Thermal expansion — linear, area, volumetric

What we are measuring

Deriving linear expansion from scratch

So define $\alpha$ as the proportionality constant:

$\frac{dL}{L} = \alpha\, dT$

Why this step? $dL/L$ is fractional change — that's the physically meaningful quantity (a fraction is the same whether the rod is in mm or miles). Integrating for constant $\alpha$:

$\int_{L_0}^{L}\frac{dL}{L}=\int_{0}^{\Delta T}\alpha\,dT \;\Rightarrow\; \ln\frac{L}{L_0}=\alpha\,\Delta T$

$L = L_0\,e^{\alpha \Delta T}$

Why this step? Exact form. But $\alpha \sim 10^{-5}\,K^{-1}$ is tiny, so $\alpha\Delta T \ll 1$. Use $e^x \approx 1+x$:

Area expansion — derive $\beta = 2\alpha$

Take a square plate, side $L_0$, area $A_0 = L_0^2$.

$A = L^2 = \big[L_0(1+\alpha\Delta T)\big]^2 = L_0^2(1+\alpha\Delta T)^2$

Why this step? Both sides expanded by the same linear factor (isotropic material). Expand the square:

$A = A_0\big(1 + 2\alpha\Delta T + \alpha^2\Delta T^2\big)$

Why this step? The term $\alpha^2\Delta T^2 \sim (10^{-5})^2 \sim 10^{-10}$ is negligible. Drop it:

Volume expansion — derive $\gamma = 3\alpha$

Cube of side $L_0$, volume $V_0=L_0^3$:

$V = L_0^3(1+\alpha\Delta T)^3 = V_0\big(1 + 3\alpha\Delta T + 3\alpha^2\Delta T^2 + \alpha^3\Delta T^3\big)$

Why this step? Three directions → cube of the linear factor. Keep only the first-order term:

The "hole expands too" idea

Worked examples

Forecast-then-Verify

Common mistakes

Active recall

Connections

• Kinetic theory of gases — same atomic-jiggling idea; gases expand far more.
• Interatomic potential energy curve — asymmetry is the root cause.
• Stress and strain — block expansion + thermal stress when constrained.
• Anomalous expansion of water — the famous exception ($\gamma<0$ from 0–4°C).
• Pendulum and simple harmonic motion — clock-error example.
• Gas laws — where $\Delta T$ must be absolute, unlike here.

Concept Map

Hinglish (regional understanding)