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 Ångström periodic heating method measures thermal diffusivity by periodically heating one end of a rod and comparing the resulting temperature oscillations at two separated thermometers. The farther signal has a smaller amplitude and a later wave phase. Their combination measures how quickly the material redistributes heat, without requiring the absolute heater power. The method bears the name of Anders Jonas Ångström, the physicist who became a professor at Uppsala in 1858.^[ref]

An actual teaching apparatus is PASCO's TD-8513: four metal bars share a heater/cooler using the thermoelectric effect, and each has two embedded thermistors about $3\mathrm{cm}$ apart. Foam covers reduce environmental heat transfer. Its manual instructs students to measure sensor spacing and connect the thermistors through a temperature interface to a computer. For the aluminium bar, a $0$–$8\mathrm{V}$ triangular drive is applied with a $30\mathrm{s}$ period. Recording begins after the amplitudes stabilize; two cycles provide amplitudes and the delay between corresponding peaks. Repeated runs at $40$ and $50\mathrm{s}$ test consistency. The manual converts the inferred thermal diffusivity to thermal conductivity using mass density and specific heat capacity.^[ref]

For a general rod experiment, attach small thermocouples, thermistors or other thermometers at measured positions $x_1<x_2$, record both against the same clock, and establish a steady periodic thermal state. A sinusoid is especially convenient. For a switched or triangular drive, isolate the same Fourier series component at both sensors: unrelated peaks in distorted waveforms need not measure that component's phase difference. Subtract each sensor's mean temperature; a steady spatial background does not belong in the oscillating amplitude ratio. Small excursions help keep thermal conductivity, mass density and specific heat capacity constant.

The appeal is that the rod supplies two complementary clocks: decreasing amplitude measures spatial damping, while increasing wave phase lag measures delay. Unlike laser flash analysis, which follows a transient after one pulse, this method repeatedly samples a steady periodic thermal state. The timing determines thermal diffusivity $\alpha$ in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. To infer thermal conductivity $k$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, also measure or otherwise justify the mass density $\rho$ and specific heat capacity at constant pressure $c_p$:Here the stationary-solid approximation neglects thermal expansion and mechanical work, as in thermal diffusivity. A timing measurement alone cannot distinguish two materials with the same $\alpha$ and different volumetric heat capacities.

A steady periodic thermal state repeats each drive period $P$ after the startup transient has decayed. It is time dependent, unlike a time-independent steady state. For angular frequency $\omega =2\pi /P$, its single-frequency part has the phasor representation $\theta (x,t)=\mathrm{Re}\{\hat{\theta }(x)e^{i\omega t}\}$. Here $\theta$ is the oscillating temperature about the mean profile. Repeating the fitted amplitudes and wave phases over successive cycles checks that startup drift is small; merely completing one drive cycle is insufficient.

A thermal diffusion wave is a periodically forced solution of the heat equation, with spatially decreasing amplitude and varying wave phase. In a semi-infinite rod $x\ge 0$, assume constant thermal diffusivity, negligible side losses, a nearly uniform temperature across each cross-section, and the boundary conditions $\theta (0,t)={\Theta }_0\cos \omega t$ and $\theta (x,t)\to 0$ as $x\to \infty$. A steady periodic thermal state then satisfiesChoose the square root in the complex numbers with $a>0$ so that the amplitude decays into the rod. Substitution of the phasor into the heat equation proves the formula: $i\omega \hat{\theta }=\alpha {\hat{\theta }}^{\prime \prime }$. Taking the real part gives the displayed cosine. In this convention $a$ is the spatial attenuation slope and $b$ is the spatial lag slope, both in ${\mathrm{m}}^{-1}$; $\gamma$ combines them into one complex number. If a complex wavenumber $K$ is defined by $e^{i\omega t-iKx}$ instead, $K=b-ia$.

With $\Delta x=x_2-x_1>0$, let $A_j=|\hat{\theta }(x_j)|$ and define a positive downstream lag $\Delta \varphi$. Thenand either measurement determines thermal diffusivity:The logarithm uses an amplitude ratio, not a ratio of squared amplitudes. The lag is in radians and must be tracked continuously across cycles: a measured phase difference modulo $2\pi$ leaves an integer-cycle ambiguity. Use closer sensor spacings or vary frequency continuously to resolve that ambiguity. Equal attenuation and lag slopes provide a useful diagnostic of the no-side-loss model.

The apparent phase velocity is $\omega /b=\sqrt{2\alpha \omega }$, which depends on frequency. A thermal diffusion wave therefore does not give a material-independent travel speed for heat; it is a forced pattern in a dissipative diffusion equation. The ideal heat equation still has infinite propagation speed, and the observed lag is not the arrival time of a sharp front.

For the no-side-loss thermal diffusion wave, the thermal penetration depth $\delta =1/a=\sqrt{2\alpha /\omega }=\sqrt{\alpha P/\pi }$ is the distance over which the amplitude falls by $e^{-1}$ and the wave phase lags by one radian. It is a periodic counterpart of the thermal diffusion length, with its numerical factor fixed by this definition. A shorter period probes a shallower region; a longer period increases sensitivity to the far-end boundary condition.

Side losses need not destroy the method. Let $\theta$ be the small temperature disturbance about a steady background in surroundings at fixed temperature. For a rod of cross-sectional area $S$ and perimeter $\mathcal{P}$, a uniform real surface heat transfer coefficient $h_s$ produces a loss per unit volume of $H\theta$, with $H=h_s\mathcal{P}/S$. Assume constant $k,\rho ,c_p,H$, a cross-section nearly uniform in temperature, and no distributed heat source. The balance becomesHere $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, while $\lambda$ has units ${\mathrm{s}}^{-1}$. This is a distributed version of Newton's law of cooling; a constant linearized contribution from thermal radiation may be included over a small temperature range. It does not represent arbitrary time-dependent coupling to a surrounding material with its own thermal storage.

For the decaying exponential branch, the phasor equation givesso thatReal positive side loss increases $a$ and decreases $b$ at fixed angular frequency and thermal diffusivity. Their product eliminates $\lambda$, yielding the combined inferenceThis result requires that attenuation and lag belong to the same exponential complex wavenumber at the same frequency, and that the loss term is real. It does not generally survive a superposition caused by a nearby finite end, spatially varying coefficients, transverse temperature gradients, or unequal sensor delays. The inferred $\lambda =\alpha (a^2-b^2)$ supplies an additional model check; a negative value signals inconsistency with this passive-loss model rather than negative physical cooling.

Consider an original stipulated model, not measured material data: thermal diffusivity $\alpha =10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, period $P=100\mathrm{s}$, driven temperature amplitude ${\Theta }_0=4\mathrm{K}$, and thermometers at $x_1=0.020\mathrm{m}$ and $x_2=0.050\mathrm{m}$. Without side losses, $a=b=17.7245{\mathrm{m}}^{-1}$ and thermal penetration depth $\delta =0.05642\mathrm{m}$. ThusThe two no-loss estimates and the combined estimate all return $10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Now retain the same thermal diffusivity and drive but introduce $\lambda =0.010{\mathrm{s}}^{-1}$. The exact complex wavenumber gives $a=19.1863{\mathrm{m}}^{-1}$ and $b=16.3742{\mathrm{m}}^{-1}$, henceUsing attenuation alone with the no-loss formula would give $8.5343\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$; using lag alone would give $1.17174\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Their disagreement is informative. If an independent volumetric heat capacity measurement gives $\rho c_p=2.0\times 10^6\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, the combined estimate implies thermal conductivity $k=200\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H=2.0\times 10^4\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$.

The exact complex amplitudes can be checked independently by discretizing $\alpha {\hat{\theta }}^{\prime \prime }-(\lambda +i\omega )\hat{\theta }=0$ and solving the resulting linear system with a tridiagonal matrix. Halving the spatial step quarters the discrepancy, as expected for centered finite differences. Separately fitting the real time traces to a constant plus sine and cosine terms recovers both amplitudes and their lag; this also checks the sign convention used in the combined estimate.

A finite rod generally contains both exponential branches. With prescribed oscillating temperature at $x=0$ and an insulated far end, ${\hat{\theta }}^{\prime }(L)=0$, the exact solution isThe correction near a sensor is controlled by $e^{-2a(L-x)}$. The important separation is between the farther sensor and the end, not merely between the heater and the end. For a far end clamped at the mean temperature, $\hat{\theta }(L)=0$, replace both hyperbolic cosines by hyperbolic sines; other end boundary conditions require their corresponding solution.

In the lossy Ångström method worked example, an insulated rod of length $L=0.080\mathrm{m}$ gives $D=0.354866$ and $\Delta \varphi =0.615651\mathrm{rad}$. Incorrectly applying the semi-infinite formula returns $1.29418\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $29.4\%$ high. At $L=0.300\mathrm{m}$, it returns $9.99877\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $0.0123\%$ low. Side loss remains real in both cases; the error arises because the measured spatial response no longer has one exponential complex wavenumber.

Choose the period and sensor spacing so that the lag and attenuation are resolvable while the downstream signal remains larger than sensor noise. Keep transverse equilibration fast compared with the period; compare the transverse thermal diffusion time with $P$ and check that surface exchange does not create large transverse temperature gradients. A short period reduces far-end influence but can expose sensor response times; a long period increases penetration and finite-end bias. Calibrate both sensors together, exchange their positions to check channel bias, compare successive cycles, and repeat at several spacings and periods. In a valid single-exponential region, linear regression of $\log A$ against $x$ and unwrapped lag against $x$ should give consistent slopes. These are proposed diagnostic controls. If they fail, fit the finite-rod or coupled-sensor model rather than interpreting every delay as intrinsic thermal diffusivity.

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.