Work in an inertial frame centred on the <centre of mass>, and follow a fixed material mass, with no <stellar wind> or accretion through its boundary. Assume nonrelativistic motion, Newtonian self-gravity, no external gravitational field, and sufficiently regular integrable fields. Define
$$
I=\int r^2\,dm,\qquad \mathcal T=\frac12\int |\mathbf v|^2\,dm,\qquad
\Omega=-\frac G2\iint\frac{dm\,dm'}{|\mathbf r-\mathbf r'|}=\frac12\int\rho\Phi\,dV.
$$
Here $I$ is the <scalar second mass moment>, rather than a moment of inertia about one axis. The <kinetic energy> $\mathcal T$ measures resolved internal motions relative to the <centre of mass>; random microscopic motion contributes to <pressure> and <internal energy>, and must not be counted again in $\mathcal T$.

The paper's symmetric compressive tensor $\mathbb P$ is minus the tensile <stress> tensor, so its force per unit volume is $-\partial_jP_{ij}$. Differentiating the material integral twice and substituting the equation of motion gives
$$
\frac12\ddot I=2\mathcal T-\int r_i\partial_jP_{ij}\,dV-\int\mathbf r\cdot\nabla\Phi\,dm.
$$
<Integration by parts> and the <divergence theorem> turn the stress contribution into
$$
-\int r_i\partial_jP_{ij}\,dV=\int\operatorname{tr}\mathbb P\,dV-\oint r_iP_{ij}n_j\,dS.
$$
For the requested scalar <stellar virial theorem>, assume isotropic <pressure>, $P_{ij}=P\delta_{ij}$, with uniform surface pressure $P_S$. Then $\operatorname{tr}\mathbb P=3P$ and the surface integral is $P_S\oint\mathbf r\cdot\mathbf n\,dS=3P_SV$. No spherical shape is needed for this step.

Finally, symmetrizing the pair integral for the <Newtonian gravitational potential> gives
$$
-\int\mathbf r\cdot\nabla\Phi\,dm
=-\frac G2\iint\frac{(\mathbf r-\mathbf r')\cdot(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}\,dm\,dm'=\Omega.
$$
Therefore
$$
\boxed{\frac12\ddot I=2\mathcal T+3\int P\,dV-3P_SV+\Omega.}
$$
Anisotropic magnetic or viscous stresses retain their full trace and boundary terms; a changing integration mass adds transport terms.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 1 / i / Solution

{title2=$I=\int r^2\,dm$}

The scalar second mass moment is the <trace> of the <second mass moment tensor>. The <moment of inertia> about a specified axis instead integrates squared distance from that axis. For an isotropic distribution each axial <moment of inertia> is $2I/3$; the sum over three orthogonal axes is $2I$.

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