Litter Turnover Time and the Steady-State Forest Floor

A forest floor at steady state has stopped changing because two flows balance: what falls onto it each year equals what rots off it. Under that condition, the standing stock divided by the annual litterfall is the mean time a leaf spends lying there, $\tau = L_{ss}/I$ — and it is the reciprocal of Olson's decay constant, $\tau = 1/k$, arrived at with a scale and a frame instead of a litterbag. Two entirely different field methods, one number. That equivalence was Olson's own point and it remains the neatest result in the subject.

The arithmetic is a mass balance. Input is $I$; loss is $kL_{ss}$, because decomposition takes a fixed proportion of whatever is lying there. Set them equal and $k = I/L_{ss}$, so $\tau = L_{ss}/I$. Turned around, it predicts the forest floor a site will eventually carry from nothing but how much falls and how fast it rots: two stands with identical litterfall can differ tenfold in floor mass purely because one is cold and wet.

The numbers span the same two orders of magnitude the decay constant does. In lowland tropical rainforest, $\tau$ is under a year — a leaf is gone before its replacement falls, the forest floor is a thin skin over mineral soil, and almost all the ecosystem's carbon is standing up in the trees rather than lying down. A temperate deciduous forest turns its floor over in two to five years, a temperate conifer stand in five to fifteen. Boreal and montane forests run 20 to 100, and there the forest floor exceeds the annual litterfall by more than an order of magnitude. That is where high-latitude soil carbon lives, and it is why fire and drainage matter so much at those latitudes: decades of accumulated residence time can be undone in an afternoon.

Two things to be honest about. First, $\tau$ is a mean residence time, not a clearance time. In an exponential model 63 % of a cohort is gone by $\tau$, not all of it, and the tail runs much longer — three $\tau$ to reach 95 %. Second, the stand has to be at steady state for any of this to hold. A plantation still building its forest floor, a stand recovering from fire or clearfell, one freshly thinned, or one in the year after a mast crop is not at equilibrium, and the ratio then measures the disequilibrium rather than the decay rate. Approach to a new equilibrium is itself exponential and takes about three $\tau$, so a site whose decomposition slows — through acidification, cooling, or a species change to lignin-rich litter — will accumulate forest floor for decades before settling at a new and much larger stock. That accumulating layer is simultaneously a carbon store and, in a fire-prone forest, a fuel load.

Measuring it well is mostly discipline. Litter traps of known area, emptied through a full twelve months — miss the autumn peak and $I$ is meaningless. Forest floor clipped from a frame down to mineral soil, which is a judgement call in any soil without a sharp boundary. And remember what traps do not catch: fine root turnover below ground is of the same order as leaf fall above it, and it enters the soil without ever passing through the forest floor at all.