Thermal Expansion: ΔL = α L₀ ΔT

Most solids get bigger when you heat them. Not by a dramatic amount — a meter-long steel bar heated by 100°C grows by just about 1 millimeter — but enough to matter in engineering. Bridges have expansion joints, railway tracks have gaps, steam pipes have bellows, and precision clocks use special alloys for their pendulums. All of these exist because thermal expansion is real and unavoidable, and you have to design around it.

But why do solids expand? It all comes back to the microscopic picture of atoms in a crystal lattice. In a perfectly harmonic solid (where the bonds between atoms act like ideal springs), there would actually be zero thermal expansion — but real crystals are anharmonic, and that asymmetry creates the effect you can measure.

The microscopic picture

In a solid, atoms sit at equilibrium positions in a crystal lattice, held there by interatomic forces. As you heat the solid, you're adding kinetic energy, which makes the atoms vibrate more vigorously around their equilibrium positions.

If the bonds were perfectly symmetric — meaning the restoring force was equally strong whether you compressed or stretched the bond by the same amount — then the average atomic position would stay the same no matter how much the atoms vibrated. The solid wouldn't expand at all. This is what you'd get from an ideal harmonic potential (parabola-shaped energy curve).

But real bonds are anharmonic. The energy curve rises more steeply on the compression side than on the extension side — it's easier to stretch a bond a bit than to compress it by the same amount (because electron clouds repel strongly at short range but attract weakly at long range). As atoms vibrate more energetically at higher temperature, they spend more time on the "easier" side — the stretched side — and their time-averaged position shifts slightly outward.

Multiply this tiny asymmetric drift over billions of atoms in a macroscopic solid, and you get measurable thermal expansion. The effect is small (typical α around $10^{-5}$ per kelvin) because the anharmonic asymmetry is small, but it adds up across temperature ranges and object sizes.

The linear expansion formula

For most engineering purposes, the formula that matters is:

$\Delta L = \alpha\,L_{0}\,\Delta T$

where:

• $\Delta L$ is the change in length (in meters)
• $L_{0}$ is the original length at the initial temperature
• $\alpha$ is the coefficient of linear expansion in units of $\text{K}^{-1}$ (per kelvin)
• $\Delta T$ is the change in temperature (in kelvin, which equals degrees Celsius for a difference — never for absolute temperatures)

The coefficient $\alpha$ is a material property. It depends weakly on temperature but is roughly constant over modest temperature ranges (say 0–500°C) for most solids. Some typical values:

• Invar (a nickel-iron alloy): $\alpha \approx 1.2 \times 10^{-6}\;\text{K}^{-1}$
• Pyrex glass: $\alpha \approx 3.2 \times 10^{-6}$
• Soda-lime glass: $\alpha \approx 9 \times 10^{-6}$
• Steel/iron: $\alpha \approx 1.1$–$1.3 \times 10^{-5}$
• Copper: $\alpha \approx 1.7 \times 10^{-5}$
• Brass: $\alpha \approx 1.9 \times 10^{-5}$
• Aluminum: $\alpha \approx 2.3 \times 10^{-5}$
• Lead: $\alpha \approx 2.9 \times 10^{-5}$

Notice that these are all of order $10^{-5}$, which is why thermal expansion feels so small in everyday life. A 1 m copper rod heated by 100°C expands by just $1.7 \times 10^{-3}$ m = 1.7 mm — barely noticeable unless you measure carefully.

Why areal expansion is $\approx 2\alpha$

If a rod of length $L_{0}$ expands to $L_{0}(1 + \alpha\Delta T)$, what happens to a square of side $L_{0}$? Both sides expand independently, so the new area is:

$A = L_{0}^{2}(1 + \alpha\Delta T)^{2}$

Expand the square:

$A = L_{0}^{2}\bigl(1 + 2\alpha\Delta T + \alpha^{2}\Delta T^{2}\bigr)$

Since $\alpha \sim 10^{-5}$ and $\Delta T$ is at most hundreds of kelvin, the product $\alpha\Delta T$ is at most $\sim 10^{-3}$, and $\alpha^{2}\Delta T^{2}$ is at most $\sim 10^{-6}$ — completely negligible compared to the linear term. So to excellent first-order approximation:

$\Delta A = A - L_{0}^{2} \approx 2\alpha\,L_{0}^{2}\,\Delta T = 2\alpha\,A_{0}\,\Delta T$

The coefficient of areal expansion is therefore approximately $2\alpha$. This isn't a new material property — it just falls out of the linear formula applied to both dimensions of the square.

Why volumetric expansion is $\approx 3\alpha$

Same derivation for a cube. Each of the three dimensions expands independently:

$V = L_{0}^{3}(1 + \alpha\Delta T)^{3} = L_{0}^{3}\bigl(1 + 3\alpha\Delta T + 3\alpha^{2}\Delta T^{2} + \alpha^{3}\Delta T^{3}\bigr)$

Keeping only the first-order term:

$\Delta V \approx 3\alpha\,V_{0}\,\Delta T$

The coefficient of volumetric expansion (for an isotropic solid) is approximately $3\alpha$, three times the linear coefficient. Again, not a new material property — it's the linear formula applied in three dimensions.

These "factor of 2" and "factor of 3" are the most-commonly-forgotten details in thermal expansion problems. A lot of students correctly write $\Delta L = \alpha L_{0}\Delta T$ but then make a mistake with area or volume.

Method 1: Linear worked example (copper rod)

A 1-meter copper rod at 20°C is heated to 100°C. How much does it expand?

Given:

• $L_{0} = 1.0\;\text{m}$
• $\alpha_{\text{Cu}} = 1.7 \times 10^{-5}\;\text{K}^{-1}$
• $\Delta T = 100 - 20 = 80\;\text{K}$ (since a 1°C difference equals a 1 K difference)

Apply the formula:

$\Delta L = \alpha\,L_{0}\,\Delta T = (1.7 \times 10^{-5})(1.0)(80) = 1.36 \times 10^{-3}\;\text{m}$

Convert to more intuitive units:

$\Delta L = 1.36\;\text{mm}$

So the rod grows by about 1.36 millimeters — roughly the thickness of a credit card. Not much, but if you were building a precision scientific instrument or a long bridge, this would matter.

Cross-check units: α is in per-kelvin, L₀ is in meters, ΔT is in kelvin. Kelvin cancels, leaving meters. ✓

Method 2: Areal worked example (steel plate)

A 1 m² steel plate is heated by 50 K. How much does its area grow?

Given:

• $A_{0} = 1.0\;\text{m}^{2}$
• $\alpha_{\text{steel}} \approx 1.2 \times 10^{-5}\;\text{K}^{-1}$
• $\Delta T = 50\;\text{K}$

Apply the areal formula:

$\Delta A \approx 2\alpha\,A_{0}\,\Delta T = 2(1.2 \times 10^{-5})(1.0)(50) = 1.2 \times 10^{-3}\;\text{m}^{2}$

Convert:

$\Delta A \approx 12\;\text{cm}^{2}$

The plate gains about 12 square centimeters. As a fraction of the original area, that's 0.12%, which sounds small but can matter for precision components where you need micron-level tolerances.

Method 3: Volumetric worked example (brass cube)

A 1-liter brass cube is heated by 100 K. How much does its volume grow?

Given:

• $V_{0} = 1.0\;\text{L} = 10^{-3}\;\text{m}^{3}$
• $\alpha_{\text{brass}} \approx 1.9 \times 10^{-5}\;\text{K}^{-1}$
• $\Delta T = 100\;\text{K}$

Apply the volumetric formula:

$\Delta V \approx 3\alpha\,V_{0}\,\Delta T = 3(1.9 \times 10^{-5})(10^{-3})(100) = 5.7 \times 10^{-6}\;\text{m}^{3}$

Convert:

$\Delta V \approx 5.7\;\text{mL}$

The cube gains about 5.7 milliliters — roughly a teaspoon. Again, small in absolute terms (0.57% of original volume), but enough that precision machined parts can seize up or become loose at temperatures different from their design point.

Why material choice matters: the range of α

The linear expansion coefficient $\alpha$ varies by about a factor of 20 across common materials, from Invar ($1.2 \times 10^{-6}$) to lead ($2.9 \times 10^{-5}$). That huge range means engineers can choose materials to either maximize expansion (for bimetallic strips in thermostats) or minimize it (for precision instruments, optical mounts, metrology equipment).

Generally, the bonding strength and symmetry of the crystal lattice determine $\alpha$:

• Strong, stiff, symmetric lattices (like diamond, silicon, or ceramic oxides) have low $\alpha$ because the atoms can't wander far from their equilibrium positions even when energized.
• Weak, loose, asymmetric lattices (like soft metals and polymers) have high $\alpha$ because the anharmonic effect is more pronounced.

Softer metals (lead, aluminum, brass) expand more than hard metals (steel, iron, tungsten). Glasses vary dramatically: soda-lime glass has $\alpha \approx 9 \times 10^{-6}$, while borosilicate "Pyrex" is $3.2 \times 10^{-6}$ — which is why Pyrex survives sudden temperature changes that would shatter ordinary glassware. Quartz glass is even lower at $5 \times 10^{-7}$, almost 20× better than soda-lime.

Invar — the famous anomaly

Invar is a nickel-iron alloy (roughly 64% iron, 36% nickel) with $\alpha \approx 1.2 \times 10^{-6}$, about an order of magnitude lower than any of its constituent metals alone. Pure nickel has $\alpha = 1.3 \times 10^{-5}$, pure iron has $\alpha = 1.2 \times 10^{-5}$, but when you mix them in just the right ratio, the expansion nearly vanishes.

The reason is a quantum-mechanical effect involving the alloy's magnetic properties. As Invar is heated, the ferromagnetic ordering of the electron spins weakens, and the resulting shift in bond geometry almost perfectly cancels the normal thermal expansion. It's called the Invar effect, and its discovery by Charles Édouard Guillaume in 1896 was considered important enough to win the 1920 Nobel Prize in Physics.

Why does it matter? Invar (and related alloys like Kovar and Superinvar) is used in:

• Precision clocks and watches — pendulum rods that don't lengthen as the temperature of the room changes, preserving timekeeping accuracy over years.
• Metrology standard bars — calibration rods for length standards where you need sub-micron stability.
• Scientific instruments — telescope supports, interferometer arms, surveying tapes.
• Shadow masks in CRT TVs — the masks had to stay at precise dimensions as the tube heated up in use (a forgotten application now that CRTs are gone).
• LNG storage tanks — liquid natural gas at $-162°\text{C}$ would warp ordinary steel tanks; Invar panels handle the enormous temperature range without significant dimensional change.

Selecting Invar in the material widget and turning on "Show Invar zoom" shows you just how minuscule its expansion is compared to copper or aluminum.

The water anomaly (brief detour)

Everything I've said applies to solids. Liquids follow a similar pattern: most liquids expand with temperature, with a volumetric coefficient roughly comparable to or a bit larger than the corresponding solid. But water is famously anomalous.

Water reaches its maximum density at about 4°C, not at its freezing point (0°C). Below 4°C, water expands as it cools toward freezing. This has huge implications for life on Earth:

• Lakes and ponds freeze top-down, not bottom-up. Ice floats on the surface because it's less dense than liquid water, and the denser 4°C water sinks to the bottom. This creates an insulating layer of ice that protects the fish and plants below, which would otherwise freeze solid in a normal "freeze from the bottom up" scenario.
• Cold tap water coming in during winter is slightly less dense than water at 10–20°C, which can create interesting thermal-stratification patterns in water tanks.
• Pipes burst when water freezes, because ice occupies about 9% more volume than liquid water at 0°C. This is the reverse of the normal thermal-contraction pattern.

The reason for water's anomaly is hydrogen bonding. As water cools, its hydrogen-bond network gradually becomes more ice-like, with a more open "tetrahedral" structure that takes up more volume. Below 4°C, this expansion outweighs normal thermal contraction, and the net density decreases toward freezing. This is a fascinating quirk of water chemistry that has profound consequences for climate and biology.

Real-world applications

• Railway tracks. Steel rails heated on a sunny day can expand by many centimeters over a kilometer of track. Old track systems left small gaps between rail segments for this reason (the "clickety-clack" sound of train wheels crossing those gaps). Modern continuous welded rail uses pre-tensioned rails that are clamped to heavy sleepers; the steel is designed to distribute thermal stress instead of buckling, but extreme heat waves can still cause "sun kinks" where the rail warps laterally.
• Bridges. Long bridges have expansion joints at regular intervals — visible gaps or comb-shaped metal teeth where one segment meets the next — so the roadway can grow and shrink without cracking. The Golden Gate Bridge expands by about 40 cm between a cold winter morning and a hot summer afternoon. Ignoring this would cause serious structural damage.
• Steam pipes. Industrial pipes carrying hot steam grow significantly as they heat up. Pipes have expansion loops or bellows (flexible accordion-like sections) to absorb the growth without breaking their supports.
• Bimetallic strips in thermostats. Two metals with different α are bonded together. As the temperature changes, the one with larger α tries to expand more, and the strip curls. The curl triggers a switch that turns on heating or cooling. This is the oldest form of automatic temperature regulation, still used today in many mechanical thermostats.
• Sticky jar lids. Metal lids expand more than glass jars when heated. Run hot water over a stuck lid and the lid loosens just enough to twist off. Thermal expansion applied to your kitchen.
• Glass thermometers. Liquid mercury (or red alcohol) has a much larger volumetric expansion coefficient than the glass bulb. As temperature rises, the liquid expands more than the bulb and rises up the capillary tube, giving you a temperature reading.
• Cracking of heated glass. Pouring boiling water into a cold ordinary glass cup creates a huge temperature gradient — the inside heats and tries to expand, but the outside is still cold and constrains it. The internal tensile stress cracks the glass. Pyrex resists this because its $\alpha$ is about 3× smaller than soda-lime glass, so the stresses are lower.

Common mistakes

• Using Celsius for $\Delta T$ — OK; using Celsius for absolute $T$ — wrong. A difference of 50°C equals a difference of 50 K, so you can freely interchange them in $\alpha L_{0} \Delta T$. But if you're computing any absolute-temperature quantity (like Stefan-Boltzmann radiation in problem 185), you MUST use kelvin.
• Forgetting the factor of 2 for area. A common slip is to write $\Delta A = \alpha A_{0}\Delta T$ without the factor of 2. The correct formula has $2\alpha$.
• Forgetting the factor of 3 for volume. Same issue. Volumetric expansion uses $3\alpha$ for isotropic solids. For liquids, the volumetric coefficient is usually given directly as $\beta$, and you don't derive it from any linear $\alpha$ (liquids don't have linear directions).
• Assuming $\alpha$ is the same for all materials. It's wildly different — from $10^{-6}$ for Invar to nearly $10^{-4}$ for some polymers. Always look up the correct value.
• Assuming $\alpha$ is constant over all temperatures. It varies slightly — usually increasing at higher temperatures. For precision work, you'd use a temperature-dependent $\alpha$ or a table of integrated expansion.
• Confusing linear with areal with volumetric. $\alpha_{L}$, $\alpha_{A} = 2\alpha_{L}$, $\alpha_{V} = 3\alpha_{L}$ for isotropic solids. Mixing them gives answers off by factors of 2 or 3.
• Ignoring anisotropy. Crystalline solids like wood or some polymers have different expansion coefficients along different axes. Wood grain expands differently parallel vs. perpendicular to the fibers. For problems involving isotropic materials (most metals and glasses), this doesn't matter, but don't assume it's always true.

Try it

• Try different materials with the dropdown. Watch how dramatically the expansion differs between Invar and lead at the same temperature change. Lead grows about 25× more than Invar.
• Slide the temperature range. $\Delta T$ appears linearly in the formula, so doubling it doubles the expansion. A $500$ K swing isn't realistic for most materials (lead would melt), but it illustrates the dependence.
• Slide the initial length $L_{0}$. Expansion scales linearly with length, so a 5 m rod expands 5× as much as a 1 m rod at the same material and $\Delta T$.
• Enable "Show areal" to see a 2D square expanding. The visual growth is subtle because α is so small, but the magnification slider (see below) makes it visible.
• Use the magnification slider. Real thermal expansion of a 1 m rod is about 1 mm — only 0.1% of the length, essentially invisible to the eye. The magnification widget exaggerates the visual difference so you can see what's happening. Set it to 100× or 1000× to see clear growth.
• Enable "Show material comparison bar" to see all the materials simultaneously at the same $L_{0}$ and $\Delta T$. The order of magnitude from Invar to lead becomes striking.
• Enable "Show railway example" for a sketch of the classic railway expansion joint — the reason "clickety-clack" sounds existed before continuously-welded rail.
• Try "Show Invar zoom" while Invar is selected. It compares Invar's tiny expansion to copper at the same conditions, showing why Invar is such a valuable engineering material.