2
$\begingroup$

Consider the following problem found in this webpage.

Consider a spherical bubble of radius $R$, of a certain fluid of density $\rho$, trapped inside of some other fluid. The bubble is stabilized by the presence of surface tension. Namely, suppose that the bubble has a nearly, but not perfectly, spherical surface, which we describe by a function $\zeta (\theta, \phi)$, denoting the difference $\zeta = r - R$ between the actual radius $r$ and the original radius $R$. One can then write the energy cost of this deformation as:

$$ E = \alpha \int d\theta d \phi \sin\theta(R+\zeta)^2\sqrt{1+ \left( \frac{1}{R+\zeta} \frac{\partial \zeta}{\partial \theta} \right)^2 + \left( \frac{1}{(R+\zeta)\sin \theta} \frac{\partial \zeta}{\partial \phi}\right)^2} .$$

The problem is then to argue that the pressure (difference from equilibrium pressure) at the surface of the bubble is:

$$ P = \frac{2 \alpha \zeta}{R^2} + \frac{\alpha}{R^2} \nabla^2 \zeta $$

where $\nabla^2$ is spherical Laplacian.

I am not sure even in general case if one is given energy cost $E$ then how one would obtain pressure $P$. Is there a general definition or procedure that one would do?

$\endgroup$
2
  • $\begingroup$ I had written an answer which I then deleted... but can't shake the feeling that something in my method might be correct. The stuff under the square root is essentially $\sqrt{\nabla \zeta \cdot \nabla \zeta} = |\nabla \zeta |$... can't relate it to $\nabla^2$ though... $\endgroup$ – SuperCiocia Sep 30 '19 at 0:06
  • $\begingroup$ Pressure is energy per unit volume, so all you need is to bring your equation to $E = \int \mathrm{d}^3\mathbf{r}\,P$. $\endgroup$ – SuperCiocia Sep 30 '19 at 0:07
2
$\begingroup$

I think one can get the expression for $P$ without that for $E$. Let $p_o$ be the pressure due to the fluid outside and $p_i$ be the pressure due to the fluid in the bubble. Since the bubble is initially spherical, \begin{equation}\tag{e1}\label{e1} p_o - p_i = \frac{2\alpha}{R}. \end{equation} When the bubble is deformed its radius is given by $r = R + \zeta(\theta, \phi)$ so that the equation of its surface if $f(r, \theta, \phi) = 0$ where $f = r - R - \zeta(\theta, \phi)$. If the pressure inside the bubble is $p_f$, \begin{equation}\tag{e2}\label{e2} p_o - p_f = \alpha\Delta f, \end{equation} where $\Delta$ is the Laplace operator in $r, \theta, \phi$. Note that $R$ is a constant, so that \begin{equation}\tag{e3}\label{e3} \Delta f = \frac{2}{r} - \frac{\nabla^2\zeta}{r^2}, \end{equation} where $\nabla^2$ is the Laplacian in $\theta, \phi$ alone. Now, \begin{equation} r = R\left(1 + \frac{\zeta(\theta,\phi)}{R}\right). \end{equation} If $\zeta(\theta,\phi) \ll R$, we can approximate \begin{equation}\tag{e4}\label{e4} \frac{1}{r} = \frac{1}{R} - \frac{\zeta}{R^2}. \end{equation} Similarly, \begin{equation}\tag{e5}\label{e5} \frac{1}{r^2} = \frac{1}{R^2} - \frac{2\zeta}{R^3}. \end{equation} Substitution equations (e5) and (e4) in (e3) we get \begin{equation}\tag{e6}\label{e6} \Delta f = \frac{2}{R} - 2\frac{\zeta}{R^2} - \frac{\nabla^2\zeta}{R^2}, \end{equation} where we have ignored the term $\zeta\nabla^2\zeta$ it being of a higher order in $\zeta$. From equation (e2) and (e6), \begin{equation}\tag{e7}\label{e7} p_o - p_f = \frac{2\alpha}{R} - \frac{2\alpha\zeta}{R^2} - \frac{\alpha\nabla^2\zeta}{R^2} \end{equation} Subtracting (e7) from (e1) we get \begin{equation}\tag{e8}\label{e8} p_f - p_i = \frac{2\alpha\zeta}{R^2} + \frac{\alpha\nabla^2\zeta}{R^2}. \end{equation} $p_f - p_i$ is the difference in the deformed bubble from its equilibrium pressure.

The integral in the expression for $E$ is just the area of a deformed sphere. You may want to refer to one of my questions for more details.

$\endgroup$
2
  • $\begingroup$ Thanks! This is a very interesting approach, can you please tell me where $p_o - p_i = \alpha \Delta f$ is coming from? Is it assumption/model/exact result/approximation? $\endgroup$ – Daniels Krimans Sep 30 '19 at 8:15
  • 1
    $\begingroup$ You may want to look at the last section of the first chapter of Batchelor's book. He uses $\zeta$ for our $f$. $\endgroup$ – Amey Joshi Sep 30 '19 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.