Ground Temperature: Seasons Underground

Pick a soil and a depth. The surface swings with the seasons and with the day; the page shows the swing that reaches your thermometer, from the exact solution of the heat equation, with the surface's own swing beside it as a ghost.

Worked example — Ground Temperature: Seasons Underground

Thermometer depth in damping depths	1.247
How late the yearly peak arrives	72.5 days
Depth where the year is exactly reversed	6.30 m
Daily damping depth	0.1049 m
How late the daily peak arrives	91.0 h

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Ground Temperature: Seasons Underground

When the temperature at the surface of uniform ground swings as a sine with period P, the swing at depth z is e^−z/d of the surface's and arrives (z/d)·P/2π later, where d = √(κP/π) is the damping depth and κ the ground's thermal diffusivity. Both effects are set by one number, z/d. At z = πd the swing is exactly half a period late.

What each symbol means

d: the damping depth √(κP/π), over which a swing falls by a factor e; about 2 m for the year in dry soil, 10 cm for the day
z: the depth below the surface
P: the period of the surface's swing: a year of 365.2422 days, or a day
κ: the thermal diffusivity of the ground, in m²/s

The swing shrinks and arrives late by one and the same number 🖖

Press Calton Hill, 24 French feet and look at three rows of the results. The depth in damping depths, z/d, is 2.776. The swing there is e^−2.776 = 6.23% of the surface's. And its peak arrives 2.776 × 58.13 = 161.4 days late, because a year is 2π radians and one radian of it is 365.2422/2π = 58.13 days. The shrink and the delay are the same number, so for every factor e the swing loses on its way down it arrives 58.1 days later. Choose another ground and d changes; the 58.1 days does not. Forbes set thermometers into the rock of Calton Hill in Edinburgh at 3, 6, 12 and 24 French feet, and in 1860 William Thomson reduced thirteen years of their readings. Between 3 and 24 feet the logarithm of the yearly swing fell by 0.1160 per French foot and its peak slipped by 0.1156 radians per French foot, the bold row of the second table. Theory said the two would be equal, and they are, to 0.3%. Nearer the surface they differ by 6.2%, and Thomson put that down to the ground, not to the readings: neither the rock nor its surface is as uniform as the theory assumes.

Six metres into dry soil, the warmest day comes in winter 🖖

Press Dry soil, 6.3 m. That depth is πd to the centimetre: the swing is down to e^−π = 4.32% of the surface's and arrives 182.7 days late, half a year to within two hours, so the two curves in the lower panel are mirror images and the ground there is at its warmest when the surface is at its coldest. It is a small warmth, 0.35 °C either side of the mean. A cellar does not reach it. Press Dry soil, 2.5 m: at cellar depth the summer arrives 72.5 days late with 28.73% of the swing, an autumn peak. The coldest soil in midsummer is not at 6.30 m either but at 4.72 m, three quarters of πd, where the swing is larger and only three eighths of a year late; the results give that depth for whatever ground you choose.

Half a metre down, the night is gone and the seasons are not 🖖

Press Half a metre, daily clock. In dry soil the daily swing falls by a factor e every 0.1049 m, the yearly one every 2.004 m: 19.11 times as deep. That ratio is √365.2422, the square root of the ratio of the two periods, and κ cancels out of it, so it is the same in every ground. Half a metre down, the day's swing is 0.85% of the surface's and peaks 18.2 hours late, while the year's is still 77.92% and only 14.5 days late. A thermometer there barely registers night and day, and follows the seasons almost in step. It is the square law of The Heat Equation again: a period 365 times as long reaches √365 times as deep.

Problem solved in full

Kelvin's age of the Earth 3 steps
In 1862 William Thomson, later Lord Kelvin, treated the Earth's crust as rock with κ = 400 square feet a year, started as one solid 7,000 °F hotter than today's surface and cooling through that surface ever since. Measured, the ground warms by about 1 °F for every 50 feet of descent. How old did that make the Earth? Press Thomson's rock, 1 m first: the page shows what that κ does to the yearly wave.

Open this problem in the tool
1. In metric units Thomson's κ is the 1.178 mm²/s in the κ field, and its yearly swing falls by e every
  
  $κ = \frac{400 \times 0 . 304 8 ^{2} m ^{2}}{3 . 15569 \times 1 0 ^{7} s} = 1.178 \times 1 0^{- 6} m^{2} /s, d_{y} = \frac{400}{π} ft = 11.28 ft = 3.439 m$
2. Rock that starts uniformly hot and is cooled through its surface from time 0 has, at the surface, the gradient G = ΔT/√(πκt), Fourier's solution as Thomson wrote it. Solve for t, in his own units of °F, feet and years:
  
  $t = \frac{Δ T ^{2}}{π κ G ^{2}} = \frac{7 , 00 0 ^{2}}{π \times 400 \times ( 1/50 ) ^{2}} = \frac{49 , 000 , 000}{0 . 50265} = 97, 480, 000 years$
3. The answer goes as one over the square of the gradient, the one number measured. Half the gradient makes the Earth four times as old, and Thomson's high estimate for melting rock, 10,000 °F, gives
  
  $G = \frac{1}{100} : t = 4 \times 97.48 = 389.9 million years; Δ T = 10, 000^{\circ} F : t = 198.9 million years$
Answer

About 97.5 million years. Thomson wrote 98,000,000, and, allowing for what nobody then knew about rock at high temperatures, put the age between 20 and 400 million years. The accepted age is 4.54 billion. The usual story blames radioactivity, and it is wrong: put the Earth's radioactive heat into this calculation and the answer barely moves. What fails is the assumption that nothing inside moves. In 1895 John Perry pointed out that an interior which flows, even as slowly as hot rock creeps, can carry heat up far faster than conduction can, and then the gradient at the surface says little about how long the Earth has been cooling.

$t \approx 97.5 \times 1 0^{6} years (Thomson: 98, 000, 000; accepted: 4.54 \times 1 0^{9})$

Learning path

The heat, wave and Laplace equations

References (3)

Fourier's solution for a periodic surface temperature stated as a law, the retardation of the epoch in radians equal to the diminution of the logarithm of the amplitude, and tested on Forbes's thermometers at 3, 6, 12 and 24 French feet in the rock of Calton Hill (Table IV, thirteen years, 1842–1854: 0.1160 and 0.1156 per French foot from 3 to 24 feet), with k/c of 235.1, 260.5 and 690.7 square French feet a year for the three Edinburgh sites (§§11, 21, 35–38); and the age of the Earth from the cooling of a solid half-space, κ = 400 square feet a year and 1 °F per 50 feet, 98,000,000 years for 7,000 °F and limits of 20 and 400 million (§§6, 10–16): W. Thomson, "On the reduction of observations of underground temperature; with application to Professor Forbes' Edinburgh observations, and the continued Calton Hill series", Transactions of the Royal Society of Edinburgh 22 (1861), 405–427, doi:10.1017/S0080456800030702; and "On the secular cooling of the Earth", same journal, 23 (1862), 157–169, doi:10.1017/S0080456800018512. Both reprinted in his Mathematical and Physical Papers, vol. III (Cambridge University Press, 1890), Arts. XCIII and XCIV.
Kelvin's calculation re-run with modern values, and the case that radioactivity does not invalidate it (adding the Earth's radioactive heat barely changes the answer) while convection in the interior, argued by John Perry in 1895, does: P. England, P. Molnar and F. Richter, "John Perry's neglected critique of Kelvin's age for the Earth: A missed opportunity in geodynamics", GSA Today 17(1) (2007), 4–9.
The periodic solution and its damping, Example 5.7 (the coldest ground in high summer at three quarters of π damping depths); the thermal diffusivities of dry mineral soil, 4 × 10⁻⁷ m²/s, and granite, about 7.4 × 10⁻⁷ (Table A.2); and the accepted age of the Earth, 4.54 billion years (Problem 5.45): J. H. Lienhard V and J. H. Lienhard IV, A Heat Transfer Textbook, 6th edition. Phlogiston Press, Cambridge, Massachusetts, 2024. §5.6 and Example 5.7, pp. 233–235; Problem 5.45; Table A.2, p. 744.

By the 42Math team · Reviewed against published sources. Corrections welcome.