Temperature waves propagating into the ground — diurnal and seasonal cycles
2026-02-26
You should know
That soil can store heat, that temperature often changes with depth, and that heat tends to move from warmer places toward cooler ones.
You will learn
How heat diffuses into soil, why subsurface temperatures lag behind surface temperatures, and how diffusion equations describe that process.
Why this matters
Soil temperature affects seeds, roots, microbes, permafrost, and snowmelt timing. This is one of the cleanest examples of a physical process becoming a computational model.
If this gets hard, focus on…
Hold onto the physical idea first: heat spreads out over time, and deeper soil responds more slowly than the surface.
Wine cellars stay cool in summer and frost-free in winter not by accident but by physics. At depths of 8–10 metres in most climates, the soil temperature barely varies across the year, hovering near the mean annual air temperature. The seasonal oscillation that sends surface temperatures swinging between −20°C and +30°C has been almost completely damped out by the time it propagates 10 metres into the ground. The same principle governs the depth of permafrost in the Arctic, the temperature of groundwater in shallow aquifers, the design of underground data centres, and the calibration of soil respiration models that depend on temperature.
The mathematics describing this behaviour is the heat diffusion equation — a partial differential equation that relates the rate of temperature change at a point to the curvature of the temperature profile around it. It is one of the most important PDEs in physics, with applications from heat treatment of metals to the cooling of planetary bodies. For soil, the key result is that a sinusoidal temperature oscillation at the surface propagates downward as a damped, phase-lagged wave: the amplitude decreases exponentially with depth and the timing of the peak is progressively delayed. The damping depth — the depth at which the amplitude has fallen to 1/e of the surface value — depends only on thermal diffusivity and the period of the oscillation. This model derives that result and shows what controls it.
Why does soil temperature lag behind air temperature, and why is the lag greater at depth?
Plant a thermometer at the surface and another at 1 meter depth. Watch them for 24 hours: - Surface: Temperature swings wildly (10°C at night, 30°C at noon) - 1 m depth: Temperature barely changes (stays near 20°C all day)
The same pattern occurs seasonally: - Surface: Cold in winter, hot in summer - 10 m depth: Constant year-round (permafrost boundary, wine cellars, geothermal)
The mathematical question: How do temperature oscillations at the surface propagate downward into the soil? How far do they penetrate, and how much are they delayed?
Heat flows from warm to cool regions. In soil, this happens by conduction — molecular-scale energy transfer through collisions.
Fick’s law analogy (from diffusion):
Heat flux is proportional to the temperature gradient:
q = -k \frac{\partial T}{\partial z}
Where: - q = heat flux (W/m²), positive downward - k = thermal conductivity (W/m/K) - \frac{\partial T}{\partial z} = temperature gradient (K/m)
The negative sign means heat flows from hot to cold (down the gradient).
The surface temperature oscillates:
T_{\text{surface}}(t) = T_{\text{mean}} + A \sin(\omega t)
Where: - T_{\text{mean}} = average temperature - A = amplitude of oscillation - \omega = 2\pi / P = angular frequency (rad/s) - P = period (86400 s for daily, ~31.5×10⁶ s for annual)
This oscillation propagates downward as a damped, delayed wave.
Damping: Amplitude decreases exponentially with depth
Delay: Peak temperature occurs later at greater depths
Conservation of energy in a thin soil layer:
\rho c \frac{\partial T}{\partial t} = \frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right)
Where: - \rho = soil density (kg/m³) - c = specific heat capacity (J/kg/K) - k = thermal conductivity (W/m/K)
For constant properties (homogeneous soil):
\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial z^2}
Where \alpha = k / (\rho c) is the thermal diffusivity (m²/s).
Physical interpretation: - Left side: Rate of temperature change - Right side: Curvature of temperature profile (how “bent” the profile is)
If the profile is concave up (\partial^2 T/\partial z^2 > 0), temperature increases. If concave down, temperature decreases.
\alpha = \frac{k}{\rho c}
Typical values:
| Material | \alpha (m²/s) |
|---|---|
| Dry sand | 0.3 × 10⁻⁶ |
| Moist sand | 0.6 × 10⁻⁶ |
| Clay (dry) | 0.25 × 10⁻⁶ |
| Clay (wet) | 0.5 × 10⁻⁶ |
| Peat | 0.1 × 10⁻⁶ |
| Rock | 1.0 × 10⁻⁶ |
| Water | 1.4 × 10⁻⁶ |
| Air | 20 × 10⁻⁶ |
Key insight: Moist soil has higher diffusivity than dry soil (water conducts heat better than air).
Boundary condition (surface oscillates):
T(z=0, t) = T_m + A_0 \sin(\omega t)
Deep boundary (constant temperature):
T(z \to \infty, t) = T_m
Solution (for z > 0):
T(z,t) = T_m + A_0 e^{-z/d} \sin\left(\omega t - \frac{z}{d}\right)
Where d is the damping depth (or e-folding depth):
d = \sqrt{\frac{2\alpha}{\omega}}
Two key features: 1. Amplitude decays exponentially: A(z) = A_0 e^{-z/d} 2. Phase lags linearly: Phase lag = z/d radians
The damping depth is where amplitude drops to $1/e %$ of surface value.
Daily cycle (P = 86400 s, \omega = 7.27 \times 10^{-5} rad/s):
For typical soil (\alpha = 0.5 \times 10^{-6} m²/s):
d_{\text{daily}} = \sqrt{\frac{2 \times 0.5 \times 10^{-6}}{7.27 \times 10^{-5}}} = 0.12 \text{ m}
At depth d = 12 cm, daily temperature swing is 37% of surface swing.
Annual cycle (P = 365.25 \times 86400 s, \omega = 1.99 \times 10^{-7} rad/s):
d_{\text{annual}} = \sqrt{\frac{2 \times 0.5 \times 10^{-6}}{1.99 \times 10^{-7}}} = 2.2 \text{ m}
Scaling rule:
\frac{d_{\text{annual}}}{d_{\text{daily}}} = \sqrt{\frac{P_{\text{annual}}}{P_{\text{daily}}}} = \sqrt{365.25} \approx 19
Annual waves penetrate ~19 times deeper than daily waves.
Time delay for temperature peak to reach depth z:
\Delta t = \frac{z}{d} \times \frac{P}{2\pi}
Example: Daily cycle, z = 0.5 m, d = 0.12 m:
\Delta t = \frac{0.5}{0.12} \times \frac{86400}{2\pi} = 4.17 \times 13750 = 57,300 \text{ s} \approx 16 \text{ hours}
Surface peaks at noon → 0.5 m depth peaks at 4 AM the next day.
Problem: A soil has thermal diffusivity \alpha = 0.4 \times 10^{-6} m²/s. Surface temperature oscillates daily with mean 15°C and amplitude 10°C.
(a) Damping depth
Daily period: P = 86400 s
\omega = \frac{2\pi}{P} = \frac{2\pi}{86400} = 7.27 \times 10^{-5} \text{ rad/s}
d = \sqrt{\frac{2\alpha}{\omega}} = \sqrt{\frac{2 \times 0.4 \times 10^{-6}}{7.27 \times 10^{-5}}}
= \sqrt{\frac{0.8 \times 10^{-6}}{7.27 \times 10^{-5}}} = \sqrt{1.1 \times 10^{-2}} = 0.105 \text{ m} = 10.5 \text{ cm}
(b) Amplitude at 20 cm
A(z) = A_0 e^{-z/d} = 10 \times e^{-0.20/0.105}
= 10 \times e^{-1.90} = 10 \times 0.150 = 1.5°\text{C}
Temperature swing at 20 cm depth is ±1.5°C (compared to ±10°C at surface).
(c) Phase lag
\phi = \frac{z}{d} = \frac{0.20}{0.105} = 1.90 \text{ radians}
Convert to time:
\Delta t = \phi \times \frac{P}{2\pi} = 1.90 \times \frac{86400}{2\pi} = 1.90 \times 13750 = 26,125 \text{ s} \approx 7.3 \text{ hours}
Surface peaks at noon (12:00) → 20 cm depth peaks at 19:18 (7:18 PM).
Below is an interactive soil temperature profile simulator.
<label>
Soil type:
<select id="soil-type">
<option value="dry-sand">Dry Sand (α=0.3)</option>
<option value="moist-sand" selected>Moist Sand (α=0.6)</option>
<option value="clay">Clay (α=0.4)</option>
<option value="peat">Peat (α=0.1)</option>
</select>
</label>
<label>
Surface temperature amplitude (°C):
<input type="range" id="amplitude-slider" min="5" max="20" step="1" value="10">
<span id="amplitude-value">10</span> °C
</label>
<label>
Mean temperature (°C):
<input type="range" id="mean-temp-slider" min="5" max="25" step="1" value="15">
<span id="mean-temp-value">15</span> °C
</label>
<label>
Time of day (hours):
<input type="range" id="time-slider" min="0" max="24" step="0.5" value="12">
<span id="time-value">12.0</span> h
</label>
<div class="button-group">
<button id="animate-toggle">Animate Daily Cycle</button>
</div><p><strong>Damping depth:</strong> <span id="damping-depth"></span> cm</p>
<p><strong>Temperature at 50 cm:</strong> <span id="temp-50cm"></span> °C</p>Try this: - Animate the daily cycle: Watch the temperature wave propagate downward - Switch to peat: Low diffusivity → shallow damping depth → surface insulation - Switch to moist sand: Higher diffusivity → deeper penetration - Observe phase lag: Surface peaks at noon, but 50 cm peaks hours later - Notice damping: Temperature swing decreases rapidly with depth
Key insight: The ground acts as a low-pass filter — high-frequency oscillations (daily) are filtered out quickly, while low-frequency (annual) penetrate deep.
At 2–3 meters depth, daily temperature oscillations are negligible (amplitude < 1% of surface).
Annual oscillations still penetrate: - Summer surface temperature: 30°C - 3 m depth: Lags by ~3 months → peaks in fall - Result: Basement cool in summer, relatively warm in winter
In Arctic regions, soil can be: - Active layer (top ~0.5–2 m): Thaws in summer, freezes in winter - Permafrost (below): Permanently frozen
Active layer thickness is approximately where annual temperature wave amplitude drops below the freezing point.
Climate warming → deeper summer thaw → thicker active layer → permafrost degradation.
Ocean has very high thermal inertia (high c, high \rho): - Slow to warm in spring - Slow to cool in fall - Moderates coastal climates
Desert soil (dry) has low thermal inertia: - Heats quickly during day - Cools quickly at night - Large diurnal temperature range
Frost depth in winter depends on: - Duration of freezing temperatures (how long the “cold wave” lasts) - Soil thermal properties - Snow cover (insulation)
Simplified estimate: Frost penetrates to depth where soil temperature stays above 0°C.
Real soil has layers with different thermal properties: - Organic litter (low diffusivity) - Topsoil (medium) - Subsoil (higher) - Rock (high)
Boundary conditions at each interface complicate the solution.
When soil freezes, latent heat of fusion is released:
L_f = 3.34 \times 10^5 \text{ J/kg}
This slows the freezing front (energy must be extracted before temperature drops further).
Phase change creates a nonlinear problem — \alpha changes abruptly at the freezing front.
Moisture moves in soil (evaporation, drainage, capillary rise). Water carries heat (advection).
Our model assumes no water movement — pure conduction only.
In reality, evaporation at the surface cools the soil (latent heat), and infiltrating rainfall heats or cools deeper layers.
The solution assumes sinusoidal forcing has been running for many periods (steady periodic state).
After a sudden change (e.g., tillage, snow removal), soil takes time to adjust (transient response, not covered by our periodic solution).
Real soil temperature is a superposition of multiple cycles:
T(z,t) = T_m + A_d e^{-z/d_d} \sin(\omega_d t - z/d_d) + A_a e^{-z/d_a} \sin(\omega_a t - z/d_a)
Where subscript d = daily, a = annual.
Daily component: Damping depth ~10 cm, negligible below 50 cm
Annual component: Damping depth ~2 m, significant to 10 m
Superposition works because the heat equation is linear.
Consider a thin layer of soil from depth z to z + \Delta z.
Energy in:
Heat flux entering from above: q(z)
Energy out:
Heat flux leaving through bottom: q(z + \Delta z)
Storage:
Energy stored in the layer: \rho c \Delta z \frac{\partial T}{\partial t}
Balance:
q(z) - q(z + \Delta z) = \rho c \Delta z \frac{\partial T}{\partial t}
Divide by \Delta z and take limit \Delta z \to 0:
-\frac{\partial q}{\partial z} = \rho c \frac{\partial T}{\partial t}
Substitute Fourier’s law q = -k \frac{\partial T}{\partial z}:
\frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right) = \rho c \frac{\partial T}{\partial t}
For constant k:
\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial z^2}
This is the diffusion equation (or heat equation).
Same equation governs: - Moisture diffusion in soil - Chemical diffusion in fluids - Pollutant spreading in groundwater - Population spread in ecology
Universal form:
\frac{\partial \phi}{\partial t} = D \frac{\partial^2 \phi}{\partial z^2}
Where \phi is the diffusing quantity and D is the diffusion coefficient.