= Solution
The <magnetic field> satisfies $\nabla\cdot\mathbf B=0$. Expanding the <divergence> of its <Maxwell stress tensor> gives
$$
\partial_jM_{ij}=\frac1{\mu_0}\left(B_j\partial_jB_i-\frac12\partial_iB^2\right)=\frac{[(\nabla\times\mathbf B)\times\mathbf B]_i}{\mu_0}.
$$
For the <Newtonian gravitational field> $\mathbf g=-\nabla\Phi$, <Poisson equation for Newtonian gravity> gives $\partial_jg_j=-4\pi G\rho$, and the <gradient> representation gives $\partial_jg_i=\partial_ig_j$. Consequently
$$
\partial_j\left(g_ig_j-\frac12g^2\delta_{ij}\right)=g_i\partial_jg_j+g_j(\partial_jg_i-\partial_ig_j)=-4\pi G\rho g_i.
$$
The negative of the <Newtonian gravitational stress tensor>, in this force-stress convention, therefore supplies $\rho g_i$. Adding the <pressure> stress supplies $-\partial_ip$, so
$$
\boxed{\rho D_tu_i=\partial_jT_{ij},\qquad T_{ij}=-p\delta_{ij}-\frac{g_ig_j-g^2\delta_{ij}/2}{4\pi G}+\frac{B_iB_j-B^2\delta_{ij}/2}{\mu_0}.}
$$
Each summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.
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