Thermal Expansion & Thermal Stress

Track a bar's length L as its temperature T changes and the natural quantity is not dL/dT, which doubles when you take a bar twice as long, but the fractional rate α = (1/L)(dL/dT) = d(ln L)/dT. Integrate it: ln(L/L₀) = ∫α dT, so L = L₀ exp(∫α dT). For constant α that is L₀ exp(αΔT) = L₀(1 + αΔT + ½(αΔT)² + …), and the linear law ΔL = αL₀ΔT is the first term. It is an excellent first term: steel has α ≈ 1.2 × 10⁻⁵ K⁻¹, so a 100 K rise gives αΔT = 1.2 × 10⁻³ and the dropped term, ½(αΔT)² = 7.2 × 10⁻⁷, is 0.06% of it. What limits the law over a wide range is not the truncation but the drift of α itself: for most metals it rises with temperature, and by the third law it falls to zero as T → 0 K. A tabulated α is a mean over the range in which it was measured, and a long excursion needs the integral, as the hard example shows.

Two neighbouring atoms sit in a potential well U(r) with its minimum at the bond length r₀. If the well were parabolic, the vibration would be symmetric about r₀ at every amplitude: the mean separation would stay r₀ however hot the solid, and α would be exactly zero. Real wells are steeper on the compressed side than on the stretched side — a cubic term in U(r) — so an atom with more energy spends longer at large r than at small r, and its mean position creeps outward. Classically that outward shift is proportional to k_BT, which is why α is roughly constant at ordinary temperatures; quantum freezing of the vibrations is what makes it vanish near absolute zero. The outward shift is proportional to the vibrational energy the bonds hold, so α rises and falls with the heat capacity C_V — the Grüneisen relation, whose constant of proportionality measures the anharmonicity. Expansion is not bonds being stretched by heat; it is the asymmetry of the bond made visible.

Every length in a body with one α scales by the factor 1 + αΔT, so for a plate ln A = 2 ln L and for a block ln V = 3 ln L. Differentiating gives an area coefficient 2α and a volume coefficient β = 3α exactly, provided α means the same thing in every direction. Expanding the product instead, (1 + αΔT)³ = 1 + 3αΔT + 3(αΔT)² + (αΔT)³, shows the same 3α with the same second-order leftovers as the linear law. Crystals need not be isotropic: β is the sum of the three principal coefficients, β = α₁ + α₂ + α₃, and the terms can have opposite signs. Calcite grows along its c-axis at about 25 × 10⁻⁶ K⁻¹ and shrinks across it at about −6 × 10⁻⁶ K⁻¹, so β ≈ (25 − 6 − 6) × 10⁻⁶ K⁻¹ = 13 × 10⁻⁶ K⁻¹, nowhere near three times either number. A polycrystalline metal averages its randomly oriented grains and behaves isotropically; a single crystal or a layered material does not.

Raise a plate uniformly by ΔT and every length in it — between any two atoms, across any hole — is multiplied by the same 1 + αΔT. That is a similarity transformation, a photographic enlargement of the whole picture, and the picture includes the holes. A steel ring of inner diameter 50.000 mm warmed by 200 K has an inner diameter of 50.000 × (1 + 1.2 × 10⁻⁵ × 200) = 50.120 mm, which is how a collar is shrink-fitted onto a shaft. The same rule governs liquids in vessels. A flask does not hold its liquid at fixed volume: its cavity grows with coefficient βᵥₑₛₛₑₗ = 3α_glass, so only the excess (β_liquid − βᵥₑₛₛₑₗ) spills. A 250.0 cm³ soda-lime flask (α ≈ 9 × 10⁻⁶ K⁻¹, so β ≈ 2.7 × 10⁻⁵ K⁻¹) brim-full of mercury (β = 1.8 × 10⁻⁴ K⁻¹) warmed by 40 K overflows (1.8 × 10⁻⁴ − 2.7 × 10⁻⁵) × 250.0 × 40 = 1.53 cm³, not the 1.80 cm³ a rigid flask would spill.

Clamp a rod between rigid walls and warm it by ΔT. Free, it would take a thermal strain αΔT; the walls demand total strain zero, so the rod must also carry a mechanical strain of −αΔT, and Hooke's law prices that at σ = Y × (−αΔT) = −YαΔT, compressive. No length appears, because the walls forbid a fraction, not a distance: a 1 m link and a 30 m rail warmed alike carry the same stress, and only the force F = σA depends on anything else. For steel, Y = 2.0 × 10¹¹ Pa and a 50 K rise give σ = 2.0 × 10¹¹ × 1.2 × 10⁻⁵ × 50 = 1.2 × 10⁸ Pa, 120 MPa — about half the yield stress of mild steel, from a temperature swing a rail sees every summer. Cooling reverses the sign, and a member cast in place can crack in tension. Leave a clearance s and the rod grows freely until αL₀ΔT = s; beyond that the blocked strain is αΔT − s/L₀, and length enters only through s/L₀, a fixed gap turned into a strain.

Nothing in the definition requires α or β to be positive. Water between 0 °C and 3.98 °C has β < 0: warm it and it contracts, so its density peaks at 999.97 kg m⁻³ at 3.98 °C. At 0 °C β ≈ −6.8 × 10⁻⁵ K⁻¹, at 20 °C it is about +2.1 × 10⁻⁴ K⁻¹, and water at 0 °C and at about 8 °C have almost the same density. A single coefficient cannot describe this, so the linear law fails outright across the range even though the changes are tiny. The cause is structure, not anharmonicity: hydrogen bonding builds an open, ice-like arrangement that collapses as the liquid warms, and the collapse outruns ordinary thermal expansion until about 4 °C. Ice itself is about 8% less dense than the water it freezes from. The consequence is a lake that cools from the surface only until the whole depth reaches 4 °C; after that, colder surface water floats, ice forms on top, and the bottom holds at 4 °C all winter.