Linear Thermal Expansion

Equivalent forms

L = L_0(1 + \alpha \Delta T)

\frac{\Delta L}{L_0} = \alpha \Delta T

A single coefficient α encodes how a material's atomic bond stiffness and anharmonicity respond to temperature.

Symbol	Name	SI unit	Dimension	Range
$Delta_L$	Change in lengthoutput Increase (or decrease) in length due to temperature change	m	L	0 – 0.1
$alpha$	Coefficient of linear expansion Fractional length change per degree temperature change	1/K	Θ⁻¹	1e-7 – 0.0001
$L_0$	Original length Length of the object at the reference temperature	m	L	0.01 – 1000
$Delta_T$	Temperature change Change in temperature from reference	K	Θ	-200 – 500

Unit systems

Symbol	SI	CGS	Imperial
$Delta_L$	m	cm	in
$alpha$	1/K	1/K	1/°F
$L_0$	m	cm	ft
$Delta_T$	K	K	°F

Where it holds

Valid for small temperature changes where expansion is approximately linear. Breaks down near phase transitions, at cryogenic temperatures, and for very large

\Delta T

where higher-order terms matter.

Dimensional analysis

$[\Delta L] = [\alpha ][L_{0}][\Delta T] \to L = (\Theta ^{-1})(L)(\Theta ) = L \checkmark$

Discovery

John Harrison · 1730

Harrison exploited differential thermal expansion in his marine chronometers using bimetallic strips, enabling accurate longitude determination at sea.

Try this

Why do bridges have expansion joints with gaps in them?

A steel bridge beam (α = 12×10⁻⁶ /K) is 100 m long at 15°C. Find how much it lengthens on a 40°C summer day.

Research status: stable

Real-world applications

Bridge expansion joints and railway rail gaps
Bimetallic strip thermostats
Shrink-fitting mechanical parts (heat the outer ring to slip it over the shaft)
Dental fillings must match the expansion coefficient of tooth enamel to prevent cracking

Common misconceptions

Holes in a material expand too — a heated ring gets larger on both the inside and outside, it does not close up
Water is anomalous: it contracts when heated from $0^{\circ}C$ to $4^{\circ}C$ , then expands normally above $4^{\circ}C$
$\alpha is$ not truly constant — it varies with temperature, especially near phase transitions and at cryogenic temperatures

Experimental verification

Harrison's bimetallic strip experiments (1730s). Modern measurement via dilatometry: a sample is heated in a furnace while a push-rod or laser interferometer tracks length changes to nanometer precision. NIST maintains reference data

for \alpha of

metals, ceramics, and polymers.

Derivation

At the atomic level, thermal expansion arises from the asymmetry (anharmonicity) of the interatomic potential energy curve.

As temperature increases, atoms vibrate with larger amplitude, but because the potential well is steeper on the compression side, the average bond length increases.

The linear coefficient

\alpha = (1/L_{0})(dL/dT)

is approximately constant over moderate temperature ranges.

Limiting cases

\Delta _T \to 0

⟶

\Delta _L \to 0

No temperature change, no expansion.

\alpha \to 0

(Invar alloy)⟶

\Delta _L \to 0

Materials with near-zero expansion coefficient barely change length.

Large Delta_T⟶ Linear approximation breaks downHigher-order terms and phase changes become significant at extreme temperatures.

What if…

What if you heat a metal ring — does the hole shrink or expand?

The hole expands. Every linear dimension scales by the same factor $(1 + \alpha \Delta T)$ , including holes. Imagine drawing the ring on a rubber sheet and stretching it uniformly — all distances grow.

What if you use Invar alloy

(\alpha \approx 1.2\times 10^{-6} /K)

?

Expansion drops by a factor of 10 compared to steel. Invar was invented specifically for precision instruments (clock pendulums, survey tapes) where thermal stability is critical.

1

Steel bridge expansion

Given ·

alpha:: 0.000012
L 0:: 100
Delta T:: 25

Find · Delta_L

Steps

$\alpha = 12\times 10^{-6} /K$ for steel
$\Delta T = 40 - 15 = 25\,\mathrm{K}$
$\Delta L = 1.2\times 10^{-5} \times 100 \times 25 = 0.03\,\mathrm{m} = 3\,\mathrm{cm}$

Result ·

\Delta L = \alpha L_{0}\Delta T = 1.2\times 10^{-5} \times 100 \times 25 = 0.03\,\mathrm{m} = 3\,\mathrm{cm}

2

Aluminum power line sag

Given ·

alpha:: 0.000023
L 0:: 200
Delta T:: 40

Find · Delta_L

Steps

$\alpha = 23\times 10^{-6} /K$ for aluminum
A 200 m span heated from $-10^{\circ}C$ to $30^{\circ}C$ : $\Delta T = 40\,\mathrm{K}$
$\Delta L = 2.3\times 10^{-5} \times 200 \times 40 = 0.184\,\mathrm{m}$ — this causes noticeable sag

Result ·

\Delta L = \alpha L_{0}\Delta T = 2.3\times 10^{-5} \times 200 \times 40 = 0.184\,\mathrm{m} \approx 18.4\,\mathrm{cm}

Fourier's Law of Heat Conduction→Ideal Gas Law→First Law of Thermodynamics→