Decomposition and Soil Carbon Turnover

From litterfall to soil organic matter — the slow carbon cycle

2026-02-26

The peat bogs of the Hudson Bay Lowlands have been accumulating organic carbon for roughly 8,000 years since the last glaciation. The reason is simple and astonishing: decomposition in waterlogged, cold, anaerobic conditions is so slow that plant material accumulates faster than it decays. In some bogs, a buried leaf 1,000 years old is still recognisable. The same organic matter in a warm, well-drained agricultural soil in southern Ontario would decompose completely within a few decades. The carbon content of a soil is not determined by how much plant material falls into it; it is determined by the balance between inputs and the rate of decomposition — and that rate varies by four orders of magnitude across Earth’s environments.

Soils hold roughly 1,500 billion tonnes of organic carbon — more than twice the amount in the atmosphere and three times the amount in all living vegetation. Small changes in decomposition rates, driven by warming temperatures or altered moisture, can therefore have large consequences for atmospheric CO₂. Modelling decomposition requires tracking carbon through multiple pools with different turnover times: fresh litter that breaks down in months, slower humus that persists for decades, and a passive fraction that can persist for centuries under the right conditions. This model builds that multi-pool structure, derives the temperature and moisture dependencies of decay rates, and calculates steady-state soil carbon storage for different climate regimes.

1. The Question

Where does the carbon go after plants die?

Leaves fall, roots die, wood decays. This dead organic matter (litter) accumulates on and in the soil, where microbes and fungi break it down, releasing CO₂ through heterotrophic respiration.

The balance between litter inputs and decomposition outputs determines: - Soil carbon storage — soils hold 2–3× more carbon than atmosphere - Net Ecosystem Productivity (NEP = NPP - R_h) - Carbon sequestration potential — can soils store more carbon? - Soil fertility — decomposition releases nutrients

The mathematical question: How do we model the flow of carbon from litter to soil organic matter to CO₂, accounting for different decay rates and environmental controls?


2. The Conceptual Model

Litter and Soil Carbon Pools

Three pools with different turnover times:

  1. Litter (fast pool):
  2. Soil Organic Matter - Active (medium pool):
  3. Soil Organic Matter - Slow/Passive (slow pool):

Decomposition as Exponential Decay

Each pool loses carbon through first-order decay:

\frac{dC}{dt} = I - k C

Where: - C = carbon in pool (kg C/m²) - I = input rate (kg C/m²/year) - k = decay constant (year⁻¹)

Turnover time (\tau):

\tau = \frac{1}{k}

Example: Litter with k = 0.5 year⁻¹ has \tau = 2 years (half decomposes each 2 years).

Environmental Controls

Temperature: Decomposition rate increases exponentially with temperature.

Q₁₀ rule: Decomposition rate doubles per 10°C warming.

k(T) = k_{20} \times Q_{10}^{(T-20)/10}

Where: - k_{20} = decay rate at 20°C - Q_{10} \approx 2 (typically 1.5–3) - T = soil temperature (°C)

Moisture: Decomposition maximized at moderate moisture.

f(W) = \begin{cases} W/W_{\text{opt}} & \text{if } W < W_{\text{opt}} \\ 1 & \text{if } W = W_{\text{opt}} \\ (1-W)/(1-W_{\text{opt}}) & \text{if } W > W_{\text{opt}} \end{cases}

Where W is relative soil moisture (0–1) and W_{\text{opt}} \approx 0.6.

Heterotrophic Respiration

Total heterotrophic respiration:

R_h = k_L C_L + k_A C_A + k_S C_S

Where subscripts L, A, S denote litter, active SOM, slow SOM.

Net Ecosystem Productivity:

\text{NEP} = \text{NPP} - R_h

Positive NEP: Ecosystem is a carbon sink
Negative NEP: Ecosystem is a carbon source


3. Building the Mathematical Model

Three-Pool Decomposition Model

State variables: - C_L = litter carbon (kg C/m²) - C_A = active SOM carbon (kg C/m²) - C_S = slow SOM carbon (kg C/m²)

Litter dynamics:

\frac{dC_L}{dt} = I_L - k_L C_L

Where I_L is litterfall from vegetation (from Model 22: turnover of leaves/roots).

Active SOM dynamics:

\frac{dC_A}{dt} = f_{LA} k_L C_L - k_A C_A

Where f_{LA} is fraction of decomposed litter entering active SOM (typically 0.3–0.5, rest respired).

Slow SOM dynamics:

\frac{dC_S}{dt} = f_{AS} k_A C_A - k_S C_S

Where f_{AS} is fraction of decomposed active SOM entering slow SOM (typically 0.2–0.3).

Decay Constants

Typical values at 20°C, optimal moisture:

Pool Decay rate (k, year⁻¹) Turnover time (\tau, years)
Litter 0.5–2.0 0.5–2
Active SOM 0.05–0.20 5–20
Slow SOM 0.001–0.01 100–1000

Variation by climate: - Tropical: High k (warm, wet) - Temperate: Moderate k - Boreal: Low k (cold) - Arid: Very low k (dry, limited decomposers)

Steady State

At equilibrium:

C_L^* = \frac{I_L}{k_L}

C_A^* = \frac{f_{LA} k_L C_L^*}{k_A} = \frac{f_{LA} I_L}{k_A}

C_S^* = \frac{f_{AS} k_A C_A^*}{k_S} = \frac{f_{AS} f_{LA} I_L}{k_S}

Total soil carbon:

C_{\text{soil}}^* = C_L^* + C_A^* + C_S^*

Key insight: Slow pool dominates total carbon despite small input fraction, because k_S \ll k_L.

Heterotrophic Respiration at Steady State

R_h^* = k_L C_L^* + k_A C_A^* + k_S C_S^*

Substituting steady-state values:

R_h^* = I_L + f_{LA} I_L + f_{AS} f_{LA} I_L

Wait, this doesn’t equal I_L. Let me recalculate correctly.

Actually, at steady state, total inputs = total outputs:

R_h^* = (1 - f_{LA}) k_L C_L^* + (1 - f_{AS}) k_A C_A^* + k_S C_S^*

After all transfers, this sums to I_L (conservation of carbon).


4. Worked Example by Hand

Problem: A temperate forest has: - Litterfall: I_L = 0.4 kg C/m²/year (from leaf/root turnover) - Litter decay rate: k_L = 1.0 year⁻¹ - Active SOM decay rate: k_A = 0.1 year⁻¹ - Slow SOM decay rate: k_S = 0.005 year⁻¹ - Transfer fractions: f_{LA} = 0.4, f_{AS} = 0.3

Calculate steady-state carbon in each pool.

Solution

Litter pool:

C_L^* = \frac{I_L}{k_L} = \frac{0.4}{1.0} = 0.4 \text{ kg C/m}^2

Active SOM:

C_A^* = \frac{f_{LA} I_L}{k_A} = \frac{0.4 \times 0.4}{0.1} = \frac{0.16}{0.1} = 1.6 \text{ kg C/m}^2

Slow SOM:

C_S^* = \frac{f_{AS} f_{LA} I_L}{k_S} = \frac{0.3 \times 0.4 \times 0.4}{0.005}

= \frac{0.048}{0.005} = 9.6 \text{ kg C/m}^2

Total soil carbon:

C_{\text{soil}}^* = 0.4 + 1.6 + 9.6 = 11.6 \text{ kg C/m}^2

Interpretation: - Slow SOM holds 83% of total soil carbon (9.6 / 11.6) - This matches observations: most soil carbon is old, stable material - Total = 11.6 kg C/m² = 116 tonnes C/ha (typical for temperate forest soil)

Heterotrophic respiration:

At steady state, R_h = I_L = 0.4 kg C/m²/year (all inputs eventually respired).


5. Computational Implementation

Below is an interactive decomposition simulator.

<label>
  Ecosystem type:
  <select id="ecosystem-decomp">
    <option value="tropical">Tropical Forest</option>
    <option value="temperate" selected>Temperate Forest</option>
    <option value="boreal">Boreal Forest</option>
    <option value="grassland">Grassland</option>
    <option value="desert">Desert</option>
  </select>
</label>
<label>
  Litterfall rate (kg C/m²/year):
  <input type="range" id="litterfall-slider" min="0.1" max="1.0" step="0.05" value="0.4">
  <span id="litterfall-value">0.4</span>
</label>
<label>
  Temperature (°C):
  <input type="range" id="temp-decomp-slider" min="0" max="30" step="1" value="15">
  <span id="temp-decomp-value">15</span> °C
</label>
<label>
  Soil moisture (0-1):
  <input type="range" id="moisture-slider" min="0" max="1" step="0.05" value="0.6">
  <span id="moisture-value">0.6</span>
</label>
<label>
  Simulation years:
  <input type="range" id="years-decomp-slider" min="50" max="500" step="50" value="200">
  <span id="years-decomp-value">200</span> years
</label>
<div class="button-group">
  <button id="run-decomp-sim">Run Simulation</button>
</div>
<p><strong>Total soil C:</strong> <span id="total-soil-c"></span> kg C/m²</p>
<p><strong>Slow SOM fraction:</strong> <span id="slow-fraction"></span>%</p>
<p><strong>R<sub>h</sub>:</strong> <span id="rh-result"></span> kg C/m²/year</p>
<p><strong>Mean residence time:</strong> <span id="mrt-result"></span> years</p>

Try this: - Tropical forest: High temperature → fast decomposition → low soil carbon accumulation - Boreal forest: Low temperature → slow decomposition → huge soil carbon stores (peat) - Desert: Dry conditions (low moisture) → very slow decomposition - Increase temperature: Speeds all decay rates → reduces equilibrium carbon - Increase litterfall: More inputs → higher steady-state carbon - Notice: Slow SOM takes centuries to equilibrate, dominates total carbon

Key insight: Cold/wet environments accumulate soil carbon because decomposition is slower than production. This is why peatlands and permafrost soils are massive carbon stores.


6. Interpretation

Global Soil Carbon Distribution

Largest soil carbon stores: 1. Boreal forests & tundra: Cold → slow decomposition 2. Peatlands: Waterlogged → anaerobic, very slow decay 3. Grasslands: Deep roots, moderate climate

Smallest stores: 1. Tropical forests: Fast turnover (hot, wet) 2. Deserts: Low productivity + dry conditions

Paradox: Tropical forests have low soil C despite high productivity (fast decomposition balances high input).

Climate Change Feedbacks

Warming accelerates decomposition:

If temperature increases 5°C and Q₁₀ = 2:

k_{\text{new}} = k_{\text{old}} \times 2^{5/10} = k_{\text{old}} \times 1.41

41% faster decomposition → releases soil carbon → positive feedback.

Permafrost thaw: - Frozen organic matter begins decomposing - Releases CO₂ and CH₄ - Major climate concern (~1000 Pg C in permafrost)

Soil Carbon Sequestration

Practices to increase soil C: 1. No-till agriculture: Reduces disturbance → less decomposition 2. Cover crops: More inputs to soil 3. Biochar: Very slow decay (k \approx 0.0001 year⁻¹) 4. Organic amendments: Compost, manure

Saturation: Soil can only hold so much carbon (determined by clay content, climate).


7. What Could Go Wrong?

Assuming Single Decay Rate

Real litter has multiple components: - Soluble sugars: k \approx 10 year⁻¹ (days to decompose) - Cellulose: k \approx 1 year⁻¹ (months) - Lignin: k \approx 0.1 year⁻¹ (years)

Better model: Multiple litter pools with different decay rates.

Ignoring Priming Effects

Priming: Fresh litter addition stimulates decomposition of old SOM.

Mechanism: New carbon provides energy for microbes → they also decompose stable SOM.

Effect: Non-linear interactions between pools.

Constant Transfer Fractions

Real f_{LA}, f_{AS} vary with: - Litter quality: High lignin → less transfer to SOM - Soil clay content: More clay → more stabilization - Oxygen availability: Anaerobic → more SOM formation

Neglecting Nutrient Limitation

Decomposition requires nitrogen (microbes need C:N ~10:1).

Low N litter (wood, C:N > 100): - Slow decomposition - Microbes immobilize N from soil - Net N immobilization

High N litter (legumes, C:N < 20): - Fast decomposition - Net N mineralization

Our model is carbon-only. Full model couples C and N cycles.


8. Extension: Century Model

The Century model (Parton et al., 1987) is the classic soil C model:

Five pools: 1. Structural litter (slow decay) 2. Metabolic litter (fast decay) 3. Active SOM (fast) 4. Slow SOM (medium) 5. Passive SOM (very slow)

Climate modifiers: - Temperature (Q₁₀) - Moisture (optimal curve) - Soil texture (clay slows decay)

Lignin effects: - High lignin → more to structural litter - Affects decay rates

Widely used in ecosystem models (CASA, TEM, BIOME-BGC).


9. Math Refresher: First-Order Decay

General Solution

For \frac{dC}{dt} = I - kC with constant input I:

Steady state:

C^* = \frac{I}{k}

Approach to steady state:

C(t) = C^* + (C_0 - C^*) e^{-kt}

Where C_0 is initial value.

Time to 95% of equilibrium:

t_{95} = \frac{-\ln(0.05)}{k} = \frac{3}{k}

Example: If k = 0.1 year⁻¹:

t_{95} = \frac{3}{0.1} = 30 \text{ years}

Takes 30 years to reach 95% of equilibrium.

For slow pool (k = 0.005 year⁻¹):

t_{95} = \frac{3}{0.005} = 600 \text{ years}

This is why soil carbon responds slowly to land use change.


Summary