11.5 The age of the Earth

In 1862 William Thomson (Lord Kelvin) used the heat equation to estimate the age of the Earth. Model the crust as a half-space \(x > 0\) (depth), initially at the temperature \(T_0\) of molten rock, with the surface held at \(0\) from \(t = 0\). The solution, by the image method of Section 9, is \(u = T_0\operatorname {erf}\bigl (x/\sqrt {4\kappa t}\bigr )\), and the temperature gradient at the surface is \[ u_x(0, t) = \frac {T_0}{\sqrt {\pi \kappa t}} . \] Measuring the gradient \(G\) in mines and solving for \(t\) gives \(t = T_0^2/\pi \kappa G^2\). With Kelvin’s values, \(T_0 \approx 3900\) K, \(G \approx 0.0365\) K/m (one degree Fahrenheit per fifty feet) and \(\kappa \approx 1.2 \times 10^{-6}\) m\(^2\)/s, this is about \(96\) million years (Software 11.6.3), and Kelvin quoted \(98\) million, with a range of \(20\) to \(400\) million. Geologists and Darwin needed far longer. The mathematics was right; the model was not. Perry (1895) pointed out that convection in a fluid interior keeps the deep temperature high and greatly lengthens the estimate, and the discovery of radioactive heating in the rocks removed the premise altogether. The modern age is \(4.5\) billion years.

The half-space is the right model because the diffusion length \(\sqrt {4\kappa t}\) for \(100\) million years is about \(120\) km, small beside the Earth’s radius of \(6400\) km; for a whole sphere one separates in spherical coordinates, and the radial modes are \(\sin (n\pi r/a)/r\) (Problem 11.7.4).

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