Thermal expansion — linear, area, volumetric: Visual walkthrough

We will only assume you can multiply, and that you know a rectangle's area is "side times side". Everything else is built here.

Step 1 — Why atoms sit farther apart when hot

WHY this matters. Everything on this page is just bookkeeping on top of this one fact: a single atom-to-atom gap grows a tiny bit when heated. If that gap did not grow, nothing would expand. See the Interatomic potential energy curve for the deep reason the spring is lopsided.

PICTURE. In the figure, the curve is the energy stored versus the gap between two atoms. The bottom of the bowl is where a cold atom rests (its comfortable spacing). A hot atom has more energy, so it rocks between the two walls of the bowl — but the left wall (squeeze) is steeper than the right wall (stretch), so the middle of its swing (its average position, the magenta dot) drifts rightward, to a bigger gap.

Figure — Thermal expansion — linear, area, volumetric

Step 2 — From one gap to a whole rod

WHY. Growth is fractional, not fixed. Ten gaps that each grow by $1%$ make a chain that grows by $1%$ — the fraction does not care how many gaps there are. This is why the rod's length and one atomic gap obey the identical rule.

PICTURE. Top row: cold rod, five gaps. Bottom row: hot rod, same five gaps each stretched by the same sliver (orange). The total stretch $Δ L$ is the sum of all the little slivers.

Step 3 — Name the growth factor once

WHY. Every side of every shape multiplies by the same $k$ (the material is isotropic — it grows equally in all directions). Writing $k$ instead of $(1 + α Δ T)$ lets us watch the geometry without drowning in symbols. At the end we substitute $k$ back and expand.

PICTURE. A single cold segment of length $L_{0}$ becomes a hot segment of length $k L_{0}$ . The green bracket is the tiny extra bit, $α Δ T \cdot L_{0}$ .

Recall Why is

$k$ only just bigger than 1? Because $α \approx 1 0^{- 5} K^{- 1}$ is tiny. For a $100 K$ rise, $α Δ T = 1 0^{- 3}$ , so $k = 1.001$ . That "barely bigger than 1" is exactly why we can later throw away $α^{2}$ terms. How big is $α Δ T$ for a $100 K$ heating of steel ( $α = 1.2 \times 1 0^{- 5}$ )? ::: $1.2 \times 1 0^{- 3}$ — about one part in a thousand.

Step 4 — Area: multiply two growing sides

WHY $k^{2}$ , not $2 k$ ? Because area is a product of two lengths, not a sum. When both factors in a product grow, the product grows by the product of their growth factors. This is the whole reason area outruns length.

PICTURE. The cold square is the magenta core. Heating widens it rightward (one orange strip) and upward (a second orange strip), and fills in the tiny violet corner where the two strips overlap. Two strips = the two $α Δ T$ contributions; the corner = the $α^{2} Δ T^{2}$ leftover.

$A = (k L_{0}) (k L_{0}) = k^{2} L_{0}^{2} = A_{0} k^{2}$

$A_{0} = L_{0}^{2}$ — cold area (the magenta core).
$k^{2}$ — the area growth factor.

Now substitute $k = 1 + α Δ T$ and multiply it out:

$A = A_{0} (1 + α Δ T)^{2} = A_{0} (core 1 + two strips 2 α Δ T + tiny corner α^{2} Δ T^{2})$

$1$ — the original square (magenta).
$2 α Δ T$ — the two orange strips (one per direction).
$α^{2} Δ T^{2}$ — the violet corner where strips overlap.

Step 5 — Why we delete the corner

WHY. If $α Δ T = 1 0^{- 3}$ , then its square is $1 0^{- 6}$ — a thousand times smaller than the strips. Keeping it is like measuring a road trip and worrying about the width of one atom. Dropping it is the "first-order approximation".

PICTURE. Same square as Step 4, but the two orange strips are bold (kept) and the corner is faded to a whisper (dropped), with the sizes labelled $1 0^{- 3}$ vs $1 0^{- 6}$ .

Step 6 — Volume: three growing sides

WHY $k^{3}$ ? Volume is length $\times$ width $\times$ height — a product of three growing lengths. Three factors, each carrying its own $α Δ T$ , so three first-order slabs appear.

PICTURE. The cold cube is the magenta core. Heating adds a slab on the right face, a slab on the top, and a slab on the front — three orange slabs, one per direction. (The thin edge-bars and the tiny corner cube are the higher-order leftovers we will drop.)

Take a cube of cold side $L_{0}$ . Its cold volume is the magenta core:

$V_{0} = L_{0}^{3}$

$V_{0}$ — the cold volume of the cube (side $\times$ side $\times$ side).
$L_{0}$ — the cold side length, same one from Step 2.

Each side grows to $k L_{0}$ , so:

$V = (k L_{0})^{3} = V_{0} k^{3} = V_{0} (1 + α Δ T)^{3}$

$V = (k L_{0})^{3}$ — hot volume, all three sides multiplied by $k$ .
$V_{0} k^{3}$ — cold volume times the volume growth factor $k^{3}$ .

Multiply out:

$V = V_{0} (core 1 + three slabs 3 α Δ T + three edge-bars 3 α^{2} Δ T^{2} + corner cube α^{3} Δ T^{3})$

$1$ — original cube $V_{0}$ .
$3 α Δ T$ — the three face-slabs (the parts that actually matter).
$3 α^{2} Δ T^{2}$ and $α^{3} Δ T^{3}$ — edge-bars ( $1 0^{- 6}$ ) and corner cube ( $1 0^{- 9}$ ): deleted, same reason as Step 5.

Step 7 — The degenerate cases (never left uncovered)

Case $Δ T = 0$ (no heating). Then $k = 1$ , so $L = L_{0}$ , $A = A_{0}$ , $V = V_{0}$ . Nothing moves. Good — the formulas do nothing when you do nothing.

Case $Δ T < 0$ (cooling). Then $α Δ T < 0$ , so $k < 1$ and everything shrinks by exactly the same rule. Expansion and contraction are one formula with two signs of $Δ T$ .

Case of a hole. A hole is just "missing metal" surrounded by real metal. Every real atom moves outward by fraction $α Δ T$ , so the boundary of the hole moves outward too — the hole grows, by the same $β = 2 α$ as solid area. It does not close up.

Case $γ < 0$ (water, 0–4 °C). For most solids $α > 0$ , but water between $0$ and $4^{\circ} C$ has $γ < 0$ : it contracts on heating. Our algebra is unchanged; only the sign of the coefficient flips. See Anomalous expansion of water.

PICTURE. Left panel: cold ring with a hole. Right panel: hot ring — the outer edge grew (orange) and the hole grew (the dashed magenta circle is bigger), arrows radiating outward from every point.

The one-picture summary

α L = L_{0} k^{1} β = 2 α A = A_{0} k^{2} γ = 3 α V = V_{0} k^{3}

Keep only the first-order term of each and the coefficients fall out clean: $α : β : γ = 1 : 2 : 3.$

Recall Feynman retelling — the walkthrough in plain words

Picture one lopsided spring between two atoms; heat it and the two atoms drift a hair apart (Step 1). A rod is a long chain of these springs, so the rod grows by the same tiny fraction — call that fraction's "keep everything and add a bit" multiplier $k$ (Steps 2–3). Now a flat plate has two sides, both stretched by $k$ , and area is side times side, so area multiplies by $k$ twice — $k^{2}$ — which splits into the old square plus two thin strips (Steps 4–5). A solid block has three sides, so its volume multiplies by $k$ three times — $k^{3}$ — giving the old block plus three thin slabs (Step 6). The extra corners and edges are so ridiculously small we rub them out. Count the strips and slabs: two for area, three for volume. That is the entire $1 : 2 : 3$ — it is just how many directions the shape can grow in. Cooling? Same story, $k$ dips below one. A hole? Everything scales outward like a zoomed photo, so the hole gets bigger too (Step 7).

Connections

Thermal expansion — linear, area, volumetric (index 1.7.7) — the parent result this page draws.
Interatomic potential energy curve — the lopsided spring of Step 1.
Kinetic theory of gases — same "jiggle harder, spread out" idea, far stronger in gases.
Anomalous expansion of water — the $γ < 0$ case from Step 7.
Stress and strain — what happens when a heated rod is prevented from growing.
Pendulum and simple harmonic motion — a lengthened rod slows a clock.
Gas laws — where $Δ T$ must be absolute, unlike here.