Integrating Young Laplace Surface Tension along two radii of curveture

Question

Consider a point P on the surface as shown in Fig. and draw a curve at a constant distance ρ from P. This curve forms the boundary of a cap for which we shall find the equilibrium condition as ρ tends to zero. Through P we draw the two principal curvature sections AB and CD on the surface. Their radii of curvature at P are R1 and R2. At the point A, an element δl of the boundary line is subjected to a force σδl whose projection along the normal PN is

$$ σδlsinφ ≈ σφδl = σ \frac{ρ}{R}δl $$

since φ by assumption is small. If we consider four elements δl of the periphery at A, B, C, and D, they will contribute with a force

$$ 2ρσδl (\frac{1}{{R}_1}+\frac{1}{{R}_2})$$

Since this expression by Euler’s theorem, Eq. (10), is independent of the choice of AB and CD, it can be integrated around the circumference. Since four orthogonal elements are considered, the integration is made over one quarter of a revolution to give

$$πρ^2σ(\frac{1}{{R}_1}+\frac{1}{{R}_2})$$

The force on the surface element caused by the pressure difference over the surface is given by (p1 − p2)πρ2,and equating the last two expressions, the Young-Laplace equation follows.

Question is I cant understand how to integrate over this "quarter" because I cant neither visualize it nor write a meaningful equation for it. Sadly this part is bypassed by writer I did understand the former part though.

Answer 1

You haven't actually included Euler's Theorem. You would do better to read a modern differential geometry text, as opposed to an old-fashioned physics treatment. You are averaging the sum of what's called the normal curvature of the surface in two orthogonal directions. Let $k_1$ and $k_2$ be the principal curvatures (the maximum and minimum normal curvatures as you look at all possible directions in the tangent plane at $P$. These occur in orthogonal directions, $e_1$ and $e_2$. Euler's Theorem says that the normal curvature in the direction of $\cos\theta\,e_1 + \sin\theta\,e_2$ is $$k_n(\theta) = k_1\cos^2\theta + k_2\sin^2\theta.$$ Integrating over $[0,\pi/2]$ (or $[0,2\pi]$) is what the author was referring to. The average value is $\frac12(k_1+k_2)$, either way. Not surprisingly, this is the mean curvature $H$ of the surface at $P$. ($k_i=1/R_i$, where the $R_i$ are the minimum and maximum radii of curvature at $P$.)

Answer 2

Please note the force is exerted on the full surface of the area ADBC containing all normal sections concurrent at P.

Euler's theorem of normal curvatures

$$k_n(\theta) = k_1\cos^2\theta + k_2\sin^2\theta$$

$$\theta= 0, k_n(\theta)== k_1 ; ~\theta= \pi/2, k_n(\theta)== k2;$$

$$\dfrac{\int _{0}^{n \pi/2} k_n d \theta }{n \pi/2}=\frac{k_1+k2}{2}=H, $$

the mean curvature which is valid only for integer $n$.

For a particular case $n=1$ the average value

$$\dfrac{\int _{0}^{ \pi/2}( k_1 \cos^2 \theta+ k_2 \sin^2 \theta ) d \theta}{\pi/2}=H $$

A graph of $(k_n- \tau_g, $ normal curvature vs. geodesic torsion in Mohr's curvature Circle) also shows that we need to consider infinitesimal segment lengths for integration $\theta $ from $0$ at A, ending at $D,B,C$...