How is pressure related to the energy cost

Question

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?

Answer 1

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.