Because $\phi$ is a conserved composition density, an infinitesimal material displacement $\mathbf r\mapsto\mathbf r+\mathbf u$ changes it at fixed position by
$$
\delta\phi=-\nabla\mathbin\cdot(\phi\mathbf u)
=-\mathbf u\mathbin\cdot\nabla\phi-\phi\nabla\mathbin\cdot\mathbf u.
$$
The second term is essential when the deformation is compressible. By the definition of the <chemical potential>, and taking $\mathbf u$ to vanish on the boundary,
$$
\delta F=\int\mu\,\delta\phi\,d\mathbf r
=-\int\mu\nabla\mathbin\cdot(\phi\mathbf u)d\mathbf r
=\int\phi\nabla_j\mu\,u_jd\mathbf r.
$$
The same free-energy change written in terms of the <stress>[stress tensor] is
$$
\delta F=\int\Sigma_{ij}\nabla_i u_jd\mathbf r
=-\int(\nabla_i\Sigma_{ij})u_jd\mathbf r.
$$
Since $\mathbf u$ is arbitrary,
$$
\boxed{\nabla_i\Sigma_{ij}=-\phi\nabla_j\mu.}
$$
This is the <Korteweg force> density of a diffuse-interface mixture.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-344/1/a/solution