Half-space cooling model

Remarks

Proof idea: the problem has no length scale, so the solution depends only on the similarity variable $\eta = z/(2\sqrt{\kappa t})$. The heat equation becomes $f'' + 2\eta f' = 0$, whose solution with $f(0) = 0$ and $f(\infty) = 1$ is $\operatorname{erf}$. If the plate spreads from the ridge at constant speed $u$, then $t = x/u$ at distance $x$ from the ridge. Let the rock have density $\rho_m$ at $T_m$ and thermal expansion coefficient $\alpha$, and let water have density $\rho_w$. If the cooled column is in isostatic balance, the seafloor sinks below the ridge depth by $$w(t) = \frac{2 \rho_m \alpha (T_m - T_0)}{\rho_m - \rho_w} \sqrt{\frac{\kappa t}{\pi}},$$ using the complementary error function integral $\int_0^\infty \operatorname{erfc}(s) \, ds = 1/\sqrt{\pi}$. So depth and heat flow vary as $\sqrt{t}$ and $1/\sqrt{t}$, which matches observations for young seafloor; older seafloor flattens, which suggests a plate of finite thickness. The surface value of the geothermal gradient is $q(t)/k$.

Going deeper

For a student

The model is the basic link between plate tectonics and heat flow. It is the classical solution of the heat equation in a half-space with a step change of surface temperature. The same solution appears in Kelvin's estimate of the age of the Earth. Fitting the $\sqrt{t}$ law to bathymetry gives an estimate of the product $\alpha (T_m - T_0)$. The departure of old seafloor from the law is explained by the plate model, which adds a bottom boundary at fixed temperature.