Cubic polytropic interior

The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

For the cubic polytropic interior, $\nabla \rho$ vanishes at each vertex, so the locally defined potential $\Phi =C-2K\rho$ predicts zero gravitational acceleration there. But at the vertex $(0,0,0)$, the actual self-gravitational acceleration of the positive mass distribution iswhose three components are strictly positive. The fields cannot match continuously. Thus the interior Helmholtz equation and zero density on the faces do not construct an isolated static cubic star. External stresses or an external gravitational field would be needed to realize such a boundary-value construction.

The sine-product cubic polytropic interior has zero mass density gradient at every corner. Its proposed hydrostatic gravitational potential ${\Phi }_{\mathrm{int}}=C-2K\rho$ therefore satisfies $\nabla {\Phi }_{\mathrm{int}}(0)=0$ at the corner $(0,0,0)$. For the actual Newtonian gravitational potential of that positive mass density in $[0,L{]}^3$,Every component is strictly negative: all source positions have positive coordinates and positive interior mass density. The gravitational acceleration $\mathbf{g}=-\nabla {\Phi }_N$ consequently points into the cube, while the potential gradient points in the opposite direction. Neither vector vanishes.

This sign distinction preserves the matching obstruction: the hydrostatic interior predicts zero corner force, but its own mass produces a nonzero corner force. Satisfying the interior Poisson equation for Newtonian gravity and zero density on the faces does not ensure a globally isolated equilibrium. An added external gravitational field or external stresses would change the boundary problem.