Adams–Williamson equation

Remarks

Hydrostatic equilibrium gives $dP/dr = -\rho g$. Under the assumptions, density changes only by adiabatic compression, so $d\rho/dP = \rho/K_S = 1/\Phi$. The chain rule $d\rho/dr = (d\rho/dP)(dP/dr)$ gives the equation. Example: $\Phi$ is known from seismology, and $g = Gm/r^2$ by Newton's law of gravitation. So the equation and $dm/dr = 4\pi r^2 \rho$ form a closed system of ordinary differential equations for $\rho$ and $m$. It is integrated inward from a starting density and the total mass. The mantle and core masses must also match the total mass and the moment of inertia of the Earth. The adiabatic and uniform-composition hypotheses are necessary: across a phase change, such as near $660$ km depth, or in a thermal boundary layer, the density does not follow the equation. The Bullen parameter measures this departure.

Going deeper

For a student

The key identity is $\alpha^2 - \frac43\beta^2 = K_S/\rho$, which follows from $\alpha^2 = (K + \frac43\mu)/\rho$ and $\beta^2 = \mu/\rho$. So seismology gives $K_S/\rho$ but not $\rho$ itself, and hydrostatics supplies the missing relation. In Earth models the equation is used piecewise within each region, with jumps in density placed at seismic discontinuities. These jumps are then fixed by the mass and the moment of inertia, and today also by the frequencies of the normal modes.