Question 1.7: Capilarity effects are taken into account by considering tha...

In order to minimise the internal energy U(h, R), use the Lagrange multipliers method to impose the condition that the volume of the drop is fixed, i.e. V (h, R) = V_0. The function F (h, R, λ) to be minimised is,

F (h, R, λ) = U(h, R) − λ \bigl(V (h, R) − V_0\bigr) = (\gamma _{s\ell } – \gamma _{sg }) π (2 R h − h^2) + \gamma _{\ell g } 2π R h − λ \bigl(\frac{π}{3} h^2(3R − h) – V_0\bigr).

where λ is the Lagrange multiplier. According to this method, the partial derivative of the function F (h, R, λ) with respect to h has to vanish,

\frac{\partial F}{\partial h} =(\gamma _{s\ell }-\gamma _{sg}) 2\pi (R-h) -\gamma _{\ell g} 2\pi R +\lambda \pi (2Rh -h^2) =0.

which yields an expression for the Lagrange multiplier,

\lambda = \biggl(\frac{2}{2Rh – h^2}\biggr)(R-h) (\gamma _{s\ell }-\gamma _{sg})+\biggl(\frac{2}{2Rh-h^2}\biggr) R \gamma _{\ell g}.

The partial derivative of the function F (h, R, λ) with respect to R has to vanish as well,

\frac{\partial F}{\partial R} = (\gamma _{s\ell } – \gamma _{sg})2\pi h + \gamma _{\ell g}2\pi h-\lambda \pi h^2 =0.

which yields another expression for the Lagrange multiplier,

\lambda = \frac{2}{h} (\gamma _{s\ell }-\gamma _{sg}) + \frac{2}{h} \gamma _{\ell g}.

Equating the two expressions for the Lagrange multiplier λ yields,

(R-h)(\gamma _{s\ell }-\gamma _{sg}) + R \gamma _{\ell g} = (2R-h) (\gamma _{s\ell }-\gamma _{sg}) +(2R-h) \gamma _{\ell g}.

which reduces to,

(\gamma _{s\ell }-\gamma _{sg}) + \Bigl(\frac{R-h}{R}\Bigr) \gamma _{\ell g} =0 .

By graphical inspection (Fig. 1.2),

\cos \theta = \frac{R-h}{R}.

Thus, we find that,

(\gamma _{s\ell }-\gamma _{sg}) + \gamma _{\ell g} \cos \theta =0.