= Solution
Use the local momentum equation in the current volume:
$$
\sigma_{jk,k}=\rho(a_j-b_j).
$$
Symmetry of the <Cauchy stress tensor> and the product rule give
$$
\partial_k(x_i\sigma_{jk})=\sigma_{ij}+\rho x_i(a_j-b_j).
$$
Apply the <divergence theorem> and use $t_j=\sigma_{jk}n_k$. The <mean stress and strain boundary identities> follow:
$$
\boxed{V\overline\sigma_{ij}=\int_Sx_it_j\,dS+\int_V\rho x_i(b_j-a_j)\,dV.}
$$
For <infinitesimal strain tensor>, $e_{ij}=(u_{i,j}+u_{j,i})/2$, and a second application of the <divergence theorem> gives
$$
\boxed{V\overline e_{ij}=\frac12\int_S(u_in_j+u_jn_i)\,dS.}
$$
To first order, evaluate self-gravity in the undeformed uniform sphere. The <shell theorem> says that only the enclosed mass $M(r)=4\pi\rho r^3/3$ contributes to the inward field. Consequently
$$
b(x)=-\frac{GM(r)}{r^2}\frac{x}{r}
=-\frac{4\pi G\rho}{3}x=-\frac ga x,\qquad
g=\frac{4\pi G\rho a}{3}.
$$
The field vanishes at the centre and increases linearly to its surface value.
The requested radius change concerns the final traction-free <static elastic equilibrium>, not the <acceleration> immediately after gravity is switched on. In equilibrium the boundary <traction> and <acceleration> vanish. Spherical symmetry gives
$$
\frac1V\int_Vx_ix_j\,dV=\frac{a^2}{5}\delta_{ij},\qquad
\overline\sigma_{ij}=-\frac{\rho ga}{5}\delta_{ij}.
$$
The homogeneous <linear elasticity> law commutes with volume averaging:
$$
\overline\sigma=\kappa\,\operatorname{tr}(\overline e)I+
2\mu\left(\overline e-\frac13\operatorname{tr}(\overline e)I\right),
\qquad \kappa=\lambda+\frac23\mu.
$$
Thus $\overline e=e_0I$ with $e_0=-\rho ga/(15\kappa)$. This conclusion is about the mean <strain>, not a claim that the local <strain> is uniform.
Let the boundary displacement be $u(a)=u_a n$. Since $\int_Sn_in_j\,dS=(4\pi a^2/3)\delta_{ij}$, the boundary identity gives $\overline e=(u_a/a)I$. Therefore the <equilibrium contraction of a self-gravitating elastic sphere> is
$$
\boxed{u_a=-\frac{\rho ga^2}{15\kappa},\qquad
\text{radius decrease}=\frac{\rho ga^2}{15\kappa}.}
$$
The <shear modulus> cancels because the mean deformation is purely volumetric.
The full radial field provides a useful check without assuming uniform deformation. For $u=f(r)e_r$, momentum balance is
$$
(\lambda+2\mu)\left(f''+\frac{2f'}r-\frac{2f}{r^2}\right)=\frac{\rho g}{a}r.
$$
Regularity at the centre gives $f=Ar+Br^3$, with
$$
B=\frac{\rho g}{10a(\lambda+2\mu)},\qquad
A=-\frac{5\lambda+6\mu}{3\lambda+2\mu}Ba^2
$$
from $\sigma_{rr}(a)=0$. Substitution reproduces $f(a)=-\rho ga^2/(15\kappa)$. During the transient, the <acceleration> term in the mean-stress identity remains; gravitational feedback from the small density and shape changes is higher order in the equilibrium calculation.
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