Thermal diffusivity ($\kappa$)

Thermal diffusivity measures how rapidly thermal conduction smooths a temperature difference inside a material. Thermal conductivity measures the heat flux driven by a specified temperature gradient; thermal diffusivity also accounts for the heat capacity that must be warmed. A material can carry substantial heat yet change temperature slowly because it stores substantial internal energy per kelvin.

Write $\alpha =\kappa$ for thermal diffusivity, reserving $k$ for thermal conductivity. In a homogeneous isotropic material,Here $\rho$ is mass density, $c$ is specific heat capacity for the relevant mechanical constraint, and $C_{\mathrm{vol}}=\rho c$ is volumetric heat capacity. The familiar $k/(\rho c_p)$ uses specific heat capacity at constant pressure. For ordinary solids with negligible thermal expansion, the difference between the two constrained specific heat capacities is often neglected. A rigid fixed-volume model instead uses specific heat capacity at constant volume; a compressible moving fluid requires the accompanying work and transport terms. Thus the constraint is part of the approximation, not a choice made solely from the units. ^[ref]

Assume a stationary material in the continuum approximation, local thermal equilibration on times shorter than those resolved, no internal heating, no phase transition, and negligible convection and mechanical work. Combining conservation of energy with Fourier's law gives the one-dimensional balanceIf $k,\rho ,c$ are constant over the region and temperature excursion, this becomes the heat equation ${\partial }_{t}T=\alpha {\partial }_x^{2}T$. If $k$ varies, retain the differentiated product; replacing it with $k{\partial }_x^{2}T$ loses a term. Anisotropic materials need a thermal conductivity tensor. The scalar constant-$\alpha$ examples below deliberately exclude these complications. ^[ref]

For an original comparison, stipulate two model materials with equal thermal conductivity $k=2\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and equal mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, but specific heat capacities $c_A=1000$ and $c_B=4000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. ThenAt the same temperature gradient, both have the same instantaneous heat flux by Fourier's law. At the same initial temperature profile and boundary conditions, material B takes four times as long to smooth a given spatial pattern. With fixed end temperatures, identical geometry and no sources, both eventually have the same steady profile and heat flux. The larger heat capacity changes the transient, not this steady state. These are stipulated parameters, not measurements of named substances.

The Prandtl number is the ratio of kinematic viscosity to thermal diffusivity. It compares the rates at which momentum and temperature variations diffuse. In thermal-diffusion units the Boussinesq equations have momentum inertia multiplied by ${\mathrm{Pr}}^{-1}$; large, rather than small, Prandtl number supplies the usual instantaneous Stokes flow limit.

The thermal diffusion time $t_D=L^2/\alpha$, with $\alpha =\kappa$ the thermal diffusivity, is a characteristic scale for smoothing a temperature variation over length $L$. It is the thermal instance of the diffusive time scale. Rescaling $x=L\hat{x}$ and $t=(L^2/\alpha )\hat{t}$ removes $L$ and $\alpha$ from the constant-coefficient heat equation. Doubling length therefore multiplies comparable internal equilibration times by four; increasing thermal diffusivity by four divides them by four.

This scale is not a universal time of complete thermal equilibrium. A nonzero Fourier mode decays exponentially and never becomes exactly zero at finite time. The geometry, boundary conditions, initial pattern and accepted residual determine the numerical factor. For example, an insulated bar's slowest nonuniform Fourier mode has decay time $L^2/({\pi }^2\alpha )$, and reducing its amplitude to a fraction $\epsilon$ requiresExternal heat transfer can impose a longer cooling time even when internal temperature differences smooth quickly. Thermal diffusivity alone does not determine cooling to the room's temperature.

With thermal diffusivity $\xi$, a Fourier mode of angular wavenumber $k$ decays at rate $\beta =\xi k^2$. To avoid confusing this $k$ with thermal conductivity, use $\alpha =\xi$ and angular wavenumber $K$ below: the same rate is $\beta =\alpha K^2$, with units ${\mathrm{s}}^{-1}$.

Consider an insulated bar $0\le x\le L$ with constant thermal diffusivity, no sources and no heat flux through either end or the sides. An original initial temperature profile and its exact evolution areThe cosine has zero temperature gradient at both ends, satisfying the insulated Neumann boundary conditions. Its spatial mean is zero, so conservation of energy leaves the mean temperature $\overline{T}$ unchanged while the nonuniform part disappears. A short derivation substitutes $T=\overline{T}+A(t)\cos (K_{n}x)$ into the heat equation: the second spatial derivative supplies $-K_n^2$, hence $A^{\prime }=-\alpha K_n^{2}A$. The exponential decay time is ${\tau }_n=1/(\alpha K_n^2)$; doubling angular wavenumber quarters this time. If cycles-per-length spatial frequency $f_x$ is used instead, $K=2\pi f_x$ and the rate is $4{\pi }^2\alpha f_x^2$.

For the two model materials in thermal diffusivity and $L=0.010\mathrm{m}$, ${\tau }_{1,A}\approx 5.07\mathrm{s}$ and ${\tau }_{1,B}\approx 20.3\mathrm{s}$. After $10\mathrm{s}$, the remaining amplitude fractions are approximately $0.139$ and $0.610$. A narrow hot-and-cold pattern disappears faster than a broad one because its temperature curvature, and hence the imbalance in conductive heat flux, is larger.

The thermal diffusion length $\ell (t)=\sqrt{\alpha t}$ is the characteristic length over which thermal conduction changes a temperature distribution during time $t$ in the constant-thermal diffusivity model. It is a scale, not a sharp front or distance travelled at a constant speed: the heat equation has infinite propagation speed mathematically. Different definitions of penetration depth introduce numerical factors. In particular, the normalized one-dimensional heat kernel has spatial variance $2\alpha t$, so its root-mean-square width is $\sqrt{2\alpha t}$.

At $t=10\mathrm{s}$, the model materials have thermal diffusion lengths $4.47$ and $2.24\mathrm{mm}$. For $L=10\mathrm{mm}$ their thermal diffusion times are $50$ and $200\mathrm{s}$. These values describe the same heat equation as the thermal diffusion rate of a Fourier mode; the different numerical factors come from the definition of the scale.

Laser flash analysis measures thermal diffusivity by heating one face of a specimen briefly and recording the opposite face's temperature rise. The original 1961 experiment used a flash tube rather than a laser: a blackened thin metal specimen sat in an opaque ceramic holder, with thermocouple wires pressed against its rear face. An amplifier, oscilloscope and camera recorded the signal. Uniform front illumination and the black coating promoted uniform absorption; the opaque low-thermal conductivity holder blocked direct rear illumination and reduced edge losses. The delayed rear response reveals internal thermal conduction rather than direct heating of the sensor. ^[ref]

A modern version uses a laser or flash lamp, a specimen of measured thickness $L$, controlled initial temperature, and a rear-facing infrared detector with a synchronized recorder. The readout is a temperature rise versus time, starting from a pre-pulse baseline. Thermal conductivity additionally requires mass density and specific heat capacity; it is not obtained from timing alone. ^[ref]

Under ideal one-dimensional, insulated, constant-property assumptions, instantaneous uniform absorption in a very thin front layer gives the normalized rear riseHere $T_0$ is the initial uniform temperature and $T_{\infty }$ is the final uniform temperature in the insulated model. Fitting this curve estimates thermal diffusivity; each term has the same $\alpha K_n^2$ decay as the insulated-bar Fourier modes. Equivalently, the half-rise time $t_{1/2}$ solves $R(t_{1/2})=1/2$. Writing $s=\alpha t/L^2$, the series gives $s_{1/2}\approx 0.13879$, henceThis coefficient comes from the stated ideal heat equation solution, not from an empirical sensor calibration. It has been independently checked by solving the series equation. Absolute absorbed pulse energy cancels from the normalized curve; determining heat capacity from its absolute height needs additional information. ^[ref]

Parker and colleagues identify finite flash duration, heat loss, nonuniform illumination, recording bandwidth and electrical interference as practical limits. A specimen too thin responds during the pulse; one too thick allows losses before the rear rise finishes. They shielded the electrical chain and recorded a baseline; their reported tests were preliminary metal measurements near ambient and moderately elevated temperatures. ^[ref]

Useful additional controls are to measure thickness at several points, verify the recorder's time scale and detector response independently, record a no-pulse baseline, and repeat at different pulse energies while avoiding phase transitions. Compare different thicknesses of the same homogeneous material: ideal half-rise times scale as $L^2$. Failure of these checks can expose finite-pulse effects, sensor delay, optical transmission, coating storage or edge losses; a thickness error is especially consequential because the inferred thermal diffusivity depends on $L^2$. These are proposed diagnostic controls, not invented measurements. When the pulse is not short or the sample exchanges appreciable heat with its surroundings, fit a model that includes those effects instead of applying the ideal half-rise coefficient unchanged.