# How is pressure related to the energy cost

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?

• 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... – SuperCiocia Sep 30 '19 at 0:06
• Pressure is energy per unit volume, so all you need is to bring your equation to $E = \int \mathrm{d}^3\mathbf{r}\,P$. – SuperCiocia Sep 30 '19 at 0:07

## 1 Answer

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.

• 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? – Daniels Krimans Sep 30 '19 at 8:15
• You may want to look at the last section of the first chapter of Batchelor's book. He uses $\zeta$ for our $f$. – Amey Joshi Sep 30 '19 at 8:43