Question

Stabilities of droplets. Certain molecular aggregates (like surfactant micelles and oil droplets) shrink owing to surface tension and expand owing to volumetric forces. Suppose the free energy \(g(r)\) as a function of the radius \(r\) of the aggregate is \(g(r)=2 r^{2}-r^{3}\). Such a system has two equilibrium radii. (a) Calculate the two radii. (b) Identify each radius as stable, unstable, neutral, or metastable.

Short answer

The equilibrium radii are 0 (stable) and 4/3 (unstable).

Step-by-step solution

- Finding the Equilibrium Points

Equilibrium points occur where the derivative of the free energy with respect to the radius is zero. First, find the derivative of the given function: \( g(r) = 2r^2 - r^3 \). Differentiate with respect to \( r \) to obtain \( g'(r) = 4r - 3r^2 \). Set this derivative equal to zero to find the critical points: \( 4r - 3r^2 = 0 \).

- Solving for the Equilibrium Radii

Solve the equation \( 4r - 3r^2 = 0 \) for \( r \). Factor the equation to get \( r(4 - 3r) = 0 \). This gives two solutions: \( r = 0 \) and \( r = \frac{4}{3} \). Therefore, the two equilibrium radii are \( r = 0 \) and \( r = \frac{4}{3} \).

- Determining the Stabilities

To determine whether each radius is stable or unstable, examine the second derivative of the free energy function: \( g''(r) \). The second derivative is \( g''(r) = 4 - 6r \). Evaluate \( g''(r) \) at the two radii. For \( r = 0 \), \( g''(0) = 4 - 6(0) = 4 \) which is positive, indicating a local minimum and thus a stable equilibrium. For \( r = \frac{4}{3} \), \( g''(\frac{4}{3}) = 4 - 6(\frac{4}{3}) = 4 - 8 = -4 \) which is negative, indicating a local maximum and thus an unstable equilibrium.

Key concepts

Free Energy Function

The free energy function, often represented as \( g(r) \), is crucial in thermodynamics to describe the stability of molecular systems. It describes how the energy in the system varies with parameters like the radius \( r \) of a droplet or aggregate. In the given problem, the free energy is defined as \( g(r) = 2r^2 - r^3 \). This means the energy depends on the surface tension, which tries to shrink the droplet, and the volumetric forces, which tend to expand it. By studying the shape of this function, we can understand when the system reaches equilibrium and whether this state is stable or unstable.

Derivatives in Physics

Derivatives are a powerful tool in physics, especially for finding equilibrium points in a system. The first derivative of the free energy function, denoted \( g'(r) \), helps identify these points. For our function, \( g(r) = 2r^2 - r^3 \), the first derivative is \( g'(r) = 4r - 3r^2 \). Setting this derivative equal to zero, \( 4r - 3r^2 = 0 \), allows us to find the critical points by solving the equation. These critical points, or equilibrium points, are where the system can potentially rest without changing the radius further.

Equilibrium Points

Equilibrium points are where the system is at rest, and these can be found by setting the first derivative \( g'(r) \) of the free energy to zero. In the example, solving \( 4r - 3r^2 = 0 \) gives the solutions \( r = 0 \) and \( r = \frac{4}{3} \).
The stability of each equilibrium point is determined by examining the second derivative \( g''(r) \). For \( g(r) = 2r^2 - r^3 \), the second derivative is \( g''(r) = 4 - 6r \). At \( r = 0 \), it is positive, indicating a stable equilibrium. At \( r = \frac{4}{3} \), it is negative, indicating an unstable equilibrium. Therefore, \( r = 0 \) is a stable point, while \( r = \frac{4}{3} \) is not.