Mean stress and strain boundary identities ($V{\overline{\sigma }}_{ij}={\int }_S{x}_i{t}_{j}dS+{\int }_V\rho x_i(b_j-a_j)dV$)

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

For a uniform traction-free sphere, first-order self-gravity is $b=-gx/a$. Static averaging gives $\overline{\sigma }=-\rho gaI/5$. Homogeneous linear elasticity then gives $\overline{e}=-\rho gaI/(15\kappa )$. The boundary strain identity identifies $\overline{e}=u(a)I/a$, proving the displayed positive radius decrease. The local strain is not uniform, and the cancellation of the shear modulus concerns only the mean contraction.