrangement is, however. $\endgroup$ – user197851 Nov 22 '18 at 20:27
  • $\begingroup$ When calculating the force due to the pressure that glues the two plates together, should a different value of $R$ be taken to reflect the surface area in contact with the plates which will have a larger radius than the $R$ of the diagram? $\endgroup$ – Nemon27 Nov 22 '18 at 20:59
  • $\begingroup$ I've added a few lines to my answer: we're discouraged from allowing the comments section to become an extended discussion! $\endgroup$ – user197851 Nov 22 '18 at 21:04

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