1.2 Ocean Basins

Step 1: Set up the heat conduction problem

New oceanic lithosphere is created at mid-ocean ridges at the mantle temperature $T_m \approx 1300°$C. As it moves away from the ridge, it cools by vertical heat conduction. We model the lithosphere as a semi-infinite half-space with the 1D heat equation:

$$\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial z^2}$$

where $\kappa \approx 10^{-6}$ m²/s is the thermal diffusivity and $z$ is depth below the seafloor.

Step 2: Boundary and initial conditions

The surface is held at the seafloor temperature $T_s \approx 0°$C, and at great depth the temperature approaches the mantle value:

$$T(z=0, t) = T_s, \quad T(z \to \infty, t) = T_m, \quad T(z, t=0) = T_m$$

Step 3: Solve using the error function

Introducing the similarity variable $\eta = z / (2\sqrt{\kappa t})$, the solution is:

$$T(z, t) = T_s + (T_m - T_s) \cdot \text{erf}\left(\frac{z}{2\sqrt{\kappa t}}\right)$$

where $\text{erf}(\eta) = \frac{2}{\sqrt{\pi}} \int_0^{\eta} e^{-u^2}\,du$ is the error function.

Step 4: Compute thermal contraction

Cooling causes the lithosphere to contract. The total thermal contraction (column shortening) is found by integrating the temperature deficit over depth using the volumetric thermal expansion coefficient $\alpha_v$:

$$\Delta h = \alpha_v \int_0^{\infty} \bigl[T_m - T(z,t)\bigr]\, dz = \alpha_v (T_m - T_s) \cdot 2\sqrt{\frac{\kappa t}{\pi}}$$

Step 5: Apply isostatic compensation

The thermal contraction causes subsidence as the denser, cooler lithosphere sinks isostatically. The water column above must compensate, so the depth increase relative to the ridge crest is:

$$\Delta d(t) = \frac{\rho_m}{\rho_m - \rho_w} \cdot \alpha_v (T_m - T_s) \cdot 2\sqrt{\frac{\kappa t}{\pi}}$$

Step 6: Substitute typical parameter values

Using $\rho_m = 3300$ kg/m³, $\rho_w = 1025$ kg/m³, $\alpha_v = 3.1 \times 10^{-5}$ K⁻¹, $T_m - T_s = 1300$ K, $\kappa = 10^{-6}$ m²/s:

$$\Delta d = \frac{3300}{2275} \times 3.1 \times 10^{-5} \times 1300 \times 2\sqrt{\frac{10^{-6} \cdot t}{\pi}}$$

Step 7: Convert to the empirical age-depth formula

Converting time from seconds to millions of years ($1 \text{ Ma} = 3.156 \times 10^{13}$ s) and evaluating the constants yields:

$$\Delta d \approx 350\sqrt{t_{\text{Ma}}} \quad \text{meters}$$

Therefore the total ocean depth at crustal age $t$ (in Ma) is:

$$d(t) = d_r + 350\sqrt{t} \approx 2600 + 350\sqrt{t} \quad \text{meters}$$

This holds well for crust younger than ~70 Ma. For older crust, the finite plate thickness (~100 km) causes the depth to flatten toward an asymptotic value of ~6400 m (plate model).

Step 1: Introduce the plate model

For old lithosphere ($t > 70$ Ma), the half-space model overpredicts subsidence because the lithosphere has a finite thickness $a \approx 125$ km. The plate model adds a bottom boundary condition $T(z = a, t) = T_m$, giving:

$$T(z, t) = T_s + (T_m - T_s)\left[\frac{z}{a} + \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{1}{n}\sin\left(\frac{n\pi z}{a}\right) e^{-n^2\pi^2\kappa t/a^2}\right]$$

Step 2: Integrate for the depth anomaly

Integrating the temperature deficit over the plate thickness and applying isostasy gives the depth:

$$d(t) = d_r + \frac{\rho_m \alpha_v (T_m - T_s) a}{2(\rho_m - \rho_w)} \left[1 - \frac{8}{\pi^2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} e^{-(2n+1)^2\pi^2\kappa t/a^2}\right]$$

Step 3: Asymptotic depth

As $t \to \infty$, the exponential terms vanish and the depth approaches a maximum:

$$d_{\infty} = d_r + \frac{\rho_m \alpha_v (T_m - T_s) a}{2(\rho_m - \rho_w)} \approx 2600 + 3800 \approx 6400 \text{ m}$$

Step 4: Young vs. old crust comparison

For young crust ($t \ll a^2/(\pi^2 \kappa)$), the plate model reduces to the half-space result $d \approx d_r + 350\sqrt{t}$. The transition occurs at the "thermal time constant":

$$\tau = \frac{a^2}{\pi^2 \kappa} \approx \frac{(125 \times 10^3)^2}{\pi^2 \times 10^{-6}} \approx 50 \text{ Ma}$$

Observationally, the $\sqrt{t}$ law works well for $t < 70$ Ma. Beyond that, the plate model with $a = 125$ km and $T_m = 1350°$C (GDH1 model of Stein & Stein, 1992) matches global bathymetry data to within ~200 m.