Between the last raindrop and the first root hair lies the busiest few meters of the water cycle: a zone that is neither dry nor full, where water hangs from mineral grains under tension, climbs against gravity, and can take a week to move the length of your hand. Rivers get the credit, but the vadose zone decides how much rain becomes river, how much becomes plant, and how much becomes groundwater. This companion teaches you to measure its water, to predict which way that water moves, and to model it with Richards’ equation — running live in this page. Click a part of the soil profile below (or a chapter card) to explore, experiment, and test yourself.
Companion to the Part 04 lecture slides · Evaristo Critical Zone Hydrology Lab, University of Georgia. Scores are self-assessment only — nothing is recorded, and progress resets if you reload.
Here is a puzzle. Water runs downhill. A soil is full of open pores connected all the way to the water table. So why, a week after the last rain, is the topsoil still damp — and why do plants survive on it? The answer is electrostatic, and it sets the rules for everything else in this Part.
Strip a soil to its essentials and two things remain: particles and the void space between them, and that void space can be filled interchangeably with air or with water. Water does not flow through soil so much as around it — along interconnected pores whose paths wind, split, and rejoin. A drop entering the top of a column may travel far more than the column’s length before it leaves the bottom. Which pores are wet decides how easily that journey goes: the largest pores are the freeways, but they are also the first to empty. Manns et al. (2024) put numbers to the classes — micropores (below about 0.05 mm) mostly store water; mesopores (0.05–0.3 mm) move it slowly; macropores (0.3–1 mm, and the larger voids to 5 mm) carry fast, bypassing flow. Drag the water content below and watch the pore network fill from the finest pores outward — and drain from the coarsest pores inward.
How can soil store water for weeks so that plants survive between storms? Look at the molecule. Water is polar: its two hydrogens sit on one side of the oxygen (the bond angle is 104.5°), so one end of the molecule carries a partial positive charge and the other a partial negative one. Molecules therefore stick to each other — cohesion, through hydrogen bonds, the source of water’s high surface tension — and to charged surfaces — adhesion. Soil particles have charged surfaces (most soil is quartz, SiO2, the same material as glass), so water is pulled onto them as films and into the narrow spaces between them. Electromagnetism, not gravity, is what stores water in soils.
The cleanest demonstration is a thin glass tube standing in a cup: water climbs the tube above the level in the cup, higher the thinner the tube. A Teflon tube of the same size does nothing, because Teflon carries no charge for the water to adhere to. The rise stops when the upward pull of surface tension around the tube’s rim balances the weight of the lifted column:
Run the capillary lab: set the tube radius and the material, watch the meniscus settle, and read the rise. The bank of tubes at the right shows how quickly the effect grows as pores shrink.
Water moves from high energy to low energy, and the energy that matters here is the sum of gravitational (potential) energy and pressure energy. Engineers long ago divided that energy by the specific weight of water, ρg, so it could be written as a length — the hydraulic head. A column of water of height h, area A, and density ρ weighs W = Ahρg; the pressure at its base is P = W/A = hρg; divide by ρg and you recover h — the pressure head. If we know the fluid density, we can describe a pressure by the depth of fluid above a point.
In a soil the pressure head ψ can be positive, zero, or negative. Below the water table molecules are pushed together by the weight of water above them: ψ > 0, growing with depth. At the water table ψ = 0 — that is one definition of the water table, the surface where pressure head vanishes; another is the surface connecting water levels in wells, which is the same thing. Above the water table water hangs from the particles by adhesion and cohesion, its molecules pulled apart: ψ < 0, a tension (also called matric potential or suction). Tension is measured with a tensiometer — a water-filled tube closed by a porous ceramic cup, with a vacuum gauge; water moves out through the cup until the tube’s pressure matches the soil’s, and the gauge reads the pull. Tensiometers cannot read very large tensions: beyond roughly −8 m (about −80 kPa) the water column cavitates and the instrument loses contact. Drag the probe up and down the column below, and move the water table.
Water is pulled onto soil particles as films, and the thinner the film, the more strongly it is held. As a soil dries the films thin, the largest pores empty first, and it takes more and more tension to pull the next drop loose. A wet soil drains its macropores under gravity alone; at about 0.1 bar (1 m of tension) rapid drainage stops and the soil is at field capacity; from there plants do the pulling, emptying the mesopores, until at 15 bars (150 m) they can pull no more and wilt — the wilting point. The soil is not dry: air-dry soil in a room at 48 % relative humidity still holds water at about 1,000 bars, and only an oven (105 °C, 48 h, by definition 10,000 bars — a suction equal to lifting a column of water 100 km tall) removes the last of it. Every rung of the ladder pairs a tension with the largest pore that can still hold water and with the relative humidity of the pore air in equilibrium with it. Slide the marker up and down the ladder.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
“The soil is 25 % water” sounds like a fact. It is not — not until you say 25 % of what. By mass? By volume? As a depth of water you could pour on the surface? Each answer is right, each is different, and a bulk density is the exchange rate between them.
Sieve out the gravel and what is left of the mineral soil sorts into three size fractions: sand (2–0.05 mm), silt (0.05–0.002 mm), and clay (below 0.002 mm). The textural class is the relative proportion, by mass, of the three — a point inside the USDA triangle. Texture is the first thing to know about a soil because pore sizes follow particle sizes: sands have big pores that drain at low tension; clays have tiny pores that hold water at tensions no plant can match. Click or drag anywhere in the triangle, or type the fractions, and read the class with the typical bulk density and pore space from the lecture table and the Clapp–Hornberger hydraulic parameters from Dingman’s Table 7.4.
Drive a steel ring (a Kopecky ring holds 100 cm3) into the soil, trim it flush, weigh it, dry it at 105–107 °C for a day, and weigh it again. Two numbers fall out. The bulk density is the dry mass over the ring volume, ρb = ms/V, typically 1.2–1.7 g cm−3; and because the particles themselves have a nearly constant particle density (about 2.65 g cm−3 for quartz), the porosity follows: n = 1 − ρb/ρs. The ring preserves the soil’s structure but may compact it while it is driven; re-packing sieved soil into a cylinder gives a density for a plowed field instead. Now the water:
The lecture’s clay loam: a 250-g sample dries to 200 g, so θg = 50/200 = 0.25; with ρb = 1.3 g cm−3 that is θv = 0.33, or 330 mm of water per meter of soil, or 231 mm in a 0.70-m root zone. Use the core lab to run the chain yourself: change the masses and the ring, watch the three-phase bar redraw, and try the presets.
Every soil has its own relationship between how much water it holds and how hard that water is held: the soil moisture characteristic (or moisture release, or water retention) curve. Its shape follows the pore-size distribution, which follows texture — and biology (ants, earthworms, roots), compaction, and organic matter. Plotted with tension on a logarithmic axis, the curve starts nearly vertical at saturation, bends once air enters the largest pores (the air-entry tension, whose magnitude equals the height of the capillary fringe), falls steeply, and flattens again toward a tightly bound residual content. From a hydrologic standpoint the whole curve matters. From a plant’s standpoint two points matter:
Explore the curve. The two soils from the lecture are here with their plotted points (a fine-textured, high-porosity soil and a coarse, low-porosity one); so are Dingman’s Clapp–Hornberger textures and a soil you can shape yourself with van Genuchten’s α and n. Switch the tension axis between meters and bars, drag the FC and WP lines, and read the plant-available water and its depth for a chosen soil thickness.
A core from an A horizon was taken with a ring of 72 cm3. In the laboratory the sample was brought to four tensions and weighed: 143 g at saturation (0 bar), 126 g at 0.1 bar, 108 g at 15 bars, and 100 g oven-dry (10,000 bars). Find the bulk density, the porosity, the water contents by mass and by volume at each tension, and the depth of plant-available water at field capacity if the horizon is 20 cm thick. Enter each value; every step checks your number before the next opens — and stay alert at the end, because this data set has a surprise in it.
Real root zones cross several horizons. The text’s Table 5.5 works a pedon with a plowed topsoil (Ap, 15 cm), a transition layer (BA, 30 cm), and a clay-rich subsoil (Bt, 45 cm), each sampled with a ring of radius 3 cm and height 5 cm (V = πr2h = 141 cm3, tare 2 g) and weighed oven-dry, at field capacity, and at the wilting point. The Ap row is worked for you. Complete the BA and Bt rows and total the pedon.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
Downhill? Too simple — it cannot explain water moving up a dry soil profile, or sideways. Toward money? It explains the large-scale plumbing of water-scarce regions surprisingly well. From high head to low head? Always works.
Picture two pools whose bottoms sit at different elevations, connected by a pipe with the valve just opened. The head in each pool is its bottom elevation plus its water depth: H = z + ψ. Water flows from the pool with the higher head to the pool with the lower one — even if that means flowing uphill through the pipe — until the heads (the water-surface elevations) are equal. Now picture a single still pool: the pressure head is zero at the surface and equals the depth at the bottom, and since depth falls exactly as fast as elevation rises, the head is the same at every level. Equal heads everywhere: the hydrostatic condition, and no flow. A wetted soil, covered against evaporation and left to drain, approaches the same state: ψ becomes more negative upward at the same rate z increases, the soil is wetter at the bottom than at the top, and nothing moves. Set the two pools (the lecture’s two cases are presets) and open the valve; then probe the still pool.
Three tensiometers sit in a profile: w in the A horizon at 19.8 m elevation, x in the E horizon at 19.6 m, y in the Bt horizon at 19.4 m. Compute the head at each and the direction of flow follows. In the lecture’s first situation w and x are both at field capacity (ψ = −1 m): Hw = 18.8, Hx = 18.6, flow is downward. In the second, w is slightly drier (−1.2 m) and the heads tie: no flow, although the two points hold different amounts of water. In the third, x is much drier than y (−4.0 vs −1.4 m): Hx = 15.6 < Hy = 18.0, and water moves up from the Bt into the E horizon — suction beating gravity, as it does when roots have dried the topsoil. Set the three pressure heads, watch the arrows, and try the challenges.
In 1856 Henry Darcy, engineer to the city of Dijon, was filtering the town’s water through sand and wanted to know what controlled the flow. He forced water at a steady rate Q through a sand-filled column with two manometers a distance Δs apart, and found that Q was proportional to the head difference Δh between them and inversely proportional to Δs. Written for a column of cross-section A:
The constant Ks is the hydraulic conductivity — the ease with which a medium transmits water, a property of the material, and one that spans twelve orders of magnitude from gravel to shale. Two velocities hide in the law. The specific discharge q is what you would measure if water could use the whole cross-section; it cannot, because most of the cross-section is solid, so the water in the pores moves faster than q, at v = q/ne. Run the column: pick a material (or set K directly), tilt the heads, and watch the tracer outrun the Darcy velocity.
Problem A. A silty sand is tested in an apparatus like Darcy’s: inside diameter 10 cm, manometers Δs = 25 cm apart, steady flow Q = 1.7 cm3 min−1, head difference Δh = 15 cm. Find Ks. Problem B. A storm has packed a culvert under a road with sand from end to end: 5 m long, 0.8 m in diameter, K = 3 m d−1, ne = 0.38, and the water level at one end 1.6 m higher than at the other, the whole culvert submerged. Find the discharge, the specific discharge, and the average linear velocity. Solve along.
Dingman’s Box 7.4: two tensiometers in an unsaturated soil, A with its cup 20 cm deep and B at 50 cm, read on five days. Which way is the water moving? Take the deeper cup as datum (zB = 0, zA = +30 cm), add elevation to each reading to get total head, and compare. Fill in the flow directions, then check the arithmetic in the book’s own table — one entry does not add up, and it changes the answer for that day. The second tab holds the text’s exercise on flux and velocity between two points.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
Darcy’s law was found in a saturated sand filter. Take the same law into a soil that is only part full and one word changes everything: the conductivity is no longer a constant. Follow that change through conservation of mass and you arrive at the one equation that describes infiltration, capillary rise, root uptake, and drainage at once.
In a saturated soil every pore carries water and K = Ks, the maximum. As the soil dries, the largest pores empty first — the interstate highways close — and flow is confined to smaller, more tortuous side streets, through a shrinking share of the cross-section. So K(θ) < Ks, always, and the drop is not gentle: between field capacity and wilting point a soil’s conductivity can fall by four to six orders of magnitude. Because ψ and K both depend on θ, the retention curve and the conductivity curve are two faces of one soil. The lecture’s MATLAB snippet used the van Genuchten–Mualem model:
Drag the water content and watch the pore network, the retention curve, and the conductivity curve respond together; switch among the lecture soil and three textures.
Write the head as H = z + ψ with ψ < 0 in the unsaturated zone (the matric potential), let the conductivity follow the water content, and Darcy’s law for vertical flow becomes:
Gravity always pulls down and always with the same strength: one meter of head per meter of height. The matric gradient can point either way and can be enormous — across a wetting front the tension changes by meters over centimeters. Flow goes up when the capillary pull beats gravity: the tension must increase upward by more than one meter per meter, so that dψ/dz + 1 turns negative. (The lecture slide states the rule as “upward if dψ/dz > 1” — read its dψ/dz as the magnitude of the suction gradient; with ψ negative and z pointing up, the strict condition is dψ/dz < −1.) The sign decides the direction; K(θ) decides how much gets through. Set the gradient dial and the water content and read the flux.
Mass conservation for a thin layer of soil says that what comes in minus what goes out is the change in storage: ∂θ/∂t = −∂q/∂z. Substitute the modified Darcy law for q and the two laws fuse into one, first written by L. A. Richards in 1931:
Read it aloud: the rate at which soil moisture changes equals the divergence of the water flux, where the flux is the soil’s ability to conduct water times the total driving force — capillary pull plus gravity. It is the master equation because all the physics of vadose-zone flow sits in that one line. It is also nonlinear — K and ψ both depend on the unknown θ — so except for highly simplified soils and boundary conditions it has no closed-form solution and is solved numerically, layer by thin layer and step by small time step. Tests since the 1960s have shown good agreement with field and laboratory measurements; the next section runs such a solver in your browser.
Richards’ equation explains infiltration after rain (a steep ∂ψ/∂z at the wetting front drives water down), capillary rise from a water table (the matric gradient overcomes gravity), root water uptake (a sink that carves local moisture gradients), and redistribution after irrigation stops (gravity drainage racing capillary retention). The simulator below solves the equation numerically for a soil column — van Genuchten soils, a mass-conservative implicit scheme, and the same boundary conditions a research code would use. Choose a scenario, set the soil and the forcing, and run it. Then scrub through time, read the water balance, and compare with the lecture’s sketches.
A pot 30 cm tall with drainage holes has sat unwatered for a week; its soil is quite dry (θ = 0.15). You water from the top until water just starts dripping from the holes, then stop. The study guide asks you to sketch what happens; here you predict, then let the equation draw. Pick a sketch for each part, check it, and press “run this part” to see the solver’s answer — then settle the argument between your two friends about little-and-often versus deep-and-infrequent watering.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.