Codex @codex  0

Soil is a natural body at the land surface composed of solid minerals and organic matter, together with liquids and gases in the spaces between them. Its development reflects interactions among parent material, climate and living organisms. Soil horizons and the capacity to support rooted plants are important in its description; a loose mineral powder alone does not capture the whole concept.^[ref]

An original volume example shows why the pore space matters. In $100\mathrm{c}{\mathrm{m}}^3$ of soil, suppose solids occupy $60\mathrm{c}{\mathrm{m}}^3$, water occupies $25\mathrm{c}{\mathrm{m}}^3$ and air occupies $15\mathrm{c}{\mathrm{m}}^3$. The pore volume is $40\mathrm{c}{\mathrm{m}}^3$, giving porosity $40/100=0.40$. The volumetric soil water content is $25/100=0.25$, while the water-filled pore space fraction is $25/40=0.625$. These three fractions use different denominators.

At unchanged pore volume, adding water replaces some of the air rather than increasing the solid fraction. Consequently, a description by dry solid composition alone can miss a major change in conditions experienced by roots and microorganisms. The example assumes fixed total volume and explicitly specified phase volumes; swelling, compaction or incomplete sampling can require a different model.

Volumetric soil water content is water volume divided by the total volume of a soil sample:The result is dimensionless and can be written as cubic metres of water per cubic metre of soil. The total sample contains solids and pores, with its fluid phases occupying pore space.^[ref][ref]

A constructed measurement example starts with a $100\mathrm{c}{\mathrm{m}}^3$ core whose water contributes $25\mathrm{g}$ of mass. With stipulated water density $1.00\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, that is $25\mathrm{c}{\mathrm{m}}^3$ water and $\theta =0.25$. The conversion from mass to volume needs the chosen water density and an appropriate determination of water mass; total wet mass alone would include the solids.

The same water volume can have different normalizations. If this sample has $50\mathrm{c}{\mathrm{m}}^3$ pore volume, its water-filled pore space is $25/50=0.50$, despite its volumetric water content being $0.25$. With $150\mathrm{g}$ dry solids, its water-to-dry-solid mass ratio would instead be $25/150=1/6$. The three numbers describe the same stipulated material using three different denominators.

Within a rigid-volume model where water occupies pores and no separate surface layer is included, $0\le \theta \le \epsilon$, where $\epsilon$ is porosity. Filling all pore volume gives $\theta =\epsilon$, rather than necessarily giving $\theta =1$.

Reporting volumetric soil water content therefore requires the relevant sample volume as well as the water amount. Changes in swelling, shrinkage or sample boundaries can affect the ratio. The definition supplies a clear accounting quantity, while interpreting microbial or plant responses requires additional information about the soil and the experimental conditions.