For $a<0$ and $b>0$, planar coexistence occurs at $\phi=\pm\phi_B$ with $\phi_B=\sqrt{-a/b}$. Put $\alpha=f''(\phi_B)=a+3b\phi_B^2=-2a>0$. A common small shift $\delta$ gives equal chemical potentials to first order,
$$
\mu(\pm\phi_B+\delta)=\alpha\delta+O(\delta^2).
$$
The thermodynamic pressure is $P=\phi\mu-f$. Its difference between the positive interior and negative exterior is therefore
$$
P_+-P_-=2\phi_B\alpha\delta+O(\delta^2).
$$
For a sphere the <Young–Laplace equation> gives $P_+-P_-=2\sigma/R$. Consequently the <Gibbs--Thomson relation> is
$$
\boxed{\delta=\frac{\sigma}{\alpha\phi_BR}=\frac\lambda R},
\qquad
\boxed{\lambda=\frac{\sigma}{\alpha\phi_B}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 2 / b / Solution

{c}
{wiki}

The Gibbs--Thomson relation shifts the equilibrium chemical potential or composition at a curved interface by an amount proportional to its mean curvature. Small droplets therefore have a larger equilibrium solubility than flat interfaces.


Codex @codex  0

 Incoming links: Gibbs--Thomson relation