1.2
What is cavitation?
§1 · The liquid state

1.2 Tensile strength in principle

How much tension can a perfectly pure, perfectly homogeneous liquid withstand before the intermolecular potential of Lesson 1.1 loses cohesion entirely? Two arguments — one mechanical, one thermodynamic — give estimates in the range of −103-10^3 atm for water at room temperature.

The mechanical (Born) estimate

The simplest argument inverts the bulk modulus. The liquid responds linearly to small pressure changes:

ΔVV=−ΔpK.\frac{\Delta V}{V} = -\frac{\Delta p}{K}.

Extrapolate this linear response outside its regime of validity, to the strain at which −∂U/∂r-\partial U / \partial r in the Lennard-Jones potential reaches its maximum. For the LJ-12-6 potential, the maximum attractive force occurs at rinfl=(26/7)1/6σ0≈1.244σ0r_\text{infl} = (26/7)^{1/6} \sigma_0 \approx 1.244 \sigma_0, which corresponds to a volumetric strain of (rinfl/req)3−1≈0.235(r_\text{infl} / r_\text{eq})^3 - 1 \approx 0.235. Apply this strain in the linear response and the implied tension is

pcoh≈−0.235⋅K  ≈  −5,000 atm.p_\text{coh} \approx -0.235 \cdot K \;\approx\; -5{,}000\ \text{atm}.

This Born estimate overshoots the true theoretical strength because the linear Δp/K\Delta p / K relation badly fails at 23.5% strain. A more careful calculation that respects the actual shape of U(r)U(r) between reqr_\text{eq} and rinflr_\text{infl} gives a tension closer to −1000-1000 to −1500-1500 atm. The order of magnitude is robust: the theoretical cohesive strength of water is somewhere between −103-10^3 and −104-10^4 atm, comparable to the bulk modulus itself.

▶ Cohesive limit from the Lennard-Jones potential

A more careful estimate proceeds as follows. The force per unit area required to hold each molecular pair at separation rr across a slice through the liquid is

σ(r)=−n2/3∂U∂r,\sigma(r) = -n^{2/3} \frac{\partial U}{\partial r},

where nn is the molecular number density (cube root of which is the inverse of the typical molecule-to-molecule distance, giving force per unit area when multiplied by the per-pair force).

Differentiating the Lennard-Jones potential:

∂U∂r=4ε[−12σ012r13+6σ06r7].\frac{\partial U}{\partial r} = 4 \varepsilon \left[-12 \frac{\sigma_0^{12}}{r^{13}} + 6 \frac{\sigma_0^{6}}{r^{7}}\right].

Setting ∂2U/∂r2=0\partial^2 U / \partial r^2 = 0 to find the strain at maximum tension gives rinfl=(26/7)1/6σ0≈1.244σ0r_\text{infl} = (26/7)^{1/6} \sigma_0 \approx 1.244 \sigma_0, where the inflection-point force per molecule is

Fmax=−∂U∂r∣rinfl≈2.69εσ0.F_\text{max} = -\frac{\partial U}{\partial r}\bigg|_{r_\text{infl}} \approx \frac{2.69 \varepsilon}{\sigma_0}.

For water, take ε≈22 meV\varepsilon \approx 22 \text{ meV} per hydrogen-bond pair, with about 4 hydrogen bonds per molecule (the tetrahedrally coordinated network). Per molecule this gives an effective bonding energy of about εeff≈4⋅22/2=44 meV\varepsilon_\text{eff} \approx 4 \cdot 22 / 2 = 44 \text{ meV} (each bond shared between two molecules, so divide by 2). Combined with σ0≈2.8 A˚\sigma_0 \approx 2.8 \text{ Å} and molecular density n≈3.3×1028 m−3n \approx 3.3 \times 10^{28}\ \text{m}^{-3}:

σmax≈n2/3Fmax≈1019 m−2⋅6.6×10−12 N≈6×107 Pa=600 atm.\sigma_\text{max} \approx n^{2/3} F_\text{max} \approx 10^{19}\ \text{m}^{-2} \cdot 6.6 \times 10^{-12}\ \text{N} \approx 6 \times 10^{7}\ \text{Pa} = 600 \text{ atm}.

The order of magnitude — a few hundred atm to a thousand atm of tension — is robust across the variations in ε\varepsilon, σ0\sigma_0, and molecular packing assumptions. The argument generalises to other simple liquids with similar predictions: ethanol ∼−800\sim -800 atm, mercury ∼−30,000\sim -30,000 atm (because mercury is held by much stronger metallic bonds).

The thermodynamic (spinodal) estimate

A second route to the cohesive limit comes from the equation of state. A real liquid at temperatures below its critical point has, in the (p,V,T)(p, V, T) space, a region of metastable states where the liquid persists below its vapour-pressure saturation curve. The boundary of metastability — the spinodal — is the locus where the isothermal compressibility diverges,

∂p∂V∣T=0,\left.\frac{\partial p}{\partial V}\right|_T = 0,

i.e., where the slope of the p(V)p(V) isotherm flattens to zero. Below this pressure the liquid cannot exist in principle: any infinitesimal perturbation lowers the local free energy by separating into a denser liquid and a vapour phase.

For water at 20 °C, the spinodal pressure computed from a realistic equation of state (the Speedy/Lemmon parametrisations) sits at about −1400-1400 atm. This is the thermodynamic cohesive limit: tensions below this value are forbidden by the equation of state alone and not merely “weakly metastable.”

The two arguments — mechanical and thermodynamic — agree in order of magnitude. The cohesive limit of water at room temperature is approximately −103-10^3 atm.

What ”−1000-1000 atm” looks like physically

To pull on a column of water with a tension of 1000 atm is to apply about 100 MPa of negative pressure — equivalent to the pressure at 10 km below the sea surface but in the opposite direction. The required mechanical work to elastically extend a litre of water to this tension is roughly

W≈p2V2K≈(108 Pa)2⋅10−3 m32⋅2.2×109 Pa≈2.3 J.W \approx \frac{p^2 V}{2 K} \approx \frac{(10^8\ \text{Pa})^2 \cdot 10^{-3}\ \text{m}^3}{2 \cdot 2.2 \times 10^9\ \text{Pa}} \approx 2.3 \text{ J}.

A few joules — pulling on a chamber with a piston, slowly. Not, on paper, an extraordinary experiment.

In practice, no such experiment succeeds. Most laboratory water samples tear at tensions a thousand times smaller. The next lesson surveys what people have actually measured.