= Solution
Apply force balance to one hemisphere. The pressure force is $\pi R^2(\Pi_{\rm in}-\Pi_{\rm out})$, while the <interfacial tension> around its equator gives $2\pi R\gamma$. The <Young–Laplace equation> therefore yields
$$
\boxed{\Pi_{\rm in}-\Pi_{\rm out}=\frac{2\gamma}{R}.}
$$
To leading order in curvature, let the two bulk compositions be $\phi_+=\phi_B+\delta_+$ and $\phi_-=-\phi_B+\delta_-$. Their curvatures agree, $\alpha=f''(\pm\phi_B)=-2a$, so equal <chemical potentials> require $\alpha\delta_+=\alpha\delta_-$ and hence a common shift $\delta$.
Since $d\Pi/d\phi=\phi f''(\phi)$, expansion about the flat coexistence compositions gives $\Pi_+-\Pi_-=2\phi_B\alpha\delta+O(\delta^2)$. Matching the capillary pressure proves the <Gibbs--Thomson relation> in these variables:
$$
\boxed{\delta=\frac{\gamma}{\alpha\phi_BR},\qquad\mu_s=\alpha\delta=\frac{\gamma}{\phi_BR}.}
$$
The pressure jump balances interfacial forces; it does not permit unequal <chemical potentials>. A common shift in both compositions supplies a common interfacial <chemical potential> while their pressures differ. The expansion assumes $R$ large compared with the <phi-four diffuse interface> width.
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