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Here thermal diffusion means the spreading of temperature by thermal conduction. In a stationary homogeneous material with constant thermal conductivity $k$, mass density $\rho$ and specific heat capacity $c$, energy balance and Fourier's law give the heat equation ${\partial }_{t}T=\alpha {\nabla }^{2}T$, where $\alpha =k/(\rho c)$ is the thermal diffusivity.

A disturbance spreads over a length of order $\sqrt{\alpha t}$, so equilibration across thickness $L$ takes a time of order $L^2/\alpha$. This is the heat-transport sense of the term; temperature-gradient-driven separation of components in a mixture is a distinct transport effect.

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.

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 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.

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.

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.

Thermal effusivity measures a material's response to transient conductive contact:where $k$ is thermal conductivity, $\rho$ is mass density and $c_p$ is specific heat capacity at constant pressure in the thermal model. Its units are $\mathrm{W}{\mathrm{s}}^{1/2}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. It combines the ability to conduct heat with the ability to store thermal energy.

Consider two homogeneous semi-infinite bodies initially at uniform temperatures $T_1>T_2$. Put them in perfect planar contact, neglecting thermal resistance at the interface and using constant material properties. The heat equation similarity solution gives the common interface temperatureTo derive it, the outgoing heat flux from body 1 is $e_1(T_1-T_i)/\sqrt{\pi t}$, while the incoming heat flux to body 2 is $e_2(T_i-T_2)/\sqrt{\pi t}$. Equating them gives both the weighted temperature andEqual effusivities give the arithmetic mean; a much larger $e_2$ holds the interface closer to the initial temperature of body 2.

This explains why objects at equal actual temperature can feel differently cold on contact: the skin's heat loss depends on the transient exchange. It does not measure an electronic Fermi temperature. ^[ref] The ideal $t^{-1/2}$ initial flux is limited in practice by imperfect contact and finite response times; finite thickness matters once the thermal diffusion length reaches an outer boundary.

A heat equation problem in adjacent materials with separate thermal conductivities and volumetric heat capacities. Perfect thermal contact imposes continuous temperature and continuous normal heat flux at the interface. Multiplying each material's heat equation by its temperature cancels the interface boundary terms in a weighted energy estimate.

Substituting $u=-2\nu {\partial }_x\log \Psi$ into the viscous Burgers equation gives ${\partial }_x[({\Psi }_t-\nu {\Psi }_{xx})/\Psi ]=0$. Integration permits the arbitrary function $C(t)$ in the title. Multiplication of $\Psi$ by $e^{-\int C(t)dt}$ preserves $u$ and reduces its equation to the heat equation. Thus the heat-equation representation uses a permissible time normalization rather than a condition uniquely imposed on every representative $\Psi$.

For a spatially constant PDE source term $Q(t)$, subtract its time integral from $u$. The result obeys the homogeneous heat equation with the same initial condition. This proves the displayed formula and shows that uniform heating changes only the zero Fourier mode. In an unnormalised $N_x\times N_y$ discrete Fourier transform, a constant increment $q$ adds $N_x{N}_{y}q$ to the zero-frequency coefficient.

The viscous Burgers equation combines a nonlinear transport equation with diffusion with constant $\nu >0$. On the infinite line, a bounded initial condition gives a smooth function solving the equation for every positive time; a jump in the initial condition is smoothed rather than retained as a discontinuity. Small $\nu$ can nevertheless produce a steep layer, as the Viscous Burgers travelling shock illustrates. The Cole-Hopf transformation $u=-2\nu {\partial }_x\log \varphi$, with positive $\varphi$, converts this equation to the heat equation ${\varphi }_t=\nu {\varphi }_{xx}$.^[ref]

Positive diffusion also explains the admissibility of the inviscid limit. For the entropy flux for a scalar conservation law associated with $\eta (u)=u^2/2$, direct multiplication by $u$ givesThe last term is nonpositive. With convergence sufficient to pass to the limit, the vanishing viscosity approximation therefore retains an entropy inequality rather than just the weak solution equation. For general bounded data, the full Kruzhkov entropy inequalities specify the entropy solution; the quadratic identity is a useful illustration of dissipation.^[ref]

For positive diffusivity, the Cole-Hopf transformation with $f=2\alpha {\partial }_{\theta }\log \psi$ converts the negative-flux equation into the heat equation. Step data give a ratio of two Gaussian-tail integrals $J$. At $\theta =-UZ/2$, $J=1$ and $f=U/2$. Its vanishing-viscosity limit selects the shock or rarefaction wave in the Burgers Riemann problem with negative flux, depending on the sign of $U$.